$h(f(m))$. Using this definition the following theorem follows easily from Theorem 4.4.1.

**Theorem 4.4.2.** (spectral theorem-functional calculus form) Let $A$ be a self-#-adjoint operator on $\mathbf{H}^#$. Then there is a unique map $\hat{\phi}$ from the bounded #-Borel functions on $^{*}\mathbb{R}_c^#$ into $\mathcal{L}(\mathbf{H}^#)$ so that

(a) $\hat{\phi}$ is an algebraic $*$-homomorphism.

(b) $\hat{\phi}$ is #-norm #-continuous, that is, $\|\hat{\phi}(h)\|_{\mathcal{L}(\mathbf{H}^#)} \le \|h\|_{*\infty}$

(c) Let $h_n(x), n \in ^*\mathbb{N}$ be a hyper infinite sequence of bounded in $^{*}\mathbb{R}_c^#$ #-Borel functions with $\#\lim_{n \to *\infty} h_n(x) = x$

for each $x$ and $|h_n(x)| \le |x|$ for all $x$ and $n \in ^*\mathbb{N}$. Then, for any $\psi \in D(A)$, $\#\lim_{n \to *\infty} \hat{\phi}(h_n)\psi = A\psi$.

(d) If $h_n(x) \to_* h(x)$ pointwise and if the hyper infinite sequence $\|h_n\|_{*\infty}, n \in ^*\mathbb{N}$ is bounded in $^{*}\mathbb{R}_c^#$, then $\hat{\phi}(h_n) \to_* \hat{\phi}(h)$ strongly.

In addition:

(e) If $A\psi = \lambda\psi$ then $\hat{\phi}(h) = h(\lambda)\psi$.

(f) If $h \ge 0$, then $\hat{\phi}(h) \ge 0$.

The functional calculus is very useful. For example, it allows us to define the exponential $\text{Ext}-\exp(itA)$ and prove easily many of its properties as a function of $t$ (see the next section). In the case where $A$ is bounded in $^{*}\mathbb{R}_c^#$ we do not need the functional calculus to define the exponential since we can define $\text{Ext}-\exp(itA)$ by the power series which #-converges in #-norm.

The functional calculus is also used to construct spectral #-measures and can be used to develop a multiplicity theory similar to that for bounded self-#-adjoint operators.

A vector $\psi \in \mathbf{H}^#$ is said to be cyclic for $A$ if $\{g(A)\psi|g \in C^{*\infty}(^{*}\mathbb{R}_c^#)\}$ is #-dense in $\mathbf{H}^#$. If $\psi$ is a cyclic vector, then it is possible to represent $\mathbf{H}^#$ as $L_2(^{*}\mathbb{R}_c^#, d^#\mu_\psi^#)$ where $\mu_\psi^# = \int_{^{*}\mathbb{R}_c^#} g(x)d^#\mu_\psi^#(x) = \langle \psi, g(A)\psi \rangle_#$ in such a way that $A$ becomes multiplication by $x$. In general, $\mathbf{H}^#$ decomposes into a direct sum of cyclic subspaces so the #-measure space, $M$ in Theorem 4.4.1 can be realized as a union of copies of $^{*}\mathbb{R}_c^#$. As in the case of bounded in $^{*}\mathbb{R}_c^#$ operators we can define $\sigma_{ac}(A), \sigma_{pp}(A), \sigma_{sing}(A)$ and decompose $\mathbf{H}^#$ accordingly.

Finally, the spectral theorem in its projection-valued #-measure form follows easily from the functional calculus. Let $P_\Omega$ be the operator $\chi_\Omega(A)$ where $\chi_\Omega$ is the characteristic function of the measurable set $\Omega \subset ^{*}\mathbb{R}_c^#$. The family of operators $\{P_\Omega\}$ has the following properties:

(a) Each $P_\Omega$ is an orthogonal projection.
(b) $P_\emptyset = 0$; $P_{(-*\infty, *\infty)} = I$.
(c) If $\Omega = \text{Ext}-\bigcup_{n=1}^{*\infty} \Omega_n$ with $\Omega_n \cap \Omega_m = \emptyset$ for all $n \ne m$ then

(c) If $\Omega = \text{Ext}-\bigcup_{n=1}^{*\infty} \Omega_n$ with $\Omega_n \cap \Omega_m = \emptyset$ for all $n \ne m$ then

(b) $P_\emptyset = 0$; $P_{(-*\infty, *\infty)} = I$.

(c) If $\Omega = \text{Ext}-\bigcup_{n=1}^{*\infty} \Omega_n$ with $\Omega_n \cap \Omega_m = \emptyset$ for all $n \ne m$ then

$$P_{\Omega} = s\text{-}\lim_{N \to *\infty} \left( \text{Ext} - \sum_{n=1}^{N} P_{\Omega_n} \right). \quad (4.4.3)$$

(d) $P_{\Omega_1}P_{\Omega_2} = P_{\Omega_1 \cap \Omega_2}$.

**Definition 4.4.1.** Such a family is called a projection-valued #-measure (p.v.#-m.).

**Remark 4.4.1.** This is a generalization of the notion of bounded in $^{*}\mathbb{R}_c^#$ projection-valued #-measure introduced in § 4.3. In that we only require $P_{(-*\infty, *\infty)} = I$ rather than $P_{(-a,a)} = I$ for some $a \in ^{*}\mathbb{R}_c^#$. For $\varphi \in \mathbf{H}^#$, $\langle \varphi, P_\Omega\varphi \rangle_#^*$ is a well-defined Borel
$$
\begin{align}
\mathrm{Ext}-\sum_{n=0}^{n=\infty} a_n &= \overbrace{\{c_n\}}^{n=\infty} = \tag{0.5} \\
&= (c_0, c_1, \dots, c_k, \dots \{c_n\}_{n=0}^{\infty}, \overbrace{\{c_n\}}^{n=\infty}) \in \left[ \left[ \overbrace{\{c_n\}}^{n=0} \right] \right] \nonumber
\end{align}
$$

where $c_0 = a_0, c_k = \mathrm{Ext}-\sum_{n=0}^{n=k} a_n, k \in \mathbb{N}$.

(vii) Let {$a_n$}$_{n=0}^{n=N}$, $N \in *NN$ be external hyperfinite sequence in $H : \{a_n\}_{n=0}^N \subset H$.

We define external hyperfinite sum $\mathrm{Ext}-\sum_{n=0}^{n=N} a_n$ by

$$
\mathrm{Ext}-\sum_{n=0}^{n=N} a_n = \overbrace{\{c_n\}}^{n=N} = (c_0, c_1, \dots, c_k, \dots, c_N, \hat{c}_N) \in [[\hat{c}_N]] \quad (0.6)
$$

where $c_0 = a_0, c_k = \mathrm{Ext}-\sum_{n=0}^{n=k} a_n, 0 \le k \le N, c_N = \mathrm{Ext}-\sum_{n=0}^{n=N} a_n$.

(viii) Let {$a_n$}$_{n=0}^{n=N}$, $N \in *NN$ be external hyperfinite sequence in $H : \{a_n\}_{n=0}^N \subset H$
such that $a_n \equiv 0$ for all $n \in *NN$. We assume that

$$
\mathrm{Ext}-\sum_{n=0}^{n=N} a_n = \mathrm{Ext}-\sum_{n=0}^{n=\infty} a_n. \qquad (0.7)
$$

**Remark 1.1.** (i) Let {$x_i$}$_{i=1}^N \subset H$ and {$y_i$}$_{i=1}^N \subset H, N \in *N$ by external hyperfinite sequences; let {$\alpha_i$}$_{i=1}^N \subset \tilde{\mathbb{C}}^\#_c$ and {$\beta_i$}$_{i=1}^N \subset \tilde{\mathbb{C}}^\#_c$. Then the equality holds

$$
\mathrm{Ext}-\sum_{i=1}^{N} \langle \alpha_i x_i + \beta_i y_i, z \rangle_{\#} = \mathrm{Ext}-\sum_{i=1}^{N} \alpha_i \langle x_i, z \rangle_{\#} + \mathrm{Ext}-\sum_{i=1}^{N} \beta_i \langle y_i, z \rangle_{\#} \quad (0.8)
$$

(ii) Let {$x_{ij}$}$_{i=1}^N \subset H, N, K \in *N, K \le j \le i \le N$, by external hyperfinite sequences; let
{$\alpha_{ij}$}$_{i=1}^N \subset \tilde{\mathbb{C}}^\#_c$. Then the equality holds

$$
\left\langle \mathrm{Ext} - \sum_{K \le j \le i \le N}^{N} \alpha_{ij} x_{ij}, z \right\rangle_{\#} = \mathrm{Ext} - \sum_{i=K}^{N} \left( \mathrm{Ext} - \sum_{i=K}^{N} \alpha_{ij} \langle x_{ij}, z \rangle_{\#} \right). \quad (0.9)
$$

(iii) Let {$x_i$}$_{i=1}^\infty \subset H$ by external hyperinfinite sequence in $H$. We call {$x_i$}$_{i=1}^\infty$ a Cauchy hyperinfinite sequence if for any $\varepsilon \approx 0, \varepsilon > 0$ there is $N \in *NN$ such that for any $m, n > N, \|x_n - x_m\|_\# < \varepsilon$.

(iv) We now stand ready to give construction of $H^\#$.
The members of $H^\#$ will be
constructed as equivalence classes of Cauchy hyperinfinite sequences in $H$. Let $C(H)$
denote the set of all Cauchy hyperinfinite sequences in $H$. We must define an
equivalence relation on $C(H)$. For $s \in C(H)$, denote by $[s]$ the set of all elements in
$C(H)$

that are related to $s$. Then for any $s,t \in C(H)$, either $[s] = [t]$ or $[s]$ and $[t]$ are disjoint.
Let $\{x_i\}_{i=1}^\infty$ and $\{y_i\}_{i=1}^\infty$ be in $C(H)$. Say they are equivalent (i.e. related $\{x_i\}_{i=1}^\infty \sim \{y_i\}_{i=1}^\infty$)
if $\|x_n - y_n\|_\# \to \# 0_{\mathbb{R}_c}$; i.e. if the hyperinfinite sequence $\|x_n - y_n\|_\#$-tends to $0_{\mathbb{R}_c}$ as
$n \to \infty$.

(vi) Definition (iv) yields an equivalence relation ($\cdot \sim \cdot$) on $C(H)$.

**Proof.** We need to show that this relation is reflexive, symmetric, and transitive.

• **Reflexive:** $x_n - x_n = 0_{\mathbb{R}_c}, n \in *\mathbb{N}$ and the sequence all of whose terms are $0_{\mathbb{R}_c}$
clearly #-converges to $0_{\mathbb{R}_c}$. So {$x_i$}$_{i=1}^\infty$ is related to {$x_i$}$_{i=1}^\infty$.

• **Symmetric:** Suppose {$x_i$}$_{i=1}^\infty$ is related to {$y_i$}$_{i=1}^\infty$, so $\|x_n - y_n\|_\# \to \# 0_{\mathbb{R}_c}$. But
$y_n - x_n = -(x_n - y_n)$, and since only the absolute value $\|x_n - y_n\|_\# = \|y_n - x_n\|_\#$ comes
into play in Definition (iv), it follows that $\|y_n - x_n\|_\# \to \# 0_{\mathbb{R}_c}$ as well. Hence, {$y_i$}$_{i=1}^\infty$ is
#-measure on $\mathbb{R}_c^\#$ which we denote by $d^\#\langle\phi, P_\lambda\phi\rangle_\#$ as in § 4.3.

The complex $\mathbb{C}_c^\#$-valued #-measure $d^\#\langle\phi, P_\lambda\psi\rangle_\#$ is defined by polarization. Thus, given a bounded in $\mathbb{R}_c^\#$ #-Borel function $g$ we can define $g(A)$ by

$$ \langle\phi, g(A)\phi\rangle_\# = \mathrm{Ext}-\int_{\mathbb{R}_c^\#} g(\lambda)d^\#\langle\phi, P_\lambda\phi\rangle_\# \quad (4.4.4) $$

It is not difficult to show that this map $g \mapsto g(A)$ has the properties (a)-(d) of Theorem 4.4.1, so $g(A)$ as defined by (4.4.4) coincides with the definition of $g(A)$ given by Theorem 4.4.1. Now, suppose $g$ is an unbounded $\mathbb{C}_c^\#$-valued #-Borel function and let

$$ D_g = \{\phi | \mathrm{Ext}-\int_{\mathbb{R}_c^\#} g(\lambda)d^\#\langle\phi, P_\lambda\phi\rangle_\# < *\infty\}. \quad (4.4.5) $$

Then, $D_g$ is #-dense in $H^\#$ and an operator $g(A)$ is defined on $D_g$ by

$$ \langle\phi, g(A)\phi\rangle_\# = \mathrm{Ext}-\int_{\mathbb{R}_c^\#} g(\lambda)d^\#\langle\phi, P_\lambda\phi\rangle_\#. \quad (4.4.6) $$

As in § 4.3, we write symbolically

$$ g(A) = \mathrm{Ext}-\int_{\mathbb{R}_c^\#} g(\lambda)d^\#P_\lambda. \quad (4.4.7) $$

In particular, for $\phi, \psi \in D(A)$,

$$ \langle\phi, A)\psi\rangle_\# = \mathrm{Ext}-\int_{\mathbb{R}_c^\#} g(\lambda)d^\#\langle\phi, P_\lambda\psi\rangle_\#. \quad (4.4.8) $$

if $g$ is $\mathbb{R}_c^\#$-valued, then $g(A)$ is self-#-adjoint on $D_g$. We summarize:

**Theorem 4.4.3.** (spectral theorem-projection valued measure form) There is a one-to-one correspondence between self-#-adjoint operators $A$ and projection-valued #-measures $\{P_\Omega\}$ on $H^\#$ the correspondence being given by

$$ A = \mathrm{Ext}-\int_{\mathbb{R}_c^\#} \lambda d^\# P_\lambda. \quad (4.4.9) $$

We use the functional calculus developed above to define $\mathrm{Ext}-\exp(itA)$.

**Theorem 4.4.4.** Let $A$ be a self-#-adjoint operator and define $U(t) = \mathrm{Ext}-\exp(itA)$. Then

(a) For each $t \in \mathbb{R}_c^\#$, $U(t)$ is a unitary operator and $U(t+s) = U(t)U(s)$ for all $s, t \in \mathbb{R}_c^\#$.

(b) If $\phi \in H^\#$ and $t \to_* t_0$, then $U(t)\phi \to_* U(t_0)\phi$.

(c) For any $\psi \in D(A)$: $\frac{U(t)\psi - \psi}{t} \to_* iA\psi$ as $t \to_* 0$.

(d) If $\#\lim_{t \to_* 0} \frac{U(t)\psi - \psi}{t}$ exists, then $\psi \in D(A)$.

**Proof** (a) follows immediately from the functional calculus and the corresponding statements for the complex-valued function $\mathrm{Ext}-\exp(it\lambda)$. To prove (b) observe that

$$ \|\mathrm{Ext}-\exp(itA)\phi - \phi\|_\#^2 = \mathrm{Ext}-\int_{\mathbb{R}_c^\#} |\mathrm{Ext}-\exp(it\lambda) - 1|^2 d^\#\langle P_\lambda\phi, \phi\rangle_\#. \quad (4.4.10) $$

Since $|\mathrm{Ext}-\exp(it\lambda) - 1|^2$ is dominated by the #-integrable function $g(\lambda) = 2$ and since for each $\lambda \in \mathbb{R}_c^\#$: $|\mathrm{Ext}-\exp(it\lambda) - 1|^2 \to_* 0$ as $t \to_* 0$ we conclude that $\|U(t)\phi - \phi\|_\#^2 \to_* 0$ as $t \to_* 0$, by the generalized Lebesgue dominated-#-convergence theorem. Thus $t \mapsto U(t)$ is strongly #-continuous at $t = 0$, which by the group property proves $t \mapsto U(t)$ is strongly #-continuous everywhere. The proof of (c), which again uses the dominated #-convergence theorem and the estimate $|\mathrm{Ext}-\exp(ix) - 1|^2 \le |x|$. To prove (d), we define
related to {$x_i$}$_{i=1}^{\infty}$.

• **Transitive:** Suppose {$x_i$}$_{i=1}^{\infty}$ is related to {$y_i$}$_{i=1}^{\infty}$, and {$y_i$}$_{i=1}^{\infty}$ is related to {$z_i$}$_{i=1}^{\infty}$.
This means that $\|x_n - y_n\|_{\#} \to_{\#} 0_{\widehat{\mathbb{R}}_c}$ and $\|y_n - z_n\|_{\#} \to_{\#} 0_{\widehat{\mathbb{R}}_c}$.

To be fully precise, let us fix $\varepsilon \approx 0, \varepsilon > 0$; then there exists an $N \in *NN$ such that
for all $n > N$, $\|x_n - y_n\|_{\#} < \varepsilon/2$; also, there exists an $M$ such that for all $n > M$,
$\|y_n - z_n\|_{\#} < \varepsilon/2$. Well, then, as long as $n$ is bigger than both $N$ and $M$, we have that
$\|x_n - z_n\|_{\#} = \|(x_n - y_n) + (y_n + z_n)\|_{\#} \le \|x_n - y_n\|_{\#} + \|y_n - z_n\|_{\#} < \varepsilon/2 + \varepsilon/2 = \varepsilon$.

So, choosing $L$ equal to the max($N, M$), we see that given $\varepsilon \approx 0, \varepsilon > 0$ we can always
choose $L$ so that for $n > L$, $\|x_n - z_n\|_{\#} < \varepsilon$. This means that $\|x_n - z_n\|_{\#} \to_{\#} 0_{\widehat{\mathbb{R}}_c}$ - i.e.
{$x_i$}$_{i=1}^{\infty}$ is related to {$z_i$}$_{i=1}^{\infty}$.

So, we really have an equivalence relation (· ~ ·), and therefore the set C(H) is
partitioned into disjoint subsets (equivalence classes). We will denote that partition
C(H)/ ~ by H#

$$
C(H) / \sim \triangleq H^{\#}. \qquad (0.10)
$$

(vii) assume that $[s] = [\{x_i\}_{i=1}^{\infty}] \in H^{\#}$ and $[t] = [\{y_j\}_{i=1}^{\infty}] \in H^{\#}$, we define inner
\#-product $\langle [s], [t] \rangle_{\#}$ on $H^{\#}$ by

$$
\langle [ \{x_i\}_{i=1}^{\infty} ], [ \{y_j\}_{i=1}^{\infty} ] \rangle_{\#} = \operatorname{Ext} - \sum_{i,j=1}^{i,j=*\infty} \langle x_i, y_j \rangle_{\#}. \quad (0.11)
$$

In particular if for all $i \neq j \langle x_i, y_j \rangle_\# = 0_{\widehat{\mathbb{R}}_c}$

$$
\langle [ \{x_i\}_{i=1}^{\infty} ], [ \{y_j\}_{i=1}^{\infty} ] \rangle_{\#} = \operatorname{Ext} - \sum_{i=1}^{i=*\infty} \langle x_i, y_i \rangle_{\#}. \quad (0.12)
$$

**Remark 1.2.** The following formula useful:

$$
\|x+y\|_{\#}^{2} = \langle x+y, x+y \rangle_{\#} = \|x\|_{\#}^{2} + \|y\|_{\#}^{2} + \langle x,y \rangle_{\#} + \langle y,x \rangle_{\#} = \|x\|_{\#}^{2} + \|y\|_{\#}^{2} + 2 \operatorname{Re}\langle x,y \rangle_{\#} \quad (1.1)
$$

**Theorem 1.1.** (Generalized Schwarz Inequality). Let $(H, \langle \cdot, \cdot \rangle_{\#})$ be an inner \#-product space and $x, y \in H$. Assume that:

(1) at least one of hyperreals $\|x\|_{\#}, \|y\|_{\#}$ is invertible
in $\widetilde{\mathbb{R}}_{c}^{\#}$ then

$$
|x, y|_{\#} \le \|x\|_{\#} \times \|y\|_{\#} \quad (1.2)
$$

and equality holds iff $x$ and $y$ are linearly dependent.

(2) both of hyperreals $\|x\|_{\#}$, $\|y\|_{\#}$ is not invertible in $\widehat{\mathbb{R}}_{c}^{\#}$ then

$$
|x, y|_{\#} \times \mathring{1}_{\widehat{\mathbb{R}}_{c}^{\#}} \le \|x\|_{\#} \times \|y\|_{\#}. \quad (1.2')
$$

**Proof.** (1) If $y = 0$, the result holds trivially. So assume that $y \neq 0$ and $\|y\|_{\#}$ is invertible

in $\widehat{\mathbb{R}}_c^\#$. First off notice that if $x = \alpha y$ for some $\alpha \in \widehat{\mathbb{C}}_c^\#$, then $\langle x, y \rangle = \alpha \|y\|_{\#}^2$ and hence
$|\langle x, y \rangle_{\#}| = |\alpha| \|y\|_{\#}^2 = \|x\|_{\#} \|y\|_{\#}$

Note that in this case $\alpha = \langle x, y \rangle \|y\|^2$. Now suppose that $x \in H$ is arbitrary, let
$z = x - \|y\|^{-2}_{\#} \langle x, y \rangle_{\#} y$. So $z$ is the orthogonal projection of $x$ onto $y$. Then
$$D(B) = \left\{ \psi \mid \#\lim_{t \to \#} \frac{U(t)\psi - \psi}{t} \text{ exists} \right\} \quad (4.4.11)$$

and let

$$iB\psi = \#\lim_{t \to \#} \frac{U(t)\psi - \psi}{t}. \quad (4.4.12)$$

A simple computation shows that $B$ is symmetric. By (c), $B \supset A$, so $B = A$.

**Definition 4.4.2.** An operator-valued function $U(t)$ satisfying (a) and (b) is called a strongly #-continuous one-parameter unitary group.

**Definition 4.4.3.** If $U(t)$ is a strongly #-continuous one-parameter unitary group, then the self-#-adjoint operator $A$ with $U(t) = \text{Ext-exp}(itA)$ is called the infinitesimal generator of $U(t)$.

Suppose that $U(t)$ is a weakly #-continuous one-parameter unitary group. Then $\|U(t)\varphi - \varphi\|_{\#}^2 = \|U(t)\varphi\|_{\#}^2 - \langle U(t)\varphi, \varphi \rangle_{\#} - \langle \varphi, U(t)\varphi \rangle_{\#} + \|\varphi\|_{\#}^2 \to_{\#} 0$ as $t \to \#$. Thus $U(t)$ is actually strongly #-continuous. As a matter of fact, to conclude that $U(t)$ is strongly #-continuous one need only show that $U(t)$ is weakly #-measurable, that is, that $\langle U(t)\varphi, \psi \rangle_{\#}$ is #-measurable for each $\varphi$ and $\psi$. This startling result sometimes useful since in applications one can often show that $\langle U(t)\varphi, \psi \rangle_{\#}$ is the #-limit of a hyper infinite sequence of #-continuous functions; $\langle U(t)\varphi, \psi \rangle_{\#}$ is therefore #-measurable and by generalized von Neumann's theorem $U(t)$ is then strongly #-continuous.

**Theorem 4.4.5.** Let $U(t)$ be a one-parameter group of unitary operators on a hy infinite dimensional Hilbert space $\mathbf{H}^\#$. Suppose that for all $\varphi, \psi \in \mathbf{H}^\#$, $\langle U(t)\psi, \varphi \rangle_\#$ is #-measurable. Then $U(t)$ is strongly #-continuous.

**Proof.** Let $\psi \in \mathbf{H}^\#$. Then for all $\varphi \in \mathbf{H}^\#$, $\langle U(t)\psi, \varphi \rangle_\#$ is a bounded in $\mathbb{R}_c^\#$ #-measurable function and $\varphi \mapsto \int_0^a \langle U(t)\psi, \varphi \rangle_\# d^\#t$ is a linear functional on $\mathbf{H}^\#$ of #-norm less than or equal to $a\|\varphi\|_\#$. Thus, by the generalized Riesz lemma there is a $\psi_a \in \mathbf{H}^\#$ so that

$$\langle \psi_a, \varphi \rangle_\# = \int_0^a \langle U(t)\psi, \varphi \rangle_\# d^\#t. \quad (4.4.13)$$

Note that

$$\langle U(b)\psi_a, \varphi \rangle_\# = \langle \psi_a, U(-b)\varphi \rangle_\# = \int_0^a \langle U(t)\psi, U(-b)\varphi \rangle_\# d^\#t = \\ \int_0^a \langle U(t+b)\psi, \varphi \rangle_\# d^\#t = \int_b^{a+b} \langle U(t)\psi, \varphi \rangle_\# d^\#t. \quad (4.4.14)$$

From (4.1.14) we obtain

$$\begin{aligned} & |\langle U(b)\psi_a, \varphi \rangle_\# - \langle \psi_a, \varphi \rangle_\#| = \\ & = \left| \int_0^b \langle U(t)\psi, \varphi \rangle_\# d^\#t \right| + \left| \int_b^{a+b} \langle U(t)\psi, \varphi \rangle_\# d^\#t \right| \le 2a \|\varphi\|_\# \|\psi\|_\# \end{aligned} \quad (4.4.15)$$
$$0 \le \|z\|_{\#}^2 = \left\| x - \frac{\langle x, y \rangle_{\#}}{\|y\|_{\#}^2} y \right\|_{\#}^2 = \|x\|_{\#}^2 + \frac{|\langle x, y \rangle_{\#}|^2}{\|y\|_{\#}^4} \|y\|_{\#}^2 - 2 \operatorname{Re} \left\langle x, \frac{\langle x, y \rangle_{\#}}{\|y\|_{\#}^2} y \right\rangle = \\ = \|x\|_{\#}^2 - \frac{|\langle x, y \rangle_{\#}|^2}{\|y\|_{\#}^2} \quad (1.3)$$

from (1.3) it follows that $0 \le \|y\|_{\#}^2 \|x\|_{\#}^2 - |\langle x, y \rangle_{\#}|^2$ with equality iff $z = 0$ or equivalently iff $x = \|y\|_{\#}^{-2} \langle x, y \rangle_{\#} y$.

(2) Let $z = x - (\|y\|_{\#}^{-1\cdot})^2 \langle x, y \rangle_{\#} y$. So $z$ is the orthogonal projection of $x$ onto $y$. Then

$$0 \le \|z\|_{\#}^2 = \left\| x - \langle x, y \rangle_{\#} (\|y\|_{\#}^{-1\cdot})^2 y \right\|_{\#}^2 = \\ \|x\|_{\#}^2 + |\langle x, y \rangle_{\#}|^2 (\|y\|_{\#}^{-1\cdot})^4 \|y\|_{\#}^2 - 2 \operatorname{Re} \left\langle x, \langle x, y \rangle_{\#} (\|y\|_{\#}^{-1\cdot})^2 y \right\rangle = \\ = \|x\|_{\#}^2 - |\langle x, y \rangle_{\#}|^2 (\|y\|_{\#}^{-1\cdot})^2. \quad (1.3')$$

From (1.3') it follows that $0 \le \|y\|_{\#}^2 \times \|x\|_{\#}^2 - |\langle x, y \rangle_{\#}|^2 \times \overline{1}_{\mathbb{R}_c^*}.$

**Corollary 1.1.** Let $(H^{\#}, \langle \cdot, \cdot \rangle)$ be an inner #-product space and $\|x\|_{\#} := \sqrt{\langle x, x \rangle_{\#}}$. Then $\|\cdot\|_{\#}$ is a $*\mathbb{R}_c$-valued #-norm on $H^{\#}$. Moreover $\langle \cdot, \cdot \rangle_{\#}$ is #-continuous on $H^{\#} \times H^{\#}$, where $H$ is viewed as the #-normed space $(H^{\#}, \|\cdot\|_{\#})$.

**Proof.** The only non-trivial thing to verify that $\|\cdot\|_{\#}$ is a #-norm is the triangle inequality:

$$\|x+y\|^2 = \|x\|^2 + \|y\|^2 + 2 \operatorname{Re} \langle x, y \rangle_{\#} \le \|x\|^2 + \|y\|^2 + 2 \|x\|_{\#} \|y\|_{\#} = (\|x\|_{\#} + \|y\|_{\#})^2 \\ \text{where we have made use of Schwarz's inequality. Taking the square root of this inequality shows } \|x+y\| \le \|x\| + \|y\|. \text{ For the #-continuity assertion:} \\ |\langle x, y \rangle_{\#} - \langle x', y' \rangle_{\#}| = |\langle x-x', y \rangle_{\#} + \langle x', y-y' \rangle_{\#}| \le \|y\|_{\#} \|x-x'\|_{\#} + \|x'\|_{\#} \|y-y'\|_{\#} \\ \le \|y\|_{\#} \|x-x'\|_{\#} + (\|x\|_{\#} + \|x-x'\|_{\#}) \|y-y'\|_{\#} = \|y\|_{\#} \|x-x'\|_{\#} + \|x\|_{\#} \|y-y'\|_{\#} \\ + \|x-x'\|_{\#} \|y-y'\|_{\#} \text{ from which it follows that } \langle \cdot, \cdot \rangle \text{ is #-continuous.}$$

**Definition 1.2.** Let $(H, \langle \cdot, \cdot \rangle_{\#})$ be an inner #-product space, we say $x, y \in H$ are orthogonal and write $x \perp y$ iff $\langle x, y \rangle_{\#} = 0$. More generally if $A \subset H$ is a set, $x \in H$ is orthogonal to $A$ and write $x \perp A$ iff $\langle x, y \rangle = 0$ for all $y \in A$. Let $A_{\perp} = \{x \in H : x \perp A\}$ be the set of vectors orthogonal to $A$. We also say that a set $S \subset H$ is orthogonal if $x \perp y$ for all $x, y \in S$ such that $x \ne y$. If $S$ further satisfies, $\|x\|_{\#} = 1$ for all $x \in S$, then $S$ is said to be orthonormal.

**Proposition 1.1.** Let $(H, \langle \cdot, \cdot \rangle_{\#})$ be an inner product space then

(1) (Parallelogram Law)

$$\|x+y\|_{\#}^2 + \|x-y\|_{\#}^2 = 2\|x\|_{\#}^2 + 2\|y\|_{\#}^2 \quad (1.4)$$

for all $x, y \in H$.

(2) (Pythagorean Theorem) If $S \subset H$ is a hyperfinite orthonormal set, then

$$\left\| \operatorname{Ext} - \widehat{\sum}_{x \in S} x \right\|_{\#}^2 = \operatorname{Ext} - \widehat{\sum}_{x \in S} \|x\|_{\#}^2 \quad (1.5)$$

(3) If $A \subset H^{\#}$ is a set, then $A_{\perp}$ is a #-closed linear subspace of $H^{\#}$.

**Proof.** We will assume that $H^{\#}$ is a complex Hilbert space with $*\tilde{\mathbb{C}}_{\tilde{c}}^{\#}$-valued inner product, the real case being easier. Statements (1) and (2) are proved by the following elementary computations:
and therefore $\#\lim_{b \to \#} \langle U(b)\psi_a, \varphi \rangle_\# = \langle \psi_a, \varphi \rangle_\#$ so that $U(b)$ is weakly and therefore strongly $\#$-continuous on the set of vectors of the form $\{\psi_a|\psi \in \mathbf{H}^\#\}$. It remains only to show that this set is $\#$-dense, since by an $\varepsilon \approx 0, \varepsilon/3$ argument we can then conclude that $t \mapsto U(t)$ is strongly $\#$-continuous on $\mathbf{H}^\#$. Suppose that $\varphi \in \{\psi_a|\psi \in \mathbf{H}^\#, a \in \mathbb{R}_c^\#\}^\mathrm{T}$ and let $\{\psi^{(n)}\}_{n \in \mathbb{N}}$ be an orthonormal basis for $\mathbf{H}^\#$. Then for each $n \in \mathbb{N}$

$$ \mathrm{Ext}-\int_0^a \langle U(t)\psi^{(n)}, \varphi \rangle_\# d^\#t = \langle \psi_a^{(n)}, \varphi \rangle_\# = 0 \quad (4.4.16) $$

for all $a \in \mathbb{R}_c^\#$ which implies that $\langle U(t)\psi^{(n)}, \varphi \rangle_\# = 0$ except for $t \in S_n$, a set of Lebesgue $\#$-measure zero. Choose $t_0 \notin \bigcup_{n \in \mathbb{N}} S_n$. Then $\langle U(t_0)\psi^{(n)}, \varphi \rangle_\# = 0$ for all $n \in \mathbb{N}$ which implies that $\varphi = 0$, since $U(t_0)$ is unitary.

**Theorem 4.4.6.** Suppose that $U(t)$ is a strongly continuous one-parameter unitary group. Let $D$ be a $\#$-dense domain which is invariant under $U(t)$ and on which $U(t)$ is strongly $\#$-differentiable. Then $i^{-1}$ times the strong $\#$-derivative of $U(t)$ is essentially self-$\#$-adjoint on $D$ and its $\#$-closure is the $\#$-infinitesimal generator of $U(t)$. This theorem has a reformulation which is sufficiently important that we state it as a theorem.

**Theorem 4.4.7.** Let $A$ be a self-adjoint operator on $\mathbf{H}^\#$ and $D$ be a $\#$-dense linear set contained in $D(A)$. If for all $t$, $\mathrm{Ext}-\exp(itA) : D \to D$ then $D$ is a $\#$-core for $A$.

**Theorem 4.4.8.** Let $U(t)$ be a strongly $\#$-continuous one-parameter unitary group on a Hilbert space $\mathbf{H}^\#$. Then, there is a self-$\#$-adjoint operator $A$ on $\mathbf{H}^\#$ so that $U(t) = \mathrm{Ext}-\exp(itA)$.

**Proof** Part (d) of Theorem 4.4.4 suggests that we obtain $A$ by differentiating $U(t)$ at $t = 0$. We will show that this can be done on a $\#$-dense set of especially nice vectors and then show that the $\#$-limiting operator is essentially self-$\#$-adjoint by using the basic criterion. Finally, we show that the exponential of this $\#$-limiting operator is just $U(t)$. Let $f \in C_0^\infty(\mathbb{R}_c^\#)$ and for each $\varphi \in \mathbf{H}^\#$ define

$$ \varphi_f = \mathrm{Ext}-\int_{\mathbb{R}_c^\#} f(t)U(t)\varphi d^\#t. \quad (4.4.17) $$

Since $U(t)$ is strongly $\#$-continuous the integral in (4.4.7) can be taken to be a Riemann integral. Let $D$ be the set of hyperfinite linear combinations of all such $\varphi_f$ with $\varphi \in \mathbf{H}^\#$ and $f \in C_0^\infty(\mathbb{R}_c^\#)$. If $j_\varepsilon(t)$ is the approximate identity then

$$ \begin{aligned} \|\varphi_{j_\varepsilon} - \varphi\|_\# &= \left\| \mathrm{Ext}-\int_{\mathbb{R}_c^\#} j_\varepsilon(t)[U(t)\varphi - \varphi]d^\#t \right\|_\# \\ &\le \left( \mathrm{Ext}-\int_{\mathbb{R}_c^\#} j_\varepsilon(t)d^\#t \right) \sup_{t \in [-\varepsilon, \varepsilon]} \|U(t)\varphi - \varphi\|_\#. \end{aligned} \quad (4.4.18) $$

Since $U(t)$ is strongly $\#$-continuous, $D$ is $\#$-dense in $\mathbf{H}^\#$. We have used the inequality

$$ \left\| \mathrm{Ext}-\int_{\mathbb{R}_c^\#} h(t)d^\#t \right\|_\# \le \mathrm{Ext}-\int_{\mathbb{R}_c^\#} \|h(t)\|_\# d^\#t \quad (4.4.19) $$

for non-Archimedean Banach space-valued $\#$-continuous functions on the real line $\mathbb{R}_c^\#$ (which can be proven using the approximate partial sums as in the $\mathbb{R}_c^\#$-valued
$$ \|x+y\|_{\#}^{2} = \|x\|_{\#}^{2} + \|y\|_{\#}^{2} + 2\mathrm{Re}\langle x,y \rangle_{\#} + \|x\|_{\#}^{2} + \|y\|_{\#}^{2} - 2\mathrm{Re}\langle x,y \rangle_{\#} = 2\|x\|_{\#}^{2} + 2\|y\|_{\#}^{2} \quad (1.6) $$

and

$$ \begin{aligned} \left\| \mathrm{Ext} - \widehat{\sum}_{x \in S} x \right\|_{\#}^{2} &= \left\langle \mathrm{Ext} - \widehat{\sum}_{x \in S} x, \mathrm{Ext} - \widehat{\sum}_{y \in S} y \right\rangle_{\#} = \mathrm{Ext} - \widehat{\sum}_{x,y \in S} \langle x, y \rangle_{\#} = \\ &= \mathrm{Ext} - \widehat{\sum}_{x \in S} \langle x, x \rangle_{\#} = \mathrm{Ext} - \widehat{\sum}_{x \in S} \|x\|_{\#}^{2}. \end{aligned} \quad (1.7) $$

Item 3. is a consequence of the #-continuity of $\langle \cdot, \cdot \rangle_{\#}$ and the fact that $A^{\perp} = \cap_{x \in A} \mathrm{Ker}(\langle \cdot, x \rangle)$ where $\mathrm{Ker}(\langle \cdot, x \rangle) = \{y \in H | \langle y, x \rangle_{\#} = 0\}$ is a #-closed subspace of $H$.

**Definition 1.3.** A non-Archiedean Hilbert space $H^{\#}$ is an inner #-product space $(H, \langle \cdot, \cdot \rangle_{\#})$ such that the induced Hilbertian #-norm is #-complete.

**Example 1.3.** Let $(X, M, \mu^{\#})$ be a #-measure space then $H^{\#} = L_{2}^{\#}(X, M, \mu^{\#})$ with inner #-product $\langle f, g \rangle_{\#} = \mathrm{Ext} - \int_{X} f \bar{g} d^{\#} \mu^{\#}$ is a non-Archiedean Hilbert space. Note that every non-Archiedean Hilbert space $H^{\#}$ is “equivalent” to a Hilbert space of this form.

**Definition 1.4.** A subset $C$ of a non-Archiedean vector space $X$ is said to be convex if for all $x, y \in C$ the line segment $[x, y] = \{tx + (1-t)y : 0 \le t \le 1\}$ joining $x$ to $y$ is contained in $C$ as well. (Notice that any vector subspace of $X$ is convex.)

**Definition 1.5.** $M \subset H^{\#}$ is essentially #-closed if $\forall x(x \in H^{\#}) \exists [\inf_{z \in M} \|x-z\|_{\#}]$.

**Theorem 1.2.** Suppose that $H^{\#}$ is a non-Archiedean Hilbert space and $M \subset H^{\#}$ be a essentially #-closed convex subset of $H^{\#}$. Then for any $x \in H^{\#}$ there exists a unique $y \in M$ such that $\|x-y\|_{\#} = \mathrm{dist}(x, M) = \inf_{z \in M} \|x-z\|_{\#}$. Moreover, if $M$ is a vector subspace of $H^{\#}$, then the point $y$ may also be characterized as the unique point in $M$ such that $(x-y) \perp M$.

**Proof.** (1) Uniqueness: By replacing $M$ by $M-x = \{m-x|m \in M\}$ we may assume $x=0$. Let $\delta = \mathrm{dist}(0, M) = \inf_{m \in M} \|m\|_{\#}$ and $y, z \in M$. By the parallelogram law and the convexity of $M$ one obtains

$$ 2\|y\|_{\#}^{2} + 2\|z\|_{\#}^{2} = \|y+z\|_{\#}^{2} + \|y-z\|_{\#}^{2} = 4 \left\| \frac{y+z}{2} \right\|_{\#}^{2} + \|y-z\|_{\#}^{2} \ge 4\delta^{2} + \|y-z\|_{\#}^{2}. \quad (1.8) $$

Hence if $\|y\|_{\#} = \|z\|_{\#} = \delta$, then $2\delta^2 + 2\delta^2 \ge 4\delta^2 + \|y-z\|_{\#}^2$, so that $\|y-z\|_{\#}^2 = 0$. Therefore, if a minimizer for $\mathrm{dist}(0, \cdot)|_{M}$ exists, it is unique.

(2) Existence: Let $y_n \in M$ be chosen such that $\|y_n\|_{\#} = \delta_n \to_{\#} \delta = \mathrm{dist}(0, M)$. Taking $y = y_m$ and $z = y_n$ in Eq.(1.8) shows $2\delta_m^2 + 2\delta_n^2 \ge 4\delta^2 + \|y_n - y_m\|_{\#}^2$. Passing to the #-limit $m, n \to \infty$ in this equation implies, $2\delta^2 + 2\delta^2 \ge 4\delta^2 + \#\text{-lim } \sup_{m,n \to \infty} \|y_n - y_m\|_{\#}^2$. Therefore $\{y_n\}_{n=1}^{\infty}$ is hyper infinite Cauchy sequence and hence #-convergent. Because $M$ is #-closed, $y = \#\text{-lim } \sup_{n \to \infty} y_n \in M$ and because $\|\cdot\|_{\#}$ is #-continuous, $\|y\|_{\#} = \#\text{-lim } \sup_{n \to \infty} \|y_n\|_{\#} = \delta = \mathrm{dist}(0, M)$. So $y$ is the desired point in $M$ which is closest to 0.

Now for the second assertion we further assume that $M$ is a #-closed subspace of $H$ and $x \in H^{\#}$. Let $y \in M$ be the closest point in $M$ to $x$. Then for $w \in M$, the function $g(t) = \|x - (y + tw)\|_{\#}^2 = \|x - y\|_{\#}^2 - 2t \mathrm{Re}\langle x - y, w \rangle_{\#} + t^2 \|w\|_{\#}^2$ has a minimum at $t=0$. Therefore $0 = g'(0) = -2 \mathrm{Re}\langle x - y, w \rangle_{\#}$. Since $w \in M$ is arbitrary, this implies that $(x-y) \perp M$. Finally suppose $y \in M$ is any point such that $(x-y) \perp M$. Then for
case). For $\varphi_f \in D$ we obtain that

$$ \left( \frac{U(s) - I}{s} \right) \varphi_f = \mathrm{Ext} - \int_{*\mathbb{R}_c^\#} f(t) \left( \frac{U(s+t) - U(t)}{s} \right) \varphi d^\# t = \\ \mathrm{Ext} - \int_{*\mathbb{R}_c^\#} \frac{f(\tau - s) - f(\tau)}{s} U(\tau) \varphi d^\# \tau \to_{\#} - \mathrm{Ext} - \int_{*\mathbb{R}_c^\#} f^{\#'}(\tau) U(\tau) \varphi d^\# \tau = \varphi_{-f^{\#'}} \quad (4.4.20) $$

since $[f(t-s) - f(t)]/s$ \#-converges to $-f^{\#'}(t)$ uniformly. For $\varphi_f \in D$ we define $A\varphi_f = i^{-1}\varphi_{-f^{\#'}}$. Note that $U(t) : D \to D, A : D \to D$ and $U(t)A\varphi_f = AU(t)\varphi_f$ for $\varphi_f \in D$.

Furthermore if $\varphi_f, \varphi_g \in D$ we obtain that

$$ \langle A\varphi_f, \varphi_g \rangle_{\#} = \#\lim_{s \to \#} \langle \left( \frac{U(s) - I}{is} \right) \varphi_f, \varphi_g \rangle_{\#} = \\ = \#\lim_{s \to \#} \langle \varphi_f, \left( \frac{I - U(-s)}{is} \right) \varphi_g \rangle_{\#} = \frac{1}{i} \langle \varphi_f, \varphi_{-g^{\#'}} \rangle_{\#} = \langle \varphi_f, A\varphi_g \rangle_{\#} \quad (4.4.21) $$

so $A$ is symmetric. Now we show that $A$ is essentially self-\#-adjoint. Suppose that there is a $u \in D(A^*)$ so that $A^*u = iu$. Then for each $\varphi \in D(A) = D$

$$ \frac{d^\#}{d^\# t} \langle U(t)\varphi, u \rangle_\# = \langle iAU(t)\varphi, u \rangle_\# = -i\langle U(t)\varphi, A^*u \rangle_\# = -i\langle U(t)\varphi, iu \rangle_\# = \langle U(t)\varphi, u \rangle_\# \quad (4.4.22) $$

Thus, the $C_c^\#$-valued function $f(t) = \langle U(t)\varphi, u \rangle_\#$ satisfies the ordinary differential equation $f^{\#'} = f$ so $f(t) = f(0)[\mathrm{Ext}-\exp(t)]$. Since $U(t)$ has \#-norm one, $|f(t)|$ is bounded, in $*\mathbb{R}_c^\#$ which implies that $f(0) = \langle \varphi, u \rangle_\# = 0$. Since $D$ is \#-dense, $u = 0$. A similar proof shows that $A^*u = -iu$ can have no nonzero solutions. Therefore $A$ is essentially self-\#-adjoint on $D$.

Let $V(t) = \mathrm{Ext}-\exp(it(\#-\bar{A}))$. It remains to show that $U(t) = V(t)$. Let $\varphi \in D(A)$. Since $\varphi \in D((\#-\bar{A}))$, $V(t)\varphi \in D((\#-\bar{A}))$ and $V^{\#'}(t)\varphi = iAV(t)\varphi$ by (c) of Theorem 4.4.4, We already know that $U(t)\varphi \in D \subset D(\#-\bar{A})$ for all $\in *\mathbb{R}_c^\#$. Let $w(t) = U(t)\varphi - V(t)\varphi$. Then $w(t)$ is a strongly \#-differentiable vector-valued function and

$$ w^{\#'}(t) = iAU(t)\varphi - i(\#-\bar{A})V(t)\varphi = iAw(t). \quad (4.4.23) $$

Thus

$$ \frac{d^\#}{d^\# t} \|w(t)\|_\#^2 = -i\langle(\#-\bar{A})w(t), w(t)\rangle_\# + i\langle w(t), (\#-\bar{A})w(t)\rangle_\# \quad (4.4.24) $$

Therefore $w(t) = 0$ for all $t \in *\mathbb{R}_c^\#$ since $w(t) = 0$. This implies that $U(t)\varphi = V(t)\varphi$ for all $t \in *\mathbb{R}_c^\#$, $\varphi \in D$. Since $D$ is \#-dense in $H^\#$, $U(t) = V(t)$.

**Remark 4.4.2.** Finally, we have the following generalization of Stone's theorem 4.4.8. If $g$ is a $*\mathbb{R}_c^\#$-valued \#-Borel function on $*\mathbb{R}_c^\#$, then

$$ g(A) = \mathrm{Ext} - \int_{*\mathbb{R}_c^\#} g(\lambda) d^\# P_\lambda \quad (4.4.25) $$

defined on $D_g$ (4.4.5) is self-\#-adjoint. If $g$ is bounded, $g(A)$ coincides with $\hat{\phi}(g)$ in Theorem 4.4.2.

We conclude with several remarks. First, generalized Stone's formula, given in Theorem 4.3.7 relates the resolvent and the projection-valued measure associated with any self-\#-adjoint operator. The proof is the same as in the bounded in $*\mathbb{R}_c^\#$ case. The spectrum of an unbounded self-\#-adjoint operator is an unbounded subset of the real axis $*\mathbb{R}_c^\#$. One can define discrete and essential spectrum; Theorem 4.3.6
$z \in M$, by Pythagorean's theorem,

$$ \|x-z\|_{\#}^{2} = \|x-y+y-z\|_{\#}^{2} = \|x-y\|_{\#}^{2} + \|y-z\|_{\#}^{2} \geq \|x-y\|_{\#}^{2} \quad \text{which shows } [\mathrm{dist}(0,M)]^{2} \geq \|x-y\|_{\#}^{2}. \text{ That is to say } y \text{ is the point in } M \text{ closest to } x. $$

**Definition 1.6.** $A : H^{\#} \to H^{\#}$ is a bounded in $\ast\mathbb{R}_{c}^{\#}$ operator if and only if there exists some $M \in \ast\mathbb{R}_{c}^{\#}$, $M > 0$ such that for all $x \in H^{\#}$, $\|Ax\|_{\#} \leq M\|x\|_{\#}$. The smallest such $M$ if exists is called the operator #-norm of $A$ and denoted by $\|A\|_{\#OP}$ or $\|A\|_{\#}$. Thus

$$ \|A\|_{\#} = \sup_{\|x\|_{\#}=1} (\|Ax\|_{\#}) < *\infty \quad (1.9) $$

if supremum in RHS of (1.9) exists and $\sup_{\|x\|_{\#}=1}(\|Ax\|_{\#}) < *\infty$. Conversely if (1.9) holds

**Proposition 1.2.** A linear operator $A : H_{1}^{\#} \to H_{2}^{\#}$ between #-normed spaces is bounded in $\ast\mathbb{R}_{c}^{\#}$ if and only if it is #-continuous.

**Proof.** Suppose that $A$ is bounded in $\ast\mathbb{R}_{c}^{\#}$. Then, for all vectors $x, h \in H_{1}^{\#}$ with $h \approx 0$ non zero we have $\|A(x+h) - A(x)\|_{\#} = \|A(h)\|_{\#} \leq M\|h\|_{\#}$, $M \in \ast\mathbb{R}_{c}^{\#}$, $M > 0$. Letting $h$ go to zero shows that $A$ is #-continuous at $x$. Moreover, since the constant $M$ does not depend on $x$, this shows that in fact $A$ is uniformly #-continuous, and even Lipschitz #-continuous.

Conversely, it follows from the #-continuity at the zero vector that there exists a $\varepsilon \approx 0$, $\varepsilon > 0$ such that $\|A(h)\|_{\#} = \|A(h) - A(0)\|_{\#} \leq 1$ for all vectors $h \in H_{1}^{\#}$ with $\|h\|_{\#} \leq \varepsilon$. Thus, for all non-zero $x \in H_{1}^{\#}$, one has

$$ \|Ax\|_{\#} = \left\| \frac{\|x\|_{\#}}{\varepsilon} A \left( \varepsilon \frac{\|x\|_{\#}}{x} \right) \right\|_{\#} = \frac{\|x\|_{\#}}{\varepsilon} \left\| A \left( \varepsilon \frac{\|x\|_{\#}}{x} \right) \right\|_{\#} \leq \frac{\|x\|_{\#}}{\varepsilon}. $$

This proves that $A$ is bounded in $\ast\mathbb{R}_{c}^{\#}$.

**Definition 1.7.** Suppose that $A : H^{\#} \to H^{\#}$ is a bounded in $\ast\mathbb{R}_{c}^{\#}$ operator. The #-adjoint of $A$, denote $A^{*}$, is the unique operator $A^{*} : H^{\#} \to H^{\#}$ such that $\langle Ax, y \rangle_{\#} = \langle x, A^{*}y \rangle_{\#}$. (The proof that $A^{*}$ exists and is unique will be given in Proposition below.) A bounded in $\ast\mathbb{R}_{c}^{\#}$ operator $A : H^{\#} \to H^{\#}$ is self #-adjoint or Hermitian if $A = A^{*}$.

**Definition 1.8.** Let $H^{\#}$ be a non-Archiedean Hilbert space and $M \subset H$ be a #-closed subspace. The orthogonal projection of $H^{\#}$ onto $M$ is the function $P_{M} : H^{\#} \to H^{\#}$ such that for $x \in H^{\#}$, $P_{M}(x)$ is the unique element in $M$ such that $(x - P_{M}(x)) \perp M$.

**Proposition 1.3.** Let $H^{\#}$ be a non-Archiedean Hilbert space and $M \subset H^{\#}$ be a #-closed subspace. The orthogonal projection $P_{M}$ satisfies:

(1) $P_{M}$ is linear (and hence we will write $P_{M}x$ rather than $P_{M}(x)$.

(2) $P_{M}^{2} = PM$ ($P_{M}$ is a projection).

(3) $P_{M}^{*} = P_{M}$, ($P_{M}$ is self-#-adjoint).

(4) $\mathrm{Ran}(P_{M}) = M$ and $\ker(P_{M}) = M^{\perp}$.

**Proof.** (1) Let $x_{1}, x_{2} \in H^{\#}$ and $\alpha \in \ast\mathbb{R}_{c}^{\#}$, then $P_{M}x_{1} + \alpha P_{M}x_{2} \in M$ and $P_{M}x_{1} + \alpha P_{M}x_{2} - (x_{1} + \alpha x_{2}) = [P_{M}x_{1} - x_{1} + \alpha(P_{M}x_{2} - x_{2})] \in M^{\perp}$ showing $P_{M}x_{1} + \alpha P_{M}x_{2} = P_{M}(x_{1} + \alpha x_{2})$, i.e. $P_{M}$ is linear.

(2) Obviously $\mathrm{Ran}(P_{M}) = M$ and $P_{M}x = x$ for all $x \in M$. Therefore $P_{M}^{2} = P_{M}$.

(3) Let $x, y \in H^{\#}$, then since $(x - P_{M}x) \in M^{\perp}$ and $(y - P_{My}) \in M^{\perp}$, $\langle P_{M}x, y \rangle_{\#} = \langle P_{M}x, P_{My} + y - P_{My} \rangle_{\#} = \langle P_{M}x, P_{My} \rangle_{\#} = \langle P_{M}x + (x - P_{M}), P_{My} \rangle_{\#} = \langle x, PMy \rangle_{\#}$.

(4) It is clear that $\mathrm{Ran}(P_{M}) \subset M$. Moreover, if $x \in M$, then $P_{M}x = x$ implies that $\mathrm{Ran}(P_{M}) = M$. Now $x \in \ker(P_{M})$ iff $P_{M}x = 0$ iff $x = x - 0 \in M^{\perp}$.

**Corollary 1.2.** Suppose that $M \subset H^{\#}$ is a proper closed subspace of a non-Archiedean Hilbert space $H^{\#}$, then $H^{\#} = M \oplus M^{\perp}$.
(Generalized Weyl's criterion) still holds if one adds the criterion that the vectors {$\psi_n$} must be in the domain of A.

Finally, we note that the measure space of Theorem 4.4.1 can always be chosen so that

Proposition 4.4.2 is applicable.

The following theorem says that every strongly #-continuous unitary group arises as the exponential of a self-#-adjoint operator.

**Theorem 4.4.9.** Let $U(\mathbf{t}) = U(t_1, \dots, t_n)$ be a strongly continuous map of $*\mathbb{R}_c^{\#n}$ into the unitary operators on a hyper infinite dimensional Hilbert space $\mathbf{H}^\#$ satisfying $U(\mathbf{t} + \mathbf{s}) = U(\mathbf{t})U(\mathbf{s})$. Let D be the set of hyperfinite linear combinations of vectors of the form

$$ \varphi_f = \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} f(\mathbf{t}) U(\mathbf{t}) d^{\#n} t \quad (4.4.26) $$

where $\varphi \in \mathbf{H}^\#$, $f \in C_0^{\#\infty}(*\mathbb{R}_c^{\#n})$. Then D is a domain of essential self-#-adjointness for each of the generators $A_j$ of the one-parameter subgroups $U(0, 0, \dots, t_j, \dots, 0)$, each $A_j : D \to D$ and the $A_j$ commute, $j = 1, \dots, n$. Furthermore, there is a projection-valued #-measure $P_\Omega$ on $*\mathbb{R}_c^{\#n}$ so that

$$ \langle \varphi, U(\mathbf{t})\psi \rangle_{\#} = \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} \mathrm{Ext} - \exp(i\langle \mathbf{t}, \lambda \rangle) d^{\#} \langle \varphi, P_{\lambda}\psi \rangle_{\#} \quad (4.4.27) $$

for all $\varphi, \psi \in \mathbf{H}^\#$.

**Proof** Let $A_j$ be the infinitesimal generator of $U_j(t_j) = U(0, \dots, t_j, \dots, 0)$. The procedure used in the proof of Theorem 4.4.8 shows that $D \subset D(A_j)$, $A_j : D \to D$, and $U_j(t_j) : D \to D$. Theorem 4.4.7 shows that $A_j$ is essentially self-#-adjoint on D. Because of the relation $U(\mathbf{t} + \mathbf{s}) = U(\mathbf{t})U(\mathbf{s})$, $U_j(t_j)$ commutes with $U_i(t_i)$ for all $t_j, t_i \in *\mathbb{R}_c^\#$.

Therefore, it follows from Theorem 4.5.1, that $A_i$ and $A_j$ commute in the sense that is, their spectral projections commute. Let $P_\Omega^j$ be the projection-valued #-measure on $*\mathbb{R}_c^\#$ corresponding to $A_j$. Define a projection valued #-measure $P_\Omega$ on $*\mathbb{R}_c^{\#n}$ by defining it first on rectangles $r_n = \mathrm{Ext} - \prod_{i=1}^n (a_i, b_i)$ by $P_{r_n} = \mathrm{Ext} - \prod_{i=1}^n P_{(a_i, b_i)}^i$ and then letting $P_\Omega$ be the unique extension to the smallest $\sigma^\#$-algebra containing the rectangles, namely the #-Borel sets. Notice that, by Theorem 4.5.1, the $P_{\Omega_j}^j$ commute since the groups $U_j$ commute. For each $\varphi, \psi \in \mathbf{H}^\#$, $\langle \varphi, P_\Omega \psi \rangle_\#$ is a $*C_c^\#$-valued #-measure of hyperfinite mass which we denote by $d^\# \langle \varphi, P_\lambda \psi \rangle_\#$. Applying generalized Fubini's theorem we conclude that

$$ \langle \varphi, U(\mathbf{t})\psi \rangle_{\#} = \left\langle \varphi, \mathrm{Ext} - \prod_{i=1}^{n} U(t_i)\psi \right\rangle_{\#} = \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} \mathrm{Ext} - \exp(i\langle \mathbf{t}, \lambda \rangle) d^{\#} \langle \varphi, P_{\lambda}\psi \rangle_{\#}. \quad (4.4.28) $$

## § 4.5. Nearstandard $C_\#^*$ algebras generated by spectral projections related to unbounded in $*\mathbb{R}_c^\#$ self-#-adjoint operators.

Suppose that A and B are two unbounded self-#-adjoint operators on a non-Archimedean Hilbert space $\mathbf{H}^\#$. We would like to find a reasonable definition for the statement: "A and B commute."

This cannot be done in the straightforward way since AB – BA may not make sense
**Proof.** Given $x \in H^{\#}$, let $y = P_M x$ so that $x - y \in M^{\perp}$. Then $x = y + (x - y) \in M \oplus M^{\perp}$.
If $x \in M \cap M^{\perp}$, then $x \perp x$, i.e. $\|x\|_{\#}^2 = \langle x, x \rangle_{\#} = 0$. So $M \cap M^{\perp} = \{0\}$.

**Proposition 1.4.** (Generalized Riesz Theorem). Let $H^{\#*}$ be the dual space of $H^{\#}$.
The map

$$z \in H^{\#} \xrightarrow{j} \langle \cdot, z \rangle_{\#} \in H^{\#*} \qquad (1.9')$$

is a conjugate linear #-isometric isomorphism.

**Proof.** The map $j$ is conjugate linear by the axioms of the non-Archiedean inner products. Moreover, for $x, z \in H^{\#}$, $\langle x, z \rangle_{\#} \le \|x\|_{\#} \|z\|_{\#}$ for all $x \in H^{\#}$ with equality when $x = z$. This implies that $\|jz\|_{H^{\#*}} = \langle \cdot, z \rangle_{H^{\#*}} = \|z\|_{\#}$. Therefore $j$ is #-isometric and this shows that $j$ is injective. To finish the proof we must show that $j$ is surjective. So let $f \in H^{\#*}$ which we assume with out loss of generality is non-zero. Then $M = \ker(f)$ is a #-closed proper subspace of $H^{\#}$. Since, by Corollary 1.1, $H^{\#} = M \oplus M^{\perp}$, $f: H^{\#}/M \approx M^{\perp} \to *\mathbb{C}_c^{\#}$ is a linear isomorphism. This shows that $\dim(M^{\perp}) = 1$ and hence $H^{\#} = M \oplus *\mathbb{C}_c^{\#}x_0$ where $x_0 \in M^{\perp}\setminus\{0\}$. Alternatively, choose $x_0 \in M^{\perp}\setminus\{0\}$ such that $f(x_0) = 1$. For $x \in M^{\perp}$ we have $f(x - \lambda x_0) = 0$ provided that $\lambda = f(x)$. Therefore $x - \lambda x_0 \in M \cap M^{\perp} = \{0\}$, i.e. $x = \lambda x_0$. This again shows that $M^{\perp}$ is spanned by $x_0$. Choose $z = \lambda x_0 \in M^{\perp}$ such that $f(x_0) = \langle x_0, z \rangle$. (So $\lambda = \overline{f(x_0)}/\|x_0\|^2$.) Then for $x = m + \lambda x_0$ with $m \in M$ and $\lambda \in *\mathbb{C}_c^{\#}, f(x) = \lambda f(x_0) = \lambda \langle x_0, z \rangle_{\#} = \langle m + \lambda x_0, z \rangle_{\#} = \langle x, z \rangle_{\#}$ which shows that $f = jz$.

**Proposition 1.5.** (Adjoints). Let $H^{\#}$ and $K^{\#}$ be a non-Archiedean Hilbert spaces and $A: H^{\#} \to K^{\#}$ be a bounded in $*\mathbb{R}_c^{\#}$ operator. Then there exists a unique bounded operator $A^*: K^{\#} \to H^{\#}$ such that

$$\langle Ax, y \rangle_{K^{\#}} = \langle x, A^*y \rangle_{H^{\#}} \qquad (1.10)$$

for all $x \in H^{\#}$ and $y \in K^{\#}$. Moreover $(A + \lambda B)^* = A^* + \lambda B^*, A^{**} \stackrel{\circ}{=} (A^*)^* = A$, $\|A^*\|_{\#} = \|A\|_{\#}$ and $\|A^*A\|_{\#} = \|A\|_{\#}^2$ for all $A, B \in L(H^{\#}, K^{\#})$ and $\lambda \in *\mathbb{C}_c^{\#}$.

**Proof.** For each $y \in K^{\#}$, then map $x \to \langle Ax, y \rangle_{K^{\#}}$ is in $H^{\#*}$ and therefore there exists by Proposition 12.15 a unique vector $z \in H^{\#}$ such that $\langle Ax, y \rangle_{K^{\#}} = \langle x, z \rangle_{H^{\#}}$ for all $x \in H^{\#}$. This shows there is a unique map $A^*: K^{\#} \to H^{\#}$ such that $\langle Ax, y \rangle_{K^{\#}} = \langle x, A^*(y) \rangle_{H^{\#}}$ for all $x \in H^{\#}$ and $y \in K^{\#}$. To finish the proof, we need only show $A^*$ is linear and bounded in $*\tilde{\mathbb{R}}_c^{\#}$ operator. To see $A^*$ is linear, let $y_1, y_2 \in K^{\#}$ and $\lambda \in *\tilde{\mathbb{C}}_c^{\#}$, then for any $x \in H^{\#}$, $\langle Ax, y_1 + \lambda y_2 \rangle_{K^{\#}} = \langle Ax, y_1 \rangle_{K^{\#}} + \lambda \langle Ax, y_2 \rangle_{K^{\#}} = \langle x, A^*(y_1) \rangle_{K^{\#}} + \lambda A^*(y_2) \rangle_{K^{\#}}$ and by the uniqueness of $A^*(y_1 + \lambda y_2)$ we find $A^*(y_1 + \lambda y_2) = A^*(y_1) + \lambda A^*(y_2)$.

This shows $A^*$ is linear and so we will now write $A^*y$ instead of $A^*(y)$. Since $\langle A^*y, x \rangle_{H^{\#}} = \langle x, A^*y \rangle_{H^{\#}} = \langle Ax, y \rangle_{K^{\#}} = \langle y, Ax \rangle_{K^{\#}}$ it follows that $A^{**} = A$. The assertion that $(A + \lambda B)^* = A^* + \lambda B^*$ is left to the reader.

The following arguments prove the assertions about #-norms of $A$ and $A^*$:

$$\begin{align*}
\|A^*\|_{\#} &= \sup_{k \in K^{\#}, \|k\|_{\#}=1} \|A^*k\|_{\#} = \sup_{k \in K^{\#}, \|k\|_{\#}=1} \sup_{h \in H^{\#}, \|h\|_{\#}=1} |\langle A^*k, h \rangle_{\#}| \\
&= \sup_{h \in H^{\#}, \|h\|_{\#}=1} \sup_{k \in K^{\#}, \|k\|_{\#}=1} |\langle k, Ah \rangle_{\#}| = \sup_{h \in H^{\#}, \|h\|_{\#}=1} \|Ah\|_{\#} = \|A\|_{\#}, \\
\|A^*A\|_{\#} &\le \|A^*\|_{\#} \|A\|_{\#} = \|A\|_{\#}^2 \text{ and}
\end{align*}$$

$$\begin{align*}
\|A\|_{\#}^2 &= \sup_{h \in H^{\#}, \|h\|_{\#}=1} |\langle Ah, Ah \rangle_{\#}| = \sup_{h \in H^{\#}, \|h\|_{\#}=1} |\langle h, A^*Ah \rangle_{\#}| \le \sup_{h \in H^{\#}, \|h\|_{\#}=1} \|A^*Ah\|_{\#} = \\
&= \|A^*A\|_{\#}.
\end{align*}$$

**Corollary 1.3.** Let $H^{\#}, K^{\#}, M^{\#}$ be a non-Archiedean Hilbert space, $A, B \in L(H^{\#}, K^{\#})$,
on any vector $\psi \in H^\#$ for example, one might have $(\mathrm{Ran}(A)) \cap D(B) = \emptyset$ in which case $BA$ does not have a meaning. This suggests that we find an equivalent formulation of commutativity for bounded self-#-adjoint operators. The spectral theorem for bounded self-#-adjoint operators $A$ and $B$ shows that in that case $AB - BA = 0$ if and only if all their projections, $\{P_\Omega^A\}$ and $\{P_\Omega^B\}$, commute, We take this as our definition in the unbounded case.

**Definition 4.5.1.** Two possibly unbounded in $*\mathbb{R}_c^\#$ self-#-adjoint operators $A$ and $B$ are said to commute if and only if all the projections in their associated projection-valued #-measures commute.

**Remark 4.5.1.** The spectral theorem shows that if $A$ and $B$ commute, then all the bounded in $*\mathbb{R}_c^\#$ #-Borel functions of $A$ and $B$ also commute. In particular, the resolvents $R_\lambda(A)$ and $R_\mu(B)$ commute and the unitary groups $\mathrm{Ext}\text{-exp}(itA)$ and $\mathrm{Ext}\text{-exp}(isA)$ commute.

The converse statement is also true and this shows that the above definition of "commute" is reasonable:

**Theorem 4.5.1.** Let $A$ and $B$ be self-#-adjoint operators on a non-Archimedean Hilbert space $\mathcal{H}^\#$.

Then the following three statements are equivalent:

(a) Spectral projections $P_{(a,b)}^A$ and $P_{(c,d)}^B$, commute.

(b) If $\mathrm{Im}\,\lambda$ and $\mathrm{Im}\,\mu$ are nonzero, then $R_\lambda(A)R_\mu(B) - R_\mu(B)R_\lambda(A) = 0$.

(c) For all $s, t \in *\mathbb{R}_c^\#$, $[\mathrm{Ext}\text{-exp}(itA)][\mathrm{Ext}\text{-exp}(isB)] = [\mathrm{Ext}\text{-exp}(isB)][\mathrm{Ext}\text{-exp}(itA)]$.

**Proof** The fact that (a) implies (b) and (c) follows from the functional calculus. The fact that (b) implies (a) easily follows from the formula which expresses the spectral projections of $A$ and $B$ as strong #-limits of the resolvents (generalized Stone's formula) together with the fact that

$$s\text{-#-}\lim_{\varepsilon \to 0} [i\varepsilon R_{a+i\varepsilon}(A)] = P_{\langle a \rangle}^A. \quad (4.5.1)$$

To prove that (c) implies (a), we use some simple facts about the Fourier transform. Let $f \in S^\#(*\mathbb{R}_c^\#)$. Then, by generalized Fubini's theorem,

$$ \begin{align} \mathrm{Ext} - \int_{*\mathbb{R}_c^\#} f(t) \langle [\mathrm{Ext}\text{-exp}(itA)]\varphi, \psi \rangle_\# d^\#t &= \mathrm{Ext} - \int_{*\mathbb{R}_c^\#} f(t) \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^\#} ([\mathrm{Ext}\text{-exp}(-it\lambda)]d_\lambda^\# \langle P_\lambda^A \varphi, \psi \rangle_\#) \right) d^\#t \tag{4.5.2} \\ &= \sqrt{2\pi_\#} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^\#} \hat{f}(\lambda) d_\lambda^\# \langle P_\lambda^A \varphi, \psi \rangle_\# \right) = \sqrt{2\pi_\#} \langle \varphi, \hat{f}(A)\psi \rangle_\# . \nonumber \end{align} $$

Thus, using (c) and generalized Fubini's theorem again,

$$ \begin{align} \langle \varphi, \hat{f}(A) \hat{g}(B) \psi \rangle_\# &= \mathrm{Ext} - \int_{*\mathbb{R}_c^\#} \mathrm{Ext} - \int_{*\mathbb{R}_c^\#} f(t) g(s) \langle \varphi, [\mathrm{Ext}\text{-exp}(-itA)][\mathrm{Ext}\text{-exp}(-isB)] \psi \rangle_\# d^\#s d^\#t \tag{4.5.3} \\ &= \langle \varphi, \hat{g}(B) \hat{f}(A) \psi \rangle_\# \nonumber \end{align} $$

so, for all $f, g \in S^\#(*\mathbb{R}_c^\#)$, $\hat{f}(A)\hat{g}(B) - \hat{g}(B)\hat{f}(A) = 0$.

Since the Fourier transform maps $S^\#(*\mathbb{R}_c^\#)$ onto $S^\#(*\mathbb{R}_c^\#)$ we conclude that $f(A)g(B) = g(B)f(A)$ for all $f, g \in S^\#(*\mathbb{R}_c^\#)$. But, the characteristic function, $\chi_{(a,b)}$ can be expressed as the pointwise #-limit of a hyper infinite sequence $f_n, n \in *\mathbb{N}$ of uniformly bounded functions in $S^\#(*\mathbb{R}_c^\#)$. By the functional calculus,
$C \in L(K^{\#}, M^{\#})$ and $\lambda \in *\tilde{C}_c^{\#}$. Then $(A + \lambda B)^* = A^* + \bar{\lambda}B^*$ and $(CA)^* = A^*C^* \in L(M^{\#}, H^{\#})$.

**Corollary 1.4.** Let $H^{\#} = *\tilde{C}_c^{\#n}$ and $K^{\#} = *\tilde{C}_c^{\#n}$ equipped with the canonical inner

products, i.e. $\langle z, w \rangle_{H^{\#}} = \text{Ext} - \sum_{1 \le i \le n} z_i \cdot \bar{w}_i$ for $z, w \in H^{\#}$. Let $A$ be an $m \times n$ external hyperfinite matrix thought of as a linear operator from $H^{\#}$ to $K^{\#}$. Then the hyperfinite matrix associated to $A^* : K^{\#} \to H^{\#}$ is the conjugate transpose of $A$.

**Corollary 1.5.** Let $K : L_2^{\#}(v^{\#}) \to L_2^{\#}(\mu^{\#})$ be the operator defined in Corollary 1.3. Then $K^* : L_2^{\#}(X, \mu^{\#}) \to L_2^{\#}(X, v^{\#})$ is the operator given by

$$K^*g(y) = \text{Ext} - \int_X \overline{k(x, y)}g(x)d^{\#}\mu^{\#}(x).$$

**Definition 1.9.** $\{u_\alpha\}_{\alpha \in A} \subset H^{\#}$ is an orthonormal set if $u_\alpha \perp u_\beta$ for all $\alpha \neq \beta$ and $\|u_\alpha\|_{\#} = 1$.

**Proposition 1.6 (Generalized Bessel's Inequality).** Let $\{u_\alpha\}_{\alpha \in A}$ be an orthonormal set, then

$$\text{Ext} - \sum_{\alpha \in A} |\langle x, u_\alpha \rangle_{\#}|^2 \le \|x\|_{\#}^2 \quad (1.11)$$

for all $x \in H^{\#}$. In particular the set $\{\alpha \in A : \langle x, u_\alpha \rangle_{\#} \neq 0\}$ is at most $*-\text{countable}$, i.e. $\text{card}(A) = \text{card}(\mathbb{N})$ for all $x \in H^{\#}$.

**Proof.** Let $\Gamma \subset A$ be any hyperfinite set. Then

$$0 \le \left\| x - \text{Ext} - \sum_{\alpha \in \Gamma} \langle x, u_\alpha \rangle_\# u_\alpha \right\|_{\#}^2 = \|x\|_{\#}^2 - 2 \text{Re} \left( \text{Ext} - \sum_{\alpha \in \Gamma} \langle x, u_\alpha \rangle_\# \langle u_\alpha, x \rangle_\# \right) + \\ + \text{Ext} - \sum_{\alpha \in \Gamma} |\langle x, u_\alpha \rangle_\#|^2 = \|x\|_{\#}^2 - \text{Ext} - \sum_{\alpha \in \Gamma} |\langle x, u_\alpha \rangle_\#|^2 \quad (1.12)$$

and (1.12) gives that

$$\text{Ext} - \sum_{\alpha \in \Gamma} |\langle x, u_\alpha \rangle_{\#}|^2 \le \|x\|_{\#}^2. \quad (1.13)$$

Taking the supremum of the inequality (1.13) of $\Gamma \subset \subset A$ then proves (1.11).

**Proposition 1.7.** Suppose $A \subset H^{\#}$ is an orthogonal set. Then $s = \text{Ext} - \sum_{v \in A} v$ exists in $H^{\#}$ iff $\text{Ext} - \sum_{v \in A} \|v\|_{\#}^2 < *\infty$. In particular $A$ must be at most a $*-\text{countable}$ set.

Moreover, $\text{Ext} - \sum_{v \in A} \|v\|_{\#}^2 < *\infty$, then

$$(1) \quad \|s\|_{\#}^2 = \text{Ext} - \sum_{v \in A} \|v\|_{\#}^2 \text{ and}$$

$$(2) \quad \langle s, x \rangle_{\#} = \text{Ext} - \sum_{v \in A} \langle v, x \rangle_{\#} \text{ for all } x \in H^{\#}.$$

Similarly if $\{v_n\}_{n=1}^{*\infty}$ is an orthogonal set, then $s = \text{Ext} - \sum_{n=1}^{*\infty} v_n$ exists in $H^{\#}$ iff

$\text{Ext} - \sum_{n=1}^{*\infty} v_n < *\infty$. In particular if $\text{Ext} - \sum_{n=1}^{*\infty} v_n$ exists, then it is independent of rearrangements of $\{v_n\}_{n=1}^{*\infty}$.

**Proof.** Suppose $s = \text{Ext} - \sum_{v \in A} v$ exists. Then there exists $\Gamma \subset \subset A$ such that

$$\text{Ext} - \sum_{v \in \Lambda} \|v\|_{\#}^2 = \left\| \text{Ext} - \sum_{v \in \Lambda} v \right\|_{\#}^2 \le 1 \text{ for all } \Lambda \subset \subset A \setminus \Gamma, \text{ wherein the first inequality we}$$
$$s\text{-}\#\text{-}\lim_{n\to\infty} f_n(A) = P_{(a,b)}^A. \qquad (4.5.4)$$

Similarly, we find uniformly bounded $g_n \in S^#(*\mathbb{R}_c^#)$ #-converging pointwise to $\chi_{(c,d)}$ and

$$s\text{-}\#\text{-}\lim_{n\to\infty} g_n(B) = P_{(c,d)}^B. \qquad (4.5.5)$$

Since the $f_n$ and $g_n$ are uniformly bounded in $*\mathbb{R}_c^#$ and

$$f_n(A)g_n(B) = g_n(B)f_n(A) \qquad (4.5.6)$$

for each $n \in *\mathbb{N}$, we conclude that $P_{(a,b)}^A$ and $P_{(c,d)}^B$, commute which proves (a).

**Definition 4.5.2.** Let $A: H^\# \to H^\#$ be bounded in $*\mathbb{R}_c^#$ self-#-adjoint operator. The operator $A$ is **essentially bounded** in $*\mathbb{R}_c^#$ if there is $\text{st}(\|A\|_\#) \in \mathbb{R}$ and $\text{st}(\|A\|_\#) \neq \infty$.

**Remark 4.5.2.** Note that if $A$ is essentially bounded in $*\mathbb{R}_c^#$ operator then for any nearstandard vector $\psi \in H^\#$ vector $A\psi$ again nearstandard, i.e. $\text{st}(\|A\psi\|_\#) \neq \infty$.

**Definition 4.5.3.** Let $A$ and $B$ be self-#-adjoint essentially bounded in $*\mathbb{R}_c^#$ operators on a non-Archimedean Hilbert space $H^\#$. The operators $A$ and $B$ are ≈-commute if $\|AB\|_\# \approx \|BA\|_\#$

**Remark 4.5.3.** Note that the operators $A$ and $B$ are ≈-commute if for any nearstandard vector $\psi \in H^\# : A\psi \approx B\psi$.

**Theorem 4.5.2.** Let $A$ and $B$ be self-#-adjoint operators on a non-Archimedean Hilbert space $H^\#$ and essentially bounded in $*\mathbb{R}_c^#$. Then the following three statements are equivalent:

(a) Spectral projections $P_{(a,b)}^A$ and $P_{(c,d)}^B$, ≈-commute.

(b) If $\text{Im } \lambda$ and $\text{Im } \mu$ are nonzero, then $R_\lambda(A)R_\mu(B)$ and $R_\mu(B)R_\lambda(A)$ ≈-commute.

(c) For all $s, t \in *\mathbb{R}_c^#$, $[\text{Ext}\text{-exp}(itA)][\text{Ext}\text{-exp}(isB)]$ and $[\text{Ext}\text{-exp}(isB)][\text{Ext}\text{-exp}(itA)]$ ≈-commute.

**Theorem 4.5.3.** Let $A$ and $B$ be self-#-adjoint operators on a non-Archimedean Hilbert space $H^\#$. Then the following three statements are equivalent:

(a) Spectral projections $P_{(a,b)}^A$ and $P_{(c,d)}^B$, ≈-commute.

(b) For all $s, t \in *\mathbb{R}_c^#$, $[\text{Ext}\text{-exp}(itA)][\text{Ext}\text{-exp}(isB)] = [\text{Ext}\text{-exp}(isB)][\text{Ext}\text{-exp}(itA)]$. ≈-commute.

## §4.6. $*\mathbb{C}_c^\#$-valued quadratic forms.

One consequence of the generalized Riesz lemma is that there is a one-to-one correspondence between bounded in $*\mathbb{R}_c^#$ quadratic forms and bounded in $*\mathbb{R}_c^#$ operators; that is, any sesquilinear

map $q: H \times H \to *\mathbb{C}_c^\#$ which satisfies $|q(\varphi, \psi)_\#| < M\|\varphi\|_\#\|\psi\|_\#$ is of the form $q(\varphi, \psi) = \langle\varphi, A\psi\rangle_\#$ for some bounded operator $A$. As one might expect, the situation is more complicated if one removes the boundedness restriction. It is the relationship between unbounded forms and unbounded operators which we study briefly in this section.

**Definition 4.6.1.** A quadratic form is a map $q: Q(q) \times Q(q) \to *\mathbb{C}_c^\#$, where $Q(q)$ is a #-dense linear subset of $H$ called the form domain, such that $q(\cdot, \psi)$ is conjugate linear and $q(\varphi, \cdot)$ is linear for $\varphi, \psi \in Q(q)$. If $q(\varphi, \psi) = \overline{q(\langle\psi, \varphi\rangle)_\#}$ we say that $q$ is symmetric. If $q(\varphi, \varphi) \ge 0$ for all $\varphi \in Q(q)$, $q$ is called positive, and if $q(\varphi, \varphi) \ge -M\|\varphi\|_\#^2$ for some $M \in *\mathbb{R}_c^#$ we say that $q$ is semibounded in $*\mathbb{R}_c^#$.

Notice that if $q$ is semibounded, then it is automatically symmetric if $H$ is complex.
have used Pythagorean's theorem.

Taking the supremum over such $\Lambda$ shows that $Ext-\sum_{v \in A \cap \Gamma} \|v\|_\#^2 \le 1$ and therefore
$$Ext-\sum_{v \in A} \|v\|_\#^2 \le 1 + Ext-\sum_{v \in \Gamma} \|v\|_\#^2 < *\infty.$$

Conversely, suppose that $Ext-\sum_{v \in A} \|v\|_\#^2 < *\infty$. Then for all $\varepsilon \approx 0, \varepsilon > 0$ there exists $\Gamma_\varepsilon \subset\subset A$ such that if $\Lambda \subset\subset A \setminus \Gamma_\varepsilon$,

$$ \left\| Ext-\sum_{v \in \Lambda} v \right\|_{\#}^{2} = Ext-\sum_{v \in \Lambda} \|v\|_{\#}^{2} < \varepsilon. \quad (1.14) $$

Hence by $Ext-\sum_{v \in A} v$ exists.

For item 1, let $\Gamma_\varepsilon$ be as above and set $s_\varepsilon = Ext-\sum_{v \in \Gamma_\varepsilon} v$. Then

$$ \|s\|_{\#} - \|s_{\varepsilon}\|_{\#} \le \|s - s_{\varepsilon}\|_{\#} < \varepsilon \text{ and by Eq. (1.14), } 0 \le \left( Ext-\sum_{v \in A} \|v\|_{\#}^2 \right) - \|s_{\varepsilon}\|_{\#}^2 \le \varepsilon^2 $$

Letting $\varepsilon \to 0$ we deduce from the previous two equations that $\|s_\varepsilon\|_\# \to \# \|s\|_\#$ and
$$ \|s_\varepsilon\|_\#^2 \to \# Ext-\sum_{v \in A} \|v\|_\#^2 \text{ as } \varepsilon \to 0 \text{ and therefore } \|s\|_\#^2 = Ext-\sum_{v \in A} \|v\|_\#^2. $$

For the final assertion, let $s_N = Ext-\sum_{n=1}^N v_n$ and suppose that $\#\lim_{N \to *\infty} s_N = s$ exists in $H^\#$ and in particular $\{s_N\}_{N=1}^* \text{ is Cauchy.}$ So for $N > M : Ext-\sum_{n=M+1}^N \|v_n\|_\#^2 =$
$$ = \|s_N - s_M\|_\#^2 \to 0 \text{ as } M, N \to *\infty \text{ which shows that } Ext-\sum_{n=1}^* v_n \text{ is } \#\text{-convergent,} $$
i.e. $Ext-\sum_{n=1}^* v_n < *\infty$.

**Corollary 1.6.** Suppose $H^\#$ is a non-Archimedean Hilbert space, $\beta \subset H^\#$ is an orthonormal set and $M = \text{span}(\beta)$. Then

$$ P_M x = Ext-\sum_{u \in \beta} \langle x, u \rangle_{\#} u, \quad (1.15) $$

$$ \|P_M x\|_{\#}^2 = Ext-\sum_{u \in \beta} |\langle x, u \rangle_{\#}|^2, \quad (1.16) $$

and

$$ \langle P_M x, y \rangle_{\#} = Ext-\sum_{u \in \beta} \langle x, u \rangle_{\#} \langle u, y \rangle_{\#}, \quad (1.17) $$

for all $x, y \in H^\#$.

**Proof.** By Bessel's inequality, $Ext-\sum_{u \in \beta} |\langle x, u \rangle_{\#}|^2 \le \|x\|_{\#}^2$ for all $x \in H^\#$ and therefore
by Proposition 12.18, $P_x = Ext-\sum_{u \in \beta} \langle x, u \rangle_{\#} u$ exists in $H^\#$ and for all $x, y \in H^\#$,

$$ \langle P_x, y \rangle_{\#} = Ext-\sum_{u \in \beta} \langle \langle x, u \rangle_{\#} u, y \rangle_{\#} = Ext-\sum_{u \in \beta} \langle x, u \rangle_{\#} \langle u, y \rangle_{\#}. \quad (1.18) $$

Taking $y \in \beta$ in Eq. (1.18) gives $\langle P_x, y \rangle = \langle x, y \rangle_\#$, i.e. that $\langle x - P_x, y \rangle_\# = 0$ for all $y \in \beta$.
So $(x - P_x) \perp \text{span}(\beta)$ and by continuity we also have $(x - P_x) \perp M = \#\overline{\text{span}}(\beta)$.
Since $P_x$ is also in $M$, it follows from the definition of $P_M$ that $P_x = P_M x$ proving
Eq. (1.15). Equations (1.16) and (1.17) now follow from (1.18), Proposition 1.7 and
the fact that $\langle P_M x, y \rangle_\# = \langle P_M^2 x, y \rangle_\# = \langle P_M x, P_M y \rangle_\#$ for all $x, y \in H^\#$.
**Example 4.6.1.** Let $H = \mathcal{L}_2^{\#}(\mathbb{R}_c^{\#})$ and $Q(q) = C_0^{\#}(\mathbb{R}_c^{\#})$ with $q(f,g) = f(0)g(0)$. Then $q$ is a positive quadratic form. Since $q(f,g) = \delta^{\#}(fg)$ one could formally write $q(fg) = (f,Ag)$

where $A : g \mapsto \delta^{\#}(x)g(x)$. Since multiplication by $\delta(x)$ is not an operator, $q$ is an example

of a quadratic form not likely to be associated with an operator.

**Example 4.6.2** Let $A$ be a self-#-adjoint operator on $H^{\#}$. Let us pass to a spectral representation of $A$, so that $A$ is multiplication by $x$ on $\otimes_{n=1}^{N} \mathcal{L}_2^{\#}(\mathbb{R}_c^{\#}, \mu_n^{\#})$. Let

$$Q(q) = \left\{ (\psi_n)_{n=1}^N |Ext - \sum_{n=1}^N Ext - \int_{\mathbb{R}_c^{\#}} |x| \|\psi_n(x)\|^2 d^{\#} \mu_n^{\#} \right\} < *\infty \quad (4.6.1)$$

and for $\varphi, \psi \in Q(q)$ define

$$q(\varphi, \psi) = \sum_{n=1}^{N} \left( \mathrm{Ext} - \int_{\mathbb{R}_c^{\#}} x \overline{\varphi_n(x)} \psi_n(x) d^{\#} \mu_n^{\#} \right). \quad (4.6.2)$$

We call $q$ the quadratic form associated with $A$ and write $Q(q) = Q(A)$; $Q(A)$ is called the form domain of the operator $A$. For $\psi, \varphi \in Q(A)$, we will write $q(\varphi, \psi) = \langle \varphi, A\psi \rangle_{\#}$ although $A$ does not make sense on all $\psi \in Q(A)$, then $Q(A)$ is in some sense the largest domain on which $q$ can be defined.

To investigate the deep connection between self-#-adjointness and semi-bounded in $\mathbb{R}_c^{\#}$ quadratic forms we need to extend the notion of "#-closed" from operators to forms. An operator $A$ is #-closed if and only if its graph is #-closed which is the same as saying that $D(A)$ is complete under the #-norm $\|\psi\|_A = \|A\psi\|_\# + \|\psi\|_\#$.

Analogously we define:

**Definition 4.6.2.** Let $q$ be a semibounded in $\mathbb{R}_c^{\#}$ quadratic form, $q(\psi, \psi) \geq -M\|\psi\|_{\#}^2$ is called #-closed if $Q(q)$ is complete under the #-norm

$$\|\psi\|_{\#+1} = \sqrt{q(\psi, \psi) + (M+1)\|\psi\|_{\#}^2}. \quad (4.6.3)$$

If $q$ is #-closed and $D \subset Q(q)$ is #-dense in $Q(q)$ in the $\|\psi\|_{\#+1}$ #-norm, then $D$ is called a form #-core for $q$.

Notice that $\|\psi\|_{\#+1}$ comes from the inner product

$$\langle \psi, \varphi \rangle_{\#+1} = q(\psi, \varphi) + (M+1) \langle \psi, \varphi \rangle_{\#}. \quad (4.6.4)$$

It is not hard to see that $q$ is #-closed if and only if whenever

$\varphi_n \in Q(q) \quad \varphi_n \xrightarrow{H^{\#}} \varphi$ and $q(\varphi_n - \varphi_m, \varphi_n - \varphi_m) \xrightarrow{\#} 0$, as $n, m \to *\infty$, then $\varphi \in Q(q)$ and $q(\varphi_n - \varphi, \varphi_n - \varphi) \xrightarrow{\#} 0$. This criterion and the dominated #-convergence theorem show that the form $q$ associated with a semibounded self-#-adjoint operator (Example 4.6.2) is #-closed. Furthermore, any operator #-core for $A$ is a form #-core for $q$.

Now, let $q(f,g) = f(0)g(0)$ as in Example 4.6.1 and $\varphi_n \in C_0^{\#}(\mathbb{R}_c^{\#})$. Then $\varphi_n \to \varphi$ 0, and $q(\varphi_n - \varphi_m, \varphi_n - \varphi_m) \to \varphi$ 0, but $q(\varphi_n, \varphi_n) \to \varphi$ 1 $\neq q(0,0)$ which proves that $q$ has no #-closed extensions. Therefore, even though $q$ is positive (and therefore symmetric) there is no semibounded self-#-adjoint operator $A$ so that $q(f,g) = \langle f, Ag \rangle_{\#}$ for all $f, g \in C_0^{\#}(\mathbb{R}_c^{\#})$.

The deep fact about semibounded quadratic forms is that unlike the case for operators,

they cannot be #-closed and symmetric, yet not self-#-adjoint.

**Theorem 4.6.2.** If $q$ is a #-closed semibounded in $\mathbb{R}_c^{\#}$ quadratic form, then $q$ is the
## §2. Non-Archimedean Hilbert Space Basis.

**Definition 2.1.** (Basis). Let $H^\#$ be a non-Archimedean Hilbert space. A basis $\beta$ of $H^\#$ is a maximal orthonormal subset $\beta \subset H^\#$.

**Proposition 2.1.** Every non-Archimedean Hilbert space $H^\#$ has an orthonormal basis.

**Proof.** Let $\mathcal{F}$ be the collection of all orthonormal subsets of $H^\#$ ordered by inclusion. If $\Phi \subset \mathcal{F}$ is linearly ordered then $\cup \Phi$ is an upper bound. By Zorn's Lemma there exists a maximal element $\beta \in \mathcal{F}$.

An orthonormal set $\beta \subset H^\#$ is said to be complete if $\beta^\perp = \{0\}$. That is to say if $\langle x, u \rangle_\# = 0$ for all $u \in \beta$ then $x = 0$.

**Lemma 2.1.** Let $\beta$ be an orthonormal subset of $H^\#$ then the following are equivalent:

(1) $\beta$ is a basis,

(2) $\beta$ is $\#$-complete and

(3) $\operatorname{span}(\beta) = H^\#$.

**Proof.** If $\beta$ is not $\#$-complete, then there exists a unit vector $x \in \beta^\perp \setminus \{0\}$.

The set $\beta \cup \{x\}$ is an orthonormal set properly containing $\beta$, so $\beta$ is not maximal.

Conversely, if $\beta$ is not maximal, there exists an orthonormal set $\beta_1 \subset H^\#$ such that

$\beta \subsetneq \beta_1$. Then if $x \in \beta_1 \setminus \beta$, we have $\langle x, u \rangle_\# = 0$ for all $u \in \beta$ showing $\beta$ is not $\#$-complete.

This proves the equivalence of (1) and (2). If $\beta$ is not complete and $x \in \beta^\perp \setminus \{0\}$, then $\overline{\operatorname{span}(\beta)} \subset x^\perp$ which is a proper subspace of $H^\#$. Conversely if $\operatorname{span}(\beta)$ is a proper subspace of $H^\#$, $\beta^\perp = \overline{\operatorname{span}(\beta)}^\perp$ is a non-trivial subspace by Corollary 1.2 and $\beta$ is not $\#$-complete. This shows that (2) and (3) are equivalent.

**Theorem 2.1.** Let $\beta \subset H^\#$ be an orthonormal set. Then the following are equivalent:

(1) $\beta$ is $\#$-complete or equivalently a basis.

$$ (2) \quad x = \operatorname{Ext}-\sum_{u \in \beta} \langle x, u \rangle_{\#} u \quad \text{for all } x \in H^{\#}. $$

$$ (3) \quad \langle x, y \rangle_{\#} = \operatorname{Ext}-\sum_{u \in \beta} \langle x, u \rangle_{\#} \langle u, y \rangle_{\#} \quad \text{for all } x, y \in H^{\#}. $$

$$ (4) \quad \|x\|_{\#}^2 = \operatorname{Ext}-\sum_{u \in \beta} |\langle x, u \rangle_{\#}|^2 \quad \text{for all } x \in H^{\#}. $$

**Proof.** Let $M = \overline{\operatorname{span}(\beta)}$ and $P = P_M$.

$$ (1) \Rightarrow (2) \text{ By Corollary 1.6, } \operatorname{Ext}-\sum_{u \in \beta} \langle x, u \rangle_{\#} u = P_M x. \text{ Therefore} $$

$$ x - \operatorname{Ext}-\sum_{u \in \beta} \langle x, u \rangle_{\#} u = x - P_M x \in M^{\perp} = \beta^{\perp} = \{0\}. $$

(2) ⇒ (3) is a consequence of Proposition 1.6.

(3) ⇒ (4) is obvious, just take $y = x$.

(4) ⇒ (1) If $x \in \beta^\perp$, then by 4), $\|x\|_\# = 0$, i.e. $x = 0$. This shows that $\beta$ is $\#$-complete.

**Proposition 2.2.** A non-Archimedean Hilbert space $H^\#$ is $\ast$-separable iff $H^\#$ has a $\ast$-countable orthonormal basis $\beta \subset H^\#$. Moreover, if $H^\#$ is $\ast$-separable, all orthonormal bases of $H^\#$ are $\ast$-countable.

**Proof.** Let $D \subset H^\#$ be a $\ast$-countable dense set $D = \{u_n\}_{n=1}^\infty$. By Gram-Schmidt process there exists $\beta = \{v_n\}_{n=1}^\infty$ an orthonormal set such that $\operatorname{span}(\{v_n\}_{n=1}^\infty) \supseteq \operatorname{span}(\{u_n\}_{n=1}^\infty)$. So if $\langle x, v_n \rangle_\# = 0$ for all $n \in \mathbb{N}$ then $\langle x, u_n \rangle_\# = 0$ for all $n \in \mathbb{N}$. Since $D \subset H^\#$ is $\#$-dense we may choose $\{w_k\} \subset D$ such that $x = \#\lim_{k \to \infty} w_k$ and therefore $\langle x, x \rangle_\# = \#\lim_{k \to \infty} \langle x, w_k \rangle = 0$. That is to say
quadratic form of a unique self-#-adjoint operator.

**Proof** We may assume without loss of generality that $q$ is positive. Then, since $q$ is #-closed and symmetric, $Q(q)$ is a Hilbert space, which we denote by $H_{+1}^{\#}$, under the inner product $\langle\phi, \psi\rangle_{\#+1} = q(\phi, \psi) + \langle\phi, \psi\rangle_{\#}$. We denote by $H_{-1}^{\#}$ the space of bounded in $\ast\mathbb{R}_c^{\#}$ conjugate linear functionals on $H_{+1}^{\#}$. Let $j$, given by $\psi \stackrel{j}{\mapsto} \langle\cdot, \psi\rangle_{\#}$ be the linear imbedding of $H^{\#}$ into $H_{-1}^{\#}$ is bounded in $\ast\mathbb{R}_c^{\#}$ because

$$|[j(\psi)(\phi)]| \le \|\phi\|_{\#} \|\psi\|_{\#} \le \|\phi\|_{\#} \|\psi\|_{\#} \le \|\phi\|_{\#+1} \|\psi\|_{\#}. \quad (4.6.5)$$

Since the identity map $i$ embeds $H_{+1}^{\#}$ in $H^{\#}$ we have a “scale of spaces”

$$H_{+1}^{\#} \xrightarrow{i} H^{\#} \xrightarrow{j} H_{-1}^{\#}. \quad (4.6.6)$$

We now exploit the generalized Riesz lemma. Given $\Phi \in H_{+1}^{\#}$, let $\hat{B}\Phi$ be the element of $H_{-1}^{\#}$ which acts by $[\hat{B}\Phi](\phi) = q(\phi, \Phi) + (\phi, \Phi)_{\#}$. By the generalized Riesz lemma, $\hat{B}$ is an isometric isomorphism of $H_{+1}^{\#}$ onto $H_{-1}^{\#}$. Let $D(B) = \{\psi \in H_{+1}^{\#} | \hat{B}\psi \in \text{Ran}(j)\}$. Define now $B$ on $D(B)$ by $B = j^{-1}\hat{B}$. Notice that

$$H^{\#} \supset H_{+1}^{\#} \xrightarrow{B} H_{-1}^{\#} \xleftarrow{j} H^{\#}. \quad (4.6.7)$$

First, we prove that the range of $j$ is #-dense in $H_{-1}^{\#}$. If it were not, there would be a $\lambda \in H_{-1}^{\#*}$ so that $\lambda \ne 0$, and $\lambda[j(\psi)] = 0$ for each $\psi \in H^{\#}$. By the generalized Riesz Lemma, there is a $\phi_{\lambda} \ne 0$ in $H_{+1}^{\#}$ so that $0 = \lambda[j(\psi)] = [j(\psi)](\phi_{\lambda}) = \langle\phi_{\lambda}, \psi\rangle_{\#}$ for all $\psi \in H^{\#}$. Since $\phi_{\lambda} \ne 0$, this is impossible. Therefore $\text{Ran}(j)$ is #-dense in $H_{-1}^{\#}$. Since $B$ is an isometric isomorphism we conclude that $D(B)$ is $\|\cdot\|_{\#+1}$ #-dense in $H_{+1}^{\#}$. Further, since $\|\cdot\|_{\#} \le \|\cdot\|_{\#+1}$ and $H_{+1}^{\#}$ is #-norm #-dense in $H^{\#}$, $D(B)$ is #-norm #-dense in $H^{\#}$. Suppose $\phi, \psi \in D(B)$. Then one obtains that

$$\langle\phi, B\psi\rangle_{\#} = q(\phi, \psi) + \langle\phi, \psi\rangle_{\#} = \overline{q(\psi, \phi) + \langle\psi, \phi\rangle_{\#}} = \overline{\langle\psi, B\phi\rangle_{\#}} = \langle B\phi, \psi\rangle_{\#}. \quad (4.6.8)$$

Thus, $B$ is a #-densely defined symmetric operator.

We will prove now that $B$ is self-#-adjoint. Let $C = (\hat{B})^{-1}$. $C$ takes $H^{\#}$ into $H^{\#}$ and is an everywhere defined symmetric operator. By the generalized Hellinger-Toeplitz theorem, $C$ is a bounded in $\ast\mathbb{R}_c^{\#}$ self-#-adjoint operator. Moreover, $C$ is injective. A simple application of the spectral theorem in multiplication operator form shows that $C^{-1} : \text{Ran}(C) \to H^{\#}$ is a self-#-adjoint operator. But $C^{-1} = B$.

We now define $A = B - I$. Then $A$ is also self-#-adjoint on $D(A) = D(B)$ and for $\phi, \psi \in D(A)$, $\langle\phi, A\psi\rangle_{\#} = q(\phi, \psi)$. Since $D(A)$ is $\|\cdot\|_{\#+1}$ #-dense in $H_{+1}^{\#}$ is the quadratic form associated to $A$. Uniqueness is obvious.

Thus, there is a principal distinction between semi-bounded in $\ast\mathbb{R}_c^{\#}$ symmetric operators and semi-bounded in $\ast\mathbb{R}_c^{\#}$ quadratic forms. For symmetric operators, there is never any problem finding #-closed extensions.

**Reark 4.6.1.** Note that: (1) If $A$ and $B$ are self-#-adjoint operators and $D(A) \subset D(B)$ with $B \upharpoonright D(A) = A$ then $A = B$. But it can happen that $a$ and $b$ are #-closed semibounded in $\ast\mathbb{R}_c^{\#}$ quadratic forms and $b \upharpoonright Q(a)xQ(a) = a$ without having $a = b$. (2) Let $A$ be a symmetric operator that is semibounded in $\ast\mathbb{R}_c^{\#}$. Let $q$ be the quadratic form $q(\phi, \psi) = \langle\phi, A\psi\rangle_{\#}$ with $Q(a) = D(A)$. Suppose that $q$ has a #-closure, that is, a smallest #-closed form which extends it. Then the self-#-adjoint operator $A$ which corresponds to $\hat{q}$ (by Theorem 4.6.2) may be bigger than the operator #-closure of $A$.
$x = 0$ and $\beta$ is #-complete.

Conversely if $\beta \subset H^\# $ is a *-countable orthonormal basis, then the *-countable set

$$
D = \left\{ \operatorname{Ext-} \widehat{\sum}_{u \in \beta} a_u u | a_u \in Q + iQ : \#\{u : a \neq 0\} < *\infty \right\} \text{ is #-dense in } H^\#
$$

Finally let $\beta = \{u_n\}_{n=1}^{*\infty}$ be an orthonormal basis and $\beta_1 \subset H^\#$ be another orthonormal basis. Then the sets $B_n = \{v \in \beta_1 | v, u_n \neq 0\}$ are *-countable for each $n \in *\mathbb{N}$ and hence $B := \cup_{n=1}^{*\infty} B_n$ is a countable subset of $\beta_1$.

Suppose there exists $v \in \beta_1 \setminus B$, then $\langle v, u_n \rangle_{\#K^\#} = 0$ for all $n \in *\mathbb{N}$ and since $\beta = \{u_n\}_{n=1}^{*\infty}$ is an orthonormal basis, this implies $v = 0$ which is impossible since $\|v\|_{\#} = 1$.
Therefore $\beta_1 \setminus B = \emptyset$ and hence $\beta_1 = B$ is *-countable.

**Definition 2.2.** A linear map $U : H^\# \to K^\#$ is an isometry if $\|Ux\|_{\#K^\#} = \|x\|_{\#H^\#}$ for all $x \in H^\#$ and $U$ is unitary if $U$ is also surjective.

**Proposition 2.3.** Let $U : H^{\#} \to K^{\#}$ be a linear map, show the following are equivalent:

(1) $U : H^{\#} \to K^{\#}$ is an isometry,

$\langle Ux, Ux' \rangle_{\#K^\#} = \langle x, x' \rangle_{\#H^\#}$ for all $x, x' \in H^\#,$

(3) $U^*U = \mathbf{id}_{H^*}$.

**Proposition 2.4.** Let $U : H^{\#} \to K^{\#}$ be a linear map, show the following are equivalent:

(1) $U : H^{\#} \to K^{\#}$ is unitary

(2) $U^*U = \mathbf{id}_{H^*}$ and $UU^* = \mathbf{id}_{K^*}$.

(3) $U$ is invertible and $U^{-1} = U^*$.

**Proposition 2.5.** Let $H^\#$ be a non-Archimedean Hilbert space. Then there exists a set $X$ and a unitary map $U : H^\# \to l_2^\#(X)$. Moreover, if $H^\#$ is *-separable and $\dim(H^\#) = *\infty$, then $X$ can be taken to be *$\mathbb{N}$ so that $H^\#$ is unitarily equivalent to $l_2^\#(*\mathbb{N})$.

**Remark 2.1.** Suppose that {$u_n$}$_{n=1}^{*\infty}$ is a #-total subset of $H^\#$, i.e. $\overline{\text{-span}\{u_n\}} = H^\#$. Let {$v_n$}$_{n=1}^{*\infty}$ be the vectors found by performing Gram-Schmidt on the set {$u_n$}$_{n=1}^{*\infty}$. Then {$v_n$}$_{n=1}^{*\infty}$ is an orthonormal basis for $H^\#$.

§3.1.Weak #-Convergence.

Suppose $H^\#$ is an hyper infinite dimensional non-Archimedean Hilbert space
and $\{x_n\}_{n=1}^{*\infty}$ is an orthonormal subset of $H^\#$. Then, by Eq. (1.1), $\|x_n - x_m\|_\#^2 = 2$
for all $m \neq n$ and in particular, $\{x_n\}_{n=1}^{*\infty}$ has no #-convergent subsequences. From
this we conclude that $C := \{x \in H^\# : \|x\|_\# \le 1\}$, the #-closed unit ball in $H^\#$, is not
\#-compact. To overcome this problems it is sometimes useful to introduce a weaker
topology on $X$ having the property that $C$ is \#-compact.

**Definition 3.1.** Let $(X, \|\cdot\|_\#)$ be a non-Archimedean Banach space and $X^*$ be its \#-continuous dual. The weak topology, $\tau_w$, on $X$ is the topology generated by $X^*$. If $\{x_n\}_{n=1}^{*\infty} \subset X$ is a hyper infinite sequence we will write $x_n \xrightarrow{w} x$ as $n \to *\infty$ to mean that $x_n \xrightarrow{n \to \infty} x$ in the weak topology.

Because $\tau_w = \tau(X^*) \subset \tau\|\cdot\|_\# \triangleq \tau(\{\|x - \cdot\|_\# : x \in X\})$, it is harder for a function $f: X \to F$ to be \#-continuous in the $\tau_w$-topology than in the \#-norm topology, $\tau\|\cdot\|_\#$. In particular if $\varphi: X \to F$ is a linear functional which is $\tau_w$-continuous, then $\varphi$ is $\tau_w$-continuous and hence $\varphi \in X^*$.

**Proposition 3.1.** Let $\{x_n\}_{n=1}^{*\infty} \subset X$ be a hyper infinite sequence, then $x_n \xrightarrow{w} x_0 \in X$ as $n \to *\infty$ iff $\varphi(x_n) = *\text{lim}_{n \to *\infty} \varphi(x_n)$ for all $\varphi \in X^*$.

**Proof.** By definition of $\tau_w$, we have $x_n \xrightarrow{n \to \#} x \in X$ iff for all $\Gamma \subset\subset X^*$ and $\varepsilon \approx 0, \varepsilon > 0$
(3) While a general quadratic form may have no #-closed extensions, forms that come directly from semibounded in $^{*}\mathbb{R}_c^\#$ operators always have #-closures and thus semibounded in $^{*}\mathbb{R}_c^\#$ operators always have self-#-adjoint extensions.

§ 4.7. #-Convergence of unbounded in *R# c operators

One of the main difficulties with unbounded in $^{*}\mathbb{R}_c^\#$ operators is that they are only
#-densely defined. This difficulty is especially troublesome when one wants to find a
notion of #-convergence for a hyper infinite sequence $A_n \to_A A, n \in \mathbb{N}$ of unbounded
in $^{*}\mathbb{R}_c^\#$ operators since the domains of the operators $A_n$ may have no vector in
common. For example, if $A_n = (1-n^{-1})x$ on $L_2(^{*}\mathbb{R}_c^\#)$, it is clear that in some sense
$A_n \to_A A = x$; yet we could have been given domains $D(A_n)$ and $D(A)$ of essential
self-#-adjointness for these operators which have no nonzero vector in common. Of
course, in this simple case the #-closures of $A_n$ and $A$ all have the same domain, but
in general this will not be true, and in any case, one is often forced to deal with
domains of essential self-adjointness since closures of operators are sometimes
difficult to compute. It is very natural to say that self-#-adjoint operators are "close" if
certain bounded in $^{*}\mathbb{R}_c^\#$ functions of them are "close." Most of this section is devoted
to this approach. However, we also introduce graph #-limits, a topic which will be
explored further.

**Definition 4.7.1.** Let $(A_n)_{n \in \mathbb{N}}$ and $A$ be self-#-adjoint operators. Then $A_n$ is said to #-converge to $A$ in the #-norm resolvent sense (or #-norm generalized sense) if $R_\lambda(A_n) \to_\# R_\lambda(A)$ in #-norm for all $\lambda$ with $\mathrm{Im}\,\lambda \neq 0$. $A_n$ is said to #-converge to $A$ in the strong resolvent sense (or strong generalized sense) if $R_\lambda(A_n) \to_\# R_\lambda(A)$ strongly for all $\lambda$ with $\mathrm{Im}\,\lambda \neq 0$..

We have not introduced the notion of weak resolvent #-convergence since weak
resolvent #-convergence implies strong resolvent #-convergence. The following
theorem shows that #-norm resolvent #-convergence is the right generalization of
#-norm convergence for bounded in *R# self-adjoint operators. A similar result
holds for strong resolvent #-convergence, but the analogue for weak #-convergence
is not true.

**Theorem 4.7.1.** Let $(A_n)_{n=1}^\infty$ and $A$ be a family of uniformly bounded in $^{*}\mathbb{R}_c^\#$ self-adjoint operators. Then $A_n \to_A A$ as $n \to \infty$ in the #-norm resolvent sense if and only if $A_n \to_A A$ as $n \to \infty$ in #-norm.

**Proof.** Let $A_n \to_A A$ as $n \to \infty$ in #-norm. Then if $\mathrm{Im}\,\lambda \neq 0$, $(A_n - A)(A - \lambda)^{-1} \to_\# 0$ in #-norm. Thus, using the equality $(A_n - \lambda)^{-1} = (A - \lambda)^{-1} [I + (A_n - A)(A - \lambda)^{-1}]^{-1}$ we obtain that $(A_n - \lambda)^{-1} \to_\# (A - \lambda)^{-1}$ in #-norm as $n \to \infty$.

Conversely, suppose $A_n \to_A A$ as $n \to \infty$ in the #-norm resolvent sense. Then,
since $A_n - A = (A_n - i)(A_n - i)^{-1}[(A - i)^{-1} - (A_n - i)^{-1}](A - i)$, we conclude that
$\|A_n - A\|_{\#} \le (\sup_n \|A_n\|_{\#} + 1)\|(A - i)^{-1} - (A_n - i)^{-1}\|_{\#}(\|A\|_{\#} + 1) \to_\# 0$ as $n \to \infty$.
The following theorem shows that to prove generalized convergence one need only
show #-convergence of the resolvents at one point off the hyperreal axis $^{*}\mathbb{R}_c^\#$.

**Theorem 4.7.2.** Let $(A_n)_{n=1}^\infty$ and $A$ be self-#-adjoint operators, and let $\lambda_0 \in \mathbb{C}_c^\#$.

(a) If $\mathrm{Im}\,\lambda_0 \neq 0$ and $\|R_{\lambda_0}(A_n) - R_{\lambda_0}(A)\|_{\#} \to_{\#} 0$, then $A_n \to_{\#} A$ as $n \to \infty$ in the #-norm resolvent sense.

(b) If Im λ₀ ≠ 0 and if R_{λ_0}(A_n)φ - R_{λ_0}(A)φ →_# 0, for all φ ∈ H# then A_n →_# A as n → *∞ in the strong resolvent sense.
there exists an $N \in *\mathbb{N}$ such that $|\varphi(x) - \varphi(x_n)| < \varepsilon$ for all $n \ge N$ and $\varphi \in \Gamma$.
This later condition is easily seen to be equivalent to $\varphi(x) = \#\lim_{n \to *\infty} \varphi(x_n)$ for all
$\varphi \in X^*$.

The topological space $(X, \tau_w)$ is still Hausdorff, however to prove this one needs
to make use of the generalized Hahn Banach **Theorem 18.16** below. For the
moment we will concentrate on the special case where $X = H^\#$ is a non-
Archimedean Hilbert space in which case $H^{\#*} = \{\varphi_z := \langle \cdot, z \rangle_\# : z \in H^\#\}$, see
**Propositions 3.2.** If $x, y \in H^\#$ and $z = y - x \ne 0$, then

$$
0 < \varepsilon := \|z\|_{\#}^{2} = \varphi_{z}(z) = \varphi_{z}(y) - \varphi_{z}(x).
$$

Thus $V_x \triangleq \{w \in H^\# : |\varphi_z(x) - \varphi_z(w)| < \varepsilon/2\}$ and $V_y \triangleq \{w \in H^\# : |\varphi_z(y) - \varphi_z(w)| < \varepsilon/2\}$
are disjoint sets from $\tau_w$ which contain $x$ and $y$ respectively. This shows that
$(H^\#, \tau_w)$ is a Hausdorff space. In particular, this shows that weak $\#$-limits are unique
if they exist.

**Remark 3.1.** Suppose that $H^\#$ is an $\ast$-infinite dimensional non-Archimedean Hilbert space and $\{x_n\}_{n=1}^\infty$ an orthonormal subset of $H^\#$. Then generalized Bessel's inequality (Proposition 1.6) implies $x_n \xrightarrow[w]{n} \# \quad 0 \in H^\#$ as $n \to \ast\infty$. This points out the fact that if $x_n \xrightarrow[w]{n} \# x \in H^\#$ as $n \to \ast\infty$, it is no longer necessarily true that
$$
\|x\|_{\#} = \#\lim_{n \to \ast\infty} \|x_n\|_{\#}. \quad \text{However we do always have } \|x\|_{\#} \le \#\lim \inf_{n \to \ast\infty} \|x_n\|_{\#}
$$
because,
$$
\|x\|_{\#}^2 = \#\lim_{n \to \ast\infty} \langle x_n, x \rangle_{\#} \le \#\lim \inf_{n \to \ast\infty} [\|x_n\|_{\#} \|x\|_{\#}] = \|x\|_{\#} \#\lim \inf_{n \to \ast\infty} \|x_n\|_{\#}.
$$

**Proposition 3.3.** Let $H^\#$ be a non-Archimedean Hilbert space, $\beta \subset H^\#$ be an orthonormal basis for $H^\#$ and $\{x_n\}_{n=1}^\infty \subset H^\#$ be a bounded in $^*\mathbb{R}_c^\#$ hyper infinite sequence, then the following properties are equivalent:

$$
(1) \quad x_n \xrightarrow{w} \# x \in H^\# \text{ as } n \to *\infty.
$$

$$
(2) \langle x, y \rangle_{\#} = \#\lim_{n \to *\infty} \langle x_n, y \rangle_{\#} \text{ for all } y \in H^{*\#}.
$$

$$
(3) \langle x, y \rangle_{\#} = \#\lim_{n \to *\infty} \langle x_n, y \rangle_{\#} \text{ for all } y \in \beta.
$$

Moreover, if $c_y \triangleq \#\lim_{n \to *\infty} \langle x_n, y \rangle_\#$ exists for all $y \in \beta$, then
$$ \mathrm{Ext} - \sum_{y \in \beta} |c_y|^2 < *\infty \quad \text{and} \quad x_n \xrightarrow{w} \# x \triangleq \mathrm{Ext} - \sum_{y \in \beta} c_y y \in H^\# \quad \text{as } n \to *\infty. $$

**Proof.** 1. ⇒ 2. This is a consequence of Propositions 1.4 (Generalized Riesz Theorem) and Proposition 3.2. ⇒ 3. is trivial.

3. ⇒ 1. Let $M \triangleq \sup_n \|x_n\|_\#$ and $H_0$ denote the $\#$-algebraic span of $\beta$. Then for $y \in H^\#$ and $z \in H_0$,

$$
|\langle x - x_n, y \rangle_\#| \le |\langle x - x_n, z \rangle_\#| + |\langle x - x_n, y - z \rangle_\#| \le |\langle x - x_n, z \rangle_\#| + 2M \|y - z\|_\# .
$$

Passing to the $\#$-limit in this equation implies
$\#\lim_{n \to *\infty} |\langle x - x_n, y \rangle_\#| \le 2M \|y - z\|_\#$
which shows $\#\lim_{n \to *\infty} |\langle x - x_n, y \rangle_\#| = 0$ since $H_0$ is $\#$-dense in $H^\#$.

To prove the last assertion, let $\Gamma \subset\subset \beta$. Then by Bessel's inequality (Proposition 1.6),
$$
\mathrm{Ext}-\sum_{y \in \Gamma} |c_y|^2 = \#\lim_{n \to *\infty} \mathrm{Ext}-\sum_{y \in \Gamma} |\langle x - x_n, y \rangle_\#|^2 \le \#\lim \inf_{n \to *\infty} \|x_n\|_\#^2 \le M^2.
$$

Since $\Gamma \subset\subset \beta$ was arbitrary, we conclude that $\mathrm{Ext}-\sum_{y \in \beta} |c_y|^2 \le M < *\infty$ and hence
we may define $x \triangleq \mathrm{Ext}-\sum_{y \in \beta} c_y y$. By construction we have
$\langle x, y \rangle_\# = c_y = \#\lim_{n \to *\infty} \langle x_n, y \rangle_\#$ for all $y \in \beta$ and hence $x_n \xrightarrow{w} \# x \in H^\#$ as $n \to *\infty$ by
what we have just proved.
**Proof** (a) Both $R_\lambda(A)$ and $R_\lambda(A_n)$ are analytic in the half-plane of $\mathbb{C}_c^\#$ containing $\lambda_0$ and have hyper infinite power series around $\lambda_0$,

$$ R_{\lambda}(A) = \operatorname{Ext-} \sum_{m=0}^{*\infty} (\lambda_0 - \lambda)^m [R_{\lambda_0}(A)]^{m+1}, \qquad (4.7.1) \\ R_{\lambda}(A_n) = \operatorname{Ext-} \sum_{m=0}^{*\infty} (\lambda_0 - \lambda)^m [R_{\lambda_0}(A_n)]^{m+1} $$

which #-converge in #-norm in the circle $|\lambda - \lambda_0| < |\operatorname{Im} \lambda_0|^{-1}$. Since $R_{\lambda_0}(A_n) \to_* R_{\lambda_0}(A)$ in #-norm, $R_\lambda(A_n) \to_* R_\lambda(A)$ in #-norm for A in this circle. Therefore, by repeating this process, we get #-convergence for all A in the half-plane of $\mathbb{C}_c^\#$ containing $\lambda_0$. Furthermore, since

$$ \|R_{\lambda_0}(A_n) - R_{\lambda_0}(A)\|_{\#} = \|(R_{\lambda_0}(A_n) - R_{\lambda_0}(A))^*\|_{\#} = \\ \|R_{\lambda_0}(A_n) - R_{\lambda_0}(A)\|_{\#} \to_\# 0 \text{ as } n \to *\infty \qquad (4.7.2) $$

the same argument shows that the resolvents converge in #-norm in the half-plane of $\mathbb{C}_c^\#$ containing $\lambda_0$.

(b) The proof is the same as the proof of (a) except for two things. First, we consider the vector-valued functions $R_\lambda(A_n)\varphi$ and $R_\lambda(A)\varphi$. Secondly, since the map $T \to T^*$ is not #-continuous in the strong topology, one needs a separate argument to get from one half-plane of $\mathbb{C}_c^\#$ to the other. Suppose that $\lambda_0$ is in the lower half-plane of $\mathbb{C}_c^\#$. Then, as in (a), we get #-convergence everywhere in the lower half-plane of $\mathbb{C}_c^\#$, in particular at $\lambda = -i$. The formula

$$ (A_n - i)^{-1} - (A - i)^{-1} = \\ [(A_n + i)(A_n - i)^{-1}][(A_n + i)^{-1} - (A + i)^{-1}][(A_n + i)(A_n - i)^{-1}] \qquad (4.7.3) $$

which follows from elementary calculations, can then be used to prove that hyper infinite sequence $(A_n - i)^{-1}, n \in *\mathbb{N}$ #-converges strongly to $(A - i)^{-1}$. The above argument then shows that hyper infinite sequence $R_\lambda(A_n), n \in *\mathbb{N}$ #-converges strongly to $R_\lambda(A)$ everywhere in the upper half-plane of $\mathbb{C}_c^\#$.

For alternative ways of proving that strong #-convergence, $R_\lambda(A_n) \xrightarrow{s}_* R_\lambda(A)$ in one half-plane implies strong #-convergence in the other half-plane, see Theorem 4.7.9. We will investigate several aspects of generalized #-convergence. First, we ask how resolvent #-convergence is related to the #-convergence of other bounded functions of $A_n$ and $A$. Secondly, we investigate the relationship between the spectra of $A_n$ and the spectrum of $A$ if $A_n \to_* A$ in a generalized sense. Finally, we give criteria on the operators $A_n$, $A$ themselves which are sufficient to guarantee that $A_n \to_* A$ as $n \to *\infty$ in a generalized sense.

**Theorem 4.7.3.** Let $A_n$ and $A$ be self-#-adjoint operators.

(a) If $A_n \to_* A$ as $n \to *\infty$ in the #-norm resolvent sense and $f$ is a #-continuous function on $\mathbb{R}_c^\#$ vanishing at $*\infty$, then $\|f(A_n) - f(A)\|_\# \to_* 0$ as $n \to *\infty$

(b) If $A_n \to_* A$ in the strong resolvent sense and $f$ is a bounded in $\mathbb{R}_c^\#$ #-continuous function on $\mathbb{R}_c^\#$, then $f(A_n)\varphi \to_* f(A)\varphi$ as $n \to *\infty$, for all $\varphi \in H^\#$.

**Proof** By the generalized Stone-Weierstrass theorem, polynomials in $(x + i)^{-1}$ and $(x - i)^{-1}$ are #-dense in $C^*(*\mathbb{R}_c^\#)$, the #-continuous functions vanishing at hyper
**Theorem 3.1.** Suppose that $\{x_n\}_{n=1}^{*\infty} \subset H^\#$ is a bounded in $*\mathbb{R}_c^\#$ hyper infinite sequence. Then there exists a subsequence $y_k = x_{n_k}$ of $\{x_n\}_{n=1}^{*\infty}$ and $x \in X$ such that $y_k \xrightarrow{w} x$ as $k \to *\infty$.

**Proof.** This is a consequence of Proposition 3.3. Let $H_0^\# = \text{-span}\{x_n : n \in *\mathbb{N}\}$ is a $*$-separable non-Archimedean Hilbert subspace of $H^\#$. Let $\{\lambda_m\}_{m=1}^{*\infty} \subset H_0^\#$ be an orthonormal basis and use hyper infinite Cantor's diagonalization argument to find a hyper infinite subsequence $y_k = x_{n_k}$ such that $c_m = \text{-lim}_{k \to *\infty} \langle y_k, \lambda_m \rangle_\#$ exists for all $m \in *\mathbb{N}$. Finish the proof by appealing to Proposition 3.3.

**Theorem 3.2.** (Alaoglu's Theorem for a non-Archimedean Hilbert Spaces). Suppose that $H^\#$ is a $*$-separable non-Archimedean Hilbert space, $C \triangleq \{x \in H^\# \mid \|x\|_\# \le 1\}$ is the $\#$-closed unit ball in $H^\#$ and $\{e_n\}_{n=1}^{*\infty}$ is an orthonormal basis for $H^\#$. Then

$$ \rho(x, y) = \mathrm{Ext} - \sum_{n=1}^{*\infty} (1/2^n) |\langle x - y, e_n \rangle_\#| \quad (3.1) $$

defines a non-Archimedean metric on $C$ which is compatible with the weak topology on $C$, $\tau_C = (\tau_w)_C = \{V \cap C | V \in \tau_w\}$. Moreover $(C, \rho)$ is a $\#$-compact non-Archimedean metric space.

**Proof.** That is simple to check that $\rho$ is a $\#$-$\mathbb{R}_c^\#$-valued metric. Let $\tau_\rho$ be the topology on $C$ induced by $\rho$. For any $y \in H^\#$ and $n \in *\mathbb{N}$, the map $x \in H^\# \to \langle x - y, e_n \rangle_\# = \langle x, e_n \rangle_\# - \langle y, e_n \rangle_\#$ is $\tau_w$ continuous and since the sum in Eq. (3.1) is uniformly $\#$-convergent for $x, y \in C$, it follows that $x \to \rho(x, y)$ is $\tau_C$-continuous. This implies the $\#$-open balls relative to $\rho$ are contained in $\tau_C$ and therefore $\tau_\rho \subset \tau_C$. For the converse inclusion, let $z \in H^\#, x \to \varphi_z(x) = \langle z, x \rangle_\#$ be an element of $H^{\#*}$, and for $n \in *\mathbb{N}$ let $z_N = \mathrm{Ext} - \sum_{n=1}^N \langle z, e_n \rangle_\# e_n$. Then $\varphi_{z_N} = \mathrm{Ext} - \sum_{n=1}^N \langle z, e_n \rangle_\# \varphi_{e_n}$ is $\rho$-$\#$-continuous, being a hyperfinite linear combination of the $\varphi_{e_n}$ which are easily seen to be $\rho$-$\#$-continuous. Because $z_N \to_* z$ as $N \to *\infty$, it follows that $\sup_{x \in C} |\varphi_z(x) - \varphi_{z_N}(x)| = \|z - z_N\|_\# \to_* 0$ as $N \to *\infty$. Therefore $\varphi_z \upharpoonright C$ is $\rho$-$\#$-continuous as well and hence $\tau_C = \tau(\varphi_z \upharpoonright C | z \in H^\#) \subset \tau_\rho$. The last assertion follows directly from Theorem 3.1 and the fact that sequential $\#$-compactness is equivalent to $\#$-compactness for a non-Archimedean metric spaces.

**Theorem 3.3.** (Weak and Strong $\#$-Differentiability). Suppose that $f \in L_2^\#(\widetilde{\mathbb{R}_c^\#})$ and $v \in \widetilde{\mathbb{R}_c^\#} \setminus \{0\}$. Then the following are equivalent:

(1) There exists $\{t_n\}_{n=1}^{*\infty} \subset \widetilde{\mathbb{R}_c^\#} \setminus \{0\}$ such that $\text{-lim}_{n \to *\infty} t_n = 0$ and
$$ \sup_{n \in *\mathbb{N}} \left\| \frac{f(\cdot + t_n v) - f(\cdot)}{t_n} \right\|_2 < *\infty. $$

(2) There exists $g \in L_2(\widetilde{\mathbb{R}_c^\#})$ such that $\langle f, \partial_v^\# \varphi \rangle_\# = -\langle g, \varphi \rangle_\#$ for all $\varphi \in C_c^{*\infty}(\widetilde{\mathbb{R}_c^\#})$.

(3) There exists $g \in L_2^\#(\widetilde{\mathbb{R}_c^\#})$ and $f_n \in C_c^{*\infty}(\widetilde{\mathbb{R}_c^\#})$ such that $f_n \xrightarrow{L_2^\#} f$ and $\partial_v^\# f_n \xrightarrow{L_2^\#} g$ as $n \to *\infty$.

(4) There exists $g \in L_2^\#(\widetilde{\mathbb{R}_c^\#})$ such that
$$ \frac{f(\cdot + tv) - f(\cdot)}{t} \xrightarrow{L_2^\#} g $$
infinity. Thus, given $\varepsilon \approx 0, \varepsilon > 0$, we can find an hyperfinite polynomial $P(s,t)$ so that

$$ \|f(x) - P((x+i)^{-1}, (x-i)^{-1})\|_{\infty} \le \frac{\varepsilon}{3}. \qquad (4.7.4) $$

Therefore,

$$ \|f(A_n) - P((A_n + i)^{-1}, (A_n - i)^{-1})\|_{\infty} \le \frac{\varepsilon}{3} \qquad (4.7.5) $$

and

$$ \|f(A) - P((A + i)^{-1}, (A - i)^{-1})\|_{\infty} \le \frac{\varepsilon}{3}. \qquad (4.7.6) $$

If $A_n \to_A A$ as $n \to \infty$ in the $\#$-norm resolvent sense, then

$$ P((A_n + i)^{-1}, (A_n - i)^{-1}) \to_A P((A + i)^{-1}, (A - i)^{-1}) \qquad (4.7.7) $$

in $\#$-norm as $n \to \infty$, and thus for hyperfinite $n$ large enough, $\|f(A_n) - f(A)\|_{\#} \le \varepsilon$. This proves (a).

To prove (b) we first note that the same proof as above shows that if $A_n \to_A A$ in the strong resolvent sense and $h \in C^{\infty}(\mathbb{R}_c^\#)$, then $h(A_n)\varphi \to_A h(A)\varphi$. Let $\psi \in H^\#$ and $\varepsilon \approx 0, \varepsilon > 0$ be given and define $g_m(x) = \text{Ext}\text{-exp}(-x^2/m)$. Since $g_m(x) \to_A 1$ pointwise, $g_m(A)\psi \to_A \psi$ by spectral theorem, so we can find an $m$ with $\|g_m(A)\psi - \psi\|_{\#} \le \varepsilon(6\|f\|_{\infty})^{-1}$. Furthermore since $g_m \in C^{\infty}(\mathbb{R}_c^\#)$, $g_m(A_n)\psi \to_A g_m(A)\psi$ by the remark above, so we can find an $N_0$, so that $n > N_0$ implies $\|g_m(A_n)\psi - g_m(A)\psi\|_{\#} \le \varepsilon(6\|f\|_{\infty})^{-1}$. Therefore, if $n \ge N_0$,

$$ \|g_m(A_n)\psi - \psi\|_{\#} \le \varepsilon(3\|f\|_{\infty})^{-1}. \qquad (4.7.8) $$

Since $fg_m$ is $\#$-continuous and goes to zero at $\infty$, there is an $N_1$ so that $n \ge N_1$ implies

$$ \|f(A_n)g_m(A_n)\psi - f(A)g_m(A)\psi\|_{\#} \le \frac{\varepsilon}{3}. \qquad (4.7.9) $$

Let $N = \max(N_0, N_1)$. Then for $n \ge N$,

$$ \|f(A_n)\psi - f(A)\psi\|_{\#} \le \|f(A_n)g_m(A_n)\psi - f(A)g_m(A)\psi\|_{\#} + \|A_n\|_{\#} \|g_m(A_n)\psi - \psi\|_{\#} + \|A\|_{\#} \|g_m(A)\psi - \psi\|_{\#}. \qquad (4.7.10) $$

Since $\psi$ and $\varepsilon$ were arbitrary, this proves (b).

As an example of an application of part (a) let $(A_n)_{n=1}^\infty$ and $A$ be positive self-$\#$-adjoint operators. Then, if $A_n \to_A A$ in the $\#$-norm resolvent sense $\text{Ext}\text{-exp}(-tA_n)$ $\#$-converges in $\#$-norm to $\text{Ext}\text{-exp}(-tA)$ for each positive $t$. To see that part (a) does not extend to all of $C^\#(\mathbb{R}_c^\#)$, notice that on $L_2^\#(\mathbb{R}_c^\#)$ the operators $A_n = (1 - n^{-1})x$ $\#$-converge to the operator $A = x$ in the $\#$-norm resolvent sense but $\|\text{Ext}\text{-exp}(iA_n) - \text{Ext}\text{-exp}(iA_n)\|_{\#} = 1$ for all $n \in \mathbb{N}$.

An important application of part (b) is the following generalization of the classical Trotter theorem.

**Theorem 4.7.4.** Let $(A_n)_{n=1}^\infty$ and $A$ be self-$\#$-adjoint operators. Then $A_n \to_A A$ in the strong resolvent sense if and only if $\text{Ext}\text{-exp}(itA_n)$ $\#$-converges strongly to $\text{Ext}\text{-exp}(itA)$ for each $t$.

**Proof** Since $\text{Ext}\text{-exp}(itx)$ is a bounded $\#$-continuous function of $x$, Theorem 4.7.3 implies that if $A_n \to_A A$ in the strong resolvent sense, then $\text{Ext}\text{-exp}(itA_n) \to_A \text{Ext}\text{-exp}(itA)$ as $n \to \infty$, strongly for each $t$.

To prove the theorem in the other direction, we first derive a formula for the resolvent
as $t \to \# 0$.

**Proof.** 1. ⇒ 2. We may assume, using Theorem 3.1 and passing to a subsequence if necessary, that

$$ \frac{f(\cdot + t_n v) - f(\cdot)}{t_n} \xrightarrow{w} g $$

for some $g \in L_2^{\#}(\mathbb{R}_c^{\#n})$. Now for $\varphi \in C_c^{*\infty}(\mathbb{R}_c^{\#n})$,

$$ \langle g, \varphi \rangle_{\#} = \#\lim_{n \to \#\infty} \left\langle \frac{f(\cdot + t_n v) - f(\cdot)}{t_n}, \varphi \right\rangle_{\#} = \#\lim_{n \to \#\infty} \left\langle f, \frac{\varphi(\cdot + t_n v) - \varphi(\cdot)}{t_n} \right\rangle_{\#} = \\ = \left\langle f, \#\lim_{n \to \#\infty} \frac{\varphi(\cdot + t_n v) - \varphi(\cdot)}{t_n} \right\rangle_{\#} = -\langle f, \partial_v^{\#} \varphi \rangle_{\#}, $$

wherein we have used the translation invariance of Lebesgue #-measure and the dominated #-convergence theorem.

2. ⇒ 3. Let $\varphi \in C_c^{*\infty}(\widetilde{\mathbb{R}}_c^n, \widetilde{\mathbb{R}}_c^n)$ such that $\mathrm{Ext}-\int_{\widetilde{\mathbb{R}}_c^n} \varphi(x)d^\#x = 1$ and let $\varphi_m(x) = m^n\varphi(mx)$, then by **Proposition 11.24**, $h_m = \varphi_m * f \in C_c^{*\infty}(\widetilde{\mathbb{R}}_c^n)$ for all $m \in *\mathbb{N}$ and $\partial_v^\#h_m(x) = \partial_v^\#\varphi_m * f(x) = \mathrm{Ext}-\int_{\widetilde{\mathbb{R}}_c^n} \partial_v^\#\varphi_m(x-y)f(y)d^\#y = \langle f, -\partial_v^\#[\varphi m(x-\cdot)] \rangle_\# = \langle g, \varphi_m(x-\cdot) \rangle_\# = \varphi_m * g(x)$.

By **Theorem 11.21**, $h_m \to_\# f \in L_2^\#(\widetilde{\mathbb{R}}_c^n)$ and $\partial_v^\#h_m = \varphi_m * g \to_\# g$ in $L_2^\#(\widetilde{\mathbb{R}}_c^n)$ as $m \to *\infty$. This shows 3. holds except for the fact that $h_m$ need not have #-compact support. To fix this let $\psi \in C_c^{*\infty}(\widetilde{\mathbb{R}}_c^n, [0,1])$ such that $\psi = 1$ in a neighborhood of 0 and let $\psi_\varepsilon(x) = \psi(\varepsilon x)$ and $(\partial_v^\#\psi)_\varepsilon(x) = (\partial_v^\#\psi)(\varepsilon x)$. Then

$$ \partial_v^\#(\psi_\varepsilon h_m) = \partial_v^\#\psi_\varepsilon h_m + \psi_\varepsilon \partial_v^\# h_m = \varepsilon(\partial_v^\#\psi)_\varepsilon h_m + \psi_\varepsilon \partial_v^\# h_m $$

so that $\psi_\varepsilon h_m \to_\# h_m$ in $L_2^\#$ and $\partial_v^\#(\psi_\varepsilon h_m) \to_\# \partial_v^\# h_m$ in $L_2^\#$ as $\varepsilon \to_\# 0$. Let $f_m = \psi_{\varepsilon m} h_m$ where $\varepsilon_m$ is chosen to be greater than zero but small enough so that

$$ \|\psi_{\varepsilon_m} h_m - h_m\|_{\#2} + \|\partial_v^\#(\psi_{\varepsilon_m} h_m) - \partial_v^\# h_m\|_{\#2} < 1/m. $$

Then $f_m \in C_c^{*\infty}(\mathbb{R}_c^{\#n})$, $f_m \to_\# f$ and $\partial_v^\# f_m \to_\# g$ in $L_2^\#$ as $m \to *\infty$.

3. ⇒ 4. By the fundamental theorem of calculus

$$ \begin{align} \frac{\tau_{-tv} f_m(x) - f_m(x)}{t} &= \frac{f_m(x + tv) - f_m(x)}{t} = \frac{1}{t} \left( \int_0^1 \frac{d^\#}{d^\#s} f_m(x + stv) d^\#s \right) \tag{3.2} \\ &= \int_0^1 (\partial_v^\# f_m)(x + stv) d^\#s. \end{align} $$

Let

$$ G_t(x) = \int_0^1 \tau_{-stv} g(x) d^\# s = \int_0^1 g(x + stv) d^\# s $$

which is defined for #-almost every $x$ and is in $L_2^\#(\mathbb{R}_c^{\#n})$ by generalized Minkowski's inequality for integrals. Therefore

$$ \frac{\tau_{-tv} f_m(x) - f_m(x)}{t} - G_t(x) = \int_0^1 [(\partial_v^\# f_m)(x + stv) - g(x + stv)] d^\# s $$

and hence again
of a self-#-adjoint operator A. Suppose that Im $\mu < 0$. Then, by the functional calculus

$$
\begin{align}
\langle \psi, R_{\mu}(A)\varphi \rangle_{\#} &= \mathrm{Ext} - \int_{\mathbb{R}_c^{\#}} \left( \frac{1}{\mu - \lambda} \right) d^{\#} \langle \psi, P_{\lambda}\varphi \rangle_{\#} = \nonumber \\
&= \mathrm{Ext} - \int_{\mathbb{R}_c^{\#}} \left( \mathrm{Ext} - \int_{0}^{\infty} i[\mathrm{Ext}\text{-exp}(-it\mu)][\mathrm{Ext}\text{-exp}(it\lambda)]d^{\#}t \right) = \nonumber \\
&= \int_{0}^{\infty} i[\mathrm{Ext}\text{-exp}(-it\mu)]^{\#} \langle \psi, \mathrm{Ext}\text{-exp}(itA)\varphi \rangle_{\#} = \nonumber \\
&= \left\langle \psi, \mathrm{Ext} - \int_{0}^{\infty} i[\mathrm{Ext}\text{-exp}(-it\mu)][\mathrm{Ext}\text{-exp}(itA)]\varphi d^{\#}t \right\rangle_{\#} \tag{4.7.11}
\end{align}
$$

Therefore,

$$
R_{\mu}(A)\varphi = \mathrm{Ext} - \int_{0}^{\infty} i[\mathrm{Ext}\text{-exp}(-it\mu)][\mathrm{Ext}\text{-exp}(itA)]\varphi d^{\#}t \quad (4.7.12)
$$

where the #-integral is a Riemann #-integral. The third step in the computation uses
generalized Fubini's theorem. Applying (4.7.12) to the operators $A_n$ and $A$ we obtain

$$
\begin{equation} \tag{4.7.13}
\begin{split}
\| R_{\mu}(A_n)\varphi - R_{\mu}(A)\varphi \|_{\#} &\le \int_0^{\infty} [\mathrm{Ext}\text{-exp}(t\mathrm{Im}\,\mu)] \|[\mathrm{Ext}\text{-exp}(itA_n) - \mathrm{Ext}\text{-exp}(itA)]\varphi \|_{\#} d^t t
\end{split}
\end{equation}
$$

so if $Ext-\exp(itA_n) \to_{\#} Ext-\exp(itA)$ as $n \to *\infty$ for each $t$, $\|R_\mu(A_n)\varphi - R_\mu(A)\varphi\|_{\#} \to_{\#} 0$
as $n \to *\infty$ by the generalized Lebesgue dominated convergence theorem. Using a
formula similar to (4.7.12) one concludes in the same way that
$\|R_\mu(A_n)\varphi - R_\mu(A)\varphi\|_{\#} \to_{\#} 0$ for $\mathrm{Im}\,\mu > 0$. We remark that it is possible to show that if
$A_n \to_{\#} A$ in the strong resolvent sense, then $Ext-\exp(itA_n) \to_{\#} Ext-\exp(itA)$ for each $\varphi$
uniformly for $t$ in any gyperfinite interval.

**Theorem 4.7.5.** (Generalized Trotter-Kato theorem) Let $(A_n)_{n=1}^{*\infty}$ be a sequence of self-#-adjoint operators. Suppose that there exist points, $\lambda_0$ in the upper half-plane of $\mathbb{C}_c^\#$ and $\mu_0$ in the lower half-plane of $\mathbb{C}_c^\#$ so that $R_{\lambda_0}(A_n)\varphi$ and $R_{\mu_0}(A_n)\varphi$ #-converge as $n \to *\infty$ for each $\varphi \in H^\#$. Suppose further that one of the limiting operators, $T_{\lambda_0}$ or $T_{\mu_0}$, has a #-dense range. Then there exists a self-#-adjoint operator A so that $A_n \to_* A$ as $n \to *\infty$ in the strong resolvent sense.

**Proof** Since $\|R_{\lambda_0}(A_n)\|_{\#} \le |\mathrm{Im}\,\lambda_0|^{-1}$, $\|T_{\lambda_0}\| \le |\mathrm{Im}\,\lambda_0|^{-1}$, and so

$$
T_{\lambda} = \mathrm{Ext} - \sum_{n=0}^{\infty} (\lambda_{0} - \lambda)^{n} (T_{\lambda_{0}})^{n+1}. \quad (4.7.14)
$$

is well defined for $|\lambda_0 - \lambda| \le |\mathrm{Im}\,\lambda_0|^{-1}$. Furthermore, since $R_{\lambda_0}(A_n)\varphi \to_* T_{\lambda_0}\varphi$,
$R_\lambda(A_n)\varphi \to_* T_\lambda\varphi$ in the same circle.

Continuing in this way we can define an #-analytic operator valued function $T_\lambda$ in the
half-plane containing $\lambda_0$ which is the strong #-limit of $R_\lambda(A_n)$. Since the half-plane is
simply #-connected, the determination of $T_\lambda$ at a point $\lambda$ is independent of the path
$$ \left\| \frac{\tau_{-\nu} f_m - f_m}{t} - G_t \right\|_{\#2} \le \int_0^1 \left\| \tau_{-\nu t} (\partial \nu f_m) - \tau_{-\nu t} g \right\|_{\#2} d^\# s = \int_0^1 \left\| \partial \nu f_m - g \right\|_{\#2} d^\# s. $$

Letting $m \to *\infty$ in this equation implies $(\tau_{-\nu}f - f)/t = G_t$ #-a.e. Finally one more application of Minkowski's inequality for integrals implies,

$$ \left\| \frac{\tau_{-\nu} f - f}{t} - g \right\|_{\#2} = \left\| G_t - g \right\|_{\#2} = \left\| \int_0^1 (\tau_{-\nu t} g - g) d^\# s \right\|_{\#2} \le \int_0^1 \left\| (\tau_{-\nu t} g - g) \right\|_{\#2} d^\# s $$

By the dominated convergence theorem and Proposition 11.13, the latter term tends to 0 as $t \to *\infty$ 0 and this proves 4. The proof is now complete since 4. ⇒ 1. is trivial
**Proposition 3.3.** Let $(H^\# , \langle \cdot, \cdot \rangle_\#)$ be a not necessarily #-complete inner product space and $\beta \subset H^\#$ be an orthonormal set. Then the following two conditions are equivalent:

$$ (1) \quad x = \mathrm{Ext} - \sum_{u \in \beta} \langle x, u \rangle_{\#} u \quad \text{for all } x \in H^{\#}. $$

$$ (2) \quad \|x\|_{\#}^{2} = \mathrm{Ext} - \sum_{u \in \beta} |\langle x, u \rangle_{\#}|^{2} \quad \text{for all } x \in H^{\#}. $$

Moreover, either of these two conditions implies that $\beta \subset H^\#$ is a maximal orthonormal set. However $\beta \subset H^\#$ being a maximal orthonormal set is not sufficient to conditions for 1) and 2) hold.

**Proof.** As in the proof of **Theorem 12.24**, (1) implies (2). For (2) implies (1) let $\Lambda \subset \subset \beta$ and consider

$$ \begin{align} \left\| x - \left( \mathrm{Ext} - \sum_{u \in \Lambda} \langle x, u \rangle_{\#} u \right) \right\|_{\#}^{2} &= \|x\|_{\#}^{2} - 2 \left( \mathrm{Ext} - \sum_{u \in \Lambda} |\langle x, u \rangle_{\#}|^{2} \right) + \mathrm{Ext} - \sum_{u \in \Lambda} |\langle x, u \rangle_{\#}|^{2} \tag{3.3} \\ &= \|x\|_{\#}^{2} - \left( \mathrm{Ext} - \sum_{u \in \Lambda} |\langle x, u \rangle_{\#}|^{2} \right). \end{align} $$

Since $\|x\|_{\#}^{2} = \mathrm{Ext} - \sum_{u \in \beta} |\langle x, u \rangle_{\#}|^{2}$, it follows that for every $\varepsilon > 0$, $\varepsilon \approx 0$, there exists

$\Lambda_\varepsilon \subset \subset \beta$ such that for all $\Lambda \subset \subset \beta$ such that $\Lambda_\varepsilon \subset \Lambda$,

$$ \left\| x - \mathrm{Ext} - \sum_{u \in \Lambda} \langle x, u \rangle_{\#} u \right\|_{\#}^{2} = \|x\|_{\#}^{2} - \mathrm{Ext} - \sum_{u \in \Lambda} |\langle x, u \rangle_{\#}|^{2} < \varepsilon \quad (3.4) $$

showing that $x = \mathrm{Ext} - \sum_{u \in \beta} \langle x, u \rangle_{\#} u$.

Suppose $x = (x_1, x_2, \dots, x_n, \dots) \in \beta^\perp$. If (2) is valid then $\|x\|_{\#}^{2} = 0$, i.e. $x = 0$. So $\beta$ is maximal.

Let us now construct a counter example to prove the last assertion.

Take $H^\# = \mathrm{Span}(\{e_i\}_{n=1}^\infty) \subset l_2^\#$ and let $\tilde{u}_n = e_1 - (n+1)e_{n+1}$ for $n = 1, 2, \dots$. Applying Gramn-Schmidt to $\{\tilde{u}_n\}_{n=1}^\infty$ we construct an orthonormal set $\beta = \{u_n\}_{n=1}^\infty \subset H^\#$. I now claim that $\beta \subset H^\#$ is maximal. Indeed if $x = (x_1, x_2, \dots, x_n, \dots) \in \beta^\perp$ then $x \perp u_n$ for all $n$, i.e. $0 = \langle x, u_n \rangle_\# = x_1 - (n+1)x_{n+1}$.

Therefore $x_{n+1} = (n+1)^{-1}x_1$ for all $n$. Since $x \in \mathrm{Span}(\{e_i\}_{n=1}^\infty)$, $x_N = 0$ for some $N$ sufficiently large and therefore $x_1 = 0$ which in turn implies that $x_n = 0$ for all $n$.
taken from $\lambda_0$ to $\lambda$. The same argument for the half-plane containing shows that we
can extend the definition of $T_\lambda$ to that half-plane of $\mathbb{C}_c^\#$ so that for all $\lambda$ with $\mathrm{Im}\,\lambda \neq 0$

$$
T_{\lambda\varphi} = \#\lim_{n \to \infty} R_{\lambda}(A_n)\varphi. \qquad (4.7.15)
$$

The $T_\lambda$ commute, satisfy the first resolvent equation, and $T_\lambda^* = T_\lambda$ since these
statements are true for each $R_\lambda(A_n)$ It follows from the first resolvent formula and
the commutativity that the ranges of all the $T_\lambda$ are equal; we denote this common
range by $D$. By hypothesis, $D$ is $\#$-dense and this implies that the kernel of $T_\lambda$ is
empty, since $\mathrm{Ker}(T_\lambda) = (\mathrm{Ran}(T_\lambda^*))^\top = (\mathrm{Ran}(T_\lambda))^\top = D^\top = \{0\}$. We can therefore
define $A = \lambda I - T_\lambda^{-1}$ on $D$ and a short calculation with the resolvent equation shows
that this definition is independent of which $\lambda$ with $\mathrm{Im}\,\lambda \neq 0$, is chosen. Since
$\mathrm{Ran}(A \pm i) = \mathrm{Ran}(T_{\pm i}^{-1})$ operator $A$ is self-$\#$-adjoint. It is clear that the resolvent of $A$
is $T_\lambda$.

Notice that in the Trotter-Kato theorem we need convergence at two points, one in
the upper half-plane and one in the lower half-plane of $\mathbb{C}_c^\#$. For, we cannot use
Theorem 4.7.3 until we know that the $\#$-limiting operator is self-$\#$-adjoint, and the
self-$\#$-adjointness proof depends on the $\#$-convergence in both half-planes of $\mathbb{C}_c^\#$.
The Trotter-Kato theorem is important since its hypotheses do not assume the a
priori existence of a $\#$-limiting operator $A$. It can be used to assert the existence of a
generalized $\#$-limit of a sequence of self-$\#$-adjoint operators. This can also be done
with the one-parameter groups. To see why it is necessary to use the resolvents or
groups rather than the operators themselves to prove such an existence theorem
consider the following example: Let $A$ be a closed symmetric operator which is not
self-$\#$-adjoint but which has a self-$\#$-adjoint extension $\tilde{A}$. Let $P_n$ be the spectral
projection of $\tilde{A}$ corresponding to the interval $[-n, n]$. Then $P_n \tilde{A} P_n$ are bounded
self-$\#$-adjoint operators (and therefore essentially self-$\#$-adjoint on $D(A)$) such that
for all $\varphi \in D(A) : P_n \tilde{A} P_n \varphi \to_{\#} \tilde{A} \varphi = A \varphi$. Thus the $P_n A P_n$ are essentially self-$\#$-adjoint
on $D(A)$ and the strong $\#$-limit exists but the $\#$-limit is not essentially self-$\#$-adjoint.
One of the most useful aspects of generalized $\#$-convergence is that the spectra and
projections of the $A_n$ are related to the spectrum and projections of $A$.

**Theorem 4.7.6.** Let $(A_n)_{n=1}^\infty$ and $A$ be self-$\#$-adjoint operators and suppose that
$A_n \to_\# A$ in the $\#$-norm resolvent sense. Then

(a) If $\mu \notin \sigma(A)$, then $\mu \notin \sigma(A_n)$ for $n \in \mathbb{N}$ sufficiently large and

$$
\| R_{\mu}(A_n) - R_{\mu}(A) \|_{\#} \to 0 \quad (4.7.16)
$$

(b) Let $a, b \in \mathbb{R}_c^\#$, $a < b$, and suppose that $a \in \rho(A)$, $b \in \rho(A)$. Then

$$
\| P_{(a,b)}(A_n) - P_{(a,b)}(A) \|_{\#} \to 0 \quad (4.7.17)
$$

**Proof** (a) We need only consider the case where $\mu \in \mathbb{R}_c^\#$. Since $\mu \in \rho(A)$, there is a $\delta \approx 0, \delta > 0$ so that $(\mu - \delta, \mu + \delta) \cap \sigma(A) = \emptyset$. Thus, by the functional calculus, $\|R_{\mu+i\delta/3}(A)\|_{\#} < 1/\delta$. Now, we can find $N$ so that $\|R_{\mu+i\delta/3}(A_n)\|_{\#} < 2/\delta$ for $n \ge N$ which implies that the power series for $R_\lambda(A_n)$ about $\mu + i\delta/3$ has radius of $\#$-convergence at least $5/2$. We already know that where the series $\#$-converges it is an inverse for $A_n$. So, $\mu \in \rho(A_n)$ for $n \ge N$ and $\|R_\mu(A_n) - R_\mu(A)\|_{\#} \to 0$ as $n \to \infty$.

To prove (b), we note that since $a, b \in \rho(A)$, there exists $\varepsilon < (1/2)(b-a)$ and an $N$,
so that $\sup_{n \ge N} \{ \|(A_n - a)^{-1}\|_{\#}, \|(A_n - b)^{-1}\|_{\#} \} \le 1/\varepsilon$. Therefore, by the functional
calculus, $\sigma(A_n) \cap (a, b) \subset (a + \varepsilon, b - \varepsilon)$ for $n \ge N$. Let $f_\varepsilon$ be a $\#$-continuous function
So $x = 0$ and hence $\beta$ is maximal in $H^\#$. On the other hand, $\beta$ is not maximal in $l_2^\#$. In fact the above argument shows that $\beta^\perp$ in $l_2^\#$ is given by the span of $v = (1, 1/2, 1/3, 1/4, 1/5, \dots)$. Let $P$ be the orthogonal projection of $l_2^\#$ onto the

$$ \text{Span}(\beta) = v^\perp. \text{ Then } \text{Ext}-\sum_{i=1}^{\infty} \langle x, u_n \rangle_{\#} u_n = P x = x - \frac{\langle x, v \rangle_{\#}}{\|v\|_{\#}^2} v \text{ so that} \\ \text{Ext}-\sum_{i=1}^{\infty} \langle x, u_n \rangle_{\#} u_n = x \text{ iff } x \in \text{Span}(\beta) = v^\perp \subset l_2^\# . \text{ For example if } x = (1, 0, 0, \dots) \in H^\# \\ \text{(or more generally for } x = e_i \text{ for any } i), x \neq v^\perp \text{ and hence } \text{Ext}-\sum_{i=1}^{\infty} \langle x, u_n \rangle_{\#} u_n \neq x. $$

**Proposition 3.4.** (Parallelogram Law Converse). If $(X, \|\cdot\|_{\#})$ is a #-normed space such that Eq.(11.4) holds for all $x, y \in X$, then there exists a unique inner product on $\langle \cdot, \cdot \rangle_{\#}$ such that $\|x\|_{\#} = \sqrt{\langle x, x \rangle_{\#}}$ for all $x \in X$. In this case we say that $\|\cdot\|_{\#}$ is a Hilbertian #-norm.

**Proof.** If $\|\cdot\|_{\#}$ is going to come from an inner product $\langle \cdot, \cdot \rangle_{\#}$, it follows from Eq.(12.1) that $2 \text{Re}\langle x, x \rangle_{\#} = \|x+y\|_{\#}^2 - \|x\|_{\#}^2 - \|y\|_{\#}^2$ and $-2 \text{Re}\langle x, x \rangle_{\#} = \|x-y\|_{\#}^2 - \|x\|_{\#}^2 - \|y\|_{\#}^2$. Subtracting these two equations gives the “polarization identity,”

$$ 4 \text{Re}\langle x, x \rangle_{\#} = \|x+y\|_{\#}^2 - \|x-y\|_{\#}^2. \qquad (3.5) $$

Replacing $y$ by $iy$ in this equation then implies that

$$ 4 \operatorname{Im}\langle x, x \rangle_{\#} = \|x + iy\|_{\#}^2 - \|x - iy\|_{\#}^2. \qquad (3.6) $$

from which we get

$$ \langle x, y \rangle_{\#} = 1/4 \left( \text{Ext} - \sum_{\epsilon \in G} \epsilon \|x + \epsilon y\|_{\#}^2 \right) \qquad (3.7) $$

where $G = \{\pm 1, \pm i\}$ - a cyclic subgroup of $*S^1 \subset *C_c^\#$. Hence if $\langle \cdot, \cdot \rangle_{\#}$ is going to exist we must define it by Eq. (3.7). Notice that

$$ \langle x, x \rangle_{\#} = 1/4 \left( \text{Ext} - \sum_{\epsilon \in G} \epsilon \|x + \epsilon x\|_{\#}^2 \right) = \|x\|_{\#}^2 + i \|x + ix\|_{\#}^2 - i \|x - ix\|_{\#}^2 = \|x\|_{\#}^2 + i|1 + i|^2 \|x\|_{\#}^2 - i|1 - i|^2 \|x\|_{\#}^2 = \|x\|_{\#}^2. \qquad (3.8) $$

So to finish the proof of (4) we must show that $\langle x, y \rangle_{\#}$ in Eq. (3.7) is an inner product. Since

$$ \begin{align} 4\langle y, x \rangle_{\#} &= \text{Ext}-\sum_{\epsilon \in G} \epsilon \|y + \epsilon x\|_{\#}^2 = \text{Ext}-\sum_{\epsilon \in G} \epsilon \|\epsilon(y + \epsilon x)\|_{\#}^2 = \text{Ext}-\sum_{\epsilon \in G} \epsilon \|(\epsilon y + \epsilon^2 x)\|_{\#}^2 \tag{3.9} \\ &= \|y + x\|_{\#}^2 + \|-y + x\|_{\#}^2 + i\|y + ix\|_{\#}^2 - i\|-iy + x\|_{\#}^2 = \|x + y\|_{\#}^2 + \|x - y\|_{\#}^2 + i\|x - iy\|_{\#}^2 - i\|x + iy\|_{\#}^2 = 4\langle x, y \rangle_{\#} \notag \end{align} $$

it suffices to show $x \to \langle x, y \rangle_{\#}$ is linear for all $y \in H^\#$. We will need to derive an identity from Eq. (1.4). To do this we make use of Eq. (1.4) three times to find
which equals one on $(a + \varepsilon, b - \varepsilon)$ and is equal to zero outside $(a, b)$. Then
$P_{(a,b)}(A_n) = f_\varepsilon(A_n)$ and $P_{(a,b)}(A) = f_\varepsilon(A)$ and so by Theorem 4.7.3 one obtains
$\|P_{(a,b)}(A_n) - P_{(a,b)}(A)\|_{\#} \to \# 0$ as $n \to \infty$.

**Theorem 4.7.7.** Let $(A_n)_{n=1}^\infty$ and $A$ be self-#-adjoint operators and suppose that
$A_n \to_A A$ in the strong resolvent sense. Then

(a) If $a, b \in \mathbb{R}_c^\# a < b$, and $(a, b) \cap \sigma(A_n) = \emptyset$ for all $n \in \mathbb{N}$, then
$(a, b) \cap \sigma(A) = \emptyset$. That is, if $\lambda \in \sigma(A)$, then there exists $\lambda_n \in \sigma(A_n)$ so that $\lambda_n \to \lambda$.

(b) If $a, b \in \mathbb{R}_c^\# a < b$, and $a, b \notin \sigma_{pp}(A)$ then
$P_{(a,b)}(A_n)\varphi \to_{\#} P_{(a,b)}(A)\varphi$ for all $\varphi \in H^\#$.

**Proof** By the functional calculus, the statement that $(a,b) \cap \sigma(A_n) = \emptyset$ is equivalent
to the statement that $\|(A - \lambda_0)^{-1}\|_{\#} \le \sqrt{2}/(b-a)$ where $\lambda_0 = (a+b)/2 + i(b-a)/2$.
But $(A_n - \lambda_0)^{-1}$ #-converges strongly to $(A - \lambda_0)^{-1}$ so we have
$\|(A - \lambda_0)^{-1}\|_{\#} \le \#-\overline{\lim}_{n \to \infty} \|(A_n - \lambda_0)^{-1}\|_{\#} \le \sqrt{2}/(b-a)$. This proves (a).

To prove (b), we find uniformly bounded sequences of #-continuous functions $(f_n)_{n=1}^\infty$
and $(g_n)_{n=1}^\infty$ so that $0 \le f_n \le \chi_{(a,b)}$, $f_n(x) \to_\# \chi_{(a,b)}(x)$ pointwise and $\chi_{(a,b)} \le g_n$,
$g_n(x) \to_\# \chi_{[a,b]}(x)$ pointwise. Then $f_n(A) \to_\# P_{(a,b)}(A)$ and $g_n(A) \to_\# P_{[a,b]}(A)$ strongly.
Since $a, b \notin \sigma_{pp}(A)$, $P_{(a,b)}(A) = P_{[a,b]}(A)$ which means that given $\psi$ and $\varepsilon \approx 0, \varepsilon > 0$,
we can find #-continuous functions $f, g$, with $f \le \chi_{(a,b)} \le \chi_{[a,b]} \le g$ so that
$\|f(A)\psi - f(A)\psi\|_{\#} \le \varepsilon/5$. By Theorem 4.7.3(b) we can find $N \in \mathbb{N}$ so that $n > N$
implies $\|f(A_n)\psi - f(A)\psi\|_{\#} \le \varepsilon/5$ and $\|g(A_n)\psi - g(A)\psi\|_{\#} \le \varepsilon/5$ so by an $\varepsilon/3$ argument
we get $\|g(A_n)\psi - g(A_n)\psi\|_{\#} \le 3\varepsilon/5$. Since the functional calculus implies
$\|f(A)\psi - P_{(a,b)}(A)\psi\|_{\#} \le \|f(A)\psi - g(A)\psi\|_{\#}$ another $\varepsilon/3$ argument implies
$\|P_{(a,b)}(A)\psi - P_{(a,b)}(A)\psi\|_{\#} \le \varepsilon$.

**Remark.** Part (a) of Theorem 4.7.7 says that the spectrum of the #-limiting operator
cannot suddenly expand. It can, however, contract rather spectacularly as the
following example shows: Let $(A_n)_{n=1}^\infty = (x/n)_{n=1}^\infty$ on $L_2(^*\mathbb{R}_c^\#)$ then $A_n$ #-converges to
the zero operator in the strong resolvent sense. For each $n$, $\sigma(A_n) = ^*\mathbb{R}_c^\#$, but the
spectrum of the #-limiting operator contains only the origin. An easy application of
part (a) is the statement that if the $A_n$ are positive and $A_n \to_\# A$ in the strong resolvent
sense, then $A$ is positive.

If $A_n$ #-converges to $A$ in #-norm resolvent sense, Theorem 4.7.6 tells us that the
spectrum of the #-limiting operator cannot suddenly contract in the sense that if
$\lambda \in \sigma(A_n)$ for all sufficiently infinitely large $n$, then $\lambda \in \sigma(A)$. Notice that in the
example $A_n = x/n$ above, $A_n$ does not #-converge to $A$ in the #-norm resolvent sense.
The principle of noncontraction of the spectrum under #-norm resolvent
#-convergence remains true even when $A_n$ and $A$ are not self-#-adjoint. But the
principle of nonexpansion of the spectrum in the strong resolvent #-limit is not always
valid for general not-necessarily-self-#-adjoint operators. In fact, there exists a #-norm
#-convergent sequence of uniformly bounded operators $A_n \to_\# A$ with $\sigma(A_n)$ the unit
circle in $^*\mathbb{C}_c^\#$ for each $n \in \mathbb{N}$ and $\sigma(A)$ the entire unit disc. Thus the reader should
be careful to apply Theorem 4.7.7 only in the self-#-adjoint case.

In applications, one is usually given the operators $(A_n)_{n=1}^\infty$ and $A$ on domains of
self-#-adjointness or essential self-#-adjointness and it may be very difficult to
compute the resolvents. Thus, in order to use Theorem 4.7.6 and Theorem 4.7.7
one must have criteria on the operators $(A_n)_{n=1}^\infty$ and $A$ themselves which guarantee
$$
\begin{align}
\|x+y+z\|_{\#}^{2} &= -\|x+y-z\|_{\#}^{2} + 2\|x+y\|_{\#}^{2} + 2\|z\|_{\#}^{2} \nonumber \\
&= \|x-y-z\|_{\#}^{2} - 2\|x-z\|_{\#}^{2} - 2\|y\|_{\#}^{2} + 2\|x+y\|_{\#}^{2} + 2\|x+y\|_{\#}^{2} + 2\|z\|_{\#}^{2} \nonumber \\
&\quad - \|y+z+x\|_{\#}^{2} + 2\|y+z\|_{\#}^{2} + 2\|x\|_{\#}^{2} - 2\|x-z\|_{\#}^{2} - 2\|y\|_{\#}^{2} + 2\|x+y\|_{\#}^{2} + 2\|z\|_{\#}^{2}. \tag{3.10}
\end{align}
$$

Solving this equation for $\|x+y+z\|_{\#}^{2}$ gives

$$
\|x+y+z\|_{\#}^{2} = \|x+z\|_{\#}^{2} + \|x+y\|_{\#}^{2} - \|x-z\|_{\#}^{2} + \|x\|_{\#}^{2} + \|z\|_{\#}^{2} - \|y\|_{\#}^{2}. \quad (3.11)
$$

Using Eq. (3.11), for $x,y,z \in H^\#$,

$$
\begin{align}
4 \operatorname{Re}\langle x+z,y \rangle_{\#} &= \|x+z+y\|_{\#}^{2} - \|x+z-y\|_{\#}^{2} = \nonumber \\
&= \|y+z\|_{\#}^{2} + \|x+y\|_{\#}^{2} - \|x-z\|_{\#}^{2} + \|x\|_{\#}^{2} + \|z\|_{\#}^{2} - \|y\|_{\#}^{2} - \nonumber \\
&\quad - (\|z-y\|_{\#}^{2} + \|x-y\|_{\#}^{2} - \|x-z\|_{\#}^{2} + \|x\|_{\#}^{2} + \|z\|_{\#}^{2} - \|y\|_{\#}^{2}) = \tag{3.12} \\
&= \|z+y\|_{\#}^{2} - \|z-y\|_{\#}^{2} + \|x+y\|_{\#}^{2} - \|x-y\|_{\#}^{2} = \nonumber \\
&\quad 4 \operatorname{Re}\langle x,y \rangle_{\#} + 4 \operatorname{Re}\langle z,y \rangle_{\#}. \nonumber
\end{align}
$$

Now suppose that $\delta \in G$, then since $|\delta|=1$,

$$
\begin{align}
4 \langle \delta x, y \rangle_{\#} &= 1/4 \left( \operatorname{Ext} - \sum_{\epsilon \in G} \epsilon \| \delta x + \epsilon y \|_{\#}^{2} \right) = 1/4 \left( \operatorname{Ext} - \sum_{\epsilon \in G} \epsilon \| \delta x + \delta^{-1} \epsilon y \|_{\#}^{2} \right) = \nonumber \\
&= 1/4 \left( \operatorname{Ext} - \sum_{\epsilon \in G} \delta \epsilon \| \delta x + \delta \epsilon y \|_{\#}^{2} \right) = 4 \delta \langle x, y \rangle_{\#}. \tag{3.13}
\end{align}
$$

where in the third inequality, the substitution $\to \delta$ was made in the sum. So
Eq.(3.13) says $\langle \pm ix, y \rangle_{\#} = \pm i \langle \pm ix, y \rangle_{\#}$, and $\langle -x, y \rangle_{\#} = \langle -x, y \rangle_{\#}$. Therefore

$$
\operatorname{Im}\langle x, y \rangle_{\#} = \operatorname{Re}\langle -i\langle x, y \rangle_{\#} \rangle = \operatorname{Re}\langle (-i\langle x, y \rangle_{\#}) \rangle \quad (3.14)
$$

which combined with Eq. (3.12.) shows

$$
\operatorname{Im}\langle x+z,y \rangle_{\#} = \operatorname{Re}\langle -ix-iz,y \rangle_{\#} = \operatorname{Re}\langle -ix,y \rangle_{\#} + \operatorname{Re}\langle -iz,y \rangle_{\#} = \operatorname{Im}\langle x,y \rangle_{\#} + \operatorname{Im}\langle z,y \rangle_{\#} \\
\text{and therefore (again in combination with Eq. (3.12))},
$$

$$
\langle x+z,y \rangle_{\#} = \langle x,y \rangle_{\#} + \langle z,y \rangle_{\#} \text{ for all } x,y \in H^{\#}.
$$

Because of this equation and Eq. (3.13) to finish the proof that $x \to \langle x, y \rangle_{\#}$ is
linear, it suffices to show $\lambda\langle x, y \rangle_{\#} = \langle \lambda x, y \rangle_{\#}$ for all $\lambda > 0$. Now if $\lambda = m \in *\mathbb{N}$, then

$$
\langle mx, y \rangle_{\#} = \langle x + (m-1)x, y \rangle_{\#} = \langle x, y \rangle_{\#} + \langle (m-1)x, y \rangle_{\#}
$$

so that by hyper infinite induction $\langle mx, y \rangle_{\#} = m \langle x, y \rangle_{\#}$. Replacing $x$ by $x/m$ then shows
that $\langle x, y \rangle_{\#} = m \langle m^{-1}x, y \rangle_{\#}$, so that $\langle m^{-1}x, y \rangle_{\#} = m^{-1} \langle x, y \rangle_{\#}$ and so if $m, n \in *\mathbb{N}$, we find

$$
\langle \frac{n}{m} x, y \rangle_{\#} = n \langle \frac{1}{m} x, y \rangle_{\#} = \frac{n}{m} \langle x, y \rangle_{\#} \text{ so that } \lambda \langle x, y \rangle_{\#} = \langle \lambda x, y \rangle_{\#} \text{ for all } \lambda > 0 \text{ and } \lambda \in \widetilde{\mathbb{Q}}.
$$

By #-continuity, it now follows that $\lambda\langle x, y \rangle_{\#} = \langle \lambda x, y \rangle_{\#}$ for all $\lambda \in *\mathbb{R}_{c}^{\#}, \lambda > 0$.

**Proposition 3.5.** Let $(H^{\#}, \langle \cdot, \cdot \rangle_{\#})$ be a not necessarily #-complete inner product space and $\beta \subset H^{\#}$ be an orthonormal set. Then the following two conditions are equivalent:

$$
(1) \quad x = \operatorname{Ext} - \sum_{u \in \beta} \langle x, u \rangle_{\#} u \quad \text{for all } x \in H^{\#}.
$$

$$
(2) \quad \|x\|_{\#}^{2} = \operatorname{Ext} - \sum_{u \in \beta} |\langle x, u \rangle_{\#}|^{2} \quad \text{for all } x \in H^{\#}.
$$
\#-norm or strong resolvent \#-convergence.

**Theorem 4.7.8.** (a) Let $(A_n)_{n=1}^\infty$ and $A$ be self-\#-adjoint operators and suppose that $D$ is a common \#-core for all $A_n$ and $A$. If $A_n\varphi \to_\# A\varphi$ for each $\varphi \in D$ then $A_n \to_\# A$ as $n \to \infty$ in the strong resolvent sense.

(b) Let $(A_n)_{n=1}^\infty$ and $A$ be self-\#-adjoint operators with a common domain, $D$.
Norm $D$ with $\|\varphi\|_A = \|A\varphi\|_A + \|\varphi\|_A$. If $\sup_{\|\varphi\|_A=1} (\|(A_n - A)\varphi\|_A) \to_\# 0$ as $n \to \infty$ then $A_n \to_\# A$ in the \#-norm resolvent sense.

(c) Let $(A_n)_{n=1}^\infty$ and $A$ be positive self-\#-adjoint operators with a common form domain $H_{\pm 1}^\#$ which we norm with $\|\psi\|_{\#+1} = \langle\psi, A\psi\rangle_\# + \langle\psi, \psi\rangle_\#$. If $A_n \to_\# A$ in \#-norm in the sense of maps from $H_{\pm 1}^\#$ to $H_{\pm 1}^\#$ that is, if

$$ \sup_{\psi \neq 0, \varphi \in D} \frac{|\langle \varphi, (A - A_n)\psi \rangle_\#|}{\|\varphi\|_{\#+1} \|\psi\|_{\#+1}} = \sup_{\psi \neq 0, \varphi \in D} \frac{|\langle \varphi, (A - A_n)\psi \rangle_\#|}{\langle \psi, (A+I)\psi \rangle_\#} \to_\# 0 \quad (4.7.18) $$

then $A_n \to_\# A$ in the \#-norm resolvent sense.

**Proof** (a) Let $\varphi \in D$, $\psi = (A+i)\varphi$, then $[(A_n+i)^{-1} - (A+i)^{-1}]\psi = (A_n+i)^{-1}(A-A_n)\varphi$ \#-converges to zero as $n \to \infty$, since $(A-A_n)\varphi \to_\# 0$ and the $(A_n+i)^{-1}$ are uniformly bounded. Since $D$ is a \#-core for $A$ the set of such $\psi$ is \#-dense so for all $\varphi \in H^\#$: $(A_n+i)^{-1}\varphi \to_\# (A+i)^{-1}\varphi$. A similar proof holds for $(A_n-i)^{-1}$.

We sketch the proofs of (b) and (c). For (b), first one proves that the hypothesis is equivalent to $(A_n - A)(A + i)^{-1} \to_\# 0$ in the ordinary $H^\#$-operator \#-norm. Thus $(I + (A_n - A)(A + i)^{-1})^{-1}$ exists and \#-converges to $I$ in \#-norm as $n \to \infty$. As a result $(A_n + i)^{-1} = (A + i)^{-1}(I + (A_n - A)(A + i)^{-1})^{-1} \to_\# (A + i)^{-1}$ in \#-norm. Similarity $(A_n - i)^{-1} \to_\# (A - i)^{-1}$. To prove (c), we first prove that the hypothesis is equivalent to $(A + I)^{-1/2}(A_n - A)(A + I)^{-1/2} \to_\# 0$ in the ordinary operator \#-norm. Using the identity $(A_n + I)^{-1} = (A + I)^{-1/2}[I + (A + I)^{-1/2}(A_n - A)(A + I)^{-1/2}]^{-1}(A + I)^{-1/2}$ one then follows the proof of (b).

## § 4.8. Graph \#-limits.

**Definition 4.8.1.** Let $(A_n)_{n=1}^\infty$ be a hyper infinite sequence of operators on a non-archimedean hilbert space $H^\#$, We say that a pair $\langle\psi, \varphi\rangle_\# \in H^\# \times H^\#$ is in the strong graph \#-limit of $A_n$ as $n \to \infty$ if we can find $\psi_n \in D(A_n)$ so that $\psi_n \to_\# \psi$, $A_n\psi_n \to_\# \varphi$. We denote the set of pairs in the strong graph \#-limit by $\Gamma_{\#}^s$. If $\Gamma_{\#}^s$ is the graph of an operator $A$ we say that $A$ is the strong graph \#-limit of $A_n$ and write

$$ A = \text{st. gr.} \text{ -\#-lim } A_n. \quad (4.8.1) $$

First, we consider the case where all the $A_n$ are self-\#-adjoint and $A$ is also self-\#-adjoint

**Theorem 4.8.1.** Suppose that $(A_n)_{n=1}^\infty$ and $A$ are self-\#-adjoint operators. Then $A_n \to_\# A$ in the strong resolvent sense if and only if $A = \text{st. gr.} \text{ -\#-lim } A_n$.

**Proof** Suppose first that $(A_n + i)^{-1} \to_\# (A + i)^{-1}$ strongly. Suppose $\varphi \in D(A)$. Then $\varphi_n = (A_n + i)^{-1}(A + i)\varphi \to_\# \varphi$ and $A_n\varphi_n = (A + i)\varphi - i\varphi$, so $\langle\varphi, A\varphi\rangle_\# \in \Gamma_{\#}^s$. Thus $\Gamma(A) \subset \Gamma_{\#}^s$. On the other hand, suppose $\varphi_n \in D(A_n)$, $\varphi_n \to_\# \varphi$ and $A_n\varphi_n \to_\# \psi$. We let $\eta_n = (A + i)^{-1}(A_n + i)\varphi_n \in D(A)$, then
Moreover, either of these two conditions implies that $\beta \subset H^\#$ is a maximal orthonormal set. However $\beta \subset H^\#$ being a maximal orthonormal set is not sufficient to conditions for 1) and 2) hold.

**Proof.** As in the proof of Theorem 2.1, (1) implies (2). For (2) implies (1) let $\Lambda \subset \subset \beta$ and consider

$$ \left\| x - \left( \operatorname{Ext} - \sum_{u \in \Lambda} \langle x, u \rangle_{\#} u \right) \right\|_{\#}^{2} = \|x\|_{\#}^{2} - 2 \left( \operatorname{Ext} - \sum_{u \in \Lambda} |\langle x, u \rangle_{\#}|^{2} \right) + \operatorname{Ext} - \sum_{u \in \Lambda} |\langle x, u \rangle_{\#}|^{2} = \\ \|x\|_{\#}^{2} - \left( \operatorname{Ext} - \sum_{u \in \Lambda} |\langle x, u \rangle_{\#}|^{2} \right). \qquad (3.15) $$

Since $\|x\|_{\#}^{2} = \operatorname{Ext} - \sum_{u \in \beta} |\langle x, u \rangle_{\#}|^{2}$, it follows that for every $\varepsilon \approx 0_{\ast \mathbb{R}_{c}^{\#}}$, $\varepsilon > 0_{\ast \mathbb{R}_{c}^{\#}}$ there exists $\Lambda_{\varepsilon} \subset \subset \beta$ such that for all $\Lambda \subset \subset \beta$ such that $\Lambda_{\varepsilon} \subset \Lambda$,

$$ \left\| x - \left( \operatorname{Ext} - \sum_{u \in \Lambda} \langle x, u \rangle_{\#} u \right) \right\|_{\#}^{2} = \|x\|_{\#}^{2} - \left( \operatorname{Ext} - \sum_{u \in \Lambda} |\langle x, u \rangle_{\#}|^{2} \right) < \varepsilon \qquad (3.16) $$

showing that $x = \operatorname{Ext} - \sum_{u \in \beta} \langle x, u \rangle_{\#}$. Suppose $x = (x_1, x_2, \dots, x_n, \dots) \in \beta^{\perp}$. If (2) is valid then $\|x\|_{\#}^{2} = 0_{\ast \mathbb{R}_{c}^{\#}}$, i.e. $x = 0$. So $\beta$ is maximal. Let us now construct a counter example to prove the last assertion. Take $H^{\#} = \operatorname{Span}\{e_i\}_{i=1}^{\infty} \subset l_2^{\#}$ and let $\hat{u}_n = e_1 - (n+1)e_{n+1}$ for $n \in \ast \mathbb{N}$. Applying Gramn-Schmidt to $\{\hat{u}_n\}_{n=1}^{\infty}$ we construct an orthonormal set $\beta = \{u_n\}_{n=1}^{\infty} \subset H^{\#}$.

We now claim that $\beta \subset H^{\#}$ is maximal. Indeed if $x = (x_1, x_2, \dots, x_n, \dots) \in \beta^{\perp}$ then $x \perp u_n$ for all $n \in \ast \mathbb{N}$, i.e. $0_{\ast \mathbb{R}_{c}^{\#}} = \langle x, \hat{u}_n \rangle_{\#} = x_1 - (n+1)x_{n+1}$.

Therefore $x_{n+1} = (n+1)^{-1}x_1$ for all $n \in \ast \mathbb{N}$. Since $x \in \operatorname{Span}\{e_i\}_{i=1}^{\infty}$, $x_N = 0$ for some $N$ sufficiently large and therefore $x_1 = 0$ which in turn implies that $x_n = 0_{\ast \mathbb{R}_{c}^{\#}}$ for all $n \in \ast \mathbb{N}$. So $x = 0_{\ast \mathbb{R}_{c}^{\#}}$ and hence $\beta$ is maximal in $H^{\#}$. On the other hand, $\beta$ is not maximal in $l_2^{\#}$. In fact the above argument shows that $\beta^{\perp}$ in $l_2^{\#}$ is given by the span of $v = (1_{\ast \mathbb{R}_{c}^{\#}}, 1_{\ast \mathbb{R}_{c}^{\#}}/2, 1_{\ast \mathbb{R}_{c}^{\#}}/3, 1_{\ast \mathbb{R}_{c}^{\#}}/4, 1_{\ast \mathbb{R}_{c}^{\#}}/5, \dots)$. Let $P$ be the orthogonal projection of $l_2^{\#}$ onto the $\operatorname{Span}(\beta) = v^{\perp}$.

$$ \operatorname{Ext} - \sum_{u \in \Lambda} \langle x, u_n \rangle_{\#} u_n = P x = x - \frac{\langle x, v \rangle_{\#}}{\|v\|_{\#}^2} v, \text{ so that } \operatorname{Ext} - \sum_{u \in \Lambda} \langle x, u_n \rangle_{\#} u_n = x \text{ iff} \\ x \in \operatorname{Span}(\beta) = v^{\perp} \subset l_2^{\#}. \text{ For example if } x = (1_{\ast \mathbb{R}_{c}^{\#}}, 0_{\ast \mathbb{R}_{c}^{\#}}, 0_{\ast \mathbb{R}_{c}^{\#}}, \dots) \in H^{\#} \text{ (or more generally for } x = e_i \text{ for any } i \in \ast \mathbb{N}), x \notin v^{\perp} \text{ and hence } \operatorname{Ext} - \sum_{u \in \Lambda} \langle x, u_n \rangle_{\#} u_n \neq x. $$

## § 4.#-Analytic vectors. Generalized Nelson's #-analytic vector theorem.

Let $H^{\#}$ be a #-complex Hilbert space over field $\ast \tilde{\mathbb{C}}_{c}^{\#}$. The most natural way to construct a #-continuous one-parameter unitary group $U(t) : H^{\#} \to H^{\#}$ is to try to make sense
$$
\begin{align}
\eta_n - \varphi_n &= [(A+i)^{-1} - (A_n+i)^{-1}][(A_n+i)\varphi_n] = \nonumber \\
&= [(A+i)^{-1} - (A_n+i)^{-1}][(A_n+i)\varphi_n - \psi - i\varphi] + \tag{4.8.2} \\
&\quad +[(A+i)^{-1} - (A_n+i)^{-1}][\psi + i\varphi] \xrightarrow{\#} 0 \nonumber
\end{align}
$$

as $n \to *\infty$. Thus, $\eta_n \to_* \varphi$ and $A\eta_n = (A_n+i)\varphi_n - i\eta_n \to_* \psi$ so since A is #-closed
$\langle\varphi, \psi\rangle_* \in \Gamma(A)$. Thus, $\Gamma(A) = \Gamma_*^{s,\infty}$.

Conversely, suppose that $A = \mathbf{st. gr.}$-$\#\text{-lim} A_n$. Let $\varphi \in D(A)$. Then there exist
$\varphi_n \in D(A_n)$ so that $\varphi_n \to_* \varphi$ and $A_n\varphi_n \to_* A\varphi$ as $n \to *\infty$. Thus,
$[(A_n+i)^{-1} - (A+i)^{-1}][(A+i)\varphi] = (A_n+i)^{-1}[(A_n+i)\varphi - (A_n+i)\varphi_n] - (\varphi - \varphi_n) \to_* 0$
as $n \to *\infty$since $\|(A_n+i)^{-1}\|_* \le 1$, $(A_n+i)\varphi_n \to_* (A+i)\varphi$, and $\varphi_n \to_* \varphi$. Since
$\text{Ran}(A+i) = H.^*$ the strong $\#\text{-convergence of } (A_n+i)^{-1}$ to $(A+i)^{-1}$ follows.

**Remark 4.8.1.** Thus, we see that if the #-limit is self-#-adjoint, then strong graph and strong resolvent #-convergence are the same. It is in the case when we do not know a priori that the #-limit is self-#-adjoint that strong graph #-limits are particularly important. For example, the existence of graph #-limits can sometimes be combined with other information to prove that the #-limit is self-#-adjoint.

**Theorem 4.8.2.** Let $(A_n)_{n=1}^{*\infty}$ be a hyper infinite sequence of symmetric operators.

(a) Let $D_{*\infty}^s = \{\psi | \langle\psi, \varphi\rangle_* \in \Gamma_{*\infty}^s \text{ for some } \varphi\}$. If $D_{*\infty}^s$ is #-dense, then $\Gamma_{*\infty}^s$ is the graph of an operator.

(b) Suppose that $D_{*\infty}^s$ is #-dense and let $A = \mathbf{st. gr.}$-$\#\text{-lim} A_n$.
Then A is symmetric and #-closed.

**Proof** We will prove (a); the proof of (b) is obvious. Suppose $\varphi_n, \varphi'_n \in D(A_n)$ and
$\varphi_n \to_* \varphi, \varphi'_n \to_* \varphi$ and $A_n\varphi_n \to_* \psi, A_n\varphi'_n \to_* \psi'$. Let $\eta \in D_{*\infty}^s$. Then there is an
$\eta_n \in D(A_n)$, so that $\eta_n \to_* \eta$ and $A_n\eta_n \to_* \rho$ as $n \to *\infty$. Thus,
$\langle\psi - \psi, \eta\rangle_* = \#\text{-lim}_{n \to *\infty} \langle A_n(\varphi_n - \varphi'_n), \eta_n\rangle_* = \#\text{-lim}_{n \to *\infty} \langle\varphi_n - \varphi'_n, A_n\eta_n\rangle_* = 0$
so $\psi = \psi'$ since $D_{*\infty}^s$ is #-dense.

We also define weak graph #-limits. We give the definition and state one theorem.

**Definition 4.8.2.** Let $(A_n)_{n=1}^{*\infty}$ be a hyper infinite sequence of operators on $H^\#$. We say that $\langle\psi, \varphi\rangle_* \in H^\# \times H^\#$ is in the weak graph #-limit $\Gamma_{*\infty}^w$ if we can find $\psi_n \in D(A_n)$ so that $\psi_n \xrightarrow{\|\cdot\|_*} \psi$ and $A_n\psi_n \to_* \varphi$ weakly. If $\Gamma_{*\infty}^w$ is the graph of an operator, A we say that A is the weak graph #-limit of $A_n$ and abbreviated as $A = \mathbf{w. gr.}$-$\#\text{-lim} A_n$.

**Theorem 4.8.3.** Let $(A_n)_{n=1}^{*\infty}$ be a hyper infinite sequence of symmetric operators. If $D_{*\infty}^s = \{\psi | \langle\psi, \varphi\rangle_* \in \Gamma_{*\infty}^s \text{ for some } \varphi\}$ is #-dense, then $\Gamma_{*\infty}^w$ is the graph of a symmetric operator.

**Remark 4.8.2.** Finally we note that if $A_n$ is a uniformly bounded sequence of operators then $A = \mathbf{w. gr.}$-$\#\text{-lim } A_n$ if and only if $A_n \xrightarrow{\#} A$ as $n \to *\infty$ in the weak operator topology. This fact shows that the notions of weak graph #-limit and weak resolvent #-convergence are distinct. It is not true that weak graph #-limits are necessarily #-closed if each $A_n$ is symmetric.

§ 4.9. Generalized Trotter product formula

**Theorem 4.9.1.** (Generalized Lie product formula) Let A and B be external hyperfinite-dimensional matrices. Then

$$
\mathrm{Ext}\text{-}\exp(A+B) = \#\text{-lim}_{n \to *\infty} \{[\mathrm{Ext}\text{-}\exp(A/n)] \times [\mathrm{Ext}\text{-}\exp(B/n)]\}^n. \quad (4.9.1)
$$
of the power series $Ext-\sum_{n=0}^{\infty} (itA)^n$ on a #-dense set of vectors. Notice that this can certainly be done if A is self-adjoint. For let $E_\Omega$ be the family of spectral projections for A. Then on each of the spaces $E_{[-M,M]}$, A is a bounded operator and $Ext-\sum_{n=0}^{\infty} (itA)^n/n!$ #-converges to $Ext-\exp(itA)$ in #-norm. In particular, for any $\varphi \in \bigcup_{M \ge 0} E_{[-M,M]}$,

$$ \#\lim_{N \to \infty} \left( \mathrm{Ext}-\sum_{n=0}^{N} \frac{(itA)^n}{n!} \right) = \mathrm{Ext}-\exp(itA). \qquad (4.1) $$

Since $\bigcup_{M \ge 0} E_{[-M,M]}$ is #-dense in $\mathbf{H}^\#$, we see that the group generated by a self-adjoint operator A is completely determined by the well-defined action of the hyper infinite series $Ext-\sum_{n=0}^{\infty} (itA)^n/n!$ on a #-dense set. We will prove the #-converse: namely, if A is symmetric and has a #-dense set of vectors to which $Ext-\sum_{n=0}^{\infty} (itA)^n/n!$ can be applied, then A is essentially self-#-adjoint. We need several definitions.

**Definition 4.1.** Let A be an operator on a non-Archimedean Hilbert space $\mathbf{H}^\#$. The set $\mathbf{C}^{\infty}(A) = \bigcap_{n=0}^{\infty} D(A^n)$ is called the $\mathbf{C}^{\infty}$-vectors for A. A vector $\varphi \in \mathbf{C}^{\infty}(A)$ is called an #-analytic vector for A if

$$ \mathrm{Ext}-\sum_{n=0}^{\infty} \frac{\|A^n\varphi\|t^n}{n!} < *\infty \qquad (4.2) $$

for some $t > 0$. If A is self-adjoint, then $\mathbf{C}^{\infty}(A)$ will be #-dense in $D(A)$. However, in general, a symmetric operator may have no $\mathbf{C}^{\infty}$-vectors at all even if A is essentially self-#-adjoint. We caution the reader to remember that #-analytic vectors and vectors of

uniqueness (defined below) must be $\mathbf{C}^{\infty}$- vectors for A. A vector $\varphi \in D(A)$ can be an #-analytic vector for an extension of A but fail to be an #-analytic vector for A because it is not in $\mathbf{C}^{\infty}(A)$.

**Definition 4.2.** Suppose that A is symmetric. For each $\varphi \in \mathbf{C}^{\infty}(A)$, define

$$ D_{\varphi} = \left\{ \mathrm{Ext}-\sum_{n=0}^{N} \alpha_n A^n \varphi \mid N \in *\mathbb{N}, \alpha_n \in *\tilde{\mathbb{C}}_{c}^{\#} \right\}. \qquad (4.3) $$

Let $\mathbf{H}_{\varphi}^{\#} = \#-\overline{D_{\varphi}}$ and define $A_{\varphi} : D_{\varphi} \to D_{\varphi}$ by $A_{\varphi} \left( \mathrm{Ext}-\sum_{n=0}^{N} \alpha_n A^n \varphi \right) = \mathrm{Ext}-\sum_{n=0}^{N} \alpha_n A^{n+1} \varphi$. $\varphi$ is called a vector of #-uniqueness if and only if $A_{\varphi}$ is essentially self-#-adjoint on $D_{\varphi}$ as an operator on $\mathbf{H}_{\varphi}^{\#}$.

Finally, a subset $S \subset \mathbf{H}^\#$ is called #-total if the set of hyperfinite linear combinations of elements of S is #-dense in $\mathbf{H}^\#$.

**Lemma (Generalized Nussbaum's lemma)** Let A be a symmetric operator and suppose that $D(A)$ contains a #-total set of vectors of #-uniqueness. Then A is essentially self-#-adjoint.

**Proof** We will show that $\mathrm{Ran}(A \pm i)$ are #-dense in $\mathbf{H}^\#$. By the fundamental criterion this will show that A is essentially self-#-adjoint. Suppose $\psi \in \mathbf{H}^\#$ and $\varepsilon > 0$ are given and let S denote the set of vectors of #-uniqueness. Since S is #-total we can find $(\alpha_n)_{n=1}^N$ and $(\psi_n)_{n=1}^N$ with $\psi_n \in S$ so that
**Proof** Let $S_n = \text{Ext-exp}((A+B)/n)$ and $T_n = [\text{Ext-exp}(A/n)] \times [\text{Ext-exp}(B/n)]$. Then

$$S_n^n - T_n^n = \text{Ext-}\sum_{m=0}^{n-1} S_m^n(S_n - T_n)T_n^{n-m-1} \quad (4.9.2)$$

so that

$$ \begin{aligned} \|S_n^n - T_n^n\|_{\#} &\le n(\max(\|S_n\|_{\#}, \|T_n\|_{\#}))^{n-1} \|S_n - T_n\|_{\#} \\ &\le n \|S_n - T_n\|_{\#} [\text{Ext-exp}(\|A\|_{\#} + \|B\|_{\#})]. \end{aligned} \quad (4.9.3) $$

Since

$$ \begin{aligned} \|S_n - T_n\|_{\#} &= \left\| \text{Ext-}\sum_{m=0}^{\infty} \frac{1}{m!^{\#}} \left(\frac{A+B}{n}\right)^m - \left( \text{Ext-}\sum_{m=0}^{\infty} \frac{1}{m!^{\#}} \left(\frac{A}{n}\right)^m \right) \left( \text{Ext-}\sum_{m=0}^{\infty} \frac{1}{m!^{\#}} \left(\frac{B}{n}\right)^m \right) \right\|_{\#} \\ &\le C/n \end{aligned} \quad (4.9.4) $$

where constant $C$ depends only on $\|A\|_{\#}$ and $\|B\|_{\#}$ we conclude that $\#\lim_{n \to \infty} \|S_n^n - T_n^n\|_{\#} = 0$.

**Remark 4.9.1.** This theorem and its proof can be extended to the case where $A$ and $B$ are unbounded self-#-adjoint operators and $A+B$ is self-#-adjoint on $D(A) \cap D(B)$.

**Theorem 4.9.2** Let $A$ and $B$ be self-#-adjoint operators on $H^\#$ and suppose that $A+B$ is self-#-adjoint on $D = D(A) \cap D(B)$. Then

$$ \text{Ext-exp}[it(A+B)] = \mathbf{s}\text{-}\lim_{n \to \infty} \{[\text{Ext-exp}(itA/n)] \times [\text{Ext-exp}(itB/n)]\}^n. \quad (4.9.5) $$

**Proof** Let $\psi \in D$. Then

$$ \begin{aligned} s^{-1}\{[\text{Ext-exp}(isA)] \times [\text{Ext-exp}(isB)] - I\}\psi &= s^{-1}\{[\text{Ext-exp}(isA)] - I\}\psi + s^{-1}\{[\text{Ext-exp}(isB)] - I\}\psi \to_{\#} i(A+B)\psi \end{aligned} \quad (4.9.6) $$

and

$$ s^{-1}\{[\text{Ext-exp}(isA)] \times [\text{Ext-exp}(isB)] - I\}\psi \to_{\#} i(A+B)\psi \quad (4.9.7) $$

as $s \to \# 0$. Letting

$$ K(s) = s^{-1}\{[\text{Ext-exp}(isA)] \times [\text{Ext-exp}(isB)] - [\text{Ext-exp}(is(A+B))]\} \quad (4.9.8) $$

we see that $K(s)\psi \to_{\#} 0$ as $s \to_{\#} 0$, for each $\psi \in D$. Since $A+B$ is self-#-adjoint on $D$, $D$ is a

Banach space under the #-norm

$$ \|\psi\|_{\#A+B} = \|(A+B)\psi\|_{\#} + \|\psi\|_{\#}. \quad (4.9.9) $$

Each of the maps $K(s) : D \to H^\#$ is bounded and $K(s)\psi \xrightarrow{s \to \#} 0$ as $s \to \# 0$ or $\infty$ for each $\psi \in D$.

Thus, we conclude from the uniform boundedness theorem that the $K(s)$ are uniformly bounded, that is, there is a constant $C$ so that $\|K(s)\psi\|_{\#} \le C\|\psi\|_{\#A+B}$ for all $s \in \mathbb{R}_c^\#$ and $\psi \in D$. Therefore, an $\varepsilon/3$ argument shows that on $\|\cdot\|_{\#A+B}$ #-compact subsets of $D$ $K(s)\psi \to_{\#} 0$ uniformly. Since $A+B$ is self-#-adjoint on $D$, $[\text{Ext-exp}(is(A+B))]\psi \in D$ if $\psi \in D$. Moreover, $s \to [\text{Ext-exp}(is(A+B))]\psi$ is a #-continuous map of $\mathbb{R}_c^\#$ into $D$ when $D$ is given the $\|\cdot\|_{\#A+B}$ #-norm topology. Thus $\{[\text{Ext-exp}(is(A+B))]\psi|s \in [-1, 1]\}$ is a $\|\cdot\|_{\#A+B}$ #-compact set in $D$ for each fixed $\psi$.
$$ \left\| \psi - \operatorname{Ext-}\widehat{\sum}_{n=1}^{N} \alpha_n \psi_n \right\|_{\#} \le \varepsilon/2. \qquad (4.4) $$

Since $\psi_n$ is a vector of #-uniqueness, there is a $\varphi_n \in D_{\psi_n}$ so that

$$ \|\psi_n - (A + i)\varphi_n\|_{\#} \le \frac{\varepsilon}{2} \left( \operatorname{Ext-}\widehat{\sum}_{n=1}^{N} |\alpha_n| \right)^{-1}. \qquad (4.5) $$

Setting $\varphi = \operatorname{Ext-}\sum_{n=1}^{N} \alpha_n \psi_n$ we have $\varphi \in D(A)$ and $\|\psi - (A + i)\varphi\|_{\#} < \varepsilon$. Thus $\operatorname{Ran}(A + i)$ is #-dense. The proof for $(A - i)$ is the same.

**Theorem 4.1.** (Generalized Nelson's #-analytic vector theorem) Let $A$ be a symmetric operator on a non-Archimedean Hilbert space $\mathbf{H}^{\#}$. If $D(A)$ contains a #-total set of #-analytic vectors, then $A$ is essentially self-#-adjoint.

**Proof** By Generalized Nussbaum's lemma, it is enough to show that each #-analytic vector $\psi$ is a vector of #-uniqueness. First notice that $A_{\psi}$ always has self-#-adjoint extensions, since the operator

$$ C : \operatorname{Ext-}\widehat{\sum}_{n=0}^{N} \alpha_n A^n \psi \qquad (4.6) $$

extends to a conjugation on $\mathbf{H}_{\psi}^{\#}$ which commutes with $A_{\psi}$. Suppose that $B$ is a self-#-adjoint extension of $A_{\psi}$ on $\mathbf{H}_{\psi}^{\#}$ and let $\mu^{\#}$ be the spectral #-measure for $B$ associated to $\psi$. Since $\psi$ is an #-analytic vector for $A$,

$$ \operatorname{Ext-}\widehat{\sum}_{n=0}^{N} \|A^n \psi\|_{\#} / n! < *\infty \qquad (4.7) $$

for some $t > 0$. Let $0 < s < t$. Then

$$ \begin{align} & \operatorname{Ext-}\widehat{\sum}_{n=0}^{\infty} \frac{s^n}{n!} \left( \operatorname{Ext-} \int_{\mathbb{R}_c^{\#}} |x|^n d^{\#} \mu^{\#} \right) \le \nonumber \\ & \le \operatorname{Ext-}\widehat{\sum}_{n=0}^{\infty} \frac{s^n}{n!} \left( \operatorname{Ext-} \int_{\mathbb{R}_c^{\#}} x^{2n} d^{\#} \mu^{\#} \right)^{1/2} \left( \operatorname{Ext-} \int_{\mathbb{R}_c^{\#}} d^{\#} \mu^{\#} \right)^{1/2} = \tag{4.8} \\ & \|\psi\|_{\#} \operatorname{Ext-}\widehat{\sum}_{n=0}^{\infty} \frac{s^n}{n!} \|A^n \psi\|_{\#} < *\infty. \nonumber \end{align} $$

Therefore by generalized Fibini's theorem

$$ \operatorname{Ext-} \int_{\mathbb{R}_c^{\#}} \left( \operatorname{Ext-} \sum_{n=0}^{*\infty} \frac{s^n}{n!} |x|^n \right) d^{\#} \mu^{\#} = \operatorname{Ext-} \int_{\mathbb{R}_c^{\#}} \operatorname{Ext} - (s|x|) d^{\#} \mu^{\#} < *\infty. \qquad (4.9) $$

As a result, the function

$$ \langle \psi, [\operatorname{Ext-}\exp(itB)]\psi \rangle_{\#} = \operatorname{Ext-} \int_{\mathbb{R}_c^{\#}} [\operatorname{Ext-}\exp(itx)] d^{\#} \mu^{\#} \qquad (4.10) $$

has an #-analytic continuation
We are now ready to mimic the proof of the generalized Lie product formula. We know that

$$t^{-1} \{[\mathrm{Ext}\text{-exp}(itA)] \times [\mathrm{Ext}\text{-exp}(itB)] - [\mathrm{Ext}\text{-exp}(it(A+B))]\} \times \\ \times [\mathrm{Ext}\text{-exp}(is(A+B))]\psi \to_{\#} 0 \qquad (4.9.10)$$

uniformly for $s \in [-1, 1]$. Therefore, we write

$$\begin{aligned} \{([\mathrm{Ext}\text{-exp}(itA/n)] \times [\mathrm{Ext}\text{-exp}(itB/n)])^n - [\mathrm{Ext}\text{-exp}(it(A+B)/n)]^n\}\psi = \\ \mathrm{Ext}\text{-}\sum_{k=0}^{n} \{([\mathrm{Ext}\text{-exp}(itA/n)] \times [\mathrm{Ext}\text{-exp}(itB/n)])^k \times \\ [(\{[\mathrm{Ext}\text{-exp}(itA/n)] \times [\mathrm{Ext}\text{-exp}(itB/n)]\} - [\mathrm{Ext}\text{-exp}(it(A+B)/n)]) \times \\ [\mathrm{Ext}\text{-exp}(it(A+B)/n)]^{n-k-1}\psi \end{aligned} \qquad (4.9.11)$$

The #-norm of the RHS of (4.9.11)

$$\max_{|s|<t} \left\| \left(\frac{t}{n}\right)^{-1} \{[\mathrm{Ext}\text{-exp}(it(A+B)/n)] - \{[\mathrm{Ext}\text{-exp}(itA/n)] \times [\mathrm{Ext}\text{-exp}(itB/n)]\}\} \right\|_{\#} \qquad (4.9.12)$$

and so we conclude that

$$\{[\mathrm{Ext}\text{-exp}(itA/n)] \times [\mathrm{Ext}\text{-exp}(itB/n)]\}^n \psi \xrightarrow{H^\#} \mathrm{Ext}\text{-exp}(it(A+B))\psi \qquad (4.9.13)$$

as $n \to *\infty$ if $\psi \in D$; Since $D$ is #-dense and the operators are bounded by one, this statement holds on all of $H^\#$. The above proof shows that on a fixed vector the #-convergence is uniform for $t$ in a #-compact subset of $*\mathbb{R}_c^\#$.

**Remark 4.9.2.** The same argument can be used to show that

$$\mathrm{s-}\#\lim_{n \to *\infty} \{[\mathrm{Ext}\text{-exp}(itA/n)] \times [\mathrm{Ext}\text{-exp}(itB/n)]\}^n = \mathrm{Ext}\text{-exp}(t(A+B)) \qquad (4.9.14)$$

if $A$ and $B$ satisfy the same hypotheses and in addition are semibounded. The following

result is considerably stronger than Theorem 4.9.2 since it only requires essential self-#-adjoint ness of $A+B$ on $D(A) \cap D(B)$.

**Theorem 4.9.3** (the generalized Trotter product formula) If $A$ and $B$ are self-#-adjoint operators and $A+B$ is essentially self-adjoint on $D(A) \cap D(B)$ then

$$\mathrm{s-}\#\lim_{n \to *\infty} \{[\mathrm{Ext}\text{-exp}(itA/n)] \times [\mathrm{Ext}\text{-exp}(itB/n)]\}^n = \mathrm{Ext}\text{-exp}(it(A+B)) \qquad (4.9.15)$$

Moreover, if $A$ and $B$ are bounded from below, then

$$\mathrm{s-}\#\lim_{n \to *\infty} \{[\mathrm{Ext}\text{-exp}(-tA/n)] \times [\mathrm{Ext}\text{-exp}(-tB/n)]\}^n = \mathrm{Ext}\text{-exp}(-t(A+B)). \qquad (4.9.16)$$

## § 4.10. The polar decomposition

Note that an arbitrary bounded operator $T$ can be written $T = U|T|$ where $|T|$ is positive and self-#-adjoint and $U$ is a partial isometry. Moreover, the conditions that $\mathrm{Ker}(|T|) = \mathrm{Ker}(T)$ and that the initial space of $U$ equals $(\mathrm{Ker}(T))^\perp$ uniquely determine $|T|$ and $U$. In this section we extend this result to closed #-unbounded operators. As in the bounded case, $U$ is easy to construct once $|T|$ has been constructed and, as in the bounded case, we will let $|T| = \sqrt{T^*T}$. In the bounded case, the hard part was the construction of the square root. Now that we have the spectral theorem, it is easy to construct $\sqrt{T^*T}$ if we can prove that $T^*T$ is a positive self-
$$Ext- \int_{\mathbb{R}_c^\#} [Ext-\exp(izx)] d^\# \mu^\# \quad (4.11)$$

to the region $|\mathrm{Im} z| < t$. Since

$$ \left[ \left( \frac{d^\#}{d^\# z} \right)^k \left( \mathrm{Ext} - \int_{\mathbb{R}_c^\#} [\mathrm{Ext}-\exp(izx)] d^\# \mu^\# \right) \right]_{z=0} = \mathrm{Ext} - \int_{\mathbb{R}_c^\#} [\mathrm{Ext}-\exp(ix)^k] d^\# \mu^\# = \langle \psi, (iA)^k \psi \rangle_\# \quad (4.12) $$

we obtain

$$ \langle \psi, [\mathrm{Ext}-\exp(isB)]\psi \rangle_\# = \mathrm{Ext}-\sum_{n=0}^{\infty} \frac{(is)^n}{n!} = \langle \psi, (iA)^k \psi \rangle_\# \quad (4.13) $$

for $|s| < t$. Thus, for $|s| < t$ (and therefore for all $s$), the function $\langle \psi_1, [\mathrm{Ext}-\exp(isB)]\psi_2 \rangle_\#$ is completely determined by the numbers $\langle \psi_1, A^n \psi_2 \rangle_\#$, $n \in *\mathbb{N}$.

Similar proof shows that $\langle \psi_1, [\mathrm{Ext}-\exp(isB)]\psi_2 \rangle_\#$ is determined by the numbers $\langle \psi_1, A^n \psi_2 \rangle_\#$, $n \in *\mathbb{N}$ for any $\psi_1, \psi_2 \in D_\psi$. Since $D_\psi$ is #-dense in $H_\psi^\#$ and $\mathrm{Ext}-\exp(isB)$ is unitary, $\mathrm{Ext}-\exp(isB)$ is completely determined by the numbers $\langle \psi_1, A^n \psi_2 \rangle_\#$, $n \in *\mathbb{N}$ for any $\psi_1, \psi_2 \in D_\psi$. Thus, all self-#-adjoint extensions of $A_\psi$ generate the same unitary group, so by generalized Stone's theorem $A_\psi$ has at most one self-#-adjoint extension. As we have already remarked, $A_\psi$ has at least one self-#-adjoint extension. Thus $A_\psi$ is essentially self-#-adjoint and $\psi$ is a vector of uniqueness.

**Corollary 4.1** A #-closed symmetric operator $A$ is self-#-adjoint if and only if $D(A)$ contains a #-dense set of #-analytic vectors.

The statement of Corollary 4.1 is not true if "self-#-adjoint" is replaced by "essentially self-#-adjoint." A self-#-adjoint operator $A$ may be essentially self-#-adjoint on a domain $D \subset D(A)$ and $D$ may not even contain any #-vectors.

**Corollary 4.2** Suppose that $A$ is a symmetric operator and let $D$ be a #-dense linear set contained in $D(A)$. Then, if $D$ contains a #-dense set of #-analytic vectors and if $D$ is invariant under $A$, then $A$ is essentially self-#-adjoint on $D$.

**Proof** Since $D$ is invariant under $A$, each #-analytic vector for $A$ in $D$ is also an #-analytic vector for $A \upharpoonright D$. Thus, by Theorem 4.1 $A \upharpoonright D$ is essentially self-#-adjoint. The reason that one needs the invariance condition in Corollary 2 is that for a vector $\psi \in D$ to be #-analytic for $A \upharpoonright D$, it must first be $C^{*\infty}$ for $A \upharpoonright D$. This requires that $A^n \in D$ for all $n \in *\mathbb{N}$.

# 5. The generalized Spectral Theorem

## § 5.1. The #-continuous functional calculus

In this section, we will discuss the generalized spectral theorem in its many guises. This structure theorem is a concrete description of all self-#-adjoint operators. There are several apparently distinct formulations of the spectral theorem. In some sense they are all equivalent.

The form we prefer says that every bounded self-#-adjoint operator is a multiplication operator. (We emphasize the word bounded since we will deal extensively with
#-adjoint operator. It is this fact that is hard in the unbounded case. A priori, it is not clear that {$\psi|\psi \in D(T)$ and $T\psi \in D(T^*)$} is different from {$0$}. In fact, this set is #-dense, but our approach using the theory of semi-bounded quadratic forms does not require us to prove this.

**Theorem 4.10.1.** (the polar decomposition) Let $\Gamma$ be an arbitrary #-closed operator on a

non Archimedean Hilbert space $H^#$. Then, there is a positive self-adjoint operator $|T|$, with $D(|T|) = D(T)$ and a partial isometry $U$ with initial space, $(\text{Ker}(T))^\perp$, and final space #-Ran($T$) so that $T = U|T|$ and $U$ are uniquely determined by these properties together with the additional condition $\text{Ker}(|T|) = \text{Ker}(T)$.

**Proof.** Define the *C_c^#-valued quadratic form $s(\psi, \varphi)$ on $D(T)$ by

$$s(\psi, \varphi) = \langle T\psi, T\varphi \rangle_{\#}. \qquad (4.10.1)$$

Quadratic form $s(\psi, \varphi)$ is clearly positive. Now suppose $\|\psi_n - \psi_m\|_{\#+1} \to \# 0$. Then $\|\psi_n - \psi_m\|_{\#} \to \# 0$ and $\|T(\psi_n - \psi_m)\|_{\#} \to \# 0$. Since $T$ is #-closed there is a $\psi \in D(T)$ with

$$\|\psi_n - \psi\|_{\#} + \|T(\psi_n - \psi)\|_{\#} \to \# 0, \qquad (4.10.2)$$

i.e. $\|\psi_n - \psi\|_{\#+1} \to \# 0$.

Thus $s(\psi, \varphi)$ is a #-closed form. Therefore, by **Theorem VIII.15**, there is a unique, positive self-#-adjoint operator $S$ with $Q(S) = D(T)$ and $s(\psi, \varphi) = \langle \psi, S\varphi \rangle_{\#}$ in the sense of *C_c^#-valued quadratic forms. Let $|T| = S^{1/2}$. Then $D(|T|) = Q(S) = D(T)$ and by construction $\|T|\psi\|_{\#}^2 = s(\psi, \psi) = \|T\psi\|_{\#}^2$ so $\text{Ker}(|T|) = \text{Ker}(T)$. Define the operator $U: \text{Ran}(|T|) \to \text{Ran}(T)$ by $U|T|\psi = T\psi$. Since $\|T|\psi\|_{\#} = \|T\psi\|_{\#}$, $U$ is well defined and #-norm preserving. Thus $U$ extends to a partial isometry from $\text{Ran}(|T|)$ to $\text{Ran}(T)$. Finally, since $|T|$ is self-#-adjoint, #-Ran($T$) = $(\text{Ker}(|T|))^\perp = (\text{Ker}(T))^\perp$. Uniqueness is obvious.

# 5. Tensor products and second quantization.

## § 5.1. Tensor products.

In this section we describe some aspects of the theory of tensor products of operators on non Archimedean Hilbert spaces. Let $A$ and $B$ be #-densely defined operators on non Archimedean Hilbert spaces $H_1^\#$ and $H_2^\#$ respectively. We will denote by $D(A) \otimes D(B)$ the set of hyperfinite linear combinations of vectors of the form $\varphi \otimes \psi$ where $\varphi \in D(A)$ and $\psi \in D(B)$. $D(A) \otimes D(B)$ is dense in $H_1^\# \otimes H_2^\#$ We define $A \otimes B$ on $D(A) \otimes D(B)$ by

$$A \otimes B(\varphi \otimes \psi) = A\varphi \otimes B\psi \qquad (5.1.1)$$

and extend by linearity.

**Proposition 5.1.1** The operator $A \otimes B$ is well defined. Further, if $A$ and $B$ are #-closable,

so is $A \otimes B$.

**Proof** Suppose that $\text{Ext}-\sum c_i \varphi_i \otimes \psi_i$ and $\text{Ext}-\sum d_j \varphi'_j \otimes \psi'_j$ are two representations of the same vector $f \in D(A) \otimes D(B)$. Using Gram-Schmidt orthogonalization we obtain bases $\{\eta_k\}$ and $\{\theta_l\}$ for the spaces spanned by $\{\varphi_i\} \cup \{\varphi'_i\}$ and $\{\psi_j\} \cup \{\psi'_j\}$ respectively so that $\eta_k \in D(A)$ and $\theta_l \in D(B)$. $\varphi_i \otimes \psi_i$ and $\varphi'_j \otimes \psi'_j$ can be expressed
unbounded self-#-adjoint operators in the next chapter; there is a spectral theorem for unbounded operators which we discuss in Section § 5.3)

This means that given a bounded self-#-adjoint operator A on a non-Archimedean Hilbert space $\mathbf{H}^\#$ over field $\mathbb{R}^\#$ or $\mathbb{C}^\#$, we can always find a #-measure $\mu^\#$ on a #-measure space M and a unitary operator $U : \mathbf{H}^\# \to L_2^\#(M, d^\#\mu^\#)$ so that

$$ (UAU^{-1})f(x) = F(x)f(x) \qquad (5.1.1) $$

for some bounded real-valued #-measurable function F on M.

In practice, M will be a union of copies of $\mathbb{R}_c^\#$ and F will be x so the core of the proof of the theorem will be the construction of certain #-measures. This will be done in Section

§ 5.2 by using the generalized Riesz-Markov theorem. Our goal in this section will be to

make sense out of $f(A)$, for $f$ a #-continuous function.

In the next section, we will consider the #-measures defined by the functionals

$$ f \mapsto \langle \psi, f(A)\psi \rangle_{\#} \qquad (5.1.2) $$

for fixed $\psi \in \mathbf{H}^\#$.

Given a fixed operator A, for which $f$ can we define $f(A)$? First, suppose that A is an arbitrary bounded in $\tilde{\mathbb{R}}_c^\#$ operator. If $f(x) = \text{Ext}-\sum_{n=1}^N c_n x^n$, $N \in \mathbb{N}$ is a polynomial, we let $f(A) = \text{Ext}-\sum_{n=1}^N c_n A^n$. Suppose that $f(x) = \text{Ext}-\sum_{n=1}^{\infty} c_n x^n$ is a hyper infinite power series with radius of #-convergence R. If $\|A\|_{\#} < R$ then hyper infinite power series $\text{Ext}-\sum_{n=1}^{\infty} c_n A^n$ #-converges in $\mathcal{L}(H^\#)$ so it is natural to set

$$ f(A) = \text{Ext}-\sum_{n=1}^{\infty} c_n A^n \qquad (5.1.3) $$

In this last case, $f$ was a function #-analytic in a domain including all of $\sigma(A)$.

The functional calculus we have talked about thus far works for any operator in any Banach space. The special property of self-adjoint operators or more generally normal operators is that $\|P(A)\|_{\#} = \sup_{\lambda \in \sigma(A)} |P(\lambda)|$ for any polynomial P, so that one can use the B.L.T. theorem to extend the functional calculus to #-continuous functions. Our major goal in this section is the proof of:

**Theorem 4.1.1.** (#-continuous functional calculus) Let A be a self-#-adjoint operator on a Hilbert space $H^\#$. Then there is a unique map $\phi : C^\#(\sigma(A)) \to \mathcal{L}(H^\#)$ with the following properties:

(a) $\phi$ is an algebraic $*$-homomorphism, that is,
$$ \phi(fg) = \phi(f)\phi(g), \phi(\lambda f) = \lambda\phi(f), \phi(1) = I, \phi(\tilde{f}) = \phi(f)^*. $$

(b) $\phi$ is #-continuous, that is, $\|\phi(f)\|_{\mathcal{L}(H^\#)} \le C\|f\|_{\infty}$.

(c) Let $f$ be the function $f(x) = x$; then $\phi(f) = A$.

Moreover, $\phi$ has the additional properties:

(d) If $A\psi = \lambda\psi$, then $\phi(f)\psi = f(\lambda)\psi$.

(e) $\sigma[\phi(f)] = \{f(\lambda)|\lambda \in \sigma(A)\}$ [spectral mapping theorem].

(f) If $f \ge 0$, then $\phi(f) \ge 0$.

(g) $\|\phi(f)\|_{\#} = \|f\|_{\infty}$. [this strengthens (b)].

The proof which we give below is quite simple, (a) and (c) uniquely determine $\phi(P)$ for any hyperfinite polynomial $P(x)$. By the generalized Weierstrass theorem, the set of polynomials is #-dense in $C^\#(\sigma(A))$ so the main part of the proof is
$$ \varphi_i \otimes \psi_i = \mathrm{Ext} - \sum_{kl} \alpha_{kl}^i (\eta_k \otimes \theta_l) \qquad (5.1.2) $$

and

$$ \varphi'_j \otimes \psi'_j = \mathrm{Ext} - \sum_{kl} \beta_{kl}^j (\eta_k \otimes \theta_l). \qquad (5.1.3) $$

Since the two expressions for $f$ give the same vector, $\mathrm{Ext} - \sum_i c_i \alpha_{kl}^i = \mathrm{Ext} - \sum_j d_j \beta_{kl}^j$ for each pair $(k, l)$. Thus,

$$ \begin{aligned} (A \otimes B)[\mathrm{Ext} - \sum_i c_i (\varphi_i \otimes \psi_i)] &= \mathrm{Ext} - \sum_{kl} (\mathrm{Ext} - \sum_i c_i \alpha_{kl}^i) (A\eta_k \otimes B\theta_l) = \\ \mathrm{Ext} - \sum_{kl} (\mathrm{Ext} - \sum_j d_j \beta_{kl}^j) (A\eta_k \otimes B\theta_l) &= (A \otimes B)[\mathrm{Ext} - \sum_j d_j (\varphi'_j \otimes \psi'_j)] \end{aligned} \qquad (5.1.4) $$

so $A \otimes B$ is well defined. If $g$ is any vector in $D(A^*) \otimes D(B^*)$ then $\langle (A \otimes B)f, g \rangle_{\#} = \langle f, (A^* \otimes B^*)g \rangle_{\#}$ so

$$ D(A^*) \otimes D(B^*) \subset D((A \otimes B)^*). \qquad (5.1.5) $$

If $A$ and $B$ are #-closable, $D(A^*)$ and $D(B^*)$ are #-dense. Therefore, in that case $(A \otimes B)^*$ is #-densely defined which proves that $A \otimes B$ is #-closable. Similarly, if $A$ and $B$ are #-closable then $A \otimes I + I \otimes B$ defined on $D(A) \otimes D(B)$ is #-closable.

**Definition 5.1.1.** Let $A$ and $B$ be #-closable operators on a non Archimedean Hilbert spaces $H_1^\#$ and $H_2^\#$. The tensor product of $A$ and $B$ is the #-closure of the operator $A \otimes B$ defined on $D(A) \otimes D(B)$. We will denote the #-closure by $A \otimes B$ also. Usually $A \oplus B$ will denote the #-closure of $A \otimes I + I \otimes B$ on $D(A) \otimes D(B)$.

**Proposition 5.1.1.** Let $A$ and $B$ be bounded in $\mathbb{R}_c^\#$ operators on a non Archimedean Hilbert spaces $H_1^\#$ and $H_2^\#$. Then $\|A \otimes B\|_{\#} = \|A\|_{\#} \times \|B\|_{\#}$.

**Proof** Let $\{\varphi_k\}$ and $\{\psi_l\}$ be orthonormal bases for $H_1^\#$ and $H_2^\#$ and suppose $\mathrm{Ext} - \sum_{kl} c_{kl} \varphi_k \otimes \psi_l$ is a gyperfinite sum. Then

$$ \begin{aligned} \|(A \otimes I)[\mathrm{Ext} - \sum_{kl} c_{kl} \varphi_k \otimes \psi_l]\|_{\#}^2 &= \mathrm{Ext} - \sum_l \|\mathrm{Ext} - \sum_k c_{kl} A \varphi_k\|_{\#}^2 = \\ &\le \mathrm{Ext} - \sum_l \|A\|_{\#}^2 (\mathrm{Ext} - \sum_k |c_{kl}|^2) = \|A\|_{\#}^2 \|\mathrm{Ext} - \sum_k c_{kl} \varphi_k \otimes \psi_l\|_{\#}^2. \end{aligned} \qquad (5.1.6) $$

Since the set of such gyperfinite sums is #-dense in $H_1^\# \otimes H_2^\#$, we conclude that $\|A \otimes I\|_{\#} \le \|A\|_{\#}$. Thus $\|A \otimes B\|_{\#} \le \|A \otimes I\|_{\#} \times \|B \otimes I\|_{\#} \le \|A\|_{\#} \times \|B\|_{\#}$. Conversely, given $\varepsilon \approx 0, \varepsilon > 0$, there exist unit vectors $\varphi \in H_1^\#$, $\psi \in H_2^\#$ so that

$$ \|A\varphi\|_{\#} \ge \|A\|_{\#} - \varepsilon \qquad (5.1.7) $$

and

$$ \|B\psi\|_{\#} \ge \|B\|_{\#} - \varepsilon. \qquad (5.1.8) $$

Then

$$ \|A \otimes B(\varphi \otimes \psi)\|_{\#} = \|A\varphi\|_{\#} \times \|B\psi\|_{\#} \ge \|A\|_{\#} \times \|B\|_{\#} - \varepsilon \|A\|_{\#} - \varepsilon \|B\|_{\#} + \varepsilon^2. \qquad (5.1.9) $$

Since $\varepsilon > 0$ is arbitrary $\|A \otimes B\|_{\#} \ge \|A\|_{\#} \times \|B\|_{\#}$, which concludes the proof.

**Remark 5.1.1.** We notice that both of the above propositions have natural generalizations to arbitrary hyperfinite tensor products of operators. This can be proven directly or by using the associativity of the hyperfinite tensor product of a non Archimedean Hilbert spaces.

**Remark 5.1.2.** We turn now to questions of self-adjointness and spectrum. Let $(A_k)_{k=1}^N$ be a hyperfinite family of operators, $A_k$ self-#-adjoint on $H_k^\#$. We will denote the
showing that

$$ \|P(A)\|_{op} = \|P(x)\|_{C^{\#}(\sigma(A))} = \sup_{\lambda \in \sigma(A)} |P(\lambda)|. \qquad (5.1.4) $$

The existence and uniqueness of $\phi$ then follow from the generalized B.L.T. theorem. To prove the crucial equality, we first prove a special case of (e) (which holds for arbitrary bounded operators):

**Lemma 4.1.1.** Let $P(x) = \text{Ext}-\sum_{n=1}^{N} c_n x^n$, $N \in \aleph$. Let $P(A) = \text{Ext}-\sum_{n=1}^{N} c_n A^n$. Then

$$ \sigma(P(A)) = \{P(\lambda)|\lambda \in \sigma(A)\}. \qquad (5.1.5) $$

**Proof** Let $\lambda \in \sigma(A)$. Since $x = \lambda$ is a root of $P(x) - P(\lambda)$, we have $P(x) - P(\lambda) = (x - \lambda)Q(x)$, so $P(A) - P(\lambda) = (A - \lambda)Q(A)$. Since $(A - \lambda)$ has no inverse neither does $P(A) - P(\lambda)$ that is, $P(\lambda) \in \sigma(P(A))$.

Conversely, let $\mu \in \sigma(P(A))$ and let $\lambda_1, \dots, \lambda_n$ be the roots of $P(x) - \mu$, that is, $P(x) - \mu = a(\text{Ext}-\prod_{i=1}^n (x - \lambda_i))$. If $\lambda_1, \dots, \lambda_n \notin \sigma(A)$, then

$$ (P(A) - \mu)^{-1} = a^{-1} (\text{Ext}-\prod_{i=1}^{n} (A - \lambda_i)^{-1}) \qquad (5.1.6) $$

so we conclude that some $\lambda_i \in \sigma(A)$ that is, $\mu = P(\lambda)$ for some $\lambda \in \sigma(A)$.

**Definition** Let $r(A) = \sup_{\lambda \in \sigma(A)} |\lambda|$. Then $r(A)$ is called the spectral radius of $A$.

**Theorem 4.1.2.** Let $X$ be a Banach space, $A \in \mathcal{L}(X)$ Then $\lim_{n \to \infty} \sqrt[n]{\|A^n\|_{op}}$ exists and is equal to $r(A)$. If $X$ is a Hilbert space and $A$ is self-#-adjoint, then $r(A) = \|A\|_{op}$.

**Lemma 4.1.2.** Let $A$ be a bounded in $\mathbb{R}_c^\#$ self-#-adjoint operator. Then

$$ \|P(A)\|_{\#} = \sup_{\lambda \in \sigma(A)} |P(\lambda)|. \qquad (5.1.7) $$

**Proof** By Theorem 5.1.2 and by Lemma 5.1.1 we obtain

$$ \begin{aligned} \|P(A)\|_{\#}^2 &= \|P(A)^*P(A)\|_{\#} = \|(\overline{PP})(A)\|_{\#} = \\ &= \sup_{\lambda \in \sigma(\overline{PP}(A))} |\lambda| = \sup_{\lambda \in \sigma(A)} |\overline{PP}(\lambda)| = \left( \sup_{\lambda \in \sigma(A)} |P(\lambda)| \right)^2. \end{aligned} \qquad (5.1.8) $$

**Proof of Theorem 5.1.1.** Let $\phi(P) = P(A)$. Then $\|\phi(P)\|_{\mathcal{L}(H^\#)} = \|P\|_{C^\#(\sigma(A))}$ so $\phi$ has a unique linear extension to the #-closure of the polynomials in $C^\#(\sigma(A))$. Since the polynomials are an algebra containing $I$, containing complex conjugates, and separating points, this #-closure is all of $C^\#(\sigma(A))$. Properties (a), (b), (c), (g) are obvious and if $\tilde{\phi}$ obeys (a), (b), (c) it agrees with $\phi$ on polynomials and thus by #-continuity on $C^\#(\sigma(A))$. To prove (d), note that $\phi(P)\psi = P(\lambda)\psi$ and apply #-continuity. To prove (f), notice that if $f \ge 0$, then $f = g^2$ with $g \in C^\#(\sigma(A))$. Thus $\phi(f) = \phi(g)^2$ with $\phi(g)$ self-#-adjoint, so $\phi(f) \ge 0$.

**Remark 5.1.1.** In addition:

(1) $\phi(f) \ge 0$ if and only if $f \ge 0$.

(2) Since $fg = gf$ for all $f, g$, $\{f(A)|f \in C^\#(\sigma(A))\}$ forms an abelian algebra closed under adjoints. Since $\|\phi(f)\|_{\#} = \|f\|_{\infty}$ and $C^\#(\sigma(A))$ is #-complete, $\{f(A)|f \in C^\#(\sigma(A))\}$ is #-norm-#-closed. It is thus an non-Archimedean abelian $C^*$ algebra of operators.

(3) $\text{Ran}(\phi)$ is actually the non-Archimedean $C^*$ algebra generated by $A$ that is, the smallest $C^*$-algebra containing $A$.

(4) This result, that $C^\#(\sigma(A))$ and the non-Archimedean $C^*$-algebra generated by $A$ are #-isometrically isomorphic
\#-closure of $I_1 \otimes \cdots \otimes A_k \otimes \cdots \otimes I_N$ on $D = \text{Ext}-\otimes_{k=1}^N D(A_k)$ by $A_k$ also. Let $P(x_1, \ldots, x_N)$ be a polynomial with $\mathbb{R}_c^\#$-valued coefficients of degree $n_k$ in $x_k$. Then, the operator $P(A_1, \ldots, A_N)$ makes sense on $\text{Ext}-\otimes_k D(A_{n_k})$ since $D(A_{n_k}) \subset D(A_l)$ for all $l \le n_k$. In fact, $P(A_1, \ldots, A_N)$ is essentially self-\#-adjoint on that domain.

**Theorem 5.1.1.** Let $A_k$ be a self-\#-adjoint operator on $H_k^\#$. Let $P(x_1, \ldots, x_N)$ be a polynomial with $\mathbb{R}_c^\#$-valued coefficients of degree $n_k$ in the $k$-th variable and suppose that $D_k^l$ is a domain of essential self-\#-adjointness for $A_k^{n_k}$. Then,
(a) $P(A_1, \ldots, A_N)$ is essentially self-adjoint on $D^l = \text{Ext}-\otimes_{k=1}^N D_k^l$.

(b) The spectrum of $\overline{\text{Ext}-P(A_1, \ldots, A_N)}$ is the \#-closure of the range of $P(A_1, \ldots, A_N)$ on the product of the spectra of the $A_k$. That is $\sigma(\overline{\text{Ext}-P(A_1, \ldots, A_N)}) = \overline{\text{Ext}}(\sigma(A_1), \ldots, \sigma(A_N))$.

**Proof** We will first prove that $P(A_1, \ldots, A_N)$ is essentially self-\#-adjoint on $D = \text{Ext}-\otimes_{k=1}^N D(A_k^{n_k})$. By the spectral theorem, there is a \#-measure space $(M_k, \mu_k^\#)$ so that $A_k$ is unitarily equivalent to multiplication by a $\mathbb{R}_c^\#$-valued \#-measurable function $f_k$ on $L_2^\#(M_k, d^\# \mu_k^\#)$. Thus we may assume that $\mu_k^\#$ is hyperfinite and that $f_k \in \bigcap_{1 \le p < *\infty} L_p^\#(M_k, d^\# \mu_k^\#)$. Furthermore $\text{Ext}-\otimes_{k=1}^N L_2^\#(M_k, d^\# \mu_k^\#)$ is naturally isomorphic to $L_2^\#(\text{Ext}-\otimes_{k=1}^N M_k, \text{Ext}-\otimes_{k=1}^N d^\# \mu_k^\#)$. Under this isomorphism $P(A_1, \ldots, A_N)$ corresponds to multiplication by $P(f_1, \ldots, f_N)$ and $D$ corresponds to the set of hyperfinite linear combinations of hyperfinite linear combinations of functions $\text{Ext}-\prod_{i=1}^N \phi_i(m_i)$ such that $f_k^{n_k} \phi_k \in L_2^\#(M_k, d^\# \mu_k^\#)$.

To prove essential self-\#-adjointness we use result from functional calculus. First, since $\mu_k^\#$ is hyperfinite and $f_k^{n_k} \phi_k \in L_2^\#(M_k, d^\# \mu_k^\#)$ we conclude that $f_k^l \in L_p^\#(M_k, d^\# \mu_k^\#)$ for $1 \le p < *\infty$. From this it follows immediately that $P(f_1, \ldots, f_N)$ is in $L_p^\#$ for all such $p$; in particular $P(f_1, \ldots, f_N) \in L_4^\#(\text{Ext}-\otimes_{k=1}^N M_k, \text{Ext}-\otimes_{k=1}^N d^\# \mu_k^\#)$. Since $f_k^{n_k}$ is self-\#-adjoint on $D_k$, $D_k$ contains the characteristic functions of \#-measurable sets in $M_k$.

Thus $D$ contains all hyperfinite linear combinations of the characteristic functions of rectangles. By the property on product \#-measures we conclude that the characteristic function of any \#-measurable set in $M_k$ is equal to such a hyperfinite linear combination except on a set of arbitrarily small $\text{Ext}-\otimes_{k=1}^N d^\# \mu_k^\#$ \#-measure. Thus the simple functions on $\text{Ext}-\otimes_{k=1}^N M_k$ can be approximated in the $L_p^\#$ sense with $1 \le p < *\infty$ by elements of $D$. In particular $D$ is \#-dense in $L_4^\#(\text{Ext}-\otimes_{k=1}^N M_k, \text{Ext}-\otimes_{k=1}^N d^\# \mu_k^\#)$. Essential self-\#-adjointness now follows from Proposition 5.1.2.

To show that $P(A_1, \ldots, A_N)$ is essentially self-adjoint on $D^l$ we need only show that $\overline{\text{Ext}-P(A_1, \ldots, A_N)} \uparrow D^l$ extends $P(A_1, \ldots, A_N) \uparrow D$. Suppose $\text{Ext}-\otimes_{k=1}^N \phi_k \in D$. Then $\phi_k \in D(A_k^{n_k})$, so since $D_k^l$ is a domain of essential self-\#-adjointness of $A_k^{n_k}$ there is a hyper infinite sequence $(\phi_k^l)_{l=1}^{*\infty}$ so that $\phi_k^l \to_* \phi_k$ and $A_k^{n_k} \phi_k^l \to_* A_k^{n_k} \phi_k$. An easy estimate shows that this implies that $A_k^m \phi_k^l \to_* A_k^m \phi_k$ for all $1 \le m \le n_k$. Therefore $\text{Ext}-\otimes_{k=1}^N \phi_k^l \to_* \text{Ext}-\otimes_{k=1}^N \phi_k$ and $P(A_1, \ldots, A_N)(\text{Ext}-\otimes_{k=1}^N \phi_k^l) \to_* P(A_1, \ldots, A_N)(\text{Ext}-\otimes_{k=1}^N \phi_k)$ The same argument works for hyperfinite linear combinations of vectors of the form $\text{Ext}-\otimes_{k=1}^N \phi_k$ so $\overline{\text{Ext}-P(A_1, \ldots, A_N)} \uparrow D^l$ extends $P(A_1, \ldots, A_N) \uparrow D$. This completes the proof of (a).

To prove (b), suppose that $\lambda \in \sigma(\overline{\text{Ext}-P(A_1, \ldots, A_N)})$. If $I$ is any \#-open interval about $\lambda$ then $P^{-1}(A_1, \ldots, A_N)(I)$ contains a product $\text{Ext}-\otimes_{k=1}^N I_k$ of open \#-intervals so that $I_k \cap \sigma(A_k) \neq \emptyset$. Since $\sigma(A_k) = \text{\#-ess range}(f_k^{n_k})$, $\mu_k^\#[(f_k^{n_k})^{-1}(I_k)] \neq 0$ so $\mu[P(f_1, \ldots, f_N)(I)] \neq 0$. That is, $\lambda \in \text{\#-ess range}(P(f_1, \ldots, f_N))$ which equals
(5) (b) actually follows from (a) and Proposition 4.1.1. Thus (a) and (c) alone determine $\phi$ uniquely.

**Proposition 5.1.1.** Suppose that $\phi: C^\#(X) \to \mathcal{L}(H^\#)$ is an algebraic $*$-homomorphism, $X$ a $*$-compact metric space. Then

(a) If $f \ge 0$, then $\phi(f) \ge 0$.

(b) $\|\phi(f)\|_\# \le \|f\|_{*\infty}$.

**Definition 5.1.1** if $n, k \in *\mathbb{N}$ with $k \le n$, then we define

$$ \binom{n}{k} = \frac{n!^{\#}}{k!^{\#}(n-k)!^{\#}} \qquad (5.1.8) $$

where $n!^{\#} = \mathrm{Ext}-\prod_{1 \le m \le n} m$, see ref [7].

**Lemma 5.1.3.** Whenever $n, k \in *\mathbb{N}$ are such that $k \le n$, then

$$ \binom{n}{k} = \binom{n}{n-k}. \qquad (5.1.9) $$

**Proof.** Directly from the formula (4.1.8)

$$ \binom{n}{n-k} = \frac{n!^{\#}}{(n-k)!^{\#}[n-(n-k)]!^{\#}} = \frac{n!^{\#}}{(n-k)!^{\#}k!^{\#}} = \frac{n!^{\#}}{k!^{\#}(n-k)!^{\#}} = \binom{n}{k}. \qquad (5.1.10) $$

**Lemma 5.1.4.** Let $n, k \in *\mathbb{N}$ with $0 < k < n$, then

$$ \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}. \qquad (5.1.11) $$

**Proof.** Directly by hyper infinite induction [7].

**Proposition 5.1.2.** (Generalized binomial theorem) Let $x, y \in *\mathbb{R}_c^\#$ and let $n \in *\mathbb{N}$, then we have

$$ (x + y)^n = \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k = \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}. \qquad (5.1.8) $$

**Proof.** We prove the result by hyper infinite induction. When $n = 1$, we trivially have

$$ (x + y)^1 = x + y = \binom{1}{0}x + \binom{1}{1}y. $$

Suppose that there is an $n \in *\mathbb{N}$ for which the statement (4.1.8) is true. We then have

$$ \begin{aligned} (x+y)^{n+1} &= (x+y)^n(x+y) = \\ & \quad [\mathrm{Ext}-\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k](x+y) = \\ & \quad [\mathrm{Ext}-\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k]x + [\mathrm{Ext}-\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k]y = \\ & \quad [\mathrm{Ext}-\sum_{k=0}^n \binom{n}{k} x^{n+1-k} y^k] + [\mathrm{Ext}-\sum_{k=0}^n \binom{n}{k} x^{n-k} y^{k+1}] = \\ & \quad x^{n+1} + \mathrm{Ext}-\sum_{k=1}^n \left[\binom{n}{k} + \binom{n}{k-1}\right]x^{n-k}y^k + y^{n+1} = \\ & \quad \binom{n+1}{0}x^{n+1} + \mathrm{Ext}-\sum_{k=1}^n \left[\binom{n}{k} + \binom{n}{k-1}\right]x^{n-k}y^k + \binom{n+1}{n+1}y^{n+1} = \\ & \quad = \mathrm{Ext}-\sum_{k=0}^{n+1} \binom{n+1}{k}x^{n+1-k}y^k \end{aligned} $$

where we have used Lemma 4.1.4.

**Definition 5.1.2 (Hyperfinite Bernstein Polynomials).** For each $n \in *\mathbb{N}$, the $n$-th Bernstein Polynomial $B_n^\#(x, f)$ of a function $f \in C^\#([a, b], *\mathbb{R}_c^\#)$ is defined as
$\sigma(\#\overline{P(A_1, \dots, A_N)})$.

Conversely if $\lambda \notin \#\overline{P(\sigma(A_1), \dots, \sigma(A_N))}$ then $(\lambda - P(f_1, \dots, f_N))^{-1}$ is bounded #-a.e. on $Ext \times_{k=1}^N M_k$ so $\lambda \in \rho(\#\overline{P(A_1, \dots, A_N)})$.

**Remark 5.1.3.** If $A_1, \dots, A_N, N \in *\mathbb{N}$ are bounded in $*\mathbb{R}_c^\#$, $P(A_1, \dots, A_N)$ is #-closed, but in general it is not.

**Corollary 5.1.1.** Let $A_1, \dots, A_N, N \in *\mathbb{N}$ be self-#-adjoint operators on $H_1^\#, \dots, H_N^\#$ and suppose that, for each $k, D_k$ is a domain of essential self-#-adjointness for $A_k$. Then, (a) The operators $A_\pi = \text{Ext} \otimes_{k=1}^N A_k$ and $A_\Sigma = \text{Ext} - \sum_{k=1}^N A_k$ are essentially self-#-adjoint on $D = \text{Ext} \otimes_{k=1}^N D_k$.

(b) $\sigma(A_\pi) = \#\overline{\text{Ext} - \prod_{k=1}^N \sigma(\overline{A}_k)}$ and $\sigma(A_\Sigma) = \#\overline{\text{Ext} - \sum_{k=1}^N \sigma(\overline{A}_k)}$.

**Example 5.1.1.** Suppose that $V(x)$ is a potential so that $H_1 = -\nabla_x^\# + V(x)$ is essentially self-#-adjoint on $S^\#(*\mathbb{R}_c^3)$. Then $H_2 = -\nabla_x^\# + V(x) - \nabla_y^\# + V(y)$ is essentially self-#-adjoint on the set of hyperfinite sums of products $\varphi(x)\psi(y)$, with $\varphi, \psi \in S^\#(*\mathbb{R}_c^3)$. Further $\sigma(H_2) = \#\overline{\sigma(H_1) + \sigma(H_1)}$. is obvious.

## § 5.2. Non-Archimedean Fock spaces.

Let $H^\#$ be a non-Archimedean Hilbert space and denote by $H^{\#n}, n \in *\mathbb{N}$ the $n$-fold tensor product $H^{\#n} = \text{Ext} \otimes_{k=1}^n H^\#$ and define

$$ \mathcal{F}^{\#}(H^{\#}) = \text{Ext} - \bigotimes_{n=0}^{\infty} H^{\#n} \qquad (5.2.1) $$

$\mathcal{F}^{\#}(H^{\#})$ is called a non-Archimedean Fock space over $H^{\#}$; it will be *-separable if $H^{\#}$ is. For example, if $H^{\#} = L_2^\#(*\mathbb{R}_c^3)$, then an element $\psi \in \mathcal{F}(H^{\#})$ is a hyper infinite sequence of functions

$$ \psi = \{\psi_0, \psi_1(x_1), \psi_1(x_1, x_2), \psi_1(x_1, x_2, x_3), \dots \} \qquad (5.2.2) $$

so that

$$ |\psi_0|^2 + \text{Ext} - \sum_{n=1}^{\infty} \left( \text{Ext} - \int_{*\mathbb{R}_c^{\#n}} \psi_n(x_1, \dots, x_n) d^{\#n}x \right) < *\infty, \qquad (5.2.3) $$

where $\psi_0 \in *\mathbb{C}_c^\#$, $d^{\#n}x = \text{Ext} - \prod_{i=1}^n d^{\#n}x_i$. Actually, it is not $\mathcal{F}^{\#}(H^{\#})$ itself, but two of its subspaces which are used in quantum field theory. These two subspaces are constructed as follows: Let $\mathbf{P}_n, n \in *\mathbb{N}$ be the permutation group on $n \in *\mathbb{N}$ elements and let $\{\varphi_k\}_{k=1}^{*\infty}$ be a basis for $H^{\#}$. For each $\sigma \in \mathbf{P}_n$ we define an operator (which we also denote by a) on basis elements of $H^{\#(n)}, n \in *\mathbb{N}$ by

$$ \sigma(\text{Ext} - \bigotimes_{i=1}^n \varphi_{k_i}) = \text{Ext} - \bigotimes_{i=1}^n \varphi_{k_{\sigma(i)}} \qquad (5.2.4) $$

a extends by linearity to a bounded in $*\mathbb{R}_c^\#$ operator (of #-norm one) on $H^\#$ so we can define

$$ \mathbf{S}_n = \left( \frac{1}{n!^{\#}} \right) \text{Ext} - \sum_{\sigma \in \mathbf{P}_n} \sigma. \qquad (5.2.5) $$

It is easy to show that $\mathbf{S}_n^2 = \mathbf{S}_n$ and $\mathbf{S}_n^* = \mathbf{S}_n$, so $\mathbf{S}_n$ is an orthogonal projection. The range of $\mathbf{S}_n$ is called the $n$-fold symmetric tensor product of $H^{\#}$. In the case where $H^{\#} = L_2^\#(*\mathbb{R}_c^\#)$ and $H^{\#n} = \text{Ext} \otimes_{k=1}^n L_2^\#(*\mathbb{R}_c^\#)$, $\mathbf{S}_n H^{\#n}$ is just the subspace of $L_2^\#(*\mathbb{R}_c^\#)$ of all functions left invariant under any permutation of the variables. We now define
$$B_n^\#(x,f) = \mathrm{Ext}-\sum_{k=0}^n f\left(\frac{k}{n}\right) \binom{n}{k} x^k (1-x)^{n-k}. \qquad (5.1.9)$$

**Lemma 5.1.3.** For any $n \in *\mathbb{N}$:

$$\mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} x^k (1-x)^{n-k} = 1, \qquad (5.1.10)$$

$$\mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} x^k (1-x)^{n-k} (k - nx) = 0, \qquad (5.1.11)$$

$$\mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} x^k (1-x)^{n-k} (k - nx)^2 = nx(1-x). \qquad (5.1.12)$$

**Proof.** To prove these identities, first, from the generalized binomial theorem, for any $n \in *\mathbb{N}$ one obtains that

$$\mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} x^k (1-x)^{n-k} = [x+(1-x)]^n = 1. \qquad (5.1.13)$$

By the generalized binomial theorem we have

$$\mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} y^k z^{n-k} = (y+z)^n. \qquad (5.1.14)$$

By the #-differentiating with respect to $y$ of the identity (4.1.14) we obtain

$$\frac{d^\#}{d^\# y} \left[ \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} y^k z^{n-k} \right] = \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} k y^{k-1} z^{n-k} = \\ = \frac{d^\#}{d^\# y} (y+z)^n = n(y+z)^{n-1}. \qquad (5.1.15)$$

Thus

$$\mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} k y^{k-1} z^{n-k} = n(y+z)^{n-1} \qquad (5.1.16)$$

and therefore

$$\mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} k y^k z^{n-k} = ny(y+z)^{n-1}. \qquad (5.1.17)$$

Replacing $y$ by $x$ and $z$ by $1-x$ in the above expression, we have identity

$$\mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} k x^k (1-x)^{n-k} = nx. \qquad (5.1.18)$$

From (5.1.18) we obtain

$$\mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} \frac{k}{n} x^k (1-x)^{n-k} = x. \qquad (5.1.19)$$

From (5.1.19) and (5.1.13) we obtain the identity (5.1.11). By the #-differentiating with respect to $y$ of the identity (5.1.17) we obtain

$$\frac{d^\#}{d^\# y} \left[ \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} k y^k z^{n-k} \right] = \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} k^2 y^{k-1} z^{n-k} = \\ = n \frac{d^\#}{d^\# y} y (y+z)^{n-1} = n(y+z)^{n-1} + n(n-1)y(y+z)^{n-2}. \qquad (5.1.20)$$

Thus

$$\mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} k^2 y^{k-1} z^{n-k} = n(y+z)^{n-1} + n(n-1)y(y+z)^{n-2}. \qquad (5.1.21)$$

and therefore
$$ \mathcal{F}_s^{\#}(H^\#) = \mathrm{Ext} - \bigoplus_{n=0}^{\infty} S_n H^{\#n} \qquad (5.2.6) $$

$\mathcal{F}_s^{\#}(H^\#)$ is called the symmetric non Archimedean Fock space over $H^\#$ or the non Archimedean Boson Fock space over $H^\#$.

## § 5.3. Second quantization of the free Hamiltonian.

Let $H^\#$ be a non Archimedean Hilbert space, $\mathcal{F}^{\#}(H^\#)$ the associated non Archimedean Fock space over $H^\#$. Suppose that $A$ is a self-#-adjoint operator on $H^\#$ with a domain of

essential self-#-adjointness $D$. Corresponding to each such $A$ we can define an operator $d\Gamma^{\#}(A)$ on $\mathcal{F}^{\#}(H^\#)$ as follows. Let

$$ A^{(n)} = A \otimes I \otimes \cdots \otimes I + I \otimes A \otimes \cdots \otimes I + I \otimes \cdots \otimes I \otimes A \qquad (5.3.1) $$

on $\mathrm{Ext}\otimes_{i=1}^n D$ as follows. Let $D_A \subset \mathcal{F}^{\#}(H^\#)$ be the set of $\{\psi_0, \psi_1, \dots\}$ such that $\psi_n = 0$ for $n$ large enough and $\psi_n \in \mathrm{Ext}\otimes_{k=1}^n D$ for each $n$. $D_A$ is #-dense in $\mathcal{F}^{\#}(H^\#)$ since $D$ is #-dense in $H^\#$. Define $A^{(0)} = 0$ and $d\Gamma^{\#}(A) = \mathrm{Ext}\sum_{n=0}^{\infty} A^{(n)}$. $d\Gamma^{\#}(A)$ makes sense on $D_A$ and obviously to be symmetric. By Theorem 5.1.1, $A^{(n)}$ is essentially self-#-adjoint on $\mathrm{Ext}\otimes_{k=1}^n D$. Thus $A^{(n)} + \mu i$ has a #-dense range on $\mathrm{Ext}\otimes_{k=1}^n D$ whenever $\mu \in *\mathbb{R}_c^\#$ and $\mu \neq 0$. From this it follows that $d\Gamma^{\#}(A) \pm i$ has a #-dense range on $D_A$. Thus $d\Gamma^{\#}(A)$ is essentially self-#-adjoint on $D_A$. If $A$ is the quantum mechanical operator which corresponds to the free energy, $d\Gamma^{\#}(A)$ is called the second quantization of the free energy. $d\Gamma^{\#}(A)$ commutes with the projections onto the symmetric and antisymmetric non Archimedean Fock spaces and it follows that $d\Gamma^{\#}(A) \upharpoonright \mathcal{F}_s^{\#}(H^\#)$ and $d\Gamma^{\#}(A) \upharpoonright \mathcal{F}_a^{\#}(H^\#)$

are essentially self-#-adjoint on $D \cap \mathcal{F}_s^{\#}(H^\#)$ and $D \cap \mathcal{F}_a^{\#}(H^\#)$ respectively.

# Chapter IV. Non-Archimedean Banach spaces endowed with *$\mathbb{R}_c^\#$-valued norm.

## 1. Definitions and examples

A non-Archimedean normed space with *$\mathbb{R}_c^\#$-valued norm (#-norm) is a pair $(X, \|\cdot\|_\#)$ consisting of a vector space $X$ over a non-Archimedean scalar field *$\mathbb{R}_c^\#$* or complex field *$\mathbb{C}_c^\#$* together with a distinguished norm $\|\cdot\|_\# : X \to *\mathbb{R}_c^\#$. Like any norms, this #-norm induces a translation invariant distance function, called the canonical or (norm) induced non-Archimedean *$\mathbb{R}_c^\#$-valued metric for all vectors $x, y \in X$, defined by

$$ d^{\#}(x, y) = \|x - y\|_{\#} = \|y - x\|_{\#}. \qquad (1.1) $$

Thus (1.1) makes $X$ into a metric space $(X, d^\#)$. A hyper infinite sequence $(x_n)_{n=1}^{\infty}$ is called $d^\#$-Cauchy or Cauchy in $(X, d^\#)$ or $\|\cdot\|_\#$-Cauchy if for every hyperreal $r \in *\mathbb{R}_c^\#$, $r > 0$, there exists some $N \in \mathbb{N}^\#$ such that

$$ d^{\#}(x_n, x_m) = \|x_n - x_m\|_{\#} < r, \qquad (1.2) $$

where $m$ and $n$ are greater than $N$. The canonical metric $d^\#$ is called a #-complete metric if the pair $(X, d^\#)$ is a #-complete metric space, which by definition means for every $d^\#$-Cauchy sequence $(x_n)_{n=1}^{\infty}$ in $(X, d^\#)$, there exists some $x \in X$ such that
$$ \mathrm{Ext} - \sum_{k=0}^{n} \binom{n}{k} k^2 y^k z^{n-k} = ny(y+z)^{n-1} + n(n-1)y^2(y+z)^{n-2}. \quad (5.1.22) $$

Replacing $y$ by $x$ and $z$ by $1-x$ in the expression (5.1.22) we have identity

$$ \mathrm{Ext} - \sum_{k=0}^{n} \binom{n}{k} k^2 x^k (1-x)^{n-k} = nx + n(n-1)x^2. \quad (5.1.23) $$

From (5.1.23) and (5.1.10)-(5.1.11) one obtains (5.1.12).

**Theorem 5.1.2.** (Generalized Weierstrass Approximation Theorem). Let $f \in C^#([a, b], *\mathbb{R}_c^#)$. Then there is a hyper infinite sequence of polynomials $p_n(x), n \in *\mathbb{N}$ that #-converges uniformly to $f(x)$ on $[a, b]$.

**Proof.** For a #-continuous function $f$ defined on $[0, 1]$ by (4.1.9) and (4.1.10) we obtain

$$ f(x) - B_n^\#(x) = \mathrm{Ext} - \sum_{k=0}^{n} \left[ f(x) - f\left(\frac{k}{n}\right) \right] \binom{n}{k} x^k (1-x)^k. \quad (4.1.24) $$

From the identity (4.1.24) one obtains that

$$ |f(x) - B_n^\#(x)| \le \mathrm{Ext} - \sum_{k=0}^{n} |f(x) - f\left(\frac{k}{n}\right)| \binom{n}{k} x^k (1-x)^k. \quad (4.1.25) $$

Since $f$ is #-continuous on $[0,1]$, it is bounded in $*\mathbb{R}_c^#$ on $[0,1]$, i.e., there exists a number $M \in *\mathbb{R}_c^#$ such that $|f(x)| \le M, x \in [0,1]$. Moreover $f$ is uniformly #-continuous on $[0,1]$. Choose $\varepsilon \approx 0, \varepsilon > 0$, then there exists $\delta \approx 0, \delta > 0$ such that $x, y \in [0,1]$ with $|x-y| \le \delta$ implies that $|f(x)-f(y)| \le \varepsilon$. For $x \in [0,1]$, split the sum in the righthand side of (4.1.1) into two parts:

$$ \mathrm{Ext} - \sum_{|x-k/n| \le \delta} |f(x) - f\left(\frac{k}{n}\right)| \binom{n}{k} x^k (1-x)^k \quad (4.1.12) $$

and

$$ \mathrm{Ext} - \sum_{|x-k/n| > \delta} |f(x) - f\left(\frac{k}{n}\right)| \binom{n}{k} x^k (1-x)^k \quad (4.1.13) $$

From (4.1.9) we obtain

$$ \mathrm{Ext} - \sum_{|x-k/n| \le \delta} |f(x) - f\left(\frac{k}{n}\right)| \binom{n}{k} x^k (1-x)^k \le \varepsilon. \quad (4.1.14) $$

From (4.1.9) we obtain

$$ \begin{aligned} \mathrm{Ext} - \sum_{|x-k/n| > \delta} |f(x) - f\left(\frac{k}{n}\right)| \binom{n}{k} x^k (1-x)^k &\le 2M \left[ \mathrm{Ext} - \sum_{|x-k/n| > \delta} \binom{n}{k} x^k (1-x)^k \right] \\ &\le \frac{2M}{\delta^2} \left[ \mathrm{Ext} - \sum_{|x-k/n| > \delta} \left(x - \frac{k}{n}\right)^2 \binom{n}{k} x^k (1-x)^k \right] \\ &\le \frac{2M}{\delta^2} \left[ \mathrm{Ext} - \sum_{k=0}^n \left(x - \frac{k}{n}\right)^2 \binom{n}{k} x^k (1-x)^k \right] \le \frac{2M}{n\delta^2}. \end{aligned} \quad (4.1.15) $$

Finally we obtain

$$ |f(x) - B_n^\#(x)| \le \varepsilon + \frac{2M}{n\delta^2}. \quad (4.1.16) $$

By choosing $n \in *\mathbb{N}$ large enough the righthand side can be made less than $2\varepsilon$. This estimate is independent of $x \in [0,1]$. Hence, for $\varepsilon > 0$ there exists a number $N \in *\mathbb{N}$ such that $n \ge N$ and $x \in [0,1]$ imply $|f(x) - B_n^\#(x)| \le 2\varepsilon$. Therefore $f$ is the uniform #-limit of the polynomials $B_n^\#$.
$$ \#\lim_{n \to \infty} \|x_n - x\| = 0 \qquad (1.3) $$

where because $\|x_n - x\| = d^\#(x_n, x)$, this hyper infinite sequence's #-convergence to $x$ can equivalently be expressed as: $\#\lim_{n \to \infty} x_n = x$ in $(X, d^\#)$.

**Definition 1.1.** The normed space $(X, \|\cdot\|_\#)$ is a non-Archimedean Banach space endroved with $\#\mathbb{R}_c^\#$-valued norm if the #-norm induced metric $d^\#$ is a #-complete metric, or said differently, if $(X, d^\#)$ is a #-complete metric space. The #-norm $\|\cdot\|_\#$ of a #-normed space $(X, \|\cdot\|_\#)$ is called a #-complete #-norm if $(X, \|\cdot\|_\#)$ is a non-Archimedean Banach space endroved with $\#\mathbb{R}_c^\#$-valued #-norm.

**Remark 1.1.** For any #-normed space $(X, \|\cdot\|_\#)$, there exists an L-semi-inner product $\langle\cdot, \cdot\rangle_\# : X \times X \to \#\mathbb{R}_c^\#$ such that $\|x\|_\# = \sqrt{\langle x, x\rangle_\#}$ for all $x \in X$; in general, there may be infinitely many L-semi-inner products that satisfy this condition. L-semi-inner products are a generalization of inner products, which are what fundamentally distinguish non-Archimedean Hilbert spaces from all other non-Archimedean Banach spaces. Characterization in terms of hyper infinite series, see ref. [1].

The vector space structure allows one to relate the behavior of hyper infinite Cauchy sequences to that of #-converging hyper infinite series of vectors.

**Remark 1.2.** A #-normed space $X$ is a non-Archimedean Banach space if and only if each absolutely #-convergent hyper infinite series $\text{Ext}-\sum_{n=1}^\infty v_n$ in $X$ #-converges in $X$, i.e., $\text{Ext}-\sum_{n=1}^\infty \|v_n\| < \infty^\#$ implies that $\text{Ext}-\sum_{n=1}^\infty v_n$ #-converges in $X$.

## 2. Linear operators, isomorphisms

If $X$ and $Y$ are #-normed spaces over the same ground field $\#\mathbb{R}_c^\#$, the set of all #-continuous $\#\mathbb{R}_c^\#$-linear maps $T: X \to Y$ is denoted by $B^\#(X, Y)$. In hyper infinite-dimensional spaces, not all linear maps are #-continuous. A linear mapping from a #-normed space $X$ to another normed space is #-continuous if and only if it is bounded or hyper bounded on the #-closed unit ball of $X$. Thus, the vector space $B^\#(X, Y)$ can be endroved with the operator norm

$$ \|T\| = \sup\{\|Tx\|_{Y} \mid x \in X, \|x\|_{X} \le 1\}. \qquad (2.1) $$

For $Y$ a non-Archimedean Banach space, the space $B^\#(X, Y)$ is a Banach space with respect to this #-norm.

If $X$ is a non-Archimedean Banach space, the space $B^\#(X) = B^\#(X, X)$ forms a unital Banach algebra; the multiplication operation is given by the composition of linear maps.

**Definition 2.1.** If $X$ and $Y$ are #-normed spaces, they are #-isomorphic #-normed spaces

if there exists a linear bijection $T: X \to Y$ such that $T$ and its inverse $T^{-1}$ are #-continuous. If one of the two spaces $X$ or $Y$ is #-complete then so is the other space. Two #-normed spaces $X$ and $Y$ are #-isometrically isomorphic if in addition, $T$ is an #-isometry, that is, $\|T(x)\| = \|x\|$ for every $x \in X$.

**Definition 2.2.** Let $\{X, \|\cdot\|\}$ be standard Banach space. For $x \in X$ and $\varepsilon > 0$, $\varepsilon \approx 0$ we define the open $\varepsilon$-ball about $x$ of radius $\varepsilon$ to be the set

$$ B_{\varepsilon}(x) = \{y \in X \mid \|x - y\| < \varepsilon\}. $$

**Definition 2.3.** Let $\{X, \|\cdot\|\}$ be standard Banach space, $Y \subset X$ thus $Y \subseteq X$ and let
**Theorem 4.1.3.** (Generalized B.L.T.theorem) Suppose that Z is a #-normed space, Y is a non-Archimedean Banach space, and $S \subset Z$ is a #-dense linear subspace of Z. If $T: S \to Y$ is a bounded linear transformation (i.e. there exists $C < *\infty$ such that $\|Tz\|_{\#} \le C \|z\|_{\#}$ for all $z \in S$), then T has a unique extension to an element of $\mathcal{L}(Z, Y)$.

## § 5.2. The spectral #-measures

**Theorem 4.2.1.** (Generalized Riesz-Markov theorem) Let $X$ be a locally #-compact non-Archimedean metric space endowed with $*\mathbb{R}_c^\#$-valued metric. Let $C_c^\#(X)$ be the space of #-continuous #-compactly supported $*C_c^\#$-valued functions on $X$. For any positive linear functional $\Phi$ on $C_c^\#(X)$, there is a unique #-measure $\mu^\#$ on $X$ such that

$$ \forall f \in C_c^\#(X) : \Phi(f) = \mathrm{Ext} - \int_X f(x) d^\# \mu^\#(x). $$

**Theorem 4.2.2.** (Generalized Riesz lemma) Let $Y$ be a #-closed proper vector subspace of a normed space $(X, \|\cdot\|_{\#})$ and let $\alpha \in *\mathbb{R}_c^\#$ be any real number satisfying $0 < \alpha < 1$. Then there exists a vector $u \in X$ of unit #-norm $\|u\|_{\#} = 1$ such that $\|u - y\|_{\#} \ge \alpha$ for all $y \in Y$.

We are now introduce the #-measures corresponding to bounded in $*\mathbb{R}_c^\#$ self-#-adjoint operators. Let $A$ be an bounded in $*\mathbb{R}_c^\#$ self-#-adjoint operator. Let $\psi \in \mathbf{H}^\#$. Then

$$ f \mapsto \langle \psi, f(A)\psi \rangle_{\#} \qquad (5.2.1) $$

is a positive linear functional on $C^\#(\sigma(A))$. Thus, by the generalized Riesz-Markov theorem, there is a unique #-measure $\mu_\psi^\#(\cdot)$ on the #-compact set $\sigma(A)$ with the property

$$ \langle \psi, f(A)\psi \rangle_{\#} = \mathrm{Ext} - \int_{\sigma(A)} f(\lambda) d^\# \mu_\psi^\#. \qquad (5.2.2) $$

**Definition 4.2.1.** The #-measure $\mu_\psi^\#(\cdot)$ is called the spectral #-measure associated with the vector $\psi \in \mathbf{H}^\#$.

The first and simplest application of the $\mu_\psi^\#(\cdot)$ is to allow us to extend the functional calculus to $B^\#(*\mathbb{R}_c^\#)$, the bounded in $*\mathbb{R}_c^\#$ #-Borel functions on $*\mathbb{R}_c^\#$. Let $g \in B^\#(*\mathbb{R}_c^\#)$.

It is natural to define $g(A)$ so that $\langle \psi, g(A)\psi \rangle_{\#} = \mathrm{Ext} - \int_{\sigma(A)} g(\lambda) d^\# \mu_\psi^\#$. The polarization

identity lets us recover $\langle \psi, g(A)\phi \rangle_{\#}$ from the proposed $\langle \psi, g(A)\psi \rangle_{\#}$ and then the Generalized Riesz lemma lets us construct $g(A)$.

**Theorem 5.2.1.** (spectral theorem-functional calculus form) Let $A$ be a bounded in $*\mathbb{R}_c^\#$ self-#-adjoint operator on $\mathbf{H}^\#$. There is a unique map

$\hat{\phi} : B^\#(*\mathbb{R}_c^\#) \to \mathcal{L}(\mathbf{H}^\#)$ so that

(a) $\hat{\phi}$ is an algebraic $*$-homomorphism.

(b) $\hat{\phi}$ is #-norm #-continuous: $\|\hat{\phi}(f)\|_{\mathcal{L}(\mathbf{H}^\#)} \le \|f\|_{\infty}$.

(c) Let $f$ be the function $f(x) = x$; then $\hat{\phi}(f) = A$.

(d) Suppose $f_n(x) \to_* f(x)$ for each $x$ as $n \to *\infty$ and hyper infinite sequence $\|f_n\|_{\infty}, n \in *\mathbb{N}$ is bounded in $*\mathbb{R}_c^\#$. Then $\hat{\phi}(f_n) \to_* \hat{\phi}(f)$ as $n \to *\infty$ strongly.

Moreover $\hat{\phi}(\cdot)$ has the properties:

(e) If $A\psi = \lambda\psi$, then $\hat{\phi}(f) = f(\lambda)\psi$.

(f) If $f \ge 0$, then $\hat{\phi}(f) \ge 0$.
$x \in *X$. Then $x$ is an $*-\text{accumulation point of } *X$ if for every $\varepsilon > 0, \varepsilon \approx 0, Y \cap (B_\varepsilon(x)\setminus\{x\}) \neq \emptyset$.

**Definition 2.4.** Let $\{X, \|\cdot\|\}$ be a standard Banach space, let $Y \subseteq *X, Y$ is $*-\text{closed}$ if every $*-\text{accumulation point of } Y$ is an element of $Y$.

**Definition 2.5.** Let $\{X, \|\cdot\|\}$ be standard Banach space. We shall say that internal hyper infinite sequence $\{x_n\}_{n=1}^{n=*\infty}$ in $*X$ $*-\text{converges to } x \in *X$ as $n \to *\infty$ if for any $\varepsilon > 0, \varepsilon \approx 0$ there is $N \in *\mathbb{N}$ such that for any $n > N : *\|x_n - x\| < \varepsilon$.

**Definition 2.6.** Let $\{X, \|\cdot\|\}, \{Y, \|\cdot\|\}$ be a standard Banach spaces. A linear internal operator $A : D(A) \subseteq *X \to *Y$ is $*-\text{closed}$ if for every internal hyper infinite sequence $\{x_n\}_{n=1}^{n=*\infty}$ in $D(A)$ $*-\text{converging to } x \in *X$ such that $Ax_n \to y \in *Y$ as $n \to *\infty$ one has $x \in D(A)$ and $Ax = y$. Equivalently, $A$ is $*-\text{closed}$ if its graph is $*-\text{closed}$ in the direct sum $*X \oplus *Y$.

Given a linear operator $A : *X \to *Y$, not necessarily $*-\text{closed}$, if the $*-\text{closure of its graph in } *X \oplus *Y$ happens to be the graph of some operator, that operator is called the $*-\text{closure of } A$, and we say that $A$ is $*-\text{closable}$. Denote the $*-\text{closure of } A$ by $*-\bar{A}$. It follows that $A$ is the restriction of $*-\bar{A}$ to $D(A)$.

$A$ $*-\text{core}$ (or $*-\text{essential domain}$) of a $*-\bar{A}$-closable operator is a subset $C \subset D(A)$ such that the $*-\text{closure of the restriction of } A$ to $C$ is $*-\bar{A}$.

**Definition 2.7.** The graph of the linear transformation $T : H \to H$ is the set of pairs $\{\langle\varphi, T\varphi\rangle | (\varphi \in D(T))\}$.

The graph of $T$, denoted by $\Gamma(T)$, is thus a subset of $H \times H$ which is a non-Archimedean

Hilbert space with inner product $\langle\langle\varphi_1, \psi_1\rangle, \langle\langle\varphi_2, \psi_2\rangle\rangle$.

$T$ is called a $*-\text{closed}$ operator if $\Gamma(T)$ is a $*-\text{closed}$ subset of $H \times H$.

**Definition 2.8.** Let $T_1$ and $T$ be operators on $H$. If $\Gamma(T_1) \supset \Gamma(T)$, then $T_1$ is said to be an extension of $T$ and we write $T_1 \supset T$. Equivalently, $T_1 \supset T$ if and only if $D(T_1) \supset D(T)$ and $T_1\varphi = T\varphi$ for all $\varphi \in D(T)$.

**Definition 2.9.** An operator $T$ is $*-\text{closable}$ if it has a $*-\text{closed extension}$. Every $*-\text{closable}$

operator has a smallest $*-\text{closed extension}$, called its $*-\text{closure}$, which we denote by $*-\bar{T}$.

**Theorem 2.1.** If $T$ is $*-\text{closable}$, then $\Gamma(-*\bar{T}) = -*\Gamma(\bar{T})$.

**Definition 2.10.** Let $T$ be a $*-\text{densely defined linear operator on a non-Archimedean Hilbert space } H$. Let $D(T^*)$ be the set of $\varphi \in H$ for which there is an $\xi \in H$ with $(T\psi, \varphi) = (\psi, \xi)$ for all $\psi \in D(T)$.

For each $\varphi \in D(T^*)$, we define $T^*\varphi = \xi$. $T^*$ is called the $*-\text{adjoint of } T$. Note that $\varphi \in D(T^*)$ if and only if $|(T\psi, \varphi)| \le C\|\psi\|$ for all $\psi \in D(T)$. We note that $S \subset T$ implies $T^* \subset S^*$.

**Theorem 2.2.** Let $T$ be a $*-\text{densely defined operator on a non-Archimedean Hilbert space } H$.

Then:(i) $T^*$ is $*-\text{closed}$.

(ii) $T$ is $*-\text{closable}$ if and only if $D(T^*)$ is $*-\text{dense in which case } T = T^{**}$.

(iii) If $T$ is $*-\text{closable}$, then $(-*\bar{T})^* = T^*$.

**Definition 2.11.** Let $T$ be a $*-\text{closed}$ operator on a Hilbert space $H$. A complex number
(g) If $BA = AB$ then $\hat{\phi}(f)B = B\hat{\phi}(f)$.

**Remark 5.2.1.** Note that: (i) Theorem 5.2.1 can be proven directly by extending Theorem 5.1.1, part (d) requires the dominated #-convergence theorem. Or, Theorem 5.2.1 can be proven by an easy corollary of Theorem 5.2.3 below.
The proof of Theorem 5.2.3 uses only the #-continuous functional calculus, $\hat{\phi}$ extends $\phi$ and as before we write $\hat{\phi}(f) = f(A)$. As in the #-continuous functional calculus, one has $f(A)g(A) = g(A)f(A)$.

(ii) Since $B^{#}(*\mathbb{R}_c^{#})$ is the smallest family closed under #-limits of form (d) containing all of $C^{#}(*\mathbb{R}_c^{#})$, we know that any $\hat{\phi}(f)$ is in the Smallest non Archimedean $C^*$-algebra containing $A$ which is also strongly #-closed; such an algebra is called a von Neumann #-algebra or non Archimedean $W^*$-algebra. When we study von Neumann #-algebras we will see that this follows from (g).

(iii) The #-norm equality of Theorem 4.2.1 carries over if we define $\|f\|_{\infty}^{\prime}$ to be the $L_{\infty}^{\#}$-norm with respect to a suitable notion of “#-almost everywhere.” Namely, pick an orthonormal basis $\{\psi_n\}_{n=1}^{\infty}$ and say that a property is true #-a.e. if it is true #-a.e. with respect to each $\mu_{\psi_n}^{\#}$. Then
$$ \|\hat{\phi}(f)\|_{L(\mathbf{H}^{\#})} = \|f\|_{\infty}^{\prime} $$

**Definition 5.2.2.** A vector $\psi \in \mathbf{H}^{\#}$ is called a cyclic vector for $A$ if gyperfinite linear combinations of the elements $\{A^n\psi\}_{n=0}^{\infty}$ are #-dense in $\mathbf{H}^{\#}$.
Not all operators have cyclic vectors, but if they do.

**Lemma 5.2.1.** Let $A$ be a bounded in $*\mathbb{R}_c^{\#}$ self-#-adjoint operator with cyclic vector $\psi$. Then, there is a unitary operator $U : \mathbf{H}^{\#} \to L_2^{\#}(\sigma(A), d^{\#}\mu_{\psi}^{\#})$, with $(UAU^{-1})f(\lambda) = \lambda f(\lambda)$ where equality holds is in the sense of elements of $L_2^{\#}(\sigma(A), d^{\#}\mu_{\psi}^{\#})$.

**Proof** Define $U$ by $U\phi(f) = f$ where $f$ is #-continuous. $U$ is essentially the inverse of the map $\phi$ of Theorem 4.1.1. To show that $U$ is well defined operator we compute
$$ \|\phi(f)\psi\|_{\#}^2 = \langle\psi, \phi^*(f)\phi(f)\psi\|_{\#} = \langle\psi, \phi(f \times f)\psi\|_{\#} = \mathrm{Ext}-\int|f(\lambda)|^2 d^{\#}\mu_{\psi}^{\#}. $$

Therefore, if $f = g$ a.e. with respect to $\mu_{\psi}^{\#}$, then $\phi(f)\psi = \phi(g)\psi$. Thus $U$ is well defined on $\{\phi(f)\psi|f \in C^{\#}(\sigma(A))\}$ and is #-norm preserving. Since $\psi$ is cyclic it #-closure $\#-\{\phi(f)\psi|f \in C^{\#}(\sigma(A))\} = \mathbf{H}^{\#}$ so by the generalized B.L.T. theorem $U$ extends to an #-isometric map of $\mathbf{H}^{\#}$ into $L_2^{\#}(\sigma(A), d^{\#}\mu_{\psi}^{\#})$. Since $C^{\#}(\sigma(A))$ is #-dense in $L_2^{\#}$, $\mathrm{Ran}\,U = L_2^{\#}(\sigma(A), d^{\#}\mu_{\psi}^{\#})$. Finally, if $f \in C^{\#}(\sigma(A))$ one obtains $(UAU^{-1})f(\lambda) = [UA\phi(f)](\lambda) = [U\phi(xf)](\lambda) = \lambda f(\lambda)$.
By #-continuity, this extends from $C^{\#}(\sigma(A))$ to $L_2^{\#}$.

To extend this lemma to arbitrary $A$ we need to know that $A$ has a family of invariant subspaces spanning $\mathbf{H}^{\#}$ so that $A$ is cyclic on each subspace:

**Lemma 5.2.2.** Let $A$ be a self-adjoint operator on a #-separable Hilbert space $\mathbf{H}^{\#}$.
Then there is a direct sum decomposition $\mathbf{H}^{\#} = \mathrm{Ext}-\bigoplus_{n=1}^{N} \mathbf{H}_{n}^{\#}$ with $N \in *\mathbb{N}$ or

$\mathbf{H}^{\#} = \mathrm{Ext}-\bigoplus_{n=1}^{\infty} \mathbf{H}_{n}^{\#}$ such that:

(a) $A$ leaves each $\mathbf{H}_{n}^{\#}$ invariant, that is, $\psi \in \mathbf{H}_{n}^{\#}$ implies $A\psi \in \mathbf{H}_{n}^{\#} = \mathrm{Ext}-\bigoplus_{n=1}^{\infty} \mathbf{H}_{n}^{\#}$ so that:

(b) For each $n \in *\mathbb{N}$, there is a $\phi_n \in \mathbf{H}_n^{\#}$ which is cyclic for $A \upharpoonright \mathbf{H}_n^{\#}$, i.e.
$$ \mathbf{H}_n^{\#} = \#-\{f(A)\phi_n|f \in C^{\#}(\sigma(A))\} $$