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

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

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

**Proof.** I will assume that $H$ is a complex Hilbert space with $\mathbb{C}_c^{\#}$-valued inner product, the real case being easier. Statements (1) and (2) are proved by the following elementary computations:

$$ \|x+y\|_{\#}^{2} = \|x\|_{\#}^{2} + \|y\|_{\#}^{2} + 2 \mathrm{Re}\langle x, y \rangle_{\#} + \|x\|_{\#}^{2} + \|y\|_{\#}^{2} - 2 \mathrm{Re}\langle x, y \rangle_{\#} = 2 \|x\|_{\#}^{2} + 2 \|y\|_{\#}^{2} \quad (1.6) $$

and

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

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 $\text{dist}(0, \cdot)|_M$ exists, it is unique.

(2) Existence: Let $y_n \in M$ be chosen such that $\|y_n\|_\# = \delta_n \to \delta = \text{dist}(0, M)$. 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, $\|y\|_\# = \#\text{-lim } \sup_{n \to \infty} \|y_n\|_\# = \delta = \text{dist}(0, M)$. 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 $g(t) = \|x - (y + tw)\|_\#^2 = \|x - y\|_\#^2 - 2t \text{Re}\langle x - y, w \rangle_\# + t^2 \|w\|_\#^2$ has a minimum at $t = 0$. Therefore $0 = g'(0) = -2 \text{Re}\langle x - y, w \rangle_\#$. 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 $z \in M$, by Pythagorean's theorem,

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

**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 $x \in H^\#$, $\|Ax\|_\# \le M\|x\|_\#$. 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 $\|A(x + h) - A(x)\|_\# = \|A(h)\|_\# \le M\|h\|_\#$, $M \in \ast\mathbb{R}_c^\#$, $M > 0$. 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 $\|A(h)\|_\# = \|A(h) - A(0)\|_\# \le 1$ for all vectors $h \in H_1^\#$ with $\|h\|_\# \le \varepsilon$. Thus, for all non-zero $x \in H_1^\#$, one has

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

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-Archimedean 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-Archimedean 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) $\text{Ran}(P_M) = M$ and $\text{ker}(P_M) = M^\perp$.

**Proof.** (1) Let $x_1, x_2 \in H^\#$ and $\alpha \in *\mathbb{R}_c^\#$, 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 $\text{Ran}(P_M) = M$ 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 $(y - P_M y) \in M^\perp$,
$\langle P_M x, y \rangle_\# = \langle P_M x, P_M y + y - P_M y \rangle_\# = \langle P_M x, P_M y \rangle_\# = \langle P_M x + (x - P_M), P_M y \rangle_\# = \langle x, P_M y \rangle_\#$.

(4) It is clear that $\text{Ran}(P_M) \subset M$. Moreover, if $x \in M$, then $P_M x = x$ implies
that $\text{Ran}(P_M) = M$. Now $x \in \text{ker}(P_M)$ 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 $H^\# = M \oplus M^\perp$.

**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 $M = \text{ker}(f)$ 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 $\lambda = \bar{f}(x_0)/\|x_0\|^2$.) Then for $x = m + \lambda x_0$ with $m \in M$ and $\lambda \in *\mathbb{C}_c^\#$, $f(x) = \lambda f(x_0) = \lambda \langle x_0, z \rangle_\# = \langle \lambda x_0, z \rangle_\# = \langle m + \lambda x_0, z \rangle_\# = \langle x, z \rangle_\#$ 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 $(A + \lambda B)^* = A^* + \lambda B^*$, $A^{**} \cong (A^*)^* = A$, $\|A^*\|_\# = \|A\|_\#$ 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 $*\mathbb{R}_c^\#$ operator. To see $A^*$ is linear, let $y_1, y_2 \in K^\#$ and $\lambda \in *\mathbb{C}_c^\#$,
then for any $x \in H^{\#}, \langle Ax, y_1 + \lambda y_2 \rangle_{K^{\#}} = \langle Ax, y_1 \rangle_{K^{\#}} + \bar{\lambda} \langle Ax, y_2 \rangle_{K^{\#}}$
$= \langle x, A^*(y_1) \rangle_{K^{\#}} + \bar{\lambda} \langle x, A^*(y_2) \rangle_{K^{\#}} = \langle x, A^*(y_1) + \lambda A^*(y_2) \rangle_{K^{\#}}$ 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
$(A + \lambda B)^* = A^* + \bar{\lambda} B^*$ is left to the reader.

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

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

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

**Corollary 1.3.** Let $H^{\#}$, $K^{\#}$, $M^{\#}$ be non-Archiedean Hilbert space, $A, B \in L(H^{\#}, K^{\#})$, $C \in L(K^{\#}, M^{\#})$ and $\lambda \in *C_c^{\#}$. Then $(A + \lambda B)^* = A^* + \bar{\lambda} B^*$ and $(CA)^* =$ $A^*C^* \in L(M^{\#}, H^{\#})$.

**Corollary 1.4.** Let $H^{\#} = *C_c^{\#n}$ and $K^{\#} = *C_c^{\#n}$ 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

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

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

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

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

and (1.12) gives that

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

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 $*$-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) $\langle s, x \rangle_{\#} = \mathrm{Ext} - \sum_{v \in A} \langle v, x \rangle_{\#}$ for all $x \in H^{\#}$.

Similarly if $\{v_n\}_{n=1}^{*\infty}$ is an orthogonal set, then $s = \mathrm{Ext} - \sum_{n=1}^{*\infty} v_n$ exists in $H^\#$ iff $\mathrm{Ext} - \sum_{n=1}^{*\infty} v_n < *\infty$. In particular if $\mathrm{Ext} - \sum_{n=1}^{*\infty} v_n$ exists, then it is independent of rearrangements of $\{v_n\}_{n=1}^{*\infty}$.

**Proof.** Suppose $s = \mathrm{Ext} - \sum_{v \in A} v$ exists. Then there exists $\Gamma \subset\subset A$ such that $\mathrm{Ext} - \sum_{v \in A} \|v\|_{\#}^2 = \|\mathrm{Ext} - \sum_{v \in A} v\|_{\#}^2 \le 1$ for all $\Lambda \subset\subset A\setminus\Gamma$, wherein the first inequality we have used Pythagorean's theorem.

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

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

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

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

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

$$ \|s\|_{\#} - \|s_{\varepsilon}\|_{\#} \le \|s - s_{\varepsilon}\|_{\#} < \varepsilon \quad \text{and by Eq. (1.14), } 0 \le (\mathrm{Ext} - \sum_{v \in A} \|v\|_{\#}^2) - \|s_{\varepsilon}\|_{\#}^2 \le \varepsilon^2 $$
Letting $\varepsilon \to 0$ we deduce from the previous two equations that $\|s_{\varepsilon}\|_{\#} \to *\infty$ and $\|s\|_{\#}^2 = \|s_{\varepsilon}\|_{\#}^2 \to *\infty$ as $\varepsilon \to 0$ and therefore $\|s\|_{\#}^2 = \|s_{\varepsilon}\|_{\#}^2$.

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

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

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

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

and

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

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

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

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

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 $(x - P_x) \perp \mathrm{span}(\beta)$ and by continuity we also have $(x - P_x) \perp M = \#\overline{\mathrm{span}}(\beta)$. 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 $x, y \in H^\#$.
## §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 $\overline{\operatorname{span}(\beta)}$ 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) $x = P$

$u \in \beta$

hx, uiu for all $x \in H$.

(3) $hx, yi = P$

$u \in \beta$

hx, ui hu, yi for all $x, y \in H$.

(4) $kxk_2 = P$

$u \in \beta$

$|hx, ui|_2$ for all $x \in H$.

**Proof.** Let $M = \operatorname{span} \beta$ and $P = \operatorname{PM}$.

(1) $\Rightarrow$ (2) By Corollary 12.20, P

$u \in \beta$

hx, uiu = PMx. Therefore

$x - X$

$u \in \beta$

$$ \text{hx, uiu} = x - \operatorname{PMx} \in M_\perp = \beta_\perp = \{0\}. $$

(2) $\Rightarrow$ (3) is a consequence of Proposition 12.19.

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

(4) $\Rightarrow$ (1) If $x \in \beta_\perp$, then by 4), $kxk = 0$, i.e. $x = 0$. This shows that $\beta$ is complete.

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

**Proof.** Let $D \subset H^#$ be a $*$-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 $\mathrm{span}(\{v_n|1 \le n \le N\}) \supseteq \mathrm{span}(\{u_n|1 \le n \le N\})$. 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 $x = \lim_{k \to \infty} wk$ and therefore $hx, xi = \lim_{k \to \infty} hx, wki = 0$. That is to say $x = 0$ and $\beta$ is $#$-complete.

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

Let $H^#$ be a $#$-complex Hilbert space over field $*\mathbb{C}^#$. The most natural way to construct a $#$-continuous one-parameter unitary group $U(t) : H^# \to H^#$ is to try to make sense of the power series $\mathrm{Ext}-\sum_{n=0}^{\infty} (itA)^n$ 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 $\mathrm{Ext}-\sum_{n=0}^{\infty} (itA)^n/n!$ $#$-converges to $\mathrm{Ext}\-\exp(itA)$ in norm. In particular, for any $\varphi \in \bigcup_{M \ge 0} E_{[-M,M]}$,

$$ \mathrm{\#-lim}_{N \to \infty} \left( \mathrm{Ext} - \sum_{n=0}^{N} \frac{(itA)^n}{n!} \right) = \mathrm{Ext}\-\exp(itA). \quad (3.1) $$

Since $\bigcup_{M \ge 0} E_{[-M,M]}$ is $#$-dense in $H^#$, 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 $\mathrm{Ext}-\sum_{n=0}^{\infty} (itA)^n/n!$ on a $#$-dense set. We will prove the $#$-converse: namely, if $A$ is symmetric and has a $#$-dense set of vectors to which $\mathrm{Ext}-\sum_{n=0}^{\infty} (itA)^n/n!$ can be applied, then $A$ is essentially self-$#$-adjoint. We need several definitions.

**Definition 1.1.** Let $A$ be an operator on a non-Archimedean Hilbert space $H^#$. The set $C^{\infty}(A) = \bigcap_{n=0}^{\infty} D(A^n)$ is called the $C^{\infty}$-vectors for $A$. A vector $\varphi \in C^{\infty}(A)$ is called an $#$-analytic vector for $A$ if

$$ \mathrm{Ext} - \sum_{n=0}^{\infty} \frac{\|A^n \varphi\| t^n}{n!} < *\infty \quad (3.2) $$

for some $t > 0$. If $A$ is self-adjoint, then $C^{\infty}(A)$ will be $#$-dense in $D(A)$. However, in general, a symmetric operator may have no $C^{\infty}$-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 $C^{\infty}$-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 $C^{\infty}(A)$.

**Definition 1.2.** Suppose that $A$ is symmetric. For each $\varphi \in C^{\infty}(A)$, define
$$D_{\phi} = \{ \mathrm{Ext} - \sum_{n=0}^{N} \alpha_n A^n \phi \mid N \in *\mathbb{N}, \alpha_n \in *\mathbb{C}_c^{\#} \}. \quad (3.3)$$

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

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

$$\| \psi - \mathrm{Ext} - \sum_{n=1}^{N} \alpha_n \psi_n \|_{\#} \le \varepsilon/2. \quad (3.4)$$

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

$$\| \psi_n - (A + i) \phi_n \|_{\#} \le \frac{\varepsilon}{2} \left( \mathrm{Ext} - \sum_{n=1}^{N} |\alpha_n| \right)^{-1}. \quad (3.5)$$

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

**Theorem 3.1.** (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

$$C : \mathrm{Ext} - \sum_{n=0}^{N} \alpha_n A^n \psi \quad (3.6)$$

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 $A$,

$$\mathrm{Ext} - \sum_{n=0}^{N} \|A^n \psi\|_{\#} / n! < *\infty \quad (3.7)$$

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

$$\begin{align}
& \mathrm{Ext} - \sum_{n=0}^{*\infty} \frac{s^n}{n!} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#}} |x|^n d^{\#} \mu^{\#} \right) \le \nonumber \\
& \le \mathrm{Ext} - \sum_{n=0}^{*\infty} \frac{s^n}{n!} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#}} x^{2n} d^{\#} \mu^{\#} \right)^{1/2} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#}} d^{\#} \mu^{\#} \right)^{1/2} = \nonumber \\
& \qquad \|\psi\|_{\#} \mathrm{Ext} - \sum_{n=0}^{*\infty} \frac{s^n}{n!} \|A^n \psi\|_{\#} < *\infty. \tag{3.8}
\end{align}$$

Therefore by generalized Fibini's theorem
$$ Ext- \int_{*\mathbb{R}_c^\#} (Ext- \sum_{n=0}^{*\infty} \frac{s^n}{n!} |x|^n) d^\# \mu^\# = Ext- \int_{*\mathbb{R}_c^\#} Ext-(s|x|) d^\# \mu^\# < *\infty. \quad (3.9) $$

As a result, the function

$$ \langle \psi, [Ext-\exp(itB)]\psi \rangle_{\#} = Ext- \int_{*\mathbb{R}_c^\#} [Ext-\exp(itx)] d^\# \mu^\# \quad (3.10) $$

has an #-analytic continuation

$$ Ext- \int_{*\mathbb{R}_c^\#} [Ext-\exp(izx)] d^\# \mu^\# \quad (3.11) $$

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

$$ \left[ \left( \frac{d^\#}{d^\# z} \right)^k \left( Ext- \int_{*\mathbb{R}_c^\#} [Ext-\exp(izx)] d^\# \mu^\# \right) \right]_{z=0} = \tag{3.12} \\ = Ext- \int_{*\mathbb{R}_c^\#} [Ext-\exp(ix)^k] d^\# \mu^\# = \langle \psi, (iA)^k \psi \rangle_{\#}, $$

we obtain

$$ \langle \psi, [Ext-\exp(isB)]\psi \rangle_{\#} = Ext- \sum_{n=0}^{*\infty} \frac{(is)^n}{n!} = \langle \psi, (iA)^k \psi \rangle_{\#} \quad (3.13) $$

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

Similar proof shows that $\langle \psi_1, [Ext-\exp(isB)]\psi_2 \rangle_{\#}$ 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 $Ext-\exp(isB)$ is unitary, $Ext-\exp(isB)$ 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 3.1** 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 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 3.2** 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 3.1 $A \upharpoonright D$ 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 $n \in *\mathbb{N}$.

## §4. The generalized Spectral Theorem
§ 4.1. 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 unbounded self-#-adjoint operators in the next chapter; there is a spectral theorem for unbounded operators which we discuss in Section § 4.3)

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

$$ (UAU^{-1}f)(x) = F(x)f(x) \qquad (4.1.1) $$

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

§ 4.2 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

$$ f \mapsto \langle \psi, f(A)\psi \rangle_{\#} \qquad (4.1.2) $$

for fixed $\psi \in H^#$.

Given a fixed operator A, for which $f$ can we define $f(A)$? First, suppose that A is an arbitrary bounded in $*\mathbb{R}_c^#$ 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

$$ f(A) = \text{Ext}-\sum_{n=1}^{\infty} c_n A^n \qquad (4.1.3) $$

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,
$$ \phi(fg) = \phi(f)\phi(g), \phi(\lambda f) = \lambda\phi(f), \phi(1) = I, \phi(\bar{f}) = \phi(f)^*. $$

(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 showing that

$$ \|P(A)\|_{\#op} = \|P(x)\|_{C^{\#}(\sigma(A))} = \sup_{\lambda \in \sigma(A)} |P(\lambda)|. \qquad (4.1.4) $$

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

$$ \sigma(P(A)) = \{P(\lambda)|\lambda \in \sigma(A)\}. \qquad (4.1.5) $$

**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

$$ (P(A) - \mu)^{-1} = a^{-1} (\text{Ext}-\prod_{i=1}^n (A - \lambda_i)^{-1}) \qquad (4.1.6) $$

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 $\aleph$ self-#-adjoint operator. Then

$$ \|P(A)\|_{\#} = \sup_{\lambda \in \sigma(A)} |P(\lambda)|. \qquad (4.1.7) $$

**Proof** By Theorem 4.1.2 and by Lemma 4.1.1 we obtain

$$ \begin{aligned} \|P(A)\|_{\#}^2 &= \|P(A)^*P(A)\|_{\#} = \|(\overline{PP})(A)\|_{\#} = \\ &= \sup_{\lambda \in \sigma((\overline{PP})(A))} |\lambda| = \sup_{\lambda \in \sigma(A)} |\overline{PP}(\lambda)| = \left( \sup_{\lambda \in \sigma(A)} |P(\lambda)| \right)^2. \end{aligned} \qquad (4.1.8) $$

**Proof of Theorem 4.1.1.** 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 $g \in \mathbb{R}^{\#}$-valued and $g \in C^{\#}(\sigma(A))$. Thus $\phi(f) = \phi(g)^2$ with $\phi(g)$ self-#-adjoint, so $\phi(f) \ge 0$.

**Remark 4.1.1.** 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 a non-Archimedean abelian $C^*$ algebra of operators.

(3) **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

(5) (b) actually follows from (a) and Proposition 4.1.1. Thus (a) and (c) alone determine $\phi$ uniquely.

**Proposition 4.1.1.** Suppose that $\phi: C^\#(X) \to \mathcal{L}(H^\#)$ is an algebraic $\ast$-homomorphism, $X$ a #-compact metric space. Then

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

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

**Definition 4.1.1.** if $n,k \in *\mathbb{N}$ with $k \le n$, then we define

$$ \binom{n}{k} = \frac{n!^{\#}}{k!^{\#}(n-k)!^{\#}} \qquad (4.1.8) $$

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

**Lemma 4.1.3.** Whenever $n,k \in *\mathbb{N}$ are such that $k \le n$, then

$$ \binom{n}{k} = \binom{n}{n-k}. \qquad (4.1.9) $$

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

$$ \binom{n}{n-k} = \frac{n!^{\#}}{(n-k)!^{\#}[n-(n-k)]!^{\#}} = \frac{n!^{\#}}{(n-k)!^{\#}k!^{\#}} = \frac{n!^{\#}}{k!^{\#}(n-k)!^{\#}} = \binom{n}{k}. \qquad (4.1.10) $$

**Lemma 4.1.4.** Let $n,k \in *\mathbb{N}$ with $0 < k < n$, then

$$ \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}. \qquad (4.1.11) $$

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

**Proposition 4.1.2.** (Generalized binomial theorem) Let $x,y \in *\mathbb{R}_c^\#$ and let $n \in *\mathbb{N}$, then we have

$$ (x+y)^n = \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} x^{n-k}y^k = \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}. \qquad (4.1.8) $$

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

where we have used Lemma 4.1.4.

**Definition 4.1.2** (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

$$ B_n^#(x,f) = \mathrm{Ext}-\sum_{k=0}^{n} f\left(\frac{k}{n}\right) \binom{n}{k} x^k (1-x)^{n-k}. \quad (4.1.9) $$

**Lemma 4.1.3.** For any $n \in *\mathbb{N}$ :

$$ \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} x^k (1-x)^{n-k} = 1, \quad (4.1.10) $$

$$ \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} x^k (1-x)^{n-k} (k - nx) = 0, \quad (4.1.11) $$

$$ \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} x^k (1-x)^{n-k} (k - nx)^2 = nx(1-x). \quad (4.1.12) $$

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

$$ \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} x^k (1-x)^{n-k} = [x+(1-x)]^n = 1. \quad (4.1.13) $$

By the generalized binomial theorem we have

$$ \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} y^k z^{n-k} = (y+z)^n. \quad (4.1.14) $$

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

$$
\begin{align}
\frac{d^\#}{d^\# y} \left[ \mathrm{Ext} - \sum_{k=0}^{n} \binom{n}{k} y^k z^{n-k} \right] &= \mathrm{Ext} - \sum_{k=0}^{n} \binom{n}{k} k y^{k-1} z^{n-k} = \nonumber \\
&= \frac{d^\#}{d^\# y} (y+z)^n = n(y+z)^{n-1}. \tag{4.1.15}
\end{align}
$$

Thus

$$ \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} k y^{k-1} z^{n-k} = n(y+z)^{n-1} \quad (4.1.16) $$

and therefore

$$ \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} k y^k z^{n-k} = ny(y+z)^{n-1}. \quad (4.1.17) $$

Replacing $y$ by $x$ and $z$ by $1-x$ in the above expression, we have identity
$$ \mathrm{Ext}-\sum_{k=0}^{n}\binom{n}{k}kx^k(1-x)^{n-k} = nx. \qquad (4.1.18) $$

From (4.1.18) we obtain

$$ \mathrm{Ext}-\sum_{k=0}^{n}\binom{n}{k}\frac{k}{n}x^k(1-x)^{n-k} = x. \qquad (4.1.19) $$

From (4.1.19) and (4.1.13) we obtain the identity (4.1.11). By the #-differentiating with respect to y of the identity (4.1.17) we obtain

$$ \begin{aligned} \frac{d^{\#}}{d^{\#}y} \left[ \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} k y^k z^{n-k} \right] &= \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} k^2 y^{k-1} z^{n-k} = \\ &= n \frac{d^{\#}}{d^{\#}y} y(y+z)^{n-1} = n(y+z)^{n-1} + n(n-1)y(y+z)^{n-2}. \end{aligned} \qquad (4.1.20) $$

Thus

$$ \mathrm{Ext}-\sum_{k=0}^{n}\binom{n}{k}k^2y^{k-1}z^{n-k} = n(y+z)^{n-1} + n(n-1)y(y+z)^{n-2}. \qquad (4.1.21) $$

and therefore

$$ \mathrm{Ext}-\sum_{k=0}^{n}\binom{n}{k}k^2y^kz^{n-k} = ny(y+z)^{n-1} + n(n-1)y^2(y+z)^{n-2}. \qquad (4.1.22) $$

Replacing y by x and z by 1-x in the expression (4.1.22) we have identity

$$ \mathrm{Ext}-\sum_{k=0}^{n}\binom{n}{k}k^2x^k(1-x)^{n-k} = nx + n(n-1)x^2. \qquad (4.1.23) $$

From (4.1.23) and (4.1.10)-(4.1.11) one obtains (4.1.12).

**Theorem 4.1.2.** (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

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

From the identity (4.1.24) one obtains that

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

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:

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

and

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

From (4.1.9) we obtain

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

From (4.1.9) we obtain
$$
\begin{align}
& \mathrm{Ext}-\sum_{|x-k/n|>\delta} |f(x)-f(\frac{k}{n})| \binom{n}{k} x^k (1-x)^k \le 2M \left[ \mathrm{Ext}-\sum_{|x-k/n|>\delta} \binom{n}{k} x^k (1-x)^k \right] \nonumber \\
& \le \frac{2M}{\delta^2} \left[ \mathrm{Ext}-\sum_{|x-k/n|>\delta} \left(x-\frac{k}{n}\right)^2 \binom{n}{k} x^k (1-x)^k \right] \le \tag{4.1.15} \\
& \le \frac{2M}{\delta^2} \left[ \mathrm{Ext}-\sum_{k=0}^n \left(x-\frac{k}{n}\right)^2 \binom{n}{k} x^k (1-x)^k \right] \le \frac{2M}{n\delta^2}. \nonumber
\end{align}
$$

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 $2\epsilon$. This estimate is independent of $x \in [0, 1]$. Hence, for $\epsilon > 0$ there exists a number $N \in *\mathbb{N}$ such that $n \ge N$ and $x \in [0, 1]$ imply $|f(x) - B_n^\#(x)| \le 2\epsilon$. Therefore $f$ is the uniform #-limit of the polynomials $B_n^\#$.

**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 $\mathcal{L}(Z, Y)$.

§ 4.2.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 $\psi \in H^#$. Then

$$
f \mapsto \langle \psi, f(A)\psi \rangle_{\#} \qquad (4.2.1)
$$

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

$$
\langle \psi, f(A) \psi \rangle_{\#} = \mathrm{Ext} - \int_{\sigma(A)} f(\lambda) d^{\#} \mu_{\psi}^{\#}. \quad (4.2.2)
$$

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

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 $\langle \psi, g(A)\psi \rangle_\# = \text{Ext}-\int_{\sigma(A)} g(\lambda)d^\#\mu_\psi^\#$. 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 4.2.1.** (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: $\|\hat{\phi}(f)\|_{\mathcal{L}(\mathbf{H}^\#)} \le \|f\|_{\cdot, \infty}$.

(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 $\|f_n\|_{\cdot, \infty}, n \in \mathbb{N}$ 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 $\hat{\phi}(f) \ge 0$.

(g) If $BA = AB$ then $\hat{\phi}(f)B = B\hat{\phi}(f)$.

**Remark 4.2.1.** Note that: (i) Theorem 4.2.1 can be proven directly by extending Theorem 4.1.1, part (d) requires the dominated $\#$-convergence theorem. Or, Theorem 4.2.1 can be proven by an easy corollary of Theorem 4.2.3 below.

The proof of Theorem 4.2.3 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 $\|f\|_{\cdot, \infty}^\#$ to be the $L^\#_{\cdot, \infty}$ $\#$-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 $\|\hat{\phi}(f)\|_{\mathcal{L}(\mathbf{H}^\#)} = \|f\|_{\cdot, \infty}^\#$.

**Definition 4.2.2.** A vector $\psi \in \mathbf{H}^\#$ is called a cyclic vector for $A$ if hyperfinite 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 4.2.1.** 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

$$ \|\phi(f)\psi\|_{\#}^2 = \langle \psi, \phi^{\#}(f)\phi(f)\psi \rangle_{\#} = \langle \psi, \phi(\tilde{f} \times f)\psi \rangle_{\#} = \text{Ext}-\int|f(\lambda)|^2 d^\#\mu_\psi^\# $$

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 $L_2^\#$, **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 4.2.2.** Let $A$ be a self-adjoint operator on a $\ast$-separable Hilbert space $\mathbf{H}^\#$.

Then there is a direct sum decomposition $\mathbf{H}^\# = \text{Ext}-\bigoplus_{n=1}^N \mathbf{H}_n^\#$ with $N \in \mathbb{N}$ or

$\mathbf{H}^\# = \text{Ext}-\bigoplus_{n=1}^\infty \mathbf{H}_n^\#$ such that:

(a) $A$ leaves each $\mathbf{H}_n^\#$ invariant, that is, $\psi \in \mathbf{H}_n^\#$ implies $A\psi \in \mathbf{H}_n^\# = \text{Ext}-\bigoplus_{n=1}^\infty \mathbf{H}_n^\#$ 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.
$$ \mathbf{H}_n^\# = \#\{f(A)\phi_n|f \in C^\#(\sigma(A))\} $$

**Theorem 4.2.3 (spectral theorem-multiplication operator form)** Let $A$ be a bounded in $\mathbb{R}_c^\#$ self-#-adjoint operator on $\mathbf{H}^\#$, a $\ast$-separable Hilbert space.

Then, there exist #-measures $\{\mu_n^\#\}_{n=1}^N$ with $N \in \mathbb{N}$ or $\{\mu_n^\#\}_{n=1}^\infty$ on $\sigma(A)$ and a unitary operator $U : \mathbf{H}^\# \to \bigoplus_{n=1}^N L_2^\#(*\mathbb{R}_c^\#, d^\#\mu_n^\#)$ or $U : \mathbf{H}^\# \to \bigoplus_{n=1}^\infty L_2^\#(*\mathbb{R}_c^\#, d^\#\mu_n^\#)$

so that $(UAU^{-1}\psi)_n(\lambda) = \lambda\psi_n(\lambda)$

where we write an element $\psi \in \bigoplus_{n=1}^N L_2^\#(*\mathbb{R}_c^\#, d^\#\mu_n^\#)$ as an $N$-tuple $(\psi_1(\lambda), \dots, \psi_N(\lambda))$ or $\ast$-tuple

This realization of $A$ is called a spectral representation.

**Proof.** Use Lemma 4.2.2 to find the decomposition and then use Lemma 4.2.1 on each component.

This theorem tells us that every bounded self-#-adjoint operator is a multiplication operator on a suitable #-measure space; what changes as the operator changes are the underlying #-measures. Explicitly:

**Corolarly 4.2.1.** Let $A$ be a bounded in $\mathbb{R}_c^\#$ self-adjoint operator on a $\ast$-separable Hilbert space $\mathbf{H}^\#$. Then there exists a finite in $\mathbb{R}_c^\#$ measure space $(M, \mu^\#)$, a bounded in $\mathbb{R}_c^\#$ function $F$ on $M$, and a unitary map, $U : \mathbf{H}^\# \to L_2^\#(M, d^\#\mu^\#)$ so that $(UAU^{-1}f)(m) = F(m)f(m)$.

**Proof** Choose the cyclic vectors $\phi_n$ so that $\|\phi_n\|_\# = 2^{-n}$. Let $M = \bigcup_{n=1}^N \mathbb{R}_c^\#$ i.e. the union of $N \in \mathbb{N}$ copies of $\mathbb{R}_c^\#$. Define $\mu$ by requiring that its restriction to the $n$-th copy of $\mathbb{R}_c^\#$ be $\mu_n$. Since $\mu(M) = \text{Ext}-\sum_{n=1}^N \mu_n^\#(*\mathbb{R}_c^\#) < \infty$, $\mu_n$ is finite in $\mathbb{R}_c^\#$. We also notice that this last theorem is essentially a rigorous form of the formaal Dirac notation. If we write $\phi_n = \phi(x; n)$, we see that in the “new representation defined by $U$” one has

$$ \langle \psi, \phi \rangle_\# = \text{Ext}-\sum_n \text{Ext}-\int d^\# \mu_n^\# \overline{\psi(\lambda; n)} \phi(\lambda; n) $$

and

$$ \langle \psi, A\phi \rangle_\# = \text{Ext}-\sum_n \text{Ext}-\int d^\# \mu_n^\# \overline{\psi(\lambda; n)} \lambda \phi(\lambda; n). $$

These are the Dirac type formulas familiar to physicists except that the formal
sums of Dirac are replaced with integrals over spectral measures, where we define:

**Definition 4.2.3.** The #-measures $d^\#\mu_n$ are called spectral measures; they are just $d^\#\mu_\psi$ for suitable $\psi$.

**Remark 4.2.2.** Notice these #-measures are not uniquely determined.
We now investigate the connection between spectral measures and the spectrum.

**Definition 4.2.3.** If $\{\mu_n^\#\}_{n=1}^N$, $N \in *\mathbb{N}$ is a family of #-measures, the support of $\{\mu_n^\#\}_{n=1}^N$ is the complement of the largest #-open set $B$ with $\mu_n^\#(B) = 0$ for all $n \in *\mathbb{N}$ so

$$ \mathrm{supp}(\{\mu_n^\#\}_{n=1}^N) = \#-\overline{\bigcup_{n=1}^N \mathrm{supp}(\mu_n^\#)}. \qquad (4.2.1) $$

**Proposition 4.2.1.** Let $A$ be a self-#-adjoint operator and $\{\mu_n^\#\}_{n=1}^N$, $N \in *\mathbb{N}$ a family of spectral #-measures. Then

$$ \sigma(A) = \mathrm{supp}(\{\mu_n^\#\}_{n=1}^N). $$

**Definition 4.2.4.** Let $F$ be a $*\mathbb{R}_c^\#$-valued function on a #-measure space $\langle M, \mu^\# \rangle$.

We say $\lambda$ is in the **essential range** of $F$ if and only if

$$ \mu^\#\{m|\lambda - \varepsilon < F(m) < \lambda + \varepsilon\} > 0. $$

for all $\varepsilon \approx 0, \varepsilon > 0$.

**Proposition 4.2.2.** Let $F$ be a bounded in $*\mathbb{R}_c^\#$ $*\mathbb{R}_c^\#$-valued function on a #-measure space $\langle M, \mu^\# \rangle$. Let $T_f$ be the operator on $L_2^\#(M, d^\#\mu^\#)$ given by $(T_F g)(m) = F(m)g(m)$
Then $\sigma(T_F)$ is the essential range of $F$.

**Definition 4.2.5.** Let $A$ be a bounded in $*\mathbb{R}_c^\#$ self-#-adjoint operator on $H^\#$

$$ \text{Let } H_{\mathrm{pp}}^{\#} = \{\psi | \mu_{\psi}^{\#} \text{ is pure point}\}, \quad H_{\mathrm{ac}}^{\#} = \{\psi | \mu_{\psi}^{\#} \text{ is absolutely #-continuous}\}, $$

$$ H_{\mathrm{sing}}^{\#} = \{\psi | \mu_{\psi}^{\#} \text{ is #-continuous singular}\}. $$

We have thus proven.

**Theorem 4.2.4.** $H^\# = H_{\mathrm{pp}}^\# \oplus H_{\mathrm{ac}}^\# \oplus H_{\mathrm{sing}}^\#$. Each of these subspaces is invariant under $A$.

$A \uparrow H_{\mathrm{pp}}^\#$ has a #-complete set of eigenvectors, $A \uparrow H_{\mathrm{ac}}^\#$ has only absolutely #-continuous spectral #-measures and $A \uparrow H_{\mathrm{sing}}^\#$ has only #-continuous singular spectral #-measures.

**Definition 4.2.6.** $\sigma_{\mathrm{pp}}(A) = \{\lambda | \lambda$ is an eigenvalue of $A\}$,

$$ \sigma_{\#\mathrm{cont}}(A) = \sigma(A \uparrow H_{\#\mathrm{cont}}^{\#}) = H_{\mathrm{sing}}^{\#} \oplus H_{\mathrm{ac}}^{\#}, $$

$$ \sigma_{\mathrm{ac}}(A) = \sigma(A \uparrow H_{\mathrm{ac}}^{\#}), $$

$$ \sigma_{\mathrm{sing}}(A) = \sigma(A \uparrow H_{\mathrm{sing}}^{\#}). $$

These sets are called the **pure point**, **#-continuous**, **absolutely #-continuous**, and **singular** (or **#-continuous singular**) spectrum respectively.

**Remark 4.2.2.** While it may happen that $\sigma_{\mathrm{ac}}(A) \cup \sigma_{\mathrm{sing}}(A) \cup \sigma_{\mathrm{pp}}(A) \neq \sigma(A)$ this is only true because we did not define $\sigma_{\mathrm{pp}}(A)$ as $\sigma(A \uparrow H_{\mathrm{pp}}^\#)$ but rather as the actual set of eigenvalues.

**Proposition 4.2.3.** $\sigma_{\#\mathrm{cont}}(A) = \sigma_{\mathrm{ac}}(A) \cup \sigma_{\mathrm{sing}}(A)$,

$$ \sigma(A) = \#-\sigma_{\mathrm{pp}}(A) \cup \sigma_{\#\mathrm{cont}}(A). $$

The sets need not be disjoint, however. The reader should be warned that $\sigma_{\mathrm{sing}}(A)$ may have nonzero #-Lebesgue measure. For many purposes, breaking up the spectrum in this way gives useful information.

Finally, we turn to the question of making canonical choices for the spectral #-measures, a subject which goes under the title of “multiplicity theory.” We will describe the basic results without proof:
§ 4.2.1. Multiplicity free operators

We must first ask when A is unitariy equivalent to multiplication by x on L₂#(*R_c#; d#μ#),
that is, when only one spectral #-measure is needed. An symple examples tells us this
happens in the finite-dimensional case only when A has no repeated eigenvalues, so
we define:

**Definition 4.2.7.** A bounded in $^{*}\mathbb{R}_c^\#\text{-}\#\text{-adjoint operator } A \text{ is called multiplicity free } \text{if and only if } A \text{ is unitarily equivalent to multiplication by } A \text{ on } L_2^\#(^{*}\mathbb{R}_c^\#, d^\#\mu^\#) \text{ for some } \#\text{-measure } \mu^\#.$

One is interested in intrinsic characterizations of “multiplicity free” and there are
several:

**Theorem 4.2.5.** The following are equivalent:

(a) A is multiplicity free.

(b) A has a cyclic vector.

(c) {B|AB = BA} is an abelian algebra.

#-Measure classes

Next we must ask about the nonuniqueness of the #-measure in the multiplicity free
case. Suppose $d^\#\mu^\#$ on $^{*}\mathbb{R}_c^\#$ is given and let $F$ be a #-measurable function which is
positive and nonzero #-a.e. with respect to $\mu^\#$ and locally $L_1^\#(^{*}\mathbb{R}_c^\#, d^\#\mu^\#)$, that is,
$\int_\Sigma |F| d^\#\mu^\# < *\infty$ for every compact set $\Sigma \subset ^{*}\mathbb{R}_c^\#$. Then $d^\#\nu = Fd^\#\mu^\#$ is a #-Borel
\#-measure and the map, $U : L_1^\#(^{*}\mathbb{R}_c^\#, d^\#\nu) \to L_1^\#(^{*}\mathbb{R}_c^\#, d^\#\mu^\#)$ given by $(Uf)(\lambda) = U(\lambda f)$
is unitary (onto since $F \neq 0$ #-a.e.) and $\lambda(Uf) = U(\lambda f)$, Thus, an operator $A$ with a
spectral representation in terms of $\pi$ could just as well be represented in terms of $\nu$.
By the generalized Radon-Nikodym theorem, $d^\#\nu = Fd^\#\mu^\#$ with $F$ #-a.e. nonzero if
and only if $\nu^\#$ and $\mu^\#$ have the same sets of #-measure zero. This suggests the
definition:

**Definition 4.2.8.** Two #-Borel #-measures $\mu^\#$ and $\nu^\#$ are called equivalent if and only if they have the same sets of #-measure zero. An equivalence class $\langle \mu^\# \rangle$ is called a **\#-measure class**.

Then, the nonuniqueness question is answered by:

**Proposition 4.2.7.** Let $\mu^\#$ and $\nu^\#$ be #-Borel #-measures on $^{*}\mathbb{R}_c^\#$ with bounded in $^{*}\mathbb{R}_c^\#$ support. Let $A_{\mu^\#}$ be the operator on $L_2^\#(^{*}\mathbb{R}_c^\#, d^\#\mu^\#)$ given by $(A_{\mu^\#}f)(\lambda) = \lambda f(\lambda)$ and similarly for $A_{\nu^\#}$ on $L_2^\#(^{*}\mathbb{R}_c^\#, d^\#\nu^\#)$. Then $A_{\mu^\#}$ and $A_{\nu^\#}$ are unitarily equivalent if and only if $\mu^\#$ and $\nu^\#$ are equivalent #-measures.

§ 4.2.2. Operators of uniform multiplicity

If one wants a canonical listing of the eigenvalues of a matrix, it is natural to list all eigenvalues of multiplicity one, all eigenvalues of multiplicity two, etc. We thus need a way of saying that $A$ is an operator of uniform multiplicity two, three, etc. It is natural to take:

**Definition 4.2.9.** A bounded self-adjoint operator *A* is said to be of uniform multiplicity *m* ∈ *N* if *A* is unitarily equivalent to multiplication by *λ* on *Ext*−⊕*L*<sup>#</sup><sub>2</sub>(*R*<sup>#</sup><sub>c</sub>, *d*<sup>#</sup>*μ*<sup>#</sup>) where there are *m* terms in the external sum and *μ*<sup>#</sup> is a fixed #-Borel #-measure.
That this is a good definition is shown by

**Proposition 4.2.8.** If *A* is unitarily equivalent to multiplication by *λ* on
$\mathrm{Ext}-\oplus L_2^\#(*\mathbb{R}_c^\#, d^\#\mu^\#)$

(m times) and on $\mathrm{Ext}-\oplus L_2^\#(*\mathbb{R}_c^\#, d^\#v)$ (n times), then $m-n$ and $\mu^\#$ and $v^\#$ are equivalent
\#-measures.

§ 4.2.3. Disjoint \#-measure classes. The multiplicity theorem

In listing eigenvalues of multiplicity one, two, three, etc. in the finitedimensional case,
we must add a requirement that prevents us from counting an eigenvalue of
multiplicity

multiplicity
three once as an eigenvalue of multiplicity one and once as an eigenvalue of

two. In the hyperfinite-dimensional case, we avoid this “error” by requiring the lists to
be

disjoint. The analogous notion for \#-measures is:

**Definition 4.2.10.** Two \#-measure classes $\langle\mu^\#\rangle$ and $\langle v^\#\rangle$ are called disjoint if any
$\mu_1^\# \in \langle\mu^\#\rangle$ and $v_1^\# \in \langle v^\#\rangle$ are mutually singular.

We can now state the basic theorem:

**Theorem 4.2.6. (commutative multiplicity theorem)** Let $A$ be abounded in
$*\mathbb{R}_c^\#$ self-\#-adjoint operator on a Hilbert space $H^\#$. Then there is a decomposition
$\mathrm{Ext}-\bigoplus_{m=1}^\infty H_m^\#$ so that

(a) $A$ leaves each $H_i^\#$ invariant.

(b) $A \upharpoonright H_m^\#$ has uniform multiplicity $m \in *\mathbb{N}$.

(c) The \#-measure classes $\langle \mu_m^\# \rangle$ associated with the spectral representation of $A \upharpoonright H_m^\#$ are mutually disjoint.

**Remark 4.2.3.** Moreover, the subspaces $\{H_m^\#\}_{m=1}^\infty$ (some of which may be zero) and
the \#-measure classes $\{\langle\mu_m^\#\rangle\}_{m=1}^\infty$ are uniquely determined by (a)-(c).

The spectral theorem with the multiplicity theory just described is thus one of those
gems of mathematics: a structure theorem, that is, a theorem that describes all
objects

of a certain sort up to a natural equivalence. Each bounded in $*\mathbb{R}_c^\#$ self-\#-adjoint
operator $A$ is described by a family of mutually disjoint \#-measure classes on
$[-\|A\|_#, \|A\|_#]$; two operators are unitarily equivalent if and only if their spectral
multiplicity \#-measure classes are identical.

§ 4.3. Spectral projections.

In the last section, we constructed a functional calculus, $f \mapsto f(A)$ for any \#-Borel
function and any bounded in $*\mathbb{R}_c^\#$ self-\#-adjoint operator $A$. The most important
functions gained in passing from the continuous functional calculus to the \#-Borel
functional calculus are the characteristic functions of sets.

**Definition 4.3.1.** Let $A$ be a bounded self-\#-adjoint operator and $\Omega$ a \#-Borel set of $*\mathbb{R}_c^\#$. $P_\Omega = \chi_\Omega(A)$ is called a spectral projection of $A$.

As the definition suggests, $P_{\Omega}$ is an orthogonal projection since $\chi_{\Omega} = \chi_{\Omega}^2 = 1$
pointwise. The properties of the family of projections $\{P_{\Omega}|_{\Omega \text{ an arbitrary \#-Borel set}}\}$
is given by the following elementary translation of the functional calculus.

**Proposition 4.3.1.** The family $\{P_{\Omega}\}$ of spectral projections of a bounded
self-#-adjoint operator A, has the following properties:

(a) Each $P_{\Omega}$ is an orthogonal projection.

(b) $P_{\emptyset} = 0$; $P_{(-a,a)} = I$ for some $a \in *\mathbb{R}_{c}^{\#}$.

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

$$P_{\Omega} = s\text{-}\lim_{N \to \infty} \left( \text{Ext}-\sum_{n=1}^{N} P_{\Omega_n} \right). \qquad (4.3.1)$$

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

**Definition 4.3.2.** A family of projections obeying (a)-(c) is called a projection-valued #-measure (p.v.#-m.).

We remark that (d) follows from (a) and (c) by abstract considerations.

As one might guess, one can integrate with respect to a p.v.#-m. If $P_{\Omega}$ is a p.v.#-m., then $\langle \phi, P_{\Omega}\phi \rangle_{\#}$ is an ordinary #-measure for any $\phi$. We will use the symbol $d^{\#}\langle \phi, P_{\lambda}\phi \rangle_{\#}$ to mean integration with respect to this #-measure. By generalized Riesz lemma methods, there is a unique operator $B$ with $\langle \phi, B\phi \rangle_{\#} = \text{Ext}-\int f(\lambda)d^{\#}\langle \phi, P_{\lambda}\phi \rangle_{\#}$.

**Theorem 4.3.1.** If $P_{\Omega}$ is a p.v.#-m. and $f$ a bounded in $*\mathbb{R}_{c}^{\#}$ #-Borel function on $\text{supp}(P_{\Omega})$, then there is a unique operator $B$ which we denote $\text{Ext}-\int f(\lambda)d^{\#}P_{\lambda}$ so that

$$\langle \phi, B\phi \rangle_{\#} = \text{Ext}-\int f(\lambda)d^{\#}\langle \phi, P_{\lambda}\phi \rangle_{\#}. \qquad (4.3.2)$$

**Theorem 4.3.2.** (spectral theorem-p.v.#-m. form) There is a one-one correspondence between (bounded) self-#-adjoint operators $A$ and (bounded) projection valued #-measures $\{P_{\Omega}\}$ given by

$$A \mapsto \{P_{\Omega}\} = \{\chi_{\Omega}(A)\} \qquad (4.3.3)$$

and

$$\{P_{\Omega}\} \mapsto A = \text{Ext}-\int \lambda d^{\#}P_{\lambda}. \qquad (4.3.4)$$

Spectral projections can be used to investigate the spectrum of $A$.

**Proposition 4.3.1.** $\lambda \in \sigma(A)$ if and only if $P_{(\lambda-\varepsilon, \lambda+\varepsilon)}(A)$ for any $\varepsilon > 0$.

The essential element of the proof is that $\|(A - \lambda)^{-1}\|_{\#} = [\text{dist}(\lambda, \sigma(A))]^{-1}$.

This suggests that we distinguish between two types of spectrum.

**Definition 4.3.3.** We say that (i) $\lambda \in \sigma_{\text{ess}}(A)$, the essential spectrum of $A$ if and only if $P_{(\lambda-\varepsilon, \lambda+\varepsilon)}(A)$ is hyper infinite dimensional for all $\varepsilon > 0$.

(ii) If $\lambda \in \sigma(A)$ but $P_{(\lambda-\varepsilon, \lambda+\varepsilon)}(A)$ is hyperfinite dimensional for some $\varepsilon > 0$, we say $\lambda \in \sigma_{\text{disc}}(A)$, the discrete spectrum of $A$. $P$ is hyper infinite dimensional means

Ran($P$) is hyper infinite dimensional.

Thus, we have a second decomposition of $\sigma(A)$. Unlike the first, it is a decomposition into two necessarily disjoint subsets. We note that $\sigma_{\text{disc}}$ is not necessarily #-closed, but notice that.

**Theorem 4.3.3** $\sigma_{\text{ess}}(A)$ is always #-closed.

**Proof** Let $\lambda_n \to_{\#} \lambda$ with each $\lambda_n \in \sigma_{\text{ess}}(A)$. Since any #-open interval $I$ about $\lambda$ contains an interval about some $\lambda_n$, $P_I(A)$ is hyper infinite dimensional.

The following three theorems give alternative descriptions of $\sigma_{\text{disc}}$ and $\sigma_{\text{ess}}$;

**Theorem 4.3.4** $\lambda \in \sigma_{\text{disc}}$ if and only if both the following hold:

(a) $\lambda$ is an #-isolated point of $\sigma(A)$ that is, for some $\varepsilon \approx 0$,

$$(\lambda - \varepsilon, \lambda + \varepsilon) \cap \sigma(A) = \{\lambda\}.$$

(b) $\lambda$ is an eigenvalue of hyperfinite multiplicity, i.e., $\{\psi|A\psi = \lambda\psi\}$ is hyperfinite
dimensional.

**Theorem 4.3.5** $\lambda \in \sigma_{ess}$ if and only if one or more of the following holds:

(a) $\lambda \in \sigma_{\#\text{cont}}(A) \leftrightarrow \sigma_{\text{ac}}(A) \cup \sigma_{\text{sing}}(A)$.

(b) $\lambda$ is a \#-limit point of $\sigma_{\text{pp}}(A)$.

(c) $\lambda$ is an eigenvalue of hyper infinite multiplicity.

**Theorem 4.3.6** (Generalized Weyl's criterion) Let $A$ be a bounded in $^{*}\mathbb{R}_c^\#$
self-\#-adjoint operator. Then (i) $\lambda \in \sigma(A)$ if and only if there exists $\{\psi_n\}_{n=1}^{\infty}$ with
$\|\psi_n\|_{\#} = 1$ and $\#\lim_{n \to \infty} \|(A - \lambda)\psi_n\|_{\#} = 0$.

(ii) $\lambda \in \sigma_{\text{ess}}(A)$ if and only if the above $\{\psi_n\}$ can be chosen to be orthogonal.
As one might guess, the essential spectrum cannot be removed by essentially
hyperfinite dimensional perturbations. In Section 4.4, we will prove a general
theorem which implies that $\sigma_{\text{ess}}(A) = \sigma_{\text{ess}}(B)$ if $A \setminus B$ is \#-compact.

Finally, we discuss one useful formula relating the resolvent and spectral projections.
It is a matter of computation to see that

$$
f_{\varepsilon}(x) \to_{\#} \begin{cases} 0 & \text{if } x \notin [a, b] \\ \frac{1}{2} & \text{if } x = a \lor x = b \\ 1 & \text{if } x \in (a, b) \end{cases}
$$

if $\varepsilon \to \# 0$, where

$$
f_{\varepsilon}(x) = (2\pi\#i)^{-1} \left( \text{Ext} - \int_{a}^{b} \left[ (x - \lambda - i\varepsilon)^{-1} - (x - \lambda + i\varepsilon)^{-1} \right] d^{\#}\lambda \right). \quad (4.3.5)
$$

Moreover, $|f_{\varepsilon}(x)|$ is bounded in $x \in * \mathbb{R}_c^\#$ uniformly in $\varepsilon \approx 0$, so by the functional calculus, one obtains that.

**Theorem 4.3.7** (Generalized Stone's formula) Let $A$ be a bounded in $^{*}\mathbb{R}_c^\#$ self-\#-adjoint operator. Then

$$
\begin{align}
\text{s-lim}_{\varepsilon \to \# 0} (2\pi \# i)^{-1} \left( \text{Ext} - \int_a^b \left[ (A - \lambda - i\varepsilon)^{-1} - (A - \lambda + i\varepsilon)^{-1} \right] d^\# \lambda \right) &= \tag{4.3.6} \\
&= \frac{1}{2} [P_{[a,b]} + P_{(a,b)}]. \nonumber
\end{align}
$$

§ 4.4. The \#-continuous functional calculus related to unbounded in $^{*}\mathbb{R}_c^\#$ self-\#-adjoint operators

In this section we will show how the spectral theorem for bounded in $^{*}\mathbb{R}_c^\#$ self-\#-adjoint operators which we developed in § 4.3 can be extended to unbounded in $^{*}\mathbb{R}_c^\#$ self-\#-adjoint operators. To indicate what we are aiming for, we first prove the following:

**Proposition 4.4.1.** Let $\langle M, \mu^\# \rangle$ be a \#-measure space with $\mu^\#$ a hyperfinite \#-measure. Suppose that $f$ is a \#-measurable, $^{*}\mathbb{R}_c^\#$-valued function on $M$ which is finite or hyperfinite a.e. $\mu^\#$. Then the operator $T_f : \varphi \to f\varphi$ on $L_2^\#(M, d^\#\mu^\#)$ with domain

$$
D(T_f) = \{\varphi | f\varphi \in L_2^\#(M, d^\#\mu^\#)\} \quad (4.4.1)
$$

is self-\#-adjoint and $\sigma(T_f)$ is the essential range of $T_f$.

**Proof** $T_f$ is clearly symmetric. Suppose that $\psi \in D(T_f^*)$ and let
$$ \chi_N = \begin{cases} 1 & \text{if } |f(m)| \le N \\ 0 & \text{otherwise} \end{cases} $$

Then, using the generalized monotone #-convergence theorem,

$$ \|T_f^* \psi\|_{\#} = \#\text{-lim}_{N \to *\infty} \|\chi_N T_f^* \psi\|_{\#} = \#\text{-lim}_{N \to *\infty} \left( \sup_{\|\varphi\|_{\#}=1} |\langle \varphi, \chi_N T_f^* \psi \rangle_{\#}| \right) = $$

$$ \#\text{-lim}_{N \to *\infty} \left( \sup_{\|\varphi\|_{\#}=1} |\langle \chi_N T_f \varphi, \psi \rangle_{\#}| \right) = \#\text{-lim}_{N \to *\infty} \left( \sup_{\|\varphi\|_{\#}=1} |\langle \varphi, \chi_N f \psi \rangle_{\#}| \right) = $$

$$ \#\text{-lim}_{N \to *\infty} \|\chi_N f \psi\|_{\#} $$

Thus, $f\psi \in L_2^\#(M, d^\#\mu^\#)$, so $\psi \in D(T_f)$ and therefore $T_f$ is self-#-adjoint. That $\sigma(T_f)$ is the essential range of $f$ follows as in the bounded case.

With more information about $f$, one can say something about the domains on which $T_f$ is essentially self-#-adjoint:

**Proposition 4.4.2.** Let $f$ and $T_f$ obey the conditions in Proposition 4.4.1. Suppose in addition that $f \in L_p^\#(M, d^\#\mu^\#)$ for $2 < p < *\infty$. Let $D$ be any #-dense set in $L_q^\#(M, d^\#\mu^\#)$ where $q^{-1} + p^{-1} = 1/2$. Then $D$ is a #-core for $T_f$.

**Proof** Let us first show that $L_q^\#$ is a #-core for $T_f$. By the generalized Holder's inequality $\|g\|_{\#2} \le \|1\|_{\#p} \cdot \|g\|_{\#q}$, and $\|fg\|_{\#2} \le \|f\|_{\#p} \cdot \|g\|_{\#q}$ so $L_p^\# \subset D(T_f)$.

Moreover, if $g \in D(T_f)$ let $g_n, n \in *\mathbb{N}$ be that function which is zero where $|g(m)| > n$ and equal to $g$ otherwise. By the generalized dominated convergence theorem, $g_n \to_* g$ and $fg_n \to_* fg$ in $L_2^\#$. Since each $g_n$ is in $L_q^\#$, we conclude that $L_q^\#$ is a #-core for $T_f$. Now let $D$ be #-dense in $L_q^\#$ and let $g \in L_q^\#$. Find $g_n \in D$ with $g_n \to_* g$ in $L_q^\#$. Since $\|g_n - g\|_{\#2} \le \|1\|_{\#p} \cdot \|g_n - g\|_{\#q}$ and $\|T_f(g_n - g)\|_{\#2} \le \|f\|_{\#p} \cdot \|g_n - g\|_{\#q}$, $g \in \#-\overline{D(T_f \upharpoonright D)}$.

Thus $L_q^\# \subset D(T_f \upharpoonright D)$ so $D$ is a #-core. Unless $f \in L_{*\infty}^\#(M, d^\#\mu^\#)$ the operator $T_f$ described in Propositions 4.4.1 and 4.4.2 will be unbounded.

Thus, we have found a large class of unbounded self-#-adjoint operators. In fact, we have found them all.

**Theorem 4.4.1.** (spectral theorem-multiplication operator form) Let $A$ be a self-adjoint operator on a $*\infty$-dimensional a non-Archimedean Hilbert space $H^\#$ with domain $D(A)$. Then there is a #-measure space $\langle M, \mu^\# \rangle$ with $\mu^\#$ a hyperfinite #-measure, a unitary operator $U: H^\# \to L_2^\#(M, d^\#\mu^\#)$, and a $*\mathbb{R}_c^\#$-valued function $f$ on $M$ which is finite or hyperfinite $\mu^\#$-a.e. so that

$$ (a) \quad \psi \in D(A) \text{ if and only if } f(\cdot)(U\psi)(\cdot) \in L_2^\#(M, d^\#\mu^\#). $$

$$ (b) \quad \text{If } \varphi \in U[D(A)], \text{ then } (UAU^{-1}\varphi)(m) = f(m)\varphi(m). $$

**Proof** It easily verify that $A + i$ and $A - i$ are one to one correspondence and $\text{Ran}(A \pm i) = H^\#$. Since $A \pm i$ are #-closed, $(A \pm i)^{-1}$ are #-closed and therefore bounded in $*\mathbb{R}_c^\#$. Note that the operators $(A + i)^{-1}$ and $(A - i)^{-1}$ commute. The equality $\langle (A - i)\psi, (A + i)^{-1}(A + i)\varphi \rangle_{\#} = \langle (A + i)^{-1}(A - i)\psi, (A + i)\varphi \rangle_{\#}$ and the fact that $\text{Ran}(A \pm i) = H^\#$ shows that $((A + i)^{-1})^* = (A - i)^{-1}$. Thus the operator $(A + i)^{-1}$ is normal.

We now use the easy extension of the spectral theorem for bounded in $*\mathbb{R}_c^\#$ self-#-adjoint operators to bounded in $*\mathbb{R}_c^\#$ normal operators. The proof of this extension is a straightforward. We conclude that there is a #-measure space $\langle M, \mu^\# \rangle$
with $\mu^\#$ a hyperfinite #-measure, a unitary operator $U : \mathbf{H}^\# \to L_2^\#(M, d^\#\mu^\#)$, and a
\#-measurable, bounded, in $*\mathbb{R}_c^\# *\mathbb{C}_c^\#$-valued function $g(m)$ so that
$U(A + i)^{-1} U^{-1} \varphi(m) = g(m)\varphi(m)$ for all $\varphi \in L_2^\#(M, d^\#\mu^\#)$. Since $\text{Ker}((A + i)^{-1})$ is empty,
$g(m) \neq 0$ a.e. $\mu^\#$, so the function $f(m) = g^{-1}(m) - i$ is hyperfinite a.e. $\mu^\#$. Now, suppose
$\psi \in D(A)$. Then $\psi = (A + i)^{-1}\varphi$ for some $\varphi \in \mathbf{H}^\#$ and $U\psi = gU\varphi$. Since $fg$ is bounded
in $*\mathbb{R}_c^\#$, we conclude that $f(U\psi) \in L_2^\#(M, d^\#\mu^\#)$. Conversely, if $f(U\psi) \in L_2^\#(M, d^\#\mu^\#)$,
then there is a $\varphi \in \mathbf{H}^\#$ so that $U\varphi = (f + i)U\psi$. Thus, $gU\varphi = g(f + i)U\psi = U\psi$, so
$\psi = (A + i)^{-1}\varphi$ which shows that $\psi \in D(A)$. This proves (a).

To prove (b) notice that if $\psi \in D(A)$ then $\psi = (A+i)^{-1}\varphi$ for some $\varphi \in \mathbf{H}^\#$ and $A\psi = \varphi - i\psi$. Therefore, $(UA\psi)(m) = (U\varphi)(m) - i(U\psi)(m) = (g^{-1}(m) - i)(U\psi)(m) = f(m)(U\psi)(m)$. Finally, if $\text{Im}(f) > 0$ on a set of nonzero Lebesgue \#-measure, there is a bounded in $*\mathbb{R}_c^\#$ set $B$ in the upper half plane so that $S = \{x|f(x) \in B\}$ has nonzero Lebesgue \#-measure. If $\chi(x)$ is the characteristic function of $S$ then $f\chi \in L_2^\#(M, d^\#\mu^\#)$ and $\text{Im}\langle\chi, f\chi\rangle > 0$. This contradicts the fact that multiplication by $i$ is self-adjoint (since it is unitarily equivalent to $A$). Thus $f$ is $*\mathbb{R}_c^\#$-valued function.

There is a natural way to define functions of a self-\#-adjoint operator by using the above theorem. Given a bounded in $*\mathbb{R}_c^\#$ \#-Borel function $h$ on $*\mathbb{R}_c^\#$ we define

$$h(A) = \mathrm{UT}_{h(f)} U^{-1} \qquad (4.4.2)$$

where $T_{h(f)}$ is the operator on $L_2^\#(M, d^\#\mu^\#)$ which acts by multiplication by the function $h(f(m))$. 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 $\text{Ext}-\text{exp}(itA)$ 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 $\text{Ext}-\text{exp}(itA)$ 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 $\mu_\psi^\#$
is the measure satisfying $\text{Ext}- \int_{*\mathbb{R}_c^\#} g(x)d^\#\mu_\psi^\#(x) = \langle \psi, g(A)\psi \rangle_\#$ 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 $n \neq m$ 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 \#-measure on $*\mathbb{R}_c^\#$ which we denote by $d^\#\langle \varphi, P_\lambda\varphi \rangle_\#$ as in § 4.3.

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

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

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

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

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

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

As in § 4.3, we write symbolically

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

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

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

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 $\mathbf{H}^\#$ the correspondence being given by

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

We use the functional calculus developed above to define $\text{Ext-exp}(itA)$.
**Theorem 4.4.4.** Let A be a self-#-adjoint operator and define $U(t) = \text{Ext-exp}(itA)$. 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 $\varphi \in \mathbf{H}^\#$ and $t \to t_0$, then $U(t)\varphi \to U(t_0)\varphi$.

(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 $\text{Ext-exp}(it\lambda)$. To prove (b) observe that

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

Since $|\text{Ext-exp}(it\lambda) - 1|^2$ is dominated by the #-integrable function $g(\lambda) = 2$ and since for each $\lambda \in *\mathbb{R}_c^\#$: $|\text{Ext-exp}(it\lambda) - 1|^2 \to_{\#} 0$ as $t \to \# 0$ we conclude that $\|U(t)\varphi - \varphi\|_{\#}^2 \to_{\#} 0$ 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 $|\text{Ext-exp}(ix) - 1|^2 \le |x|$. To prove (d), we define

$$ D(B) = \left\{ \psi \mid \#\lim_{t \to \# 0} \frac{U(t)\psi - \psi}{t} \text{ exists} \right\} \quad (4.4.11) $$

and let

$$ iB\psi = \#\lim_{t \to \# 0} \frac{U(t)\psi - \psi}{t}. \quad (4.4.12) $$

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 $t \to \# 0$. 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

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

Note that

$$ \begin{aligned} \langle U(b)\psi_a, \varphi \rangle_{\#} &= \langle \psi_a, U(-b)\varphi \rangle_{\#} = \int_0^a \langle U(t)\psi, U(-b)\varphi \rangle_{\#} d^\#t = \\ & \int_0^a \int_0^b \langle U(t+b)\psi, \varphi \rangle_{\#} d^\#t = \int_0^a \int_0^{a+b} \langle U(t)\psi, \varphi \rangle_{\#} d^\#t. \end{aligned} \qquad (4.4.14) $$

From (4.1.14) we obtain

$$ \begin{aligned} |\langle U(b)\psi_a, \varphi \rangle_{\#} - \langle \psi_a, \varphi \rangle_{\#}| = & \\ = \left| \int_0^b \langle U(t)\psi, \varphi \rangle_{\#} d^\#t \right| + \left| \int_0^{a+b} \langle U(t)\psi, \varphi \rangle_{\#} d^\#t \right| \le 2a \|\varphi\|_{\#} \|\psi\|_{\#} \end{aligned} \qquad (4.4.15) $$

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 $\varphi \in \{\psi_a|\psi \in \mathbf{H}^\#, a \in \mathbb{R}_c^\#\}^\top$ and let $\{\psi^{(n)}\}_{n \in \mathbb{N}}$ be an orthonormal basis for $\mathbf{H}^\#$.

Then for each $n \in \mathbb{N}$

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

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 $t_0 \notin \cup_{n \in \mathbb{N}} S_n$. 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

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

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

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

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

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

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 case). For $\varphi_f \in D$ we obtain that

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

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

$$ \begin{aligned} \langle A\varphi_f, \varphi_g \rangle_{\#} &= \#\mathrm{lim}_{s \to \#} \langle \left(\frac{U(s) - I}{is}\right) \varphi_f, \varphi_g \rangle_{\#} = \\ &= \#\mathrm{lim}_{s \to \#} \langle \varphi_f, \left(\frac{I - U(-s)}{is}\right) \varphi_g \rangle_{\#} = \frac{1}{i} \langle \varphi_f, \varphi_{-g^{\#'}} \rangle_{\#} = \langle \varphi_f, A\varphi_g \rangle_{\#} \end{aligned} \qquad (4.4.21) $$

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 $*\mathbb{R}_c^\#$-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

$$w^{\#}(t) = iAU(t)\varphi - i(\#-\bar{A})V(t)\varphi = iAw(t). \qquad (4.4.23)$$

Thus

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

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 $\mathbf{H}^\#$, $U(t) = V(t)$.

**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

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

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 (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

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

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

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

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 $r_n = \text{Ext}-\prod_{i=1}^n (a_i, b_i)$ by $P_{r_n} = \text{Ext}-\prod_{i=1}^n P_{(a_i, b_i)}^i$
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
\#\mathbb{C}^\#\text{-valued \#-measure of hyperfinite mass which we denote by } d^\#\langle \varphi, P_{\lambda\psi} \rangle_\#.
\text{ Applying generalized Fubini's theorem we conclude that}$

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

§ 4.5. Nearstandard $C_{\#}^*$ algebras generated by spectral
productions 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 $H^{\#}$. 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 on any vector $\psi \in H^{\#}$ for example, one might have $(\text{Ran}(A)) \cap D(B) = \emptyset$ 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 $\text{Ext}\text{-exp}(itA)$ and $\text{Ext}\text{-exp}(isA)$ 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 $H^{\#}$.

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

**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
$$s\text{-}\#\text{-}\lim_{\epsilon\to\#} \mathcal{O}[i\epsilon R_{a+i\epsilon}(A)] = P_{\{a\}}^{A}. \quad (4.5.1)$$

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,

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

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

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

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 hyperinfinite sequence $f_n, n \in \mathbb{N}$ of uniformly bounded functions in $S^{\#}(\mathbb{R}_c^\#)$. By the functional calculus,

$$ s\text{-}\#\text{-}\lim_{n\to\infty} f_n(A) = P_{(a,b)}^A. \quad (4.5.4) $$

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

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

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

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

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 $\mathrm{st}(\|A\|_\#) \in \mathbb{R}$ and $\mathrm{st}(\|A\|_\#) \neq \infty$.

**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. $\mathrm{st}(\|A\psi\|_\#) \neq \infty$.

**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 $\approx$-commute if $\|AB\|_\# \approx \|BA\|_\#$

**Remark 4.5.3.** Note that the operators $A$ and $B$ are $\approx$-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 $P_{(c,d)}^B$, $\approx$-commute.

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

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

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

**Example 4.6.1.** Let $H = \mathcal{L}_2^\#(*\mathbb{R}_c^\#)$ and $Q(q) = C_0^\infty(*\mathbb{R}_c^\#)$ 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

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

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, $q(\psi, \psi) \ge -M\|\psi\|_\#^2$ is called #-closed if $Q(q)$ is complete under the #-norm
$$ \|\psi\|_{\#+1} = \sqrt{q(\psi, \psi) + (M+1)\|\psi\|_{\#}^2}. \qquad (4.6.3) $$

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

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

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

$\varphi_n \in Q(q) \varphi_n \xrightarrow{H^\#} \varphi$ 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 $\varphi_n \in C_0^\infty(*\mathbb{R}_c^\#)$. Then $\varphi_n \xrightarrow{\#} 0$, and $q(\varphi_n - \varphi_m, \varphi_n - \varphi_m) \xrightarrow{\#} 0$, but $q(\varphi_n, \varphi_n) \xrightarrow{\#} 1 \neq q(0,0)$ 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 $f,g \in C_0^\infty(*\mathbb{R}_c^\#)$.

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 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 $\langle\varphi, \psi\rangle_{\#+1} = q(\varphi, \psi) + \langle\varphi, \psi\rangle_\#$. We denote by $H_{-1}^\#$ the space of bounded in $*\mathbb{R}_c^\#$ 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 $*\mathbb{R}_c^\#$ because

$$ [j(\psi)(\varphi)] \leq \|\varphi\|_{\#} \|\psi\|_{\#} \leq \|\varphi\|_{\#} \|\psi\|_{\#} \leq \|\varphi\|_{\#+1} \|\psi\|_{\#}. \qquad (4.6.5) $$

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

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

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 $[\hat{B}\Phi](\varphi) = q(\varphi, \Phi) + (\varphi, \Phi)_\#$. 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

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

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 $\lambda \neq 0$, and $\lambda[j(\psi)] = 0$ for each $\psi \in H^\#$. By the generalized Riesz Lemma, there is a $\varphi_\lambda \neq 0$ in $H_{+1}^\#$ so that $0 = \lambda[j(\psi)] = [j(\psi)](\varphi_\lambda) = \langle\varphi_\lambda, \psi\rangle_\#$ for all $\psi \in H^\#$. Since $\varphi_\lambda \neq 0$, 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 $\|\cdot\|_{\#} \leq \|\cdot\|_{\#+1}$ and $H_{+1}^\#$ is #-norm #-dense in $H^\#$, $D(B)$ is #-norm #-dense in $H^\#$. Suppose $\varphi, \psi \in D(B)$. Then one obtains that
$$ \langle \varphi, B\psi \rangle_{\#} = q(\varphi, \psi) + \langle \varphi, \psi \rangle_{\#} = \overline{q(\psi, \varphi) + \langle \psi, \varphi \rangle_{\#}} = \overline{\langle \psi, B\varphi \rangle_{\#}} = \langle B\varphi, \psi \rangle_{\#}. \quad (4.6.8) $$

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 $*\mathbb{R}_c^\#$ 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 $\varphi, \psi \in D(A)$, $\langle \varphi, A\psi \rangle_{\#} = q(\varphi, \psi)$. 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 $*\mathbb{R}_c^\#$ symmetric operators and semi-bounded in $*\mathbb{R}_c^\#$ 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 $B \uparrow D(A) = A$ then $A = B$. But it can happen that a and b are #-closed semibounded in $*\mathbb{R}_c^\#$ quadratic forms and $b \uparrow Q(a)xQ(a) = a$ without having $a = b$.

(2) Let A be a symmetric operator that is semibounded in $*\mathbb{R}_c^\#$. Let q be the quadratic form $q(\varphi, \psi) = \langle \varphi, A\psi \rangle_{\#}$ 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.

(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 $*\mathbb{R}_c^\#$ 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 $R_\lambda(A_n) \to_A R_\lambda(A)$ 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 $R_\lambda(A_n) \to_A R_\lambda(A)$ 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 *$\mathbb{R}^\#_c$ 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 $A_n \to_\# A$ as $n \to *\infty$ in the #-norm resolvent sense if and only if $A_n \to_\# A$ as $n \to *\infty$ in #-norm.

**Proof.** Let $A_n \to_\# A$ 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 $A_n \to_\# A$ 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 $\mathrm{Im}\,\lambda_0 \neq 0$ and if $R_{\lambda_0}(A_n)\varphi - R_{\lambda_0}(A)\varphi \to_\# 0$, for all $\varphi \in H^\#$ then $A_n \to_\# A$ as $n \to *\infty$ in the strong resolvent sense.

**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 $\lambda_0$,

$$ 
\begin{align} 
R_{\lambda}(A) &= \mathrm{Ext} - \sum_{m=0}^{*\infty} (\lambda_0 - \lambda)^m [R_{\lambda_0}(A)]^{m+1}, \tag{4.7.1} \\ 
R_{\lambda}(A_n) &= \mathrm{Ext} - \sum_{m=0}^{*\infty} (\lambda_0 - \lambda)^m [R_{\lambda_0}(A_n)]^{m+1} \notag 
\end{align} 
$$

which #-converge in #-norm in the circle $|\lambda - \lambda_0| < |\mathrm{Im}\,\lambda_0|^{-1}$. 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

$$ 
\begin{align} 
\|R_{\bar{\lambda}_0}(A_n) - R_{\bar{\lambda}_0}(A)\|_{\#} &= \|(R_{\lambda_0}(A_n) - R_{\lambda_0}(A))^*\|_{\#} = \tag{4.7.2} \\ 
&\quad \|R_{\lambda_0}(A_n) - R_{\lambda_0}(A)\|_{\#} \to_{\#} 0 \text{ as } n \to *\infty \notag 
\end{align} 
$$

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

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 $C^{*\infty}(*\mathbb{R}_c^\#)$, the #-continuous functions vanishing at hyper infinity. Thus, given $\varepsilon \approx 0, \varepsilon > 0$, we can find an hyperfinite polynomial $P(s,t)$ so that

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

Therefore,

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

and

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

If $A_n \to_{\#} A$ as $n \to *\infty$ in the #-norm resolvent sense, then

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

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 $A_n \to_{\#} A$ in the strong resolvent sense and $h \in C^{*\infty}(*\mathbb{R}_c^\#)$, then $h(A_n)\varphi \to_{\#} h(A)\varphi$. 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 $g_m(x) \to_{\#} 1$ pointwise, $g_m(A)\psi \to_{\#} \psi$ by spectral theorem, so we can find an $m$ with $\|g_m(A)\psi - \psi\|_{\#} \le \varepsilon(6\|f\|_{s,\infty})^{-1}$. Furthermore since $g_m \in C^{*\infty}(*\mathbb{R}_c^\#)$, $g_m(A_n)\psi \to_{\#} g_m(A)\psi$ by the remark above, so we can find an $N_0$, so that $n > N_0$ implies $\|g_m(A_n)\psi - g_m(A)\psi\|_{\#} \le \varepsilon(6\|f\|_{s,\infty})^{-1}$. Therefore, if $n \ge N_0$,

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

Since $fg_m$ is #-continuous and goes to zero at $*\infty$, there is an $N_1$ so that $n \ge N_1$ implies
$$ \|f(A_n)g_m(A_n)\psi - f(A)g_m(A)\psi\|_{\#} \le \frac{\epsilon}{3}. \qquad (4.7.9) $$

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

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

Since $\psi$ and $\epsilon$ 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 $A_n \to_{\#} A$ in the #-norm resolvent sense $Ext-\exp(-tA_n)$ #-converges in #-norm to $Ext-\exp(-tA)$ 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 $\|Ext-\exp(iA_n) - Ext-\exp(iA_n)\|_{\#} = 1$ 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 $A_n \to_{\#} A$ in the strong resolvent sense if and only if $Ext-\exp(itA_n)$ #-converges strongly to $Ext-\exp(itA)$ for each $t$.

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

To prove the theorem in the other direction, we first derive a formula for the resolvent of a self-#-adjoint operator $A$. Suppose that $\mathrm{Im}\,\mu < 0$. Then, by the functional calculus

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

Therefore,

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

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
$$ \|R_{\mu}(A_n)\varphi - R_{\mu}(A)\varphi\|_{\#} \le \text{Ext-} \int_{0}^{*\infty} [\text{Ext-exp}(t \text{Im } \mu)] \|[\text{Ext-exp}(itA_n) - \text{Ext-exp}(itA)]\varphi\|_{\#} d^{\#}t \quad (4.7.13) $$

so if $\text{Ext-exp}(itA_n) \to_{\#} \text{Ext-exp}(itA)$ 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 $\text{Im } \mu > 0$. We remark that it is possible to show that if $A_n \to_{\#} A$ in the strong resolvent sense, then $\text{Ext-exp}(itA_n) \to_{\#} \text{Ext-exp}(itA)$ 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 $\|R_{\lambda_0}(A_n)\|_{\#} \le |\text{Im } \lambda_0|^{-1}$, $\|T_{\lambda_0}\| \le |\text{Im } \lambda_0|^{-1}$, and so

$$ T_{\lambda} = \text{Ext-} \sum_{n=0}^{*\infty} (\lambda_0 - \lambda)^n (T_{\lambda_0})^{n+1}. \quad (4.7.14) $$

is well defined for $|\lambda_0 - \lambda| \le |\text{Im } \lambda_0|^{-1}$. 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 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 $\text{Im } \lambda \ne 0$

$$ T_{\lambda}\varphi = \#\text{-lim}_{n \to *\infty} R_{\lambda}(A_n)\varphi. \quad (4.7.15) $$

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 $\text{Ker}(T_{\lambda}) = (\text{Ran}(T_{\lambda}^*))^\top = (\text{Ran}(T_{\lambda}))^\top = D^\top = \{0\}$. 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 $\text{Im } \lambda \ne 0$, is chosen. Since $\text{Ran}(A \pm i) = \text{Ran}(T_{\pm i}^*)$ 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

$$
\| R_{\mu}(A_{n}) - R_{\mu}(A) \|_{\#} \to_{\#} 0 \qquad (4.7.16)
$$

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

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

**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, $\sigma(A_n) \cap (a, b) \subset (a \pm \varepsilon, b - \varepsilon)$ for $n \ge N$. Let $f_\varepsilon$ be a #-continuous function
which equals one on $(a \pm \varepsilon, b - \varepsilon)$ 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
$A_n \to_\# A$ 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

$$
\|(A - \lambda_0)^{-1}\|_{\#} \le \# \lim_{n \to *\infty} \|(A_n - \lambda_0)^{-1}\|_{\#} \le \sqrt{2}/(b-a). \text{ This proves (a).}
$$

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 $L_2^s(*\mathbb{R}_c^s)$ then $A_n$ #-converges to the zero operator in the strong resolvent sense. For each $n$, $\sigma(A_n) = *\mathbb{R}_c^s$, 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 $*\mathbb{C}_c^s$ 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 #-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 $\|\varphi\|_{\#A} = \|A\varphi\|_{\#} + \|\varphi\|_{\#}$. If $\sup_{\varphi \neq 0} \|\varphi\|_{\#A} = 1$ ($\|(A_n - A)\varphi\|_{\#} \to 0$ 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 $H_{+1}^s$ 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 $H_{+1}^s$ to $H_{-1}^s$ 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 $\varphi \in H_{-1}^s : (A_n + i)^{-1}\varphi \to_{\#} (A + i)^{-1}\varphi$. 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 $\Gamma_{s_\infty}^*$. If $\Gamma_{s_\infty}^*$ is the graph of an operator $A$ we say that $A$ is the strong graph $\#$-limit of $A_n$ and write

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

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 $\langle \varphi, A\varphi \rangle_\# \in \Gamma_{s_\infty}^*$. Thus $\Gamma(A) \subset \Gamma_{s_\infty}^*$. 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

$$ \begin{aligned} \eta_n - \varphi_n &= [(A + i)^{-1} - (A_n + i)^{-1}][(A_n + i)\varphi_n] = \\ &= [(A + i)^{-1} - (A_n + i)^{-1}][(A_n + i)\varphi_n - \psi - i\varphi] + \\ &\quad +[(A + i)^{-1} - (A_n + i)^{-1}][\psi + i\varphi] \to_{\#} 0 \end{aligned} \qquad (4.8.2) $$

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 $A = \text{st. gr. -}\#\text{-lim } A_n$. 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 $\#$-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 $D_{s_\infty}^* = \{\psi | \langle \psi, \varphi \rangle_\# \in \Gamma_{s_\infty}^* \text{ for some } \varphi\}$. If $D_{s_\infty}^*$ is $\#$-dense, then $\Gamma_{s_\infty}^*$ is the graph of an operator.

(b) Suppose that $D_{s_\infty}^*$ is $\#$-dense and let $A = \text{st. gr. -}\#\text{-lim } A_n$. 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 $\eta \in D^s_{\infty}$. 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, $\langle \psi - \psi, \eta \rangle_{\#} = \#\lim_{n \to *\infty} \langle A_n(\varphi_n - \varphi'_n), \eta_n \rangle_{\#} = \#\lim_{n \to *\infty} \langle \varphi_n - \varphi'_n, A_n\eta_n \rangle_{\#} = 0$ so $\psi = \psi'$ since $D^s_{\infty}$ 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 $\Gamma^w_{\infty}$ if we can find $\psi_n \in D(A_n)$ so that $\psi_n \underset{\#}{\stackrel{\|\cdot\|_\#}{\to}} \psi$ and $A_n\psi_n \to_{\#} \varphi$ weakly. If $\Gamma^w_{\infty}$ is the graph of an operator, A we say that A is the weak graph \#-limit of $A_n$ and abbreviated as $A = \mathbf{w}.\mathbf{gr.}$. \#-lim $A_n$.

**Theorem 4.8.3.** Let $(A_n)_{n=1}^{*\infty}$ be a hyper infinite sequence of symmetric operators. If $D^s_{\infty} = \{\psi | \langle \psi, \varphi \rangle_{\#} \in \Gamma^s_{\infty} \text{ for some } \varphi\}$ is \#-dense, then $\Gamma^w_{\infty}$ 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 $A = \mathbf{w}.\mathbf{gr.}$ \#-lim $A_n$ if and only if $A_n \to_{\#} A$ 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

$$ \mathrm{Ext}\text{-exp}(A + B) = \#\lim_{n \to *\infty} \{[\mathrm{Ext}\text{-exp}(A/n)] \times [\mathrm{Ext}\text{-exp}(B/n)]\}^n. \quad (4.9.1) $$

**Proof** Let $S_n = \mathrm{Ext}\text{-exp}((A+B)/n)$ and $T_n = [\mathrm{Ext}\text{-exp}(A/n)] \times [\mathrm{Ext}\text{-exp}(B/n)]$. Then

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

so that

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

Since

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

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

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

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

and

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

as $s \to \# 0$. Letting

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

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 $K(s)\psi \xrightarrow{H^\#} 0$ as $s \to \# 0$ or $\ast\infty$ 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 $s \in \ast\mathbb{R}_c^{\#}$
and $\psi \in D$. Therefore, an $\epsilon/3$ argument shows that on $\|\cdot\|_{\#A+B}$ #-compact subsets of $D$
$K(s)\psi \to_{\#} 0$ uniformly. Since $A+B$ is self-#-adjoint on $D$, $[\mathrm{Ext}\text{-exp}(is(A+B))]\psi \in D$ if
$\psi \in D$. Moreover, $s \to [\mathrm{Ext}\text{-exp}(is(A+B))]\psi$ is a #-continuous map of $\ast\mathbb{R}_c^{\#}$ into $D$
when $D$ is given the $\|\cdot\|_{\#A+B}$ #-norm topology. Thus $\{[\mathrm{Ext}\text{-exp}(is(A+B))]\psi|_{s \in [-1,1]}\}$
is a $\|\cdot\|_{\#A+B}$ #-compact set in $D$ for each fixed $\psi$.

We are now ready to mimic the proof of the generalized Lie product formula. We
know that

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

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

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

The #-norm of the RHS of (4.9.11)

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

and so we conclude that

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

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

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

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

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

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

## § 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 $\mathbf{Ker}(|T|) = \mathbf{Ker}(T)$ and that the initial space of $U$ equals $(\mathbf{Ker}(T))^{\perp}$ 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-#-adjoint operator. It is this fact that is hard in the unbounded case. A priori, it is not clear that $\{\psi|\psi \in D(T) \text{ and } T\psi \in D(T^*)\}$ 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, $(\mathbf{Ker}(T))^{\perp}$, and final space #-Ran($T$) so that $T = U|T|$ and $U$ are uniquely determined by these properties together with the additional condition $\mathbf{Ker}(|T|) = \mathbf{Ker}(T)$.

**Proof.** Define the $*\mathbb{C}_c^#$-valued quadratic form $s(\psi, \varphi)$ on $D(T)$ by

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

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

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

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 $\mathbb{C}_c^\#$-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 $\mathrm{Ker}(|T|) = \mathrm{Ker}(T)$. Define the operator $U: \mathrm{Ran}(|T|) \to \mathrm{Ran}(T)$ 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 $\mathrm{Ran}(|T|)$ to $\mathrm{Ran}(T)$. Finally, since $|T|$ is self-\#-adjoint, $\#\mathrm{Ran}(T) = (\mathrm{Ker}(|T|))^\perp = (\mathrm{Ker}(T))^\perp$. 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 $\mathrm{Ext}-\sum c_i \varphi_i \otimes \psi_i$ and $\mathrm{Ext}-\sum d_j \varphi'_j \otimes \psi'_j$ 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

$$\varphi_i \otimes \psi_i = \mathrm{Ext}-\sum_{kl} \alpha_{kl}^i (\eta_k \otimes \theta_l) \qquad (5.1.2)$$

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,

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

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

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

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

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

and

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

Then

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

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 \#-closure of $I_1 \otimes \cdots \otimes A_k \otimes \cdots \otimes I_N$ on $D = \mathrm{Ext}-\otimes_{k=1}^N D(A_k)$ by $A_k$ also. Let $P(x_1, \dots, x_N)$ be a polynomial with $*\mathbb{R}_c^\#$-valued coefficients of degree $n_k$ in $x_k$. Then, the operator $P(A_1, \dots, A_N)$ makes sense on $\mathrm{Ext}-\otimes_k D(A_{n_k})$ since $D(A_{n_k}) \subset D(A_l)$ for all $l \le n_k$. In fact, $P(A_1, \dots, A_N)$ is essentially self-\#-adjoint on that domain.

**Theorem 5.1.1.** Let $A_k$ be a self-\#-adjoint operator on $H_k^\#$. Let $P(x_1, \dots, x_N)$ 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) $P(A_1, \dots, A_N)$ is essentially self-adjoint on $D_k^l = \mathrm{Ext}-\otimes_{k=1}^N D_k^l$.

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

**Proof** We will first prove that $P(A_1, \dots, A_N)$ is essentially self-\#-adjoint on $D = \mathrm{Ext}-\otimes_{k=1}^N D(A_k^{n_k})$. By the spectral theorem, there is a \#-measure space $\langle M_k, \mu_k^\# \rangle$ 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 $\mathrm{Ext}-\otimes_{k=1}^N L_2^\#(M_k, d^\#\mu_k^\#)$ is naturally isomorphic to $L_2^\#(\mathrm{Ext}-\otimes_{k=1}^N M_k, \mathrm{Ext}-\otimes_{k=1}^N d^\#\mu_k^\#)$. Under this isomorphism $P(A_1, \dots, A_N)$ corresponds to multiplication by $P(f_1, \dots, f_N)$ and $D$ corresponds to the set of hyperfinite linear combinations of hyperfinite linear combinations of functions $\mathrm{Ext}-\prod_{i=1}^N \phi_i(m_i)$ 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 $P(f_1, \dots, f_N)$ is in $L_p^\#$ for all such $p$; in particular $P(f_1, \dots, f_N) \in L_4^\#(\text{Ext}-\otimes_{k=1}^N M_k, \text{Ext}-\otimes_{k=1}^N d^\# \mu_k^\#)$. 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 $P(A_1, \dots, A_N)$ is essentially self-adjoint on $D^l$ we need only show that $\#\overline{P}(A_1, \dots, A_N) \uparrow D^l$ extends $P(A_1, \dots, A_N) \uparrow D$. 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 $P(A_1, \dots, A_N)(\text{Ext}-\otimes_{k=1}^N \phi_k^l) \to_* P(A_1, \dots, A_N)(\text{Ext}-\otimes_{k=1}^N \phi_k)$ The same argument works for hyperfinite linear combinations of vectors of the form $\text{Ext}-\otimes_{k=1}^N \phi_k$ so $\#\overline{P}(A_1, \dots, A_N) \uparrow D^l$ extends $P(A_1, \dots, A_N) \uparrow D$. This completes the proof of (a).

To prove (b), suppose that $\lambda \in \sigma(\#\overline{P}(A_1, \dots, A_N))$. If $I$ is any #-open interval about $\lambda$ then $P^{-1}(A_1, \dots, A_N)(I)$ 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 $\sigma(A_k) = \#\text{-ess range}(f_k^{n_k})$, $\mu_k^l[(f_k^{n_k})^{-1}(I_k)] \neq 0$ so $\mu[P(f_1, \dots, f_N)(I)] \neq 0$. That is, $\lambda \in \#\text{-ess range}(P(f_1, \dots, f_N))$ which equals $\sigma(\#\overline{P}(A_1, \dots, A_N))$.

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

**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

$$ \mathcal{F}^{\#}(H^{\#}) = \text{Ext}-\otimes_{n=0}^{*\infty} H^{\#n} \qquad (5.2.1) $$
$\mathcal{F}^{\#}(H^{\#})$ is called a non-Archimedean Fock space over $H^{\#}$; it will be $*$-separable if $H^{\#}$ is. For example, if $H^{\#} = L_2^{\#}(\mathbb{R}_c^{\#})$, 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

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

where $\psi_0 \in \mathbb{C}_c^\#$, $d^{\#n}x = \mathrm{Ext} - \prod_{i=1}^n d^{\#n}x_i$. 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

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

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

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

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

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

$\mathcal{F}_s^{\#}(H^{\#})$ 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 endroved 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

$$\#\lim_{n \to \infty} \|x_n - x\|_# = 0 \qquad (1.3)$$

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 $\#\mathbb{R}^\#$, the set of all $\#$-continuous $\#\mathbb{R}^\#$-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

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

**Definition 2.3.** Let $\{X, \|\cdot\|\}$ be standard Banach space, $Y \subset X$ thus $\#Y \subset \#X$ and let $x \in \#X$. Then $x$ is an $\#$-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 $Y \subset \#X$, $Y$ is $\#$-closed if every $\#$-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 $\#X$ 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 $A: D(A) \subset \#X \to \#Y$ is $\#$-closed if for every internal hyper infinite sequence $\{x_n\}_{n=1}^{n=\infty}$ in $D(A)$ 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 $\#$-closed if its graph is $\#$-closed

in the direct sum $\#X \oplus \#Y$.

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

$A$ $\#$-core (or $\#$-essential domain) of a $\#\bar{A}$-closable operator is a subset $C \subset D(A)$ such that the $\#$-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 $\{(\varphi, T\varphi)|(\varphi \in D(T))\}$.

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 #-closed operator if $\Gamma(T)$ is a #-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 #-closable if it has a #-closed extension. Every #-closable

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

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

**Definition 2.10.** Let $T$ be a #-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 #-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 #-densely defined operator on a non-Archimedean Hilbert space $H$.

Then:(i) $T^*$ is #-closed.

(ii) $T$ is #-closable if and only if $D(T^*)$ is #-dense in which case $T = T^{**}$.

(iii) If $T$ is #-closable, then $(\#-\bar{T})^* = T^*$.

**Definition 2.11.** Let $T$ be a #-closed operator on a Hilbert space $H$. A complex number $\lambda \in \mathbb{C}^\#_c$ is in the resolvent set, $\rho(T)$, if $\lambda I - T$ is a bijection of $D(T)$ onto $H$ with a finitely or hyper finitely bounded inverse. If $\lambda \in \rho(T)$, $R_\lambda(T) = (\lambda I - T)^{-1}$ is called the resolvent of $T$ at $\lambda$.

The definitions of spectrum, point spectrum, and residual spectrum are the same for unbounded operators as they are for bounded operators. We will sometimes refer to the spectrum of nonclosed, but closable operators. In this case we always mean the spectrum of the closure.

### 3. Symmetric and self-#-adjoint operators: the basic criterion for self-#-adjointness.

**Definition 3.1.** A #-densely defined operator $T$ on a non-Archimedean Hilbert space is called symmetric (or Hermitian) if $T \subset T^*$, that is, if $D(T) \subset D(T^*)$ and $T\varphi = T^*\varphi$ for all $\varphi \in D(T)$.

Equivalently, $T$ is symmetric if and only if $(T\varphi, \psi) = (\varphi, T\psi)$ for all $\varphi, \psi \in D(T)$.

**Definition 3.2.** $T$ is called self-adjoint if $T = T^*$, that is, if and only if $T$ is symmetric and $D(T) = D(T^*)$.

A symmetric operator is always #-closable, since $D(T^*) \supset D(T)$ is #-dense in $H$. If $T$ is symmetric, $T^*$ is a closed extension of $T$ so the smallest #-closed extension $T^{**}$ of $T$ must be contained in $T^*$. Thus for symmetric operators, we have $T \subset T^{**} \subset T^*$. For #-closed symmetric operators, $T = T^{**} \subset T^*$ and, for self-adjoint operators, $T = T^{**} = T^*$.

From this one can easily see that a #-closed symmetric operator $T$ is self-adjoint if
and only if $T^*$ is symmetric.

The distinction between #-closed symmetric operators and self-adjoint operators is
very

important. It is only for self-adjoint operators that the spectral theorem holds
and it is only self-adjoint operators that may be #-exponentiated to
give the one-parameter unitary groups which give the dynamics in

QFT. Chapter X is mainly devoted to studying methods for proving that operators are
self-adjoint. We content ourselves here with proving the basic criterion for
selfadjointness.

First, we introduce the useful notion of essential self-adjointness.

**Definition 3.3** A symmetric operator $T$ is called essentially self- #-adjoint if its
\#-closure \#-T is self- \#-adjoint. If $T$ is \#-closed, a subset $D \subset D(T)$ is called a core for $T$
if

$$
\overline{\# - T \uparrow D} = T.
$$

If $T$ is essentially self-\#-adjoint, then it has one and only one self-\#-adjoint extension.

The importance of essential self-\#-adjointness is that one is often given a nonclosed
symmetric operator $T$. If $T$ can be shown to be essentially self-\#-adjoint, then there is
uniquely associated to $T$ a self-adjoint operator $T = T^{**}$. Another way of saying this is
that if $A$ is a self-\#-adjoint operator, then to specify $A$ uniquely one need not give the
exact domain of $A$ (which is often difficult), but just some \#-core for $A$

Chapter V. Semigroups of operators on a
non-Archimedean Banach spaces.

§1.Semigroups on non-Archimedean Banach spaces and
their generators.

A family of #-bounded operators {$T(t)|0 < t < \infty$#} on external hyper infinite
dimensional

non-Archimedean Banach space X endowed with $ *\mathbb{R}^{\#}_{c,+} $ - valued norm $ \|\cdot\|_{\#} $ is called a
strongly #-continuous semigroup if:

$$
(a) \quad T(0) = I
$$

$$
(b) \quad T(s)T(t) = T(s+t) \text{ for all } s,t \in *\mathbb{R}^{\#}_{c,+}
$$

(c) For each $\varphi \in X, t \mapsto T(t)$ is \#-continuous mapping.

We will see that strongly continuous semigroups are the “exponentials,”

$$
T(t) = \operatorname{Ext-}\exp(-tA), \text{ of a certain class of operators. .}
$$

We begin by studying a special class of semigroups:

**Definition 1.1.** A family {$T(t)|0 < t < \infty$#} of bounded or hyper bounded operators on external hyper infinite dimensional Banach space $X$ is called a contraction semigroup if it is a strongly \#-continuous semigroup and moreover $||T(t)||_\# < 1$ for all $t \in [0, \infty)$.

Note that the all theorems about general strongly #-continuous semigroups are easy generalizations of the corresponding theorems for #-contraction semigroups. Thus, we study the special case first. We then briefly discuss the general theory and conclude the section by studying another special class, #-holomorphic semigroups.

**Proposition 1.1.** Let $T(t)$ be a strongly \#-continuous semigroup on a non-Archimedean Banach space $X$ and set $A\varphi = \#\text{lim}_{r\to\#} A_r\varphi$ where

$$
D(A) = \{ \varphi | \#\text{lim}_{r \to \#} A_r \varphi \text{ exists} \}. \text{ Then } A \text{ is}
$$
\#-closed and \#-densely defined. A is called the infinitesimal generator of $T(t)$. We will also say that A generates $T(t)$ and write $T(t) = \text{Ext}-\exp(-tA)$.

**Proof.** Let $T(t)$ be a contraction semigroup on a Banach space $X$. We obtain the generator of $T(t)$ by \#-differentiation. Set $A_t = t^{-1}(I - T(t))$ and define

$$D(A) = \{\varphi | \#\text{-lim}_{t \to \#} 0 A_t \varphi \text{ exists}\}.$$

For $\varphi \in D(A)$, we define $A\varphi = \#\text{-lim}_{t \to \#} 0 A_t \varphi$. Our first goal is to show that $D(A)$ is \#-dense. For $\varphi \in X$, we set

$$\varphi_s = \text{Ext}-\int_0^s T(t)\varphi d^\#t. \qquad (2.1)$$

For any $r > 0$, we get

$$T(r)\varphi_s = \text{Ext}-\int_0^s T(t+r)\varphi d^\#t \qquad (2.2)$$

thus

$$A_r \varphi_s = -\frac{1}{r} \left( \text{Ext}-\int_0^s [T(t+r)\varphi - T(t)\varphi] d^\#t \right) = -\frac{1}{r} \left( \text{Ext}-\int_s^{r+s} T(t)\varphi d^\#t \right) + \frac{1}{r} \left( \text{Ext}-\int_s^r T(t)\varphi d^\#t \right). \qquad (2.3)$$

From Eq. (2.3) one obtains $\#\text{-lim}_{r \to \#} 0 A_r \varphi_s = -T(s)\varphi + \varphi$. Therefore, for each $\varphi \in X$

and $s > 0$, $\varphi_s \in D(A)$. Since $s^{-1}\varphi_s \to \# \varphi$ as $\to \# 0$, A is \#-densely defined.

Furthermore, if $\varphi \in D(A)$, then $A_r T(t)\varphi = T(t)A_r \varphi$, so $T(t) : D(A) \to D(A)$ and

$$\frac{d^\#}{d^\# t} T(t)\varphi = -AT(t)\varphi = -T(t)A\varphi \qquad (2.4)$$

A is also \#-closed, for if $\varphi_n \in D(A)$, $\#\text{-lim}_{n \to \infty \#} \varphi_n = \varphi$, and $\#\text{-lim}_{n \to \infty \#} A\varphi_n = \psi$, then

$$\begin{aligned} \#\text{-lim}_{r \to \#} 0 A_r \varphi &= \#\text{-lim}_{r \to \#} 0 \#\text{-lim}_{n \to \infty \#} \left[ -\frac{1}{r} (T(r)\varphi_n - \varphi_n) \right] = \\ &\quad \#\text{-lim}_{r \to \#} 0 \#\text{-lim}_{n \to \infty \#} \frac{1}{r} \left( \text{Ext}-\int_s^r T(t)A\varphi_n d^\#t \right) = \\ &\quad \#\text{-lim}_{r \to \#} 0 \frac{1}{r} \left( \text{Ext}-\int_s^r T(t)\psi d^\#t \right) \end{aligned} \qquad (2.5)$$

so $\varphi \in D(A)$ and $A\varphi = \psi$.

The formal Laplace transform

$$\frac{1}{\lambda + A} = - \left( \text{Ext}-\int_0^{\infty} (\text{Ext}-\exp(-\lambda t))(\text{Ext}-\exp(-tA))d^\#t \right) \qquad (2.6)$$

suggests that all $\mu \in \mathbb{C}_c^\#$ with $\text{Re}\,\mu < 0$ are in $\rho(A)$. This is in fact true and the formula (2.6) holds in the strong sense. For suppose that $\text{Re}\,\lambda > 0$. Then, since
$\|Ext\text{-}exp(-tA)\| < 1$, the formula (2.7)

$$R\varphi = Ext\text{-}\int_{0}^{\infty} (Ext\text{-}exp(-\lambda t))(Ext\text{-}exp(-tA)\varphi)d^{\#}t \quad (2.7)$$

defines a hyper bounded linear operator of #-norm less than or equal to $(\text{Re } \lambda)^{-1}$. Moreover, for $r > 0$,

$$A_r R\varphi = -\frac{1}{r} \left( \text{Ext}\text{-}\int_{0}^{\infty} (\text{Ext}\text{-}exp(-\lambda t))(\text{Ext}\text{-}exp(-(t+r)A) - \text{Ext}\text{-}exp(-tA))\varphi d^{\#}t \right) = \\ \frac{1 - \text{Ext}\text{-}exp(\lambda r)}{r} \left( \text{Ext}\text{-}\int_{0}^{\infty} (\text{Ext}\text{-}exp(-\lambda t))(\text{Ext}\text{-}exp(-tA))\varphi d^{\#}t \right) + \\ \frac{\text{Ext}\text{-}exp(\lambda r)}{r} \left( \text{Ext}\text{-}\int_{0}^{r} (\text{Ext}\text{-}exp(-\lambda t))(\text{Ext}\text{-}exp(-tA))\varphi d^{\#}t \right) \quad (2.8)$$

so as $r \to \# 0$, $A_r R\varphi \to \# (\varphi - \lambda R\varphi)$. Thus $R\varphi \in D(A)$ and $AR\varphi = \varphi - \lambda R\varphi$ which implies $(\lambda + A)R\varphi = \varphi$. In addition, for $\varphi \in D(A)$ we have $AR\varphi = RA\varphi$ since

$$A \left( \text{Ext}\text{-}\int_{0}^{\infty} (\text{Ext}\text{-}exp(-\lambda t))(\text{Ext}\text{-}exp(-tA))\varphi d^{\#}t \right) = \\ \text{Ext}\text{-}\int_{0}^{\infty} (\text{Ext}\text{-}exp(-\lambda t))A(\text{Ext}\text{-}exp(-tA))\varphi d^{\#}t = \\ \text{Ext}\text{-}\int_{0}^{\infty} (\text{Ext}\text{-}exp(-\lambda t))(\text{Ext}\text{-}exp(-tA))A\varphi d^{\#}t. \quad (2.9)$$

The first equality follows by approximation with external hyperfinite Riemann sums (see [1]) from the facts that $(\text{Ext}\text{-}exp(-\lambda t))(\text{Ext}\text{-}exp(-tA))\varphi$ and $A(\text{Ext}\text{-}exp(-\lambda t))(\text{Ext}\text{-}exp(-tA))$ are #-integrable, $A$ is #-closed. Thus, for $\varphi \in D(A)$, $R(\lambda + A)\varphi = \varphi = (\lambda + A)R\varphi$ which implies that

$$R = (\lambda + A)^{-1}. \quad (2.10)$$

The properties of $A$ which we have derived are also sufficient to guarantee that $A$ generates a contraction semigroup. In fact, we only need information about real positive $A$.

**Theorem 1.1.** (Generalized Hille-Yosida theorem) A necessary and sufficient condition that a #-closed linear operator $A$ on a Banach space $X$ generate a contraction semigroup is that

(i) $(-\infty^{\#}, 0) \subset \rho(A)$

(ii) $\|(\lambda + A)^{-1}\|_{\#}$ for all $\lambda > 0$.

Furthermore, if $A$ satisfies (i) and (ii), then the entire #-open left half-plane is contained in $\rho(A)$ and
$$ (\lambda + A)^{-1}\varphi = -\mathrm{Ext}-\int_{0}^{\infty} (\mathrm{Ext}\text{-exp}(-\lambda t))(\mathrm{Ext}\text{-exp}(-tA))d^{\#}t \quad (2.11) $$

for all $\varphi \in X$ and $\lambda$ with $\mathrm{Re}\,\lambda > 0$. Finally, if $T_1(t)$ and $T_2(t)$ are contraction semigroups generated by $A_1$ and $A_2$ respectively, then $T_2(t) \neq T_1(t)$ for some $t$ implies that $A_1 \neq A_2$.

**Proof.** Since we showed above that conditions (i) and (ii) are necessary and that (2.11)

holds, we need only show sufficiency. So, suppose that $A$ is a \#-closed operator on $X$ satisfying (i) and (ii). For $\lambda > 0$, define $A^{(\lambda)} = \lambda - \lambda^2(\lambda + A)^{-1}$. We will show that as $\lambda \to \infty^{\#}$, $A^{(\lambda)} \to_{\#} A$ strongly on $D(A)$ and then construct $\mathrm{Ext}\text{-exp}(-tA)$ as the strong \#-limit of the semigroups $\mathrm{Ext}\text{-exp}(-tA^{(\lambda)})$. For $\varphi \in D(A)$, $A^{(\lambda)}\varphi = \lambda(\lambda + A)^{-1}A\varphi$. Moreover, by (ii),

$$ \#\text{-lim}_{\lambda \to \infty^{\#}} [\lambda(\lambda + A)^{-1}\varphi - \varphi] = \#\text{-lim}_{\lambda \to \infty^{\#}} [-(\lambda + A)^{-1}A\varphi] = 0. \quad (2.12) $$

By condition (ii) the family $\{\lambda(\lambda + A)^{-1}|\lambda > 0\}$ is \#-uniformly hyperfinitely bounded in \#-norm, so since $D(A)$ is \#-dense, $\#\text{-lim}_{\lambda \to \infty^{\#}} [\lambda(\lambda + A)^{-1}\psi] = \psi$ for all $\psi \in X$. Thus $\#\text{-lim}_{\lambda \to \infty^{\#}} A^{(\lambda)}\varphi = A\varphi$ for all $\varphi \in D(A)$. Since $A$ is hyperfinitely bounded, the semigroups $\mathrm{Ext}\text{-exp}(-tA^{(\lambda)})$ can be defined by hyper infinite power series. Since

$$ \begin{aligned} \|\mathrm{Ext}\text{-exp}(-tA^{(\lambda)})\|_{\#} &= \|(\mathrm{Ext}\text{-exp}(-\lambda t))(\mathrm{Ext}\text{-exp}(t\lambda^2(\lambda + A)^{-1}))\|_{\#} \le \\ &\le (\mathrm{Ext}\text{-exp}(-\lambda t)) \left( \mathrm{Ext}-\sum_{n=0}^{\infty} \frac{t^n \lambda^{2n}}{n!} \|(\lambda + A)^{-1}\|_{\#}^n \right) \le 1 \end{aligned} \quad (2.13) $$

they are contraction semigroups. For all $\mu, \lambda, t > 0$, and all $\varphi \in D(A)$, we have

$$ [\mathrm{Ext}\text{-exp}(-tA^{(\lambda)})]\varphi - [\mathrm{Ext}\text{-exp}(-tA^{(\mu)})]\varphi = \mathrm{Ext}-\int_{0}^{t} \frac{d^{\#}}{d^{\#}s} (\mathrm{Ext}\text{-exp}(-sA^{(\lambda)}))((\mathrm{Ext}\text{-exp}(-(t-s)A^{(\lambda)}))\varphi)d^{\#}s \quad (2.14) $$

so,

$$ \begin{aligned} & \|[\mathrm{Ext}\text{-exp}(-tA^{(\lambda)})]\varphi - [\mathrm{Ext}\text{-exp}(-tA^{(\mu)})]\varphi\|_{\#} \le \\ & \mathrm{Ext}-\int_{0}^{t} \|(\mathrm{Ext}\text{-exp}(-sA^{(\lambda)}))((\mathrm{Ext}\text{-exp}(-(t-s)A^{(\lambda)}))\|_{\#} \|A^{(\mu)}\varphi - A^{(\lambda)}\varphi\|_{\#} d^{\#}s \le \\ & \le t \|A^{(\mu)}\varphi - A^{(\lambda)}\varphi\|_{\#}. \end{aligned} \quad (2.15) $$

We have used the fact that $\mathrm{Ext}\text{-exp}(-tA^{(\lambda)})$ and $[\mathrm{Ext}\text{-exp}(-(t-s)A^{(\mu)})]$ commute since $\{A^{(\lambda)}|\lambda > 0\}$ is a commuting family. Since we have proven above that $\#\text{-lim}_{\lambda \to \infty^{\#}} A^{(\lambda)}\varphi = A\varphi$, $\{\mathrm{Ext}\text{-exp}(-tA^{(\lambda)})\}$ is Cauchy as $\lambda \to \infty^{\#}$ for each $t > 0$ and $\varphi \in D(A)$. Since $D(A)$ is \#-dense and the $\mathrm{Ext}\text{-exp}(-tA^{(\lambda)})$ are uniformly hyperfinitely bounded, the same statement holds for all $\varphi \in X$. Now, define

$$ T(t)\varphi = \#\text{-lim}_{\lambda \to \infty^{\#}} [\mathrm{Ext}\text{-exp}(-tA^{(\lambda)})\varphi]. \quad (2.16) $$

$T(t)$ is a semigroup of contraction operators since these properties are preserved under strong \#-limits. The above inequality shows that the \#-convergence in Eq.(2.16)