The common way to obtain an outer #-measure is to start with a family G of “elementary sets” on which a notion of measure is defined (such as rectangles or cubes in $\mathbb{R}^n$ and then approximate arbitrary sets from the outside by hyper infinite unions of members of G.

**Proposition 5.1.1** Let $G \subseteq \tilde{P}(X)$ be a set such that $\emptyset \in G, X \in G$ and let $\rho: G \to [0, \infty]$ be a function such that $\rho(\emptyset) = 0$. For any $A \subseteq X$, define

$$ \zeta^{\#}(A) = \rho^{*}(A) = \inf\left\{\text{Ext}-\sum_{j=1}^{\infty} \rho(G_j) : G_j \in G \text{ and } A \subseteq \bigcup_{j=1}^{\infty} G_j\right\}. \quad (5.1.1) $$

if $\rho^*(A)$ exists. Then $\zeta^\#$ is an outer #-measure.

**Definition 5.1.2.** We will say that $A \subseteq X$ is admissible if $\rho^*(A)$ exists.

**Proof.** For any admissible $A \subseteq X$, $\zeta^\#(A)$ is well defined. Obviously $\zeta^\#(\emptyset) = 0$. To prove $*$-countable subadditivity, suppose $\{A_j\}_{j=1}^\infty \subseteq \tilde{P}(X)$ and $\varepsilon \approx, \varepsilon > 0$.

For each $j \in \mathbb{N}$, there exists $\{G_k^j\}_{k=1}^\infty \subseteq G$ such that $A_j \subseteq \bigcup_{k=1}^\infty G_k^j$ and

$$ \text{Ext}-\sum_{k=1}^{\infty} \rho(G_k^j) \le \zeta^\#(A_j) + \varepsilon 2^{-j}. \text{ Then if } A = \bigcup_{j=1}^{\infty} A_j, \text{ we have } A \subseteq \bigcup_{j,k=1}^{\infty} G_k^j \text{ and} $$

$$ \text{Ext}-\sum_{j,k=1}^{\infty} \rho(G_k^j) \le \sum_{j=1}^{\infty} \zeta(A_j) + \varepsilon, \text{ whence } \zeta^\#(A) \le \text{Ext}-\sum_{j=1}^{\infty} \zeta^\#(A_j) + \varepsilon. \text{ Since } \varepsilon > 0 \text{ is} $$

arbitrary, we have done.

**Definition 5.1.3.** A set $A \subseteq X$ is called $\zeta^\#$-measurable if $\rho^*(A)$ exists and $\forall B \subseteq X$ such that $\rho^*(B)$ exists the equality (5.1.2) holds

$$ \zeta^\#(B) = \zeta^\#(B \cap A) + \zeta^\#(B \cap (X \setminus A)). \quad (5.1.2) $$

Of course, the inequality $\zeta^\#(B) \le \zeta^\#(B \cap A) + \zeta^\#(B \cap (X \setminus A))$ holds for any (admissible) set A and B.

So, to prove that A is $\zeta^\#$-measurable, it suffices to prove the reverse inequality, which is trivial if $\zeta^\#(B) = \infty$. Thus, we see that A is $\zeta^\#$-measurable iff for any admissible $B \subseteq X, \zeta^\#(B) < \infty$

$$ \zeta^\#(B) \ge \zeta^\#(B \cap A) + \zeta^\#(B \cap (X \setminus A)). \quad (5.1.3) $$

**Theorem 5.1.1** (Generalized Caratheodory's theorem) Let $\zeta^\#$ be an outer #-measure on X. Then the family $\Sigma$ of all $\zeta^\#$-measurable sets is a $\sigma^\#$-algebra, and the restriction of $\zeta^\#$ to $\Sigma$ is a complete #-measure.

**Proof:** First, we observe that $\Sigma$ is closed under complements, since the definition of $\zeta^\#$-measurability of A is symmetric in A and $A^c \triangleq X \setminus A$. Next, if $A, B \in \Sigma$ and $E \subseteq X$,

$$ \zeta^\#(E) = \zeta^\#(E \cap A) + \zeta^\#(E \cap A^c) = \zeta^\#(E \cap A \cap B) + \zeta^\#(E \cap A \cap B^c) + \zeta^\#(E \cap A^c \cap B) + \\ + \zeta^\#(E \cap A^c \cap B^c). $$

But $(A \cup B) = (A \cap B) \cup (A \cap B^c) \cup (A^c \cap B)$ so, by subadditivity, $\zeta^\#(E \cap A \cap B) + \zeta^\#(E \cap A \cap B^c) + \zeta^\#(E \cap A^c \cap B) \ge \zeta^\#(E \cap (A \cup B))$, and hence $\zeta^\#(E) \ge \zeta^\#(E \cap (A \cup B)) + \zeta^\#(E \cap (A \cup B)^c)$.

It follows that $A \cup B \in \Sigma$, so $\Sigma$ is an algebra. Moreover, if $A, B \in \Sigma$ and $A \cap B = \emptyset, \zeta^\#(A \cup B) = \zeta^\#((A \cup B) \cap A) + \zeta^\#((A \cup B) \cap A^c) = \zeta^\#(A) + \zeta^\#(B)$, so $\zeta^\#$ is hyperfinitely additive on $\Sigma$.

To show that $\Sigma$ is a $\sigma^\#$-algebra, it suffices to show that $\Sigma$ is closed under $*$-countable disjoint unions. If $(A_j)_{j=1}^\infty$ is a sequence of disjoint sets in $\Sigma$, set
**Proof.** Suppose $s = \text{Ext}-\sum_{v \in A} v$ exists. Then there exists $\Gamma \subset\subset A$ such that $\text{Ext}-\sum_{v \in A} \|v\|_\#^2 = \left\| \text{Ext}-\sum_{v \in A} v \right\|_\#^2 \le 1$ for all $\Lambda \subset\subset A\Gamma$, wherein the first inequality we have used Pythagorean's theorem.

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

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

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

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

For item 1, let $\Gamma_\varepsilon$ be as above and set $s_\varepsilon = \text{Ext}-\sum_{v \in \Gamma_\varepsilon} v$. Then
$$ \|s\|_{\#} - \|s_\varepsilon\|_{\#} \le \|s - s_\varepsilon\|_{\#} < \varepsilon \quad \text{and by Eq. (1.14), } 0 \le \left( \text{Ext}-\sum_{v \in A} \|v\|_{\#}^2 \right) - \|s_\varepsilon\|_{\#}^2 \le \varepsilon^2 $$

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

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

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

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

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

and

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

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

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

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

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 $y \in \beta$.
$B_n = \bigcup_{j=1}^{n} A_j \wedge B = \bigcup_{j=1}^{*\infty} A_j$. Then, for any admissible $E \subseteq X$,

$\zeta^{\#}(E \cap B_n) = \zeta^{\#}(E \cap B_n \cap A_n) + \zeta^{\#}(E \cap B_n \cap A_n^c) = \zeta^{\#}(E \cap A_n) + \zeta^{\#}(E \cap B_{n-1})$,
so a hyperfinite induction shows that $\zeta^{\#}(E \cap B_n) = \text{Ext}-\sum_{j=1}^{n} \zeta^{\#}(E \cap A_j)$. Therefore

$$ \zeta^{\#}(E) = \zeta^{\#}(E \cap B_n) + \zeta^{\#}(E \cap B_n^c) \geq \text{Ext}-\sum_{j=1}^{n} \zeta^{\#}(E \cap A_j) + \zeta(E \cap B^c) $$

and, letting $n \to *\infty$, we obtain

$$ \zeta^{\#}(E) \geq \text{Ext}-\sum_{j=1}^{*\infty} \zeta^{\#}(E \cap A_j) + \zeta^{\#}(E \cap B^c) \geq \zeta^{\#}\left(\bigcup_{j=1}^{*\infty} E \cap A_j\right) + \zeta^{\#}(E \cap B^c) = \zeta^{\#}(E \cap B) + \\ +\zeta^{\#}(E \cap B^c) \geq \zeta^{\#}(E). $$

Thus the inequalities in this last calculation become equalities. It follows $B \in \Sigma$.

Taking $E = B$ we have $\zeta^{\#}(B) = \text{Ext}-\sum_{j=1}^{*\infty} \zeta^{\#}(A_j)$, so $\zeta^{\#}$ is $\sigma^{\#}$-additive on $\Sigma$. Finally, if

$\zeta^{\#}(A) = 0$ then we have for any admissible set $E \subseteq X$

$$ \zeta^{\#}(E) \leq \zeta^{\#}(E \cap A) + \zeta^{\#}(E \cap A^c) = \zeta^{\#}(E \cap A^c) \leq \zeta^{\#}(E), \text{ so } A \in \Sigma. $$

Therefore $\zeta^{\#}(E \cap A) = 0$ and $\zeta^{\#}|_{\Sigma}$ is a complete $\#$-measure.

Combination of Proposition 5.1.1 and Theorem 5.1.1 gives the following corollary which is also called generalized Caratheodory's theorem.

**Corollary 5.1.1** Let $G \subseteq P(X)$ be a set such that $\emptyset \in G, X \in G$, and let

$\rho: G \to [0, *\infty]$ satisfy $\rho(\emptyset) = 0$. Then the family $\Sigma$ of all $\rho^* \#$-measurable sets (where $\rho^*$ is given by (5.1.1)) is a $\sigma^*\#$-algebra, and the restriction $\rho^*|_{\Sigma}$ of $\rho^*$ to $\Sigma$ is a complete $\#$-measure.

**Definition 5.1.4** Let $\bar{A}$ be an algebra of subsets of $X$, i.e. $\bar{A}$ contains $\emptyset$ and $X$, and $\bar{A}$ is closed under hyperfinite intersections and complements. A function

$\zeta: \bar{A} \to [0, *\infty]$ is called a $\#$-premeasure if $\zeta(\emptyset) = 0$ and $\zeta\left(\bigcup_{j=1}^{*\infty} A_j\right) = \text{Ext}-\sum_{j=1}^{*\infty} \zeta(A_j)$ for any disjoint sequence $(A_j)_{j \in *\mathbb{N}}$ of elements of $\bar{A}$ such that $\bigcup_{j=1}^{*\infty} A_j \in \bar{A}$.

**Theorem 5.1.2** If $\zeta$ is a $\#$-premeasure on an algebra $\bar{A} \subseteq P(X)$ and $\zeta^*: P(X) \to [0, *\infty]$ is given by (5.1.1) then $\zeta^*|_{A} = \zeta$ and every $A \in \bar{A}$ is $\zeta^* \#$-measurable.

# § 5.2. The Lebesgue and Lebesgue – Stieltjes $\#$-measure on $*\mathbb{R}_c^\#$.

The most important application of generalized Caratheodory's theorem is the construction of the Lebesgue $\#$-measure on $*\mathbb{R}_c^\#$. Take $G$ as the set of all intervals $[a, b]$, where $a, b \in *\mathbb{R}_c^\# \cup \{-*\infty, +*\infty\}$ and $[a, b] = \emptyset$ if $a > b$. Define the function $\rho: G \to *\mathbb{R}_c^\# \cup \{*\infty\}$ by

$$ \forall a \forall b (a \leq b) [\rho([a, b]) = b - a] \text{ and } \forall a \forall b (a > b) [\rho([a, b]) = 0]. \quad (5.2.1) $$

The function $\rho$ has the obvious extension (which we denote also by $\rho$) to the algebra $A$ generated by all intervals, and this extension is a $\#$-premeasure on $A$. The $\sigma^*$-algebra $\Sigma$ given by Corollary 5.1.1 is called the Lebesgue $\sigma^*$-algebra in $\mathbb{R}$, and the restriction of $\rho^*$ to $\Sigma = \Sigma(*\mathbb{R}_c^\#)$ is called the Lebesgue $\#$-measure
So $(x - P_x) \perp \operatorname{span}(\beta)$ and by continuity we also have $(x - P_x) \perp M = \#\overline{\operatorname{span}(\beta)}$.

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

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

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

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

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

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

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

(1) ⇒ (2) By Corollary 1.6, $\operatorname{Ext}-\sum_{u \in \beta} \langle x, u \rangle_\# u = P_M x$. Therefore
$x - \operatorname{Ext}-\sum_{u \in \beta} \langle x, u \rangle_\# u = x - P_M x \in M^\perp = \beta^\perp = \{0\}$.

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

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

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

**Proposition 2.2.** A non-Archimedean Hilbert space $H^\#$ is \*-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
on $\mathbb{R}_c^\#$ and is denoted by $\mu^\#$. By Theorem 5.1.2, $\mu^\#$ is the unique extension of $\rho$. By the construction, $B^#(*\mathbb{R}_c^\#) \subseteq \Sigma(*\mathbb{R}_c^\#)$. Hence the Lebesgue #-measure is a Borel #-measure. It can be shown that $B^#(*\mathbb{R}_c^\#) \neq \Sigma(*\mathbb{R}_c^\#)$ and that the Lebesgue #-measure can be obtained also as the completion of any Borel #-measure $\omega^\#$ such that $\omega^#([a, b]) = b - a(\forall a \le b)$.

The notion of the Lebesgue measure on $*\mathbb{R}_c^\#$ has the following generalization. Suppose that $\mu^\#$ is a $\sigma^\#$-finite Borel measure on $*\mathbb{R}_c^\#$, and let $\forall x \in *\mathbb{R}_c^\#$

$$F(x) = \mu^#((-\infty, x]) \qquad (5.2.2)$$

Then $F$ is increasing and right #-continuous. Moreover, if $b > a$, $(-\infty, b] = (-\infty, a] \cup (a, b]$, so $\mu^#((a, b]) = F(b) - F(a)$.

Our procedure used above can be to turn this process around and construct a measure $\mu$ starting from an increasing, right-continuous function $F$. The special case $F(x) = x$ will yield the usual Lebesgue #-measure. As building blocks we can use the left-#-open, right-#-closed intervals in $*\mathbb{R}_c^\#$ i.e. sets of the form $(a, b]$ or $(a, *\infty)$ or $\emptyset$, where $-*\infty \le a < b < *\infty$. We call such sets $h$-intervals. The family $A$ of all finite disjoint unions of $h$-intervals is an algebra, moreover, the $\sigma^\#$-algebra generated by $A$ is the #-Borel algebra $B^#(*\mathbb{R}_c^\#)$.

**Lemma 5.2.1.** Given an increasing and right #-continuous function $F : *\mathbb{R}_c^\# \to *\mathbb{R}_c^\#$, if $(a_j, b_j](j = 1, \dots, n), n \in *\mathbb{N}$ are disjoint $h$-intervals, let

$$\mu_0^\#\left(\bigcup_{j=1}^n (a_j, b_j]\right) = \mathrm{Ext}-\sum_{j=1}^n [F(b_j) - F(a_j)], \qquad (5.2.3)$$

and let $\mu_0^\#(\emptyset) = 0$. Then $\mu_0^\#$ is a #-premeasure.

**Lemma 5.2.2.** Assume that $\{(a_a, b_a)|a \in G\}$ is a hyperfinite or $*$-countable family of intervals in $*\mathbb{R}_c^\#$ such that $[0, 1] \subseteq \cup_{a \in G} (a_a, b_a)$ then $\mathrm{Ext}-\sum_{a \in G} |a_a - b_a| > 1$.

**Theorem 5.2.1** If $F : *\mathbb{R}_c^\# \to *\mathbb{R}_c^\#$ is any increasing, right #-continuous function, there is a unique Borel #-measure $\mu_F^\#$ on $*\mathbb{R}_c^\#$ such that $\forall a \forall b (a, b \in *\mathbb{R}_c^\#)$

$$\mu_F^#((a, b]) = F(b) - F(a).$$

If $G$ is another such function then $\mu_F^\# = \mu_G^\#$ iff $F - G$ is constant.

Conversely, if $\mu^\#$ is a Borel #-measure on $*\mathbb{R}_c^\#$ that is hyperfinite on all #-bounded #-Borel sets, and we define $F(x) = \mu^#((0, x])$ if $x > 0$, $F(x) = 0$ if $x = 0$, $F(x) = -\mu^#((x, 0])$ if $x < 0$,

then $F$ is increasing and right #-continuous function, and $\mu^\# = \mu_F^\#$.

**Proof:** Each $F$ induces a #-premeasure on $B^#(*\mathbb{R}_c^\#)$ by Lemma 5.1.1. It is clear that $F$ and $G$ induce the same #-premeasure iff $F - G$ is constant, and that these

#-premeasures are $\sigma^\#$-finite (since $*\mathbb{R}_c^\# = \bigcup_{-\infty}^{*\infty} (j, j+1]$). The first two assertions

follow now from Lemma 5.2.2. As for the last one, the monotonicity of $\mu^\#$ implies the monotonicity of $F$, and the #-continuity of $\mu^\#$ from above and from below implies the right #-continuity of $F$ for $x \ge 0$ and $x < 0$. It is evident that $\mu^\# = \mu_F^\#$ on algebra $A$, and hence $\mu^\# = \mu_F^\#$ on $B^#(*\mathbb{R}_c^\#)$ (accordingly to Lemma 5.2.4).

Lebesgue – Stieltjes #-measures possess some important and useful regularity properties.

Let us fix a complete Lebesgue – Stieltjes #-measure $\mu^\#$ on $*\mathbb{R}_c^\#$ associated to an increasing, right #-continuous function $F$. We denote by $\Sigma_{\mu^\#}$ the Lebesgue algebra
process there exists $\beta = \{v_n\}_{n=1}^{*\infty}$ an orthonormal set such that $\text{span}(\{v_n|1 \le n \le N\}) \supseteq \text{span}(\{u_n|1 \le n \le N\})$. 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 $x = *\text{-} \lim_{k \to *\infty} w_k$ and therefore $\langle x, x \rangle_{\#} = *\text{-} \lim_{k \to *\infty} \langle x, w_k \rangle = 0$. That is to say $x = 0$ and $\beta$ is #-complete.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## §3.1. Weak #-Convergence.

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

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

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

$$ 
\begin{aligned}
\mu^#(E) &= \inf \left\{ \mathrm{Ext} - \sum_{j=1}^{*\infty} [F(b_j) - F(a_j)] \middle| E \subseteq \bigcup_{j=1}^{*\infty} (a_j, b_j) \right\} = \\
&= \inf \left\{ \mathrm{Ext} - \sum_{j=1}^{*\infty} \mu_F^#((a_j, b_j]) \middle| E \subseteq \bigcup_{j=1}^{*\infty} (a_j, b_j) \right\}
\end{aligned}
\qquad (5.2.4) 
$$

if infinum in RHS of (5.2.4) exists. Since $B^#(*\mathbb{R}_c^#) \subseteq \Sigma_{\mu^#}$, we may replace in the second formula for $\mu^#(E)$ h-intervals by #-open intervals, namely

**Lemma 5.2.3** For any $E \in \Sigma_{\mu^#}$,

$$ \mu^#(E) = \inf \left\{ \mathrm{Ext} - \sum_{j=1}^{*\infty} \mu_F^#((a_j, b_j)) \middle| E \subseteq \bigcup_{j=1}^{*\infty} (a_j, b_j) \right\}. \qquad (5.2.5) $$

**Theorem 5.2.2** If $E \in \Sigma_{\mu^#}$ then

$$ 
\begin{aligned}
E \in \Sigma_{\mu^#} &= \inf\{\mu^#(U) : U \supseteq E \text{ and } U \text{ is #-open}\} = \\
&= \sup\{\mu^#(K) : K \subseteq E \text{ and } K \text{ is #-compact}\}.
\end{aligned}
\qquad (5.2.6) 
$$

**Proof.** By Lemma 5.2.2, for any $\varepsilon \approx, \varepsilon > 0$, there exist intervals $(a_j, b_j)$ such that $E \subseteq \bigcup_{j=1}^{*\infty} (a_j, b_j)$ and $\mu^#(E) \le \mathrm{Ext} - \sum_{j=1}^{*\infty} \mu^#((a_j, b_j)) + \varepsilon$. If $U = \bigcup_{j=1}^{*\infty} (a_j, b_j)$ then $U$ is #-open,

$E \subseteq U$, and $\mu^#(U) \le \mu^#(E) + \varepsilon$. On the other hand, $\mu^#(U) \ge \mu^#(E)$ whenever $E \subseteq U$ so the first equality is valid.

For the second one, suppose first that $E$ is bounded in $*\mathbb{R}_c^#$. If $E$ is #-closed then $E$ is #-compact and the equality is obvious. Otherwise, given $\varepsilon \approx, \varepsilon > 0$, we can choose an #-open $U$, $(\#\bar{E})\setminus E \subseteq U$, such that $\mu^#(U) \le \mu^#((\#\bar{E})\setminus E) + \varepsilon$.

Let $K = (\#\bar{E})\setminus U$. Then $K$ is #-compact, $K \subseteq E$, and

$$ 
\begin{aligned}
\mu^#(K) &= \mu^#(E) - \mu^#(E \cap U) = \mu^#(E) - [\mu^#(U) - \mu^#(U\setminus E)] \ge \\
&\ge \mu^#(E) - \mu^#(U) + \mu^#((\#\bar{E})\setminus E) \ge \mu^#(E) - \varepsilon.
\end{aligned}
$$

If $E$ is unbounded in $*\mathbb{R}_c^#$, let $E_j = E \cap (j, j+1]$. By the preceding argument, for any $\varepsilon \approx, \varepsilon > 0$, there exist a #-compact $K_j \subseteq E_j$ with $\mu^#(K_j) \ge \mu^#(E_j) - \varepsilon 2^{-j}$. Let

$$ H_n = \bigcup_{j=-n}^{j=n} K_j. \text{ Then } H_n \text{ is #-compact, } H_n \subseteq E, \text{ and } \mu^#(H_n) \ge \mu^# \bigcup_{j=-n}^{j=n} (E_j) - \varepsilon. $$

Since $\mu^#(E) = \#\lim_{n \to *\infty} \mu^#\left(\bigcup_{j=-n}^{j=n} E_j\right)$, the result follows.

**Theorem 5.2.3.** If $E \subseteq *\mathbb{R}_c^#$, the following are equivalent:

(a) $E \in \Sigma_{\mu^#}$;

(b) $E = V \cap W_1$, where $V$ is a $G_{\delta^#}$-set and $\mu^#(N_1) = 0$;

(c) $E = H \cup N_2$, where $H$ is an $F_{\sigma^#}$-set and $\mu^#(N_2) = 0$.

**Theorem 5.2.4.** If $E \in \Sigma_{\mu^#}$ and $\mu^#(E) < *\infty$ then, for every $\varepsilon \approx, \varepsilon > 0$, there is a set $A$ that is a hyperfinite union of #-open intervals such that $\mu^#(E \Delta A) < \varepsilon$.

**Lemma 5.2.4** Let $A \subseteq P(X)$ be an algebra, let $\mu_0^#$ be a $\sigma^#$-finite #-premeasure on $A$, and let $\Omega$ be the $\sigma^#$-algebra generated by $A$. Then there exists a unique extension of $\mu_0^#$ to a #-measure $\mu^#$ on $\Omega$.
$\tau_w$ -continuous and hence $\varphi \in X^*$.

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

**Proof.** By definition of $\tau_w$, we have $x_n \xrightarrow{w} \# x \in X$ iff for all $\Gamma \subset\subset X^*$ and $\varepsilon \approx 0, \varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that $|\varphi(x) - \varphi(x_n)| < \varepsilon$ for all $n \ge N$ and $\varphi \in \Gamma$.

This later condition is easily seen to be equivalent to $\varphi(x) = \#\lim_{n \to \infty} \varphi(x_n)$ for all $\varphi \in X^*$.

The topological space $(X, \tau_w)$ is still Hausdorff, however to prove this one needs to make use of the generalized Hahn Banach **Theorem 18.16** below. For the moment we will concentrate on the special case where $X = H^\#$ is a non-

Archimedean Hilbert space in which case $H^{\#\#} = \{\varphi_z := \langle \cdot, z \rangle_\# : z \in H^\#\}$, see

**Proposition 3.2.** If $x, y \in H^\#$ and $z = y - x \ne 0$, then

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

Thus $V_x \triangleq \{w \in H^\# : |\varphi_z(x) - \varphi_z(w)| < \varepsilon/2\}$ and $V_y \triangleq \{w \in H^\# : |\varphi_z(y) - \varphi_z(w)| < \varepsilon/2\}$

are disjoint sets from $\tau_w$ which contain $x$ and $y$ respectively. This shows that $(H^\#, \tau_w)$ is a Hausdorff space. In particular, this shows that weak $\#$-limits are unique if they exist.

**Remark 3.1.** Suppose that $H^\#$ is an $\#\-$infinite dimensional non-Archimedean Hilbert space and {$x_n$}$_{n=1}^{\infty}$ an orthonormal subset of $H^\#$. Then generalized Bessel's inequality (Proposition 1.6) implies $x_n \xrightarrow{w} \# 0 \in H^\#$ as $n \to \infty$. This points out the fact that if $x_n \xrightarrow{w} \# x \in H^\#$ as $n \to \infty$, it is no longer necessarily true that

$$\|x\|_{\#} = \#\lim_{n \to \infty} \|x_n\|_{\#}.$$

$$\text{However we do always have } \|x\|_{\#} \le \#\lim \inf_{n \to \infty} \|x_n\|_{\#} \\ \text{because, } \|x\|_{\#}^2 = \#\lim_{n \to \infty} \langle x_n, x \rangle_{\#} \le \#\lim \inf_{n \to \infty} [\|x_n\|_{\#} \|x\|_{\#}] = \\ = \|x\|_{\#} \#\lim \inf_{n \to \infty} \|x_n\|_{\#}.$$

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

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

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

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

Moreover, if $c_y \triangleq \#\lim_{n \to \infty} \langle x_n, y \rangle_{\#}$ exists for all $y \in \beta$, then

$$\mathrm{Ext}-\sum_{y \in \beta} |c_y|^2 < \infty \quad \text{and} \quad x_n \xrightarrow{w} \# x \triangleq \mathrm{Ext}-\sum_{y \in \beta} c_y y \in H^\# \text{ as } n \to \infty.$$

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

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

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

Passing to the $\#$-limit in this equation implies

$$\#\lim \sup_{n \to \infty} |\langle x - x_n, y \rangle_\#| \le 2M \|y - z\|_\#$$

$$\text{which shows } \#\lim \sup_{n \to \infty} |\langle x - x_n, y \rangle_\#| = 0 \text{ since } H_0 \text{ is } \#\text{-dense in } H^\#.$$

To prove the last assertion, let $\Gamma \subset\subset \beta$. Then by Bessel's inequality (Proposition

$$1.6), \mathrm{Ext}-\sum_{y \in \Gamma} |c_y|^2 = \#\lim_{n \to \infty} \mathrm{Ext}-\sum_{y \in \Gamma} |\langle x - x_n, y \rangle_\#|^2 \le \#\lim \inf_{n \to \infty} \|x_n\|_\#^2 \le M^2.$$

Since $\Gamma \subset\subset \beta$ was arbitrary, we conclude that $\mathrm{Ext}-\sum_{y \in \beta} |c_y|^2 \le M < \infty$ and hence
§ 5.3. Product #-measures.

**Definition 5.3.1.** Let $\{(X_a, \mathcal{F}_a, \mu_a^#)\}_{a \in \Delta}$ be a nonempty family of #-measure spaces. We define the family $\Omega$ of blocks:

$$A(A_{a_1}, A_{a_2}, \dots, A_{a_n}) := \\ = A_{a_1} \times A_{a_2} \times \dots \times A_{a_n} \times \text{Ext-} \prod_{a \in \Delta, a \neq a_k, 1 \le k \le n} X_a, \qquad (5.3.1)$$

where $A_{a_k} \in \mathcal{F}_{a_k}$ and define a function

$$\mu_{\Omega}^{\#} : \Omega \to *\mathbb{R}_{c}^{\#} \cup \{ *\infty \} := \\ \mu^{\#}(A_{a_1}) \times \mu^{\#}(A_{a_2}) \times \dots \times \mu^{\#}(A_{a_n}) \times \left[ \text{Ext-} \prod_{a \in \Delta, a \neq a_k, 1 \le k \le n} \mu^{\#}(X_a) \right]. \qquad (5.3.2)$$

This function possesses an extension (by #-additivity) on the #-algebra A generated by $\Omega$. It is easily to show that $\mu_{\Omega}^{\#}$ is a #-premeasure on A.

**Definition 5.3.2** The #-measure $\hat{\mu}^{\#}$ on the $\sigma^{\#}$-algebra $\Sigma$ generated by A accordingly to **Theorem 2.1.3** is called the product #-measure of $\{\mu_{\alpha}^{\#}\}_{\alpha \in \Delta}$, and the triple

$(\prod_{a \in \Delta} X_a, \Sigma, \hat{\mu}^{\#})$ is called the product of #-measure spaces $(X_a, \Sigma_a, \mu_a^{\#})$.

We denote the $\sigma^{\#}$-algebra $\Sigma$ by $\bigotimes_{a \in \Delta} \Sigma_a$, and the #-measure $\hat{\mu}^{\#}$ by $\bigotimes_{a \in \Delta} \mu_a^{\#}$.

**Definition 5.3.3.** If $E \subseteq X_1 \times X_2$ and $x_1 \in X_1, x_2 \in X_2$, we define

$$E_{x_1} = \{x \in X_2 : (x_1, x) \in E\} \text{ and } E^{x_2} = \{x \in X_1 : (x, x_2) \in E\}.$$

If $f : X_1 \times X_2 \to *\mathbb{R}_{c}^{\#}$ is a function, we define $f_{X_1} : X_2 \to *\mathbb{R}_{c}^{\#}$ and $f^{x_2} : X_1 \to *\mathbb{R}_{c}^{\#}$ by $f_{X_1}(x) = f(x_1, x)$ and $f^{x_2}(x) = f(x, x_2)$.

**Theorem 5.3.1.** (The generalized Fubini's theorem) Let $\mu_1^{\#}, \mu_2^{\#}$ be $\sigma^{\#}$-hyperfinite #-measures on $(X_1, \mathcal{F}_1)$ and $(X_2, \mathcal{F}_2)$,

$$(X_1 \times X_2, \mathcal{F}_1 \otimes \mathcal{F}_2, \mu_1^{\#} \otimes \mu_2^{\#}) = (X_1, \mathcal{F}_1, \mu_1^{\#}) \times (X_2, \mathcal{F}_2, \mu_2^{\#}), \qquad (5.3.3)$$

and let $f \in L_1^{\#}(X_1 \times X_2, \mathcal{F}_1 \otimes \mathcal{F}_2, \mu_1^{\#} \otimes \mu_2^{\#})$. Then $f_{X_1} \in L_1^{\#}(X_2, \mathcal{F}_2, \mu_2^{\#})$, $\mu_1^{\#}$-#-a.e., and $f^{x_2} \in L_1^{\#}(X_1, \mathcal{F}_1, \mu_1^{\#})$, $\mu_2^{\#}$-#-a.e., and

$$\text{Ext-} \int_{X_1 \times X_2} f d^{\#}(\mu_1^{\#} \otimes \mu_2^{\#}) = \text{Ext-} \int_{X_2} \left[ \text{Ext-} \int_{X_1} f^{x_2} d^{\#} \mu_1^{\#} \right] d^{\#} \mu_2^{\#} = \\ = \text{Ext-} \int_{X_1} \left[ \text{Ext-} \int_{X_2} f_{X_1} d^{\#} \mu_2^{\#} \right] \qquad (5.3.4)$$

**Lemma 5.3.1.** Let $(X_1, \Sigma_1, \mu_1^{\#})$ and $(X_2, \Sigma_2, \mu_2^{\#})$ be #-measure spaces, $E \in \Sigma_1 \otimes \Sigma_2$, and let $f$ be a $\Sigma_1 \otimes \Sigma_2$-measurable function on $X_1 \times X_2$, then:

(a) $E_{X_1} \in \Sigma_2$ for all $x_1 \in X_1$ and $E_{X_2} \in \Sigma_1$ for all $x_2 \in X_2$;

(b) $f_{X_1}$ is $\Sigma_2$-measurable and $f_{X_2}$ is $\Sigma_1$-measurable for all $x_1 \in X_1$ and $x_2 \in X_2$.

**Proof.** Denote by A the collection of all $A \subseteq X_1 \times X_2$ such that $A_{X_1} \in \Sigma_2$ and $A^{x_2} \in \Sigma_1(\forall x_1 \in X_1, x_2 \in X_2)$.

The family A contains all rectangles. Thus, since

$$\left[ \bigcup_{n=1}^{\infty} A_n \right]_{x_1} = \bigcup_{n=1}^{\infty} [A_n]_{x_1}, [B_n]^{x_2} = [B_n]^{x_2} \qquad (5.3.5)$$

and
we may define $x \triangleq \mathrm{Ext}-\sum_{y \in \beta} c_y y$. By construction we have
$\langle x, y \rangle_{\#} = c_y = \#\mathrm{lim}_{n \to *\infty} \langle x_n, y \rangle_{\#}$ for all $y \in \beta$ and hence $x_n \xrightarrow{w} x \in H^{\#}$ as $n \to *\infty$ by
what we have just proved.

**Theorem 3.1.** Suppose that ${x_n}^*_{n=1} \subset H^\# $ is a bounded in $ *\mathbb{R}_c^\# $ hyper infinite sequence. Then there exists a subsequence $y_k = x_{n_k}$ of ${x_n}^*_{n=1}$ and $x \in X$ such that $y_k \xrightarrow{w} x$ as $k \to *\infty$.

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

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

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

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

**Proof.** That is simple to check that $\rho$ is a $\#-*\mathbb{R}_c^\#$-valued metric. Let $\tau_\rho$ be the topology on $\mathbb{C}$ induced by $\rho$. For any $y \in H^\#$ and $n \in *\mathbb{N}$, the map $x \in H^\# \to \langle x-y, e_n \rangle_\# = \langle x, e_n \rangle_\# - \langle y, e_n \rangle_\#$ is $\tau_w$ continuous and since the sum in Eq. (3.1) is uniformly $\#$-convergent for $x,y \in \mathbb{C}$, it follows that $x \to \rho(x,y)$ is $\tau_\mathbb{C}$-continuous. This implies the $\#$-open balls relative to $\rho$ are contained in $\tau_\mathbb{C}$ and therefore $\tau_\rho \subset \tau_\mathbb{C}$. For the converse inclusion, let $z \in H^\#$, $x \to \varphi_z(x) = \langle z, x \rangle_\#$ be an element of $H^{\#*}$, and

for $n \in *\mathbb{N}$ let $z_N = \mathrm{Ext} - \sum_{n=1}^N \langle z, e_n \rangle_\# e_n$. Then $\varphi_{z_N} = \mathrm{Ext} - \sum_{n=1}^N \langle z, e_n \rangle_\# \varphi_{e_n}$ is
$\rho$-$\#$-continuous, being a hyperfinite linear combination of the $\varphi_{e_n}$ which are easily
seen to be $\rho$-$\#$-continuous. Because $z_N \to_\# z$ as $N \to *\infty$. it follows that
$\sup_{x \in \mathbb{C}} |\varphi_z(x) - \varphi_{z_N}(x)| = \|z - z_N\|_\# \to_\# 0$ as $N \to *\infty$.

Therefore $\varphi_z \upharpoonright \mathbb{C}$ is $\rho$-$\#$-continuous as well and hence $\tau_{\mathbb{C}} = \tau(\varphi_z \upharpoonright \mathbb{C}|_{z \in H^{\#}}) \subset \tau_{\rho}$.
The last assertion follows directly from Theorem 3.1 and the fact that sequential
$\#$-compactness is equivalent to $\#$-compactness for a non-Archimedean metric
spaces.

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

(1) There exists $\{t_n\}^*_{n=1} \subset \widetilde{\mathbb{R}_c^\#} \setminus \{0\}$ such that $\#\mathrm{lim}_{n \to *\infty} t_n = 0$ and
$$ \sup_{n \in *\mathbb{N}} \left\| \frac{f(\cdot + t_n v) - f(\cdot)}{t_n} \right\|_{\#2} < *\infty. $$

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

(3) There exists $g \in L_2^\#(\widetilde{\mathbb{R}_c^\#})$ and $f_n \in C_c^{*\infty}(\widetilde{\mathbb{R}_c^\#})$ such that $f_n \xrightarrow{L_2^\#} f$ and
$$[X_1 \times X_2 \setminus A]_{x_1} = X_2 \setminus A_{x_1}, [X_1 \times X_2 \setminus A]_{x_2} = X_1 \setminus A^{x_2}, \qquad (5.3.6)$$

A is a $\sigma^{\#}$-algebra. So $\Sigma_1 \otimes \Sigma_2 \subseteq A$, and (a) is proved. Now the part (b) follows from (a) due to $f_{x_1}^{-1}(A) = [f^{-1}(A)]_{x_1}$ and $[f^{x_2}]^{-1}(A) = [f^{-1}(A)]^{x_2}(\forall A \subseteq *\mathbb{R}_c^{\#})$.

**Definition 5.3.4** A family $M \subseteq P(X)$ is called a monotone class if $M$ is closed under $*$-countable increasing unions and $*$-countable decreasing intersections.

**Lemma 5.3.2.** If $A \subseteq P(X)$ is an algebra then the monotone class generated by $A$ coincides with the $\sigma^{\#}$-algebra generated by $A$.

**Lemma 5.3.3.** Let $(X_1, \Sigma_1, \mu_1^{\#})$ and $(X_2, \Sigma_2, \mu_2^{\#})$ be $*$-measure spaces, $E \in \Sigma_1 \otimes \Sigma_2$. Then the functions $x_1 \to \mu_2^{\#}(E_{x_1})$ and $x_2 \to \mu_1^{\#}(E^{x_2})$ are $*$-measurable on $(X_1, \Sigma_1)$ and $(X_2, \Sigma_2)$, and

$$\mu_1^{\#} \otimes \mu_2^{\#}(E) = \mathrm{Ext} - \int_{X_2} \mu_1^{\#}(E^{x_2}) d^{\#} \mu_2^{\#} = \mathrm{Ext} - \int_{X_1} \mu_2^{\#}(E_{x_1}) d^{\#} \mu_1^{\#}. \qquad (5.3.7)$$

**Proof.** First we consider the case when $\mu_1^{\#}$ and $\mu_2^{\#}$ are finite. Let $A$ be the family of all $E \in \Sigma_1 \otimes \Sigma_2$ for which (5.3.7) is true. If $E = A \times B$, then $\mu_1^{\#}(E^{x_2}) = \mu_1^{\#}(A)\chi_B(x_2)$ and $\mu_2^{\#}(E_{x_1}) = \mu_2^{\#}(B)\chi_A(x_1)$, so $E \in A$. By additivity, it follows that hyperfinite disjoint unions of rectangles are in $A$ so, by Lemma 5.3.2, bit will suffice to show that $A$ is a monotone class. If $(E_n)_{n=1}^{*\infty}$ is an increasing hyper infinite sequence in $A$ and $E = \bigcup_{n=1}^{*\infty} E_n$, then the function $f_n(x_2) = \mu_1^{\#}((E_n)^{x_2})$

are $*$-measurable and increase pointwise to $f(y) = \mu_1^{\#}(E^{x_2})$. Hence $f$ is $*$-measurable and, by the monotone convergence theorem,

$$\mathrm{Ext} - \int_{X_2} \mu_1^{\#}(E^{x_2}) d\mu_2^{\#} = \mathrm{\#-lim}_{n \to *\infty} \mathrm{Ext} X_1 \int_{X_2} \mu_1^{\#}((E_n)^{x_2}) d\mu_2^{\#} = \\ \mathrm{\#-lim}_{n \to *\infty} \mu_1^{\#} \times \mu_2^{\#}(E_n) = \mu_1^{\#} \times \mu_2^{\#}(E). \qquad (5.3.8)$$

Likewise $\mu_1^{\#} \times \mu_2^{\#}(E) = \mathrm{Ext} - \int_{X_1} \mu_2^{\#}(E_x) d\mu_1^{\#}$, so $E \in A$. Similarly, if $(E_n)_{n=1}^{*\infty}$ is a decreasing hyper infinite sequence in $A$ and $E = \bigcap_{n=1}^{*\infty} E_n$, the function $x_2 \to \mu_1^{\#}((E_1)^{x_2})$ is in $L_1^{\#}(\mu_2^{\#})$ because $\mu_1^{\#}((E_1)x_2) \le \mu_1^{\#}(X_1) < *\infty$ and $\mu_2^{\#}(X_2) < *\infty$, so the dominated convergence theorem can be applied to show that $E \in A$. Thus, $A$ is a monotone class, and the proof is complete for the case of finite $*$-measure spaces. Finally, if $\mu_1^{\#}$ and $\mu_2^{\#}$ are $\sigma^{\#}$-finite, we can write $X_1 \times X_2$ as the union of an increasing hyper infinite sequence $(X_1^j \times X_2^j)_{j=1}^{*\infty}$ of rectangles of finite or hyperfinite $*$-measure. If $E \in \Sigma_1 \otimes \Sigma_2$, the preceding argument applies to $E \cap (X_1^j \times X_2^j)$ for each $j$ gives us

$$\mu_1^{\#} \times \mu_2^{\#}(E \cap (X_1^j \times X_2^j)) = \mathrm{Ext} - \int_{X_2} \mu_1^{\#}(E^{x_2} \cap X_1^j) \mu_2^{\#} = \mathrm{Ext} - \int_{X_1} \mu_2^{\#}(E_{x_1} \cap X_2^j) \mu_1^{\#}. \qquad (5.3.9)$$

The application of the monotone convergence theorem then yields the desired result.

**Lemma 5.3.3.** (Generalized Tonelli's theorem) Let $(X_1, \Sigma_1, \mu_1^{\#})$ and $(X_2, \Sigma_2, \mu_2^{\#})$ be $*$-measure spaces, and $f: X_1 \times X_2 \to *\mathbb{R}_c^{*\#}$ be a $\Sigma_1 \otimes \Sigma_2$-$*$-measurable function. Then the functions
$\partial_{\nu}^{\#} f_n \xrightarrow{L_2^{\#}} g$ as $\to \infty$.

(4) There exists $g \in L_2^{\#}(\widetilde{\mathbb{R}}_c^{\#n})$ such that

$$ \frac{f(\cdot + tv) - f(\cdot)}{t} \xrightarrow{L_2^{\#}} g $$

as $t \to \# 0$.

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

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

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

$$ \langle g, \varphi \rangle_{\#} = \#\text{-lim}_{n \to \infty} \left\langle \frac{f(\cdot + t_n v) - f(\cdot)}{t_n}, \varphi \right\rangle_{\#} = \#\text{-lim}_{n \to \infty} \left\langle f, \frac{\varphi(\cdot + t_n v) - \varphi(\cdot)}{t_n} \right\rangle_{\#} = \\ = \left\langle f, \#\text{-lim}_{n \to \infty} \frac{\varphi(\cdot + t_n v) - \varphi(\cdot)}{t_n} \right\rangle_{\#} = -\langle f, \partial_{\nu}^{\#} \varphi \rangle_{\#}, $$

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

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

By **Theorem 11.21**, $h_m \to_m f \in L_2^{\#}(\widetilde{\mathbb{R}}_c^{\#n})$ and $\partial_{\nu}^{\#} h_m = \varphi_m * g \to_m g$ in $L_2^{\#}(\widetilde{\mathbb{R}}_c^{\#n})$ as $m \to \infty$. This shows 3. holds except for the fact that $h_m$ need not have #-compact support. To fix this let $\psi \in C_c^{*\infty}(\widetilde{\mathbb{R}}_c^{\#n}, [0,1])$ such that $\psi = 1$ in a neighborhood of 0 and let $\psi_\varepsilon(x) = \psi(\varepsilon x)$ and $(\partial_\nu^\# \psi)_\varepsilon(x) = (\partial_\nu^\# \psi)(\varepsilon x)$. Then $\partial_\nu^\# (\psi_\varepsilon h_m) = \partial_\nu^\# \psi_\varepsilon h_m + \psi_\varepsilon \partial_\nu^\# h_m = \varepsilon(\partial_\nu^\# \psi)_\varepsilon h_m + \psi_\varepsilon \partial_\nu^\# h_m$ so that $\psi_\varepsilon h_m \to_m h_m$ in $L_2^{\#}$ and $\partial_\nu^\# (\psi_\varepsilon h_m) \to_m \partial_\nu^\# h_m$ in $L_2^{\#}$ as $\varepsilon \to 0$. Let $f_m = \psi_{\varepsilon m} h_m$ where $\varepsilon_m$ is chosen to be greater than zero but small enough so that $\|\psi_{\varepsilon m} h_m - h_m\|_{\#2} + \|\partial_\nu^\# (\psi_{\varepsilon m} h_m) - \partial_\nu^\# h_m\|_{\#2} < 1/m$.

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

3. $\Rightarrow$ 4. By the fundamental theorem of calculus

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

Let

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

which is defined for #-almost every $x$ and is in $L_2^{\#}(\widetilde{\mathbb{R}}_c^{\#n})$ by generalized Minkowski's inequality for integrals. Therefore
$$f_{\mu_2^{\#}}(x_1) = \mathrm{Ext}- \int_{X_2} f_{x_1} d^{\#} \mu_2^{\#} \quad \text{and} \quad f_{\mu_1^{\#}}(x_2) = \mathrm{Ext}- \int_{X_1} f^{x_2} d^{\#} \mu_1^{\#} \qquad (5.3.10)$$

are $\Sigma_1$-#-measurable and $\Sigma_2$-#-measurable, respectively, and

$$ \begin{aligned} \mathrm{Ext}- \int_{X_1 \times X_2} f d^{\#} \mu_1^{\#} \otimes \mu_2^{\#} &= \mathrm{Ext}- \int_{X_2} \left[ \mathrm{Ext}- \int_{X_1} f^{x_2} d^{\#} \mu_1^{\#} \right] d^{\#} \mu_2^{\#} = \\ &= \mathrm{Ext}- \int_{X_1} \left[ \mathrm{Ext}- \int_{X_2} f_{x_1} d^{\#} \mu_2^{\#} \right] d^{\#} \mu_1^{\#}. \end{aligned} \qquad (5.3.11) $$

**Proof:** In the case when $f$ is a characteristic function, the statement of this lemma follows from Lemma 5.3.3. Therefore, by linearity, it holds also for nonnegative simple functions. If a nonnegative #-measurable function $f$ is arbitrary, there exists a sequence of nonnegative simple functions which increase pointwise to $f$, say $(f_n)_{n=1}^{\infty}$. By the monotone convergence theorem,

$$ \begin{aligned} \mathrm{Ext}- \int_{X_1} f_{\mu_2^{\#}} d^{\#} \mu_1^{\#} &= \#\lim_{n \to \infty} \left[ \mathrm{Ext}- \int_{X_1} f_{\mu_2^{\#}}^{n} d^{\#} \mu_1^{\#} \right] = \\ &= \#\lim_{n \to \infty} \left[ \mathrm{Ext}- \int_{X_1 \times X_2} f_n d^{\#} \mu_1^{\#} \otimes \mu_2^{\#} \right] \end{aligned} \qquad (5.3.12) $$

and

$$ \begin{aligned} \mathrm{Ext}- \int_{X_2} f_{\mu_1^{\#}} d^{\#} \mu_2^{\#} &= \#\lim_{n \to \infty} \left[ \mathrm{Ext}- \int_{X_2} f_{\mu_1^{\#}}^{n} d^{\#} \mu_2^{\#} \right] = \\ &= \#\lim_{n \to \infty} \left[ \mathrm{Ext}- \int_{X_1 \times X_2} f_n d^{\#} \mu_1^{\#} \otimes \mu_2^{\#} \right], \end{aligned} \qquad (5.3.13) $$

where

$$f_{\mu_2^{\#}}^{n}(x_1) = \mathrm{Ext}- \int_{X_1} [f_n]_{x_1} d^{\#} \mu_2^{\#}, f_{\mu_1^{\#}}^{n}(x_2) = \mathrm{Ext}- \int_{X_2} [f_n]^{x_2} d^{\#} \mu_1^{\#}. \qquad (5.3.14)$$

This proves (5.3.11) and the lemma.

**Proof** of Theorem 5.3.1. Since an $\mathbb{R}_c^{\#n}$-valued function $f$ is Lebesgue #-integrable iff its positive $f^+$ and negative $f^-$ parts are #-integrable, it is sufficient to prove the theorem only for nonnegative function $f \in L_1^r(X_1 \times X_2, \Sigma, \mu_1^\# \otimes \mu_2^\#)$. But this was exactly done in Lemma 5.3.3.

## § 5.4. Lebesgue #-measure and integral in $\mathbb{R}_c^{\#n}$.

In this section, we study $\mathbb{R}_c^{\#n}$, $n \in \mathbb{N}$ and functions from $\mathbb{R}_c^{\#n}$ to $\mathbb{R}_c^\#$ from the point of view of the Lebesgue #-measure and Lebesgue integration. All results presented below possess obvious $\mathbb{C}_c^{\#n}$-valued analogs. Then we define and study generalized Cantor sets which are interesting from the point of view of the set
$$ \frac{\tau_{-tv}f_m(x) - f_m(x)}{t} - G_t(x) = \int_0^1 [(\partial_{\nu}^{\#}f_m)(x + stv) - g(x + stv)]d^\#s $$

and hence again

$$ \left\| \frac{\tau_{-tv}f_m - f_m}{t} - G_t \right\|_{\#2} \le \int_0^1 \| \tau_{-stv}(\partial v f_m) - \tau_{-stv}g \|_{\#2} d^\#s = \int_0^1 \| \partial v f_m - g \|_{\#2} d^\#s. $$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Take $H^{\#} = \operatorname{Span}(\{e_i\}_{i=1}^{*\infty}) \subset I_{2}^{\#}$ and let $\tilde{u}_n = e_1 - (n+1)e_{n+1}$ for $n = 1, 2, \dots$. Applying Gramn-Schmidt to $\{\tilde{u}_n\}_{n=1}^{*\infty}$ we construct an orthonormal set $\beta = \{u_n\}_{n=1}^{*\infty} \subset H^{\#}$.
topology and the #-measure theory. Cantor sets are #-closed #-Borel nowhere
\#-dense subsets of the interval $[0, 1]$ or, more generally, of a Hausdorff \#-space.

**Definition 5.4.1.** The Lebesgue \#-measure $\mu^{\#n}$ on $\mathbb{R}_c^{\#n}$ is the \#-completion of the product of the Lebesgue \#-measure on $\mathbb{R}_c^{\#n}$ according to Definition 5.3.1. The domain $\Sigma^n$ of $\mu^\#$ (of course, $B^\#(\mathbb{R}_c^{\#n}) \subseteq \Sigma^n$) is the class of Lebesgue \#-measurable sets in $\mathbb{R}_c^{\#n}$. We write $d^\#x^n$ for $d^\#\mu^{\#n}$ and

$$
\mathrm{Ext} - \int f(x) d^\# x^n = \mathrm{Ext} - \int f d^\# \mu^{\#n}.
$$

We extend some of the results of previous section to the $n$-dimensional case with
$n \in \mathbb{N}$. If $E = \mathrm{Ext} - \prod_{j=1}^{n} E_j$ is a block in $\mathbb{R}_c^{\#n}$, we call sets $E_j \subseteq \mathbb{R}_c^{\#n}$ the sides of
the block $E$.

**Theorem 5.4.1.** Let $E \in \Sigma^n$. Then

(a) $\mu^{\#n}(E) = \inf\{\mu^{\#n}(U) : E \subseteq U, U \text{ #-open}\} = \sup\{\mu^{\#n}(K) : K \subseteq E, K \text{ #-compact}\}$;

(b) $E = A_1 \cup N_1 = A_2 \cup N_2$, where $A_1$ is an $F_{\sigma}^{\#}$ set, $A_2$ is a $G_{\delta}^{\#}$ set, and
$\mu^{\#n}(N_1) = \mu^{\#n}(N_2) = 0$;

(c) If $\mu^{\#n}(E) < \infty$ then, for any $\varepsilon \approx 0, \varepsilon > 0$, there is a hyperfinite family $\{R_j\}_{j=1}^N$ of disjoint blocks, whose sides are intervals such that $\mu^{\#n}(E\Delta \cup_{j=1}^N R_j) < \varepsilon$.

**Proof:** By the definition of product \#-measures, if $E \in \Sigma^n$ and $\varepsilon \approx 0, \varepsilon > 0$, there is a \*-countable family $\{T_j\}_{j=1}^{\infty}$ of blocks such that $E \subseteq \cup_{j=1}^{\infty} T_j$ and

$$
\mathrm{Ext} - \sum_{j=1}^{\infty} \mu^{\#n}(T_j) \leq \mu^{\#n}(E) + \varepsilon.
$$

For each $j$, by applying Theorem 5.2.3 to the sides of $R_j$, we can find blocks
$U_j \supseteq F_j$ whose sides are \#-open sets such that $\mu^{\#n}(U_j) < \mu(T_j) + \varepsilon 2^{-j}$.

If $U = \cup_{j=1}^{\infty} U_j$ then $U$ is \#-open and

$$
\mu^{\#n}(U) \leq \mathrm{Ext} - \sum_{j=1}^{\infty} \mu^{\#n}(U_j) \leq \mu^{\#n}(E) + 2\varepsilon.
$$

This proves the first equation in part (a). The second equation and part (b)
follow as in the proofs of Theorems **2.1.6** and **2.1.7**.

Next, if $\mu^{\#n}(E) < \infty$ then $\mu^{\#n}(U_j) < \infty$ for all $j$. Since the sides of $U_j$ are
\*-countable unions of \#-open intervals, by taking suitable hyperfinite subunions,
we obtain blocks $V_j \subseteq U_j$ whose sides are hyperfinite unions of intervals such that
$\mu^{\#n}(V_j) \ge \mu^{\#n}(U_j) - \varepsilon 2^{-j}$. If $N \in \mathbb{N}$ is sufficiently hyperfinite large, we have

$$
\mu^{\#n} \left( E \setminus \bigcup_{j=1}^{N} V_j \right) \leq \mu^{\#n} \left( \bigcup_{j=1}^{N} U_j \setminus V_j \right) + \mu^{\#n} \left( \bigcup_{j=N+1}^{\infty} U_j \right) < 2\varepsilon
$$

and

$$
\mu^{\#n} \left( \bigcup_{j=1}^{N} V_j \setminus E \right) \leq \mu^{\#n} \left( \bigcup_{j=1}^{\infty} U_j E \right) < \varepsilon,
$$

so $\mu^{\#n}(E\Delta \cup_{j=1}^N V_j) < 3\varepsilon$. Since $\cup_{j=1}^N V_j$ can be expressed as a hyperfinite disjoint union of rectangles whose sides are intervals, we have proved (c).

§ 5.5. Lebesgue \#-integrable functions on \*\mathbb{R}^{\#n}_c

Let $\mu^{\#n}$ be the Lebesgue \#-measure in $\mathbb{R}_c^{\#n}$. The set $M(\mathbb{R}_c^{\#n}, \mu^{\#n})$ of all $\mathbb{R}_c^{\#n}$-valued
$\mu^{\#n}$-measurable functions on $\mathbb{R}_c^{\#n}$ is a vector space (addition and scalar multiplication
are pointwise). By $L_1^\#(\mathbb{R}_c^{\#n}, \mu^{\#n})$ we denote its subspace of all Lebesgue
\#-integrable functions (with finite in $\mathbb{R}_c^\#$ \#-integral). Now write $f \approx g$ for $f$ and $g$ in
I now claim that $\beta \subset H^\#$ is maximal. Indeed if $x = (x_1, x_2, \dots, x_n, \dots) \in \beta^\perp$ then $x \perp u_n$ for all $n$, i.e. $0 = \langle x, u_n \rangle_\# = x_1 - (n+1)x_{n+1}$.

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

$$ \operatorname{Span}(\beta) = v^\perp. \text{ Then } \operatorname{Ext}-\sum_{i=1}^{*\infty} \langle x, u_n \rangle_\# u_n = P x = x - \frac{\langle x, v \rangle_\#}{\|v\|_\#^2} v \text{ so that} $$

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

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

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

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

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

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

from which we get

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

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

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

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

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

it suffices to show $x \to \langle x, y \rangle_\#$ is linear for all $y \in H^\#$. We will need to derive an identity from Eq. (1.4). To do this we make use of Eq. (1.4) three times to find
$M(*\mathbb{R}_c^{\#n}, \mu^{\#n})$, whenever $f$ and $g$ differ only on a $\mu^{\#n}$-null set (a set of $\mu^{\#n}$-measure zero). It is easily seen that $\approx$ is an equivalencerelation. Let $L_0 = L_0(*\mathbb{R}_c^{\#n}, \mu^{\#n})$ be the set of equivalence classes of functions in $M(*\mathbb{R}_c^{\#n}, \mu^{\#n})$. We denote the equivalence classes of $f, g, \dots, by[f], [g], \dots$. The set $L_0$ becomes a vector space over field $*\mathbb{R}_c^\#$ by defining $[f] + [g] = [f + g]$ and $\alpha[f] = [\alpha f]$ for a real $\alpha \in *\mathbb{R}_c^\#$. Observe that these definitions do not depend on the choice of $f$ and $g$ in their equivalence classes. The same is true for the partial order in $L_0$, if we define $[f] \le [g]$ to mean $f(x) \le g(x)$ for all $x \in *\mathbb{R}_c^{\#n}$ except a null set. In practice, the elements of $L_0 = L_0(*\mathbb{R}_c^{\#n}, \mu^{\#n})$ are usually denoted by $f, g, \dots$ and treated as if they were functions instead of equivalence classes of functions.

**Definition 5.5.1.**

**Theorem 5.5.1.** If $f \in L_1^\#(\mu^{\#n})$ and $\varepsilon \approx 0, \varepsilon > 0$, there is a simple function $\varphi = \text{Ext}-\sum_{j=1}^N \alpha_j \chi_{R_j}$, where each $R_j$ is a product of intervals such that $\text{Ext}-\int|f-\varphi|d^\#\mu^{\#n} < \varepsilon$, and there is a #-continuous function $g$ vanishing outside of a bounded in $*\mathbb{R}_c^{\#n}$ set such that $\text{Ext}-\int|f-g|d^\#\mu^{\#n} < \varepsilon < \varepsilon$.

**Proof.** By the definition of Lebesgue #-integrable functions, we can approximate $f$ by simple functions in $L_1^#$-#-norm. Then use Theorem 5.4.1 to approximate a simple function by a function $\varphi$ of the desired form. Finally, use the generalized Urysohn Lemma to approximate such $\varphi$ by a #-continuous function.

**Theorem 5.5.2.** The Lebesgue #-measure on $*\mathbb{R}_c^{\#n}$ is translation-invariant. Namely, let $a \in *\mathbb{R}_c^{\#n}$. Define the shift $\tau_a : *\mathbb{R}_c^{\#n} \to *\mathbb{R}_c^{\#n}$ by $\tau_a(x) = x + a$.

(a) If $E \in \mathcal{L}^{\#n}$ then $\tau_a(E) \in \mathcal{L}^{\#n}$ and $\mu^{\#n}(\tau_a(E)) = \mu^{\#n}(E)$;

(b) If $f : *\mathbb{R}_c^{\#n} \to *\mathbb{R}_c^\#$ is Lebesgue #-measurable then so is $f \circ \tau_a$. Moreover, if either $f \ge 0$ or $f \in L_1^\#(\mu^{\#n})$ then

$$ \text{Ext}-\int (f \circ \tau_a) d^\# \mu^{\#n} = \text{Ext}-\int f d^\# \mu^{\#n}. \qquad (5.5.1) $$

**Proof.** Since $\tau_a$ and its inverse $\tau_{-a}$ are #-continuous, they preserve the class of #-Borel sets. The formula $\mu^{\#n}(\tau_a(E)) = \mu^{\#n}(E)$ follows easily from the trivial one dimensional variant of this result if $E$ is a block. For a general #-Borel set $E$, the formula $\mu^{\#n}(\tau_a(E)) = \mu^{\#n}(E)$ follows from the previous step, since $\mu^{\#n}$ is determined by its action on blocks. Assertion (a) now follows immediately.

If $f$ is Lebesgue #-measurable and $B$ is a #-Borel set in $*\mathbb{R}_c^\#$, we have $f^{-1}(B) = E \cup N$, where $E$ is #-Borel and $\mu^\#(N) = 0$. But $\tau_a^{-1}(E)$ is Borel and $\mu^\#(\tau_a^{-1}(N)) = 0$, so $(f \circ \tau_a)^{-1}(B) \in \Sigma^n$ and $f$ is Lebesgue #-measurable. The equality (5.5.1) reduces to the equality $\mu^{\#n}(\tau_{-a}(E)) = \mu^{\#n}(E)$ when $f = \chi_E$. It is true for simple functions by linearity, and hence for nonnegative #-measurable functions by the definition of #-integral. Taking positive and negative parts of real and imaginary parts, we obtain the result for $f \in L_1^\#(\mu^{\#n})$.

**Theorem 5.5.3.** (Generalized Lusin's theorem) If $f$ is a Lebesgue #-measurable function on $*\mathbb{R}_c^{\#n}$ and $\varepsilon \approx 0, \varepsilon > 0$ then there exist a #-measurable set $A \subseteq *\mathbb{R}_c^{\#n}$ such that $\mu^{\#n}(*\mathbb{R}_c^{\#n} \setminus A) \le \varepsilon$ and the restriction of $f$ onto $A$ is #-continuous.

## §6. *$\tilde{\mathbb{R}}_c^\#$-Valued measurable functions

**Definition 6.1** ($\sigma^\#$-algebra). Let $X$ be any set. We denote by $2^X = P(X) = \{A : A \subset X\}$ the set of all subsets of $X$. A family $\mathcal{F} \subset 2^X$ is called a $\sigma^\#$-algebra (on $X$) if:
$$
\begin{align}
\|x+y+z\|_{\#}^{2} &= -\|x+y-z\|_{\#}^{2} + 2\|x+y\|_{\#}^{2} + 2\|z\|_{\#}^{2} \nonumber \\
&= \|x-y-z\|_{\#}^{2} - 2\|x-z\|_{\#}^{2} - 2\|y\|_{\#}^{2} + 2\|x+y\|_{\#}^{2} + 2\|x+y\|_{\#}^{2} + 2\|z\|_{\#}^{2} \nonumber \\
&\qquad + \|y+z-x\|_{\#}^{2} - 2\|x-z\|_{\#}^{2} - 2\|y\|_{\#}^{2} + 2\|x+y\|_{\#}^{2} + 2\|z\|_{\#}^{2} \nonumber \\
&\qquad - \|y+z+x\|_{\#}^{2} + 2\|y+z\|_{\#}^{2} + 2\|x\|_{\#}^{2} - 2\|x-z\|_{\#}^{2} - 2\|y\|_{\#}^{2} + 2\|x+y\|_{\#}^{2} + 2\|z\|_{\#}^{2}. \tag{3.10}
\end{align}
$$

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

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

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

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

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

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

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

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

which combined with Eq. (3.12.) shows

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

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

Because of this equation and Eq. (3.13) to finish the proof that $x \to \langle x, y \rangle_{\#}$ is
linear, it suffices to show $\lambda\langle x, y \rangle_{\#} = \langle \lambda x, y \rangle_{\#}$ for all $\lambda > 0$. Now if $\lambda = m \in *\mathbb{N}$, then
$\langle mx, y \rangle_{\#} = \langle x + (m-1)x, y \rangle_{\#} = \langle x, y \rangle_{\#} + \langle (m-1)x, y \rangle_{\#}$

so that by hyper infinite induction $\langle mx, y \rangle_{\#} = m \langle x, y \rangle_{\#}$. Replacing $x$ by $x/m$ then shows
that $\langle x, y \rangle_{\#} = m \langle m^{-1}x, y \rangle_{\#}$, so that $\langle m^{-1}x, y \rangle_{\#} = m^{-1} \langle x, y \rangle_{\#}$ and so if $m, n \in *\mathbb{N}$, we find
$\langle \frac{n}{m}x, y \rangle_{\#} = n \langle \frac{1}{m}x, y \rangle_{\#} = \frac{n}{m} \langle x, y \rangle_{\#}$ so that $\lambda \langle x, y \rangle_{\#} = \langle \lambda x, y \rangle_{\#}$ for all $\lambda > 0$ and $\lambda \in *\tilde{\mathbb{Q}}$.

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

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

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

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

(ii) $\mathcal{F}$ is closed under complements, i.e. $A \in \mathcal{F}$ implies $X \Lambda A \in \mathcal{F}$;

(iii) $\mathcal{F}$ is closed under hypercountable unions, i.e. if $(A_n)_{n \in \mathbb{N}^\#}$ is a hyper infinite sequence in $\mathcal{F}$ then $\bigcup_{n \in \mathbb{N}^\#} A_n \in \mathcal{F}$.

**Proposition 6.1.** If $\mathcal{F}$ is a $\sigma^\#$-algebra on $X$ then:

1. $\mathcal{F}$ is closed under hypercountable intersections, i.e. if $(A_n)_{n \in \mathbb{N}^\#}$ is a hyper infinite sequence in $\mathcal{F}$ then $\bigcap_{n \in \mathbb{N}^\#} A_n \in \mathcal{F}$.

2. $X \in \mathcal{F}$.

3. $\mathcal{F}$ is closed under hyperfinite unions and hyperfinite intersections.

4. $\mathcal{F}$ is closed under set differences.

5. $\mathcal{F}$ is closed under symmetric differences.

**Proposition 6.2.** Suppose $\mathcal{F} \subset 2^X$ is a family of subsets satisfying the following:

1. $\emptyset \in \mathcal{F}$;

2. $\mathcal{F}$ is closed under complements;

3. $\mathcal{F}$ is closed under hyperinfinite intersections.

Then $\mathcal{F}$ is a $\sigma^\#$-algebra.

**Proposition 6.3.** If $(\mathcal{F}_\alpha)_{\alpha \in I}$ is a collection of $\sigma^\#$-algebras on $X$, then $\bigcap_\alpha \mathcal{F}_\alpha$ is also a $\sigma^\#$-algebra on $X$.

**Proposition 6.4.** ($\sigma^\#$-algebra generated by subsets). Let $K$ be a collection of subsets of $X$. There exists a $\sigma^\#$-algebra, denoted $\sigma^\#(K)$ such that $K \subset \sigma^\#(K)$ and for every other $\sigma^\#$ algebra $\mathcal{F}$ such that $K \subset \mathcal{F}$ we have that $\sigma^\#(K) \subset \mathcal{F}$
We call $\sigma^\#(K)$ the $\sigma^\#$-algebra generated by $K$.

**Proof.** Define $\sigma^\#(K) \triangleq \bigcap \{\mathcal{F} | \mathcal{F} \text{ is a } \sigma^\#$-algebra on } X, K \subset \mathcal{F}\}$. This is a $\sigma^\#$-algebra with the required properties.

**Proposition 6.5.** If $K \subset \mathcal{L}$ then $\sigma^\#(K) \subset \sigma^\#(\mathcal{L})$. Also, if $K \subset \mathcal{F}$ and $\mathcal{F}$ is a $\sigma^\#$-algebra, then $\sigma^\#(K) \subset \mathcal{F}$.

**Definition 6.2.** (Borel $\sigma^\#$-algebra). Given a topological space $X$, the Borel $\sigma^\#$-algebra is the $\sigma^\#$-algebra generated by the open sets. It is denoted $B^\#(X)$.
Specifically in the case $X = *\mathbb{R}_c^{\#d}, d \in \mathbb{N}^\#$ we have that

$$ \tilde{B}_d^\# \triangleq B^\#(*\tilde{\mathbb{R}}_c^{\#d}) = \sigma^\#(U|U \text{ is an } \#\text{-open set}). $$

A Borel-#-measurable set, i.e. a set in $B^\#(X)$, is called a #-Borel set.

**Measurable functions.** Let $f$ be a $*\tilde{\mathbb{R}}_c^\#$-valued function defined on a set $X$. We suppose that some $\sigma^\#$-algebra $\Omega \subseteq P(X)$ is fixed.

**Definition 6.3.** We say that $f$ is #-measurable, if $f^{-1}([a, b]) \in \Omega$ for any hyperreals $a, b \in *\tilde{\mathbb{R}}_c^\#$ such that $a < b$.

The following three propositions are obvious.

**Proposition 6.7.** Let $f: X \to *\tilde{\mathbb{R}}_c^\#$ be a function. Then the following conditions are equivalent:

(a) $f$ is #-measurable;

(b) $f^{-1}([0, b)) \in \Omega$ for any hyperreal $b \in *\tilde{\mathbb{R}}_c^\#$;

(c) $f^{-1}((b, \infty)) \in \Omega$ for any hyperreal $b \in *\tilde{\mathbb{R}}_c^\#$;

(d) $f^{-1}(B) \in \Omega$ for any $B \in B(R)$.

**Proposition 6.8** Let $f$ and $g$ be #-measurable functions, then

(a) $\alpha \times f + \beta \times g$ is #-measurable for any $\alpha, \beta \in *\tilde{\mathbb{R}}_c^\#$;
Moreover, either of these two conditions implies that $\beta \subset H^\#$ is a maximal orthonormal set. However $\beta \subset H^\#$ being a maximal orthonormal set is not sufficient to conditions for 1) and 2) hold.

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

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

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

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

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

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

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

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

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

Let $H^{\#}$ be a #-complex Hilbert space over field $\ast \tilde{\mathbb{C}}_{c}^{\#}$. The most natural way to construct a #-continuous one-parameter unitary group $U(t) : H^{\#} \to H^{\#}$ is to try to make sense
(b) functions $\max\{f, g\}$ and $f \times g$ are #-measurable.

In particular, functions $f^+ := \max\{f, 0\}$, $f^- := (-f)^+$, and $|f| := f^+ + f^-$ are
\#-measurable. .

Chapter IV. $\mathbb{R}_c^\#$-valued distributions.

§1. $\mathbb{R}_c^\#$-valued test functions and distributions

Definitions and theorems appropriate to analysis on non-Archemedean field $*\mathbb{R}_c^\#$ and on complex field $*\mathbb{C}_c^\# = *\mathbb{R}_c^\# + i*\mathbb{R}_c^\#$ are given in [1]-[2].

**Definition 1.1.**[3].(i) Let $U$ be a free ultrafilters on $\mathbb{N}$ and introduce an equivalence relation on sequences in $\mathbb{R}^\mathbb{N}$ as $f_1 \sim_U f_2$ iff $\{i \in \mathbb{N} | f_1(i) = f_2(i)\} \in U$.

(ii) $\mathbb{R}^\mathbb{N}$ divided out by the equivalence relation $\sim_U$ gives us the nonstandard extension $*\mathbb{R}$, the hyperreals; in symbols, $*\mathbb{R} = \mathbb{R}^\mathbb{N}/\sim_U$ and similarly $\mathbb{N}^\mathbb{N}$ divided out by the equivalence relation $\sim_U$ gives us the nonstandard extension $*\mathbb{N}$, the hyperintegers; in symbols, $*\mathbb{N} = \mathbb{N}^\mathbb{N}/\sim_U$.

**Abbreviation 1.1.** If $f \in \mathbb{R}^\mathbb{N}$, we denote its image in $*\mathbb{R}$ by $[f]$, i.e., $[f] = \{g \in \mathbb{R}^\mathbb{N} | g \sim_U f\}$.

**Remark 1.1.** For any real number $r \in \mathbb{R}$ let $\mathbf{r}$ denote the constant function $\mathbf{r} : \mathbb{N} \to \mathbb{R}$ with value $r$, i.e., $\mathbf{r}(n) = r$, for all $n \in \mathbb{N}$. We then have a natural embedding

$$ *\langle \cdot \rangle : \mathbb{R} \hookrightarrow *\mathbb{R} $$

by setting $*r = [\mathbf{r}(n)]$ for all $r \in \mathbb{R}$. We denote its image $*(\mathbb{R})$ in $*\mathbb{R}$ by $*\mathbb{R}_{\text{st}}$.

**Definition 1.2.**[3]. An element $x \in *\mathbb{R}$ is called finite if $|x| < r$ for some $r \in \mathbb{Q}, r > 0$.

**Abbreviation 1.2.** For $x \in *\mathbb{R}$ we abbreviate $x \in *\mathbb{R}_{\text{fin}}$ if $x$ is finite.

**Remark 1.2.**[3]. Let $x \in *\mathbb{R}_{\text{fin}}$ be finite. Let $D_1$, be the set of $r \in \mathbb{Q}$ such that $r < x$ and $D_2$ the set of $r' \in \mathbb{Q}$ such that $x < r'$. The pair $(D_1, D_2)$ forms a Dedekind cut in $\mathbb{R}$, hence determines a unique $r_0 \in \mathbb{R}$. A simple argument shows that $|x - r_0|$ is infinitesimal, i.e., $|x - r_0| \approx 0$.

**Definition 1.3.**[1]. This unique $r_0$ is called the standard part of $x$ and is denoted by $\circ x$ or $st(x)$.

The following notation will be used throughout this paper.

$n \in \mathbb{N}^\#$ is a fixed positive integer and $U \subset *\mathbb{R}_c^{\#n}$ is a fixed non-empty #-open subset of linear space $*\mathbb{R}_c^{\#n}$ over non Archemedean field $*\mathbb{R}_c^\#$.

$\mathbb{N} = \{0, 1, 2, \dots\}$ denotes the standard natural numbers.

$k$ will denote a non-negative integer or $\infty^\#$.

If $f$ is a function then $\text{Dom}(f)$ will denote its domain and the support of $f$, denoted by $\text{supp}(f)$, is defined to be the closure of the set $\{x \in \text{Dom}(f) : f(x) \neq 0\}$ in $\text{Dom}(f)$.

For two functions $f, g : U \to *\mathbb{C}_c^\#$, the following notation defines external canonical pairing:

$$ \langle f, g \rangle = \text{Ext-} \int_U f(x)g(x)d^\#x. \qquad (1.1) $$

A multi-index of size $n \in \mathbb{N}^\#$ is an element in $\mathbb{N}^{\#n}$, if the size of multi-indices is omitted then the size should be assumed to be $n$. The length of a multi-index $\alpha = (\alpha_1, \dots, \alpha_n) \in \mathbb{N}^{\#n}$ is defined as $\text{Ext}-\sum_{i=1}^n \alpha_i$ and denoted by $|\alpha|$. Multi-indices are particularly useful when
of the power series $Ext-\sum_{n=0}^{\infty} (itA)^n$ on a #-dense set of vectors. Notice that this can certainly be done if A is self-adjoint. For let $E_\Omega$ be the family of spectral projections for A. Then on each of the spaces $E_{[-M,M]}$, A is a bounded operator and $Ext-\sum_{n=0}^{\infty} (itA)^n/n!$ #-converges to $Ext-\exp(itA)$ in #-norm. In particular, for any $\varphi \in \bigcup_{M \ge 0} E_{[-M,M]}$,

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

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

**Definition1.1.** 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

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

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 $C^{\infty}(A)$.

**Definition1.2.** Suppose that A is symmetric. For each $\varphi \in C^{\infty}(A)$, define

$$ D_{\varphi} = \left\{ \mathrm{Ext}-\sum_{n=0}^{N} \alpha_n A^n \varphi \mid N \in *\mathbb{N}, \alpha_n \in \widetilde{\mathbb{C}}_c^{\#} \right\}. \qquad (3.3) $$

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

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

**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 $H^\#$. By the fundamental criterion this will show that A is essentially self-#-adjoint. Suppose $\psi \in H^\#$ 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
dealing with functions of several variables, in particular we introduce the following canonical notations for a given multi-index $\alpha = (\alpha_1, \dots, \alpha_n) \in \mathbb{N}^{\#n}$,

$$x^{\alpha} = x_1^{\alpha_1} \cdots x_n^{\alpha_n}, \\ \partial^{\#\alpha} = \frac{\partial^{\#|\alpha|}}{\partial x_1^{\alpha_1} \cdots \partial x_n^{\alpha_n}} \qquad (1.2)$$

We also introduce a partial order of all multi-indices by $\beta \ge \alpha$ if and only if $\beta_i \ge \alpha_i$ for all $1 \le i \le n$. When $\beta \ge \alpha$ we define their multi-index binomial coefficient as:

$$\binom{\beta}{\alpha} = \binom{\beta_1}{\alpha_1} \cdots \binom{\beta_n}{\alpha_n}.$$

1. Let $k \in \mathbb{N}^{\#} \cup \infty^{\#}$.

2. Let $C^{\#k}(U)$ denote the vector space of all $k$-times #-continuously #-differentiable $\mathbb{R}_c^\#$-valued or $\mathbb{C}_c^\#$-valued functions on $U$.

For any #-compact subset $K \subseteq U$, let $C^{\#k}(K)$ and $C^{\#k}(K; U)$ both denote the vector space of all those functions $f \in C^{\#k}(U)$ such that $\text{supp}(f) \subseteq K$.

Note that $C^{\#k}(K)$ depends on both $K$ and $U$ but we will only indicate $K$, where in particular, if $f \in C^{\#k}(K)$ then the domain of $f$ is $U$ rather than $K$. We will use the notation

$C^{\#k}(K; U)$ only when the notation $C^{\#k}(K)$ risks being ambiguous.

Every $C^{\#k}(K)$ contains the constant 0 map, even if $K = \emptyset$.

Let $C_c^{\#k}(U)$ denote the set of all $f \in C^{\#k}(U)$ such that $f \in C^{\#k}(K)$ for some #-compact subset $K$ of $U$.

Equivalently, $C_c^{\#k}(U)$ is the set of all $f \in C^{\#k}(U)$ such that $f$ has #-compact support.

$C_c^{\#k}(U)$ is equal to the union of all $C^{\#k}(K)$ as $K \subseteq U$ ranges over all #-compact subsets of $U$. If $f$ is a $\mathbb{R}_c^\#$-valued function on $U$, then $f$ is an element of $C_c^{\#k}(U)$ if and only if $f$ is a $C^{\#k}$ bump function. Every $\mathbb{R}_c^\#$-valued test function on $U$ is always also a $\mathbb{C}_c^\#$-valued test function on $U$.

For all $j, k \in \mathbb{N}$ and any #-compact subsets $K$ and $L$ of $U$, we have:

$$C_c^{\#k}(K) \subseteq C_c^{\#k}(U) \subseteq C_c^{\#k}(U);$$

$$C_c^{\#k}(K) \subseteq C_c^{\#k}(L) \text{ if } K \subseteq LC^{\#k}(K) \subseteq C_c^{\#j}(K) \text{ if } j \le k;$$

$$C_c^{\#k}(U) \subseteq C_c^{\#j}(U) \text{ if } j \le k;$$

$$C_c^{\#k}(U) \subseteq C_c^{\#j}(U) \text{ if } j \le k.$$

**Definition1.1.** Elements of $C_c^{\#\infty}(U)$ are called $\mathbb{R}_c^\#$-valued test functions on $U$ and $C_c^{\#\infty}(U)$ is

called the space of $\mathbb{R}_c^\#$-valued test functions on $U$. We will use both $D(U)$ and $C_c^{\#\infty}(U)$ to denote this space.

**Definition1.2.** Distributions on $U$ are #-continuous $\mathbb{R}_c^\#$-valued linear functionals on $C_c^{\#\infty}(U)$ when this vector space is endowed with a particular topology called the canonical

LF-topology.

The following proposition states two necessary and sufficient conditions for the #-continuity of a linear functional on $C_c^{\#\infty}(U)$ that are often straightforward to verify.

**Proposition1.1.** A linear functional $T$ on $C_c^{\#\infty}(U)$ is #-continuous, and therefore a distribution, if and only if either of the following equivalent conditions are satisfied:

1. For every #-compact subset $K \subseteq U$ there exist constants $C > 0$ and $N \in \mathbb{N}$ dependent on $K$ such that for all $f \in C_c^{\#\infty}(U)$ with support contained in $K$
$$ \left\| \psi - \operatorname{Ext-}\sum_{n=1}^{N} \alpha_n \psi_n \right\|_{\#} \le \varepsilon/2. \qquad (3.4) $$

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

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

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

**Theorem 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

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

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 $A$,

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

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

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

Therefore by generalized Fibini's theorem

$$ \operatorname{Ext-} \int_{s}^{x} \left( \operatorname{Ext-} \sum_{n=0}^{*\infty} \frac{s^n}{n!} |x|^n \right) d^\# \mu^\# = \operatorname{Ext-} \int_{s}^{x} \operatorname{Ext-} (s|x|) d^\# \mu^\# < *\infty. \qquad (3.9) $$

As a result, the function

$$ \langle \psi, [\operatorname{Ext-}\exp(itB)]\psi \rangle_{\#} = \operatorname{Ext-} \int_{s}^{x} [\operatorname{Ext-}\exp(itx)] d^{\#} \mu^{\#} \qquad (3.10) $$

has an #-analytic continuation

$$ \operatorname{Ext-} \int_{s}^{x} [\operatorname{Ext-}\exp(izx)] d^{\#} \mu^{\#} \qquad (3.11) $$
$|T(f)| \le C \sup\{|\partial^{\#a}f(x)|: x \in U, |a| \le N\}.$

2. For every #-compact subset $K \subseteq U$ and every sequence $\{f_i\}_{i=1}^{\infty}$ in $C_c^{\#\infty}(U)$ whose supports are contained in $K$, if $\{\partial^{\#a}f_i\}_{i=1}^{\infty}$ #-converges uniformly to zero on $U$ for every multi-index $\alpha$, then $\#\lim_{i \to \infty} T(f_i) = 0$.

## § 2. The non-Archimedean external $\mathbb{R}_c^\#$-Valued Schwartz distributions.

Defined below are the tempered distributions, which form a subspace of $D^{\#'}(*\mathbb{R}_c^{\#n})$, the space of distributions on $*\mathbb{R}_c^{\#n}$. This is a proper subspace: while every tempered distribution is a distribution and an element of $D^{\#'}(*\mathbb{R}_c^{\#n})$ the converse is not true. Tempered distributions are useful if one studies the Fourier transform since all tempered distributions have a Fourier transform, which is not true for an arbitrary distribution in $D^{\#'}(*\mathbb{R}_c^{\#n})$.

### § 2.1. Schwartz space $S^#(*\mathbb{R}_c^{\#n})$.

**Definition 2.1.** A function $f: X \to *\mathbb{R}_c^\#$ defined on some set $X$ is called **finitely bounded** (or **bounded**) if the set of its values is finitely bounded, i.e., $f(X) \subset [a, b]$ where $a, b \in *\mathbb{R}_{c, \text{fin}}^\#$. In other words, there exists a finite hyperreal number $M \in *\mathbb{R}_{c, \text{fin}}^\#$ such that

$$|f(X)| \le M. \qquad (2.1)$$

**Definition 2.2.** A function $f: X \to *\mathbb{R}_c^\#$ defined on some set $X$ is called **hyper finitely bounded** (or **hyper bounded**) if the set of its values is hyper finitely bounded, i.e., $f(X) \subset [a, b]$ where $a, b \in *\mathbb{R}_c^{\#} \setminus *\mathbb{R}_{c, \text{fin}}^\#$. In other words, there exists a hyperfinite hyperreal number $M \in *\mathbb{R}_c^{\#} \setminus *\mathbb{R}_{c, \text{fin}}^\#$ such that $|f(X)| \le M$.

**Definition 2.3.** For $n \in \mathbb{N}^\#$, an #-integrable function $\phi: *\mathbb{R}_c^{\#n} \to *\mathbb{R}_c^\#$ is called #-rapidly decreasing if for all $\alpha \in \mathbb{N}^{\#n}$ the product function $x \mapsto x^\alpha \phi(x)$ is a finitely bounded or hyper finitely bounded function.

**Remark 2.1.** If $\phi$ is a #-rapidly decreasing function, then its integral exists

$$\mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} \phi(x) d^{\#n} x < \infty \qquad (2.2)$$

In fact for all $\alpha \in \mathbb{N}^{\#n}$ the integral of $x \mapsto x^\alpha \phi(x)$ exists

$$\mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} x^\alpha \phi(x) d^{\#n} x < \infty. \qquad (2.3)$$

**Definition 2.4.** The Schwartz space, $S^#(*\mathbb{R}_c^{\#n})$, is the space of all #-smooth functions in $C_c^{\#\infty}(*\mathbb{R}_c^{\#n})$ that are rapidly decreasing at #-infinity along with all partial #-derivatives. Thus

$\phi: *\mathbb{R}_c^{\#n} \to *\mathbb{R}_c^\#$ is in the Schwartz space provided that any #-derivative of $\phi$, multiplied with any power of $|x|$, #-converges to 0 as $|x| \to \infty$. These functions form a #-complete TVS with a suitably defined family of seminorms. More precisely, for any multi-indices $\alpha$ and $\beta$ define:

$$p_{\alpha, \beta}(\phi) = \sup_{x \in *\mathbb{R}_c^{\#n}} |x^\alpha \partial^{\#\beta} \phi(x)|. \qquad (2.1)$$

Then $\phi$ is in the Schwartz space $S^#(*\mathbb{R}_c^{\#n})$ if all the values satisfy: $p_{\alpha, \beta}(\phi) < \infty$.
to the region $|\mathrm{Im}z|< t$. Since

$$ 
\begin{aligned}
& \left[ \left( \frac{d^{\#}}{d^{\#}z} \right)^k \left( \mathrm{Ext} - \int_{\mathbb{R}_c^{\#}} [\mathrm{Ext}\text{-exp}(izx)] d^{\#} \mu^{\#} \right) \right]_{z=0} \\
& = \mathrm{Ext} - \int_{\mathbb{R}_c^{\#}} [\mathrm{Ext}\text{-exp}(ix)^k] d^{\#} \mu^{\#} = \langle \psi, (iA)^k \psi \rangle_{\#},
\end{aligned}
\quad (3.12) $$

we obtain

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

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

Similar proof shows that $\langle \psi_1, [\mathrm{Ext}\text{-exp}(isB)]\psi_2 \rangle_{\#}$ is determined by the numbers $\langle \psi_1, A^n \psi_2 \rangle_{\#}, n \in *\mathbb{N}$ for any $\psi_1, \psi_2 \in D_\psi$. Since $D_\psi$ is #-dense in $H_\psi^\#$ and $\mathrm{Ext}\text{-exp}(isB)$ is unitary, $\mathrm{Ext}\text{-exp}(isB)$ 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 $A \uparrow D$. Thus, by Theorem 3.1 $A \uparrow D$ 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 $A \uparrow D$, it must first be $C^{*\infty}$ for $A \uparrow D$. 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
Thus

$$
S^{#}(\mathbb{R}_{c}^{\#n}, \mathbb{R}_{c}^{\#}) \triangleq \{\phi \in C^{\infty}(\mathbb{R}_{c}^{\#n}, \mathbb{R}_{c}^{\#}) | \forall \alpha, \beta \in \mathbb{N}^{\#n}(p_{\alpha, \beta}(\phi) < \infty^{\#})\}.
$$

Similarly

$$
S^{#}(\mathbb{R}_{c}^{\#n}, \mathbb{C}^{\#}) \triangleq \{\phi \in C^{\infty}(\mathbb{R}_{c}^{\#n}, \mathbb{C}^{\#}) | \forall \alpha, \beta \in \mathbb{N}^{\#n}(p_{\alpha, \beta}(\phi) < \infty^{\#})\}
$$

The family of seminorms $p_{\alpha,\beta}(\cdot)$ defines a locally convex topology on the Schwartz space $S^{#}(\mathbb{R}_{c}^{\#n})$.

For $n = 1$, the seminorms are norms on the Schwartz space $S^{#}(\mathbb{R}_{c}^{\#})$. One can also use the following family of seminorms to define the topology:

$$
\|f\|_{m,k} = \sup_{|p| \le m} \left( \sup_{x \in \mathbb{R}_c^{\#n}} \{ (1 + |x|)^k |(\hat{\partial}^{\#\alpha} f)(x)| \} \right), k, m \in \mathbb{N}^n. \quad (2.2)
$$

Otherwise, one can define a norm on $S^{#}(\mathbb{R}_{c}^{\#n})$ by

$$
\|\phi\|_k = \max_{|\alpha|+|\beta| \le k} \sup_{x \in \mathbb{R}_c^{\#n}} |x^\alpha \partial^\beta \phi(x)|, k \ge 1. \quad (2.3)
$$

The Schwartz space $S^{#}(\mathbb{R}_{c}^{\#n})$ is a Fréchet space (that is, a #-complete metrizable locally convex space). Because the Fourier transform changes $\hat{\partial}^{\#\alpha}$ into multiplication by $x^\alpha$ and vice versa, this symmetry implies that the Fourier transform of a Schwartz function is also a Schwartz function.

**Definition 2.5.** A sequence ${f_i}_{i=1}^{\infty}$ #-converges to 0 in $S^{#}(\mathbb{R}_c^{\#n})$ if and only if the functions $(1 + |x|)^k (\hat{\partial}^{\#p} f_i)(x)$ #-converge to 0 uniformly in the whole of $\mathbb{R}_c^{\#n}$, which implies that such a sequence must converge to zero in $C^{\infty}(\mathbb{R}_c^{\#n})$.

The subset of all #-analytic Schwartz functions is #-dense in $S^{#}(\mathbb{R}_{c}^{\#n})$

The Schwartz space is nuclear and the tensor product of two maps induces a canonical surjective TVS-isomorphisms $S^{#}(\mathbb{R}_{c}^{\#m}) \widehat{\otimes} S^{#}(\mathbb{R}_{c}^{\#n}) \rightarrow S^{#}(\mathbb{R}_{c}^{\#m+n})$,
where $\widehat{\otimes}$ represents the #-completion of the injective tensor product

§ 2.2.Schwartz space $S_{\text{fin}}^{\#}(\mathbb{R}_{c, \text{fin}}^{\#n})$

**Definition 2.6.** For $n \in \mathbb{N}$, an $\mathbb{R}_{c, \text{fin}}^{\#n}$-valued and #-integrable function

$\phi : \mathbb{R}_c^{\#n} \to \mathbb{R}_{c, \text{fin}}^{\#}$ is called #-rapidly decreasing if for all $\alpha \in \mathbb{N}^n$ the product function $x \mapsto x^\alpha \phi(x)$ is a finitely bounded function.

**Remark 2.2.** If $\phi$ is a #-rapidly decreasing $\mathbb{R}_{c, \text{fin}}^{\#n}$-valued function, then its integral exists and finite, i.e.,

$$
\mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} \phi(x) d^{\#n} x \in \mathbb{R}_{c, \mathrm{fin}}^{\#} \qquad (2.2)
$$

In fact for all $\alpha \in \mathbb{N}^n$ the integral of $x \mapsto x^\alpha \phi(x)$ exists and finite, i.e.,

$$
\mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} x^\alpha \phi(x) d^{\#n} x \in \mathbb{R}_{c, \mathrm{fin}}^{\#} \qquad (2.3)
$$

It follows from () that for all $\alpha \in \mathbb{N}^n$ and for any $R \in \mathbb{R}_c^\# \setminus \mathbb{R}_{c, \text{fin}}^\#$

$$
\mathrm{Ext} - \int_{\mathbb{R}_c^{\#n} \setminus B(R)} x^\alpha \phi(x) d^{\#n} x \approx 0 \quad (2.3)
$$

where $B(R) \triangleq \{x \in \mathbb{R}_c^\# | x| \le R\}$
Hilbert space $\mathbf{H}^\#$ 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 $U: \mathbf{H}^\# \to L_2^\#(M, d^\#\mu^\#)$ 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 $\psi \in \mathbf{H}^\#$.

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 $f(x) = \mathrm{Ext}-\sum_{n=1}^N c_n x^n$, $N \in \mathbb{N}$ is a polynomial, we let $f(A) = \mathrm{Ext}-\sum_{n=1}^N c_n A^n$. Suppose that $f(x) = \mathrm{Ext}-\sum_{n=1}^{\infty} c_n x^n$ is a hyper infinite power series with radius of \#-convergence $R$. If $\|A\|_{\#} < R$ then hyper infinite power series $\mathrm{Ext}-\sum_{n=1}^{\infty} c_n A^n$ \#-converges in $\mathcal{L}(H^\#)$ so it is natural to set

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

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, $\|\phi(f)\|_{\mathcal{L}(H^\#)} \le C\|f\|_{\cdot\infty}$.

(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) $\|\phi(f)\|_{\#} = \|f\|_{\cdot\infty}$. [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

$$ \|P(A)\|_{\#op} = \|P(x)\|_{C^\#(\sigma(A))} = \sup_{\lambda \in \sigma(A)} |P(\lambda)|. \qquad (4.1.4) $$
**Definition 2.7.** The Schwartz space, $S_{fin}^{\#}(*\mathbb{R}_{c,fin}^{\#n})$, is the space of all $*\mathbb{R}_{c,fin}^{\#n}$- valued #-smooth functions that are rapidly decreasing at #-infinity along with all partial #-derivatives any finite order $1 \le m < \infty$.

Thus

$\phi : *\mathbb{R}_{c}^{\#n} \to \mathbb{R}_{c}^{\#}$ is in the Schwartz space provided that any #-derivative of $\phi$, multiplied with any power of $|x|$, #-converges to 0 as $|x| \to \infty$. These functions form a #-complete TVS with a suitably defined family of seminorms. More precisely, for any multi-indices $\alpha$ and $\beta$ define:

$$p_{\alpha,\beta}(\phi) = \sup_{x \in *\mathbb{R}_{c}^{\#n}} |x^{\alpha} \partial^{\#\beta} \phi(x)|. \qquad (2.1)$$

## § 2.3. Non-Archimedean tempered distributions $S^{\#}(*\mathbb{R}_{c}^{\#n})$.

A non-Archimedean tempered distribution is a distribution $u \in D'(*\mathbb{R}_{c}^{\#n})$ that does not "grow too fast" – at most polynomial (or tempered) growth – at #-infinity in all directions; in particular it is only defined on $*\mathbb{R}_{c}^{\#n}$, not on any #-open subset. Formally, a tempered distribution is a #-continuous linear functional on the Schwartz space $S^{\#}(*\mathbb{R}_{c}^{\#n})$ of smooth functions with #-rapidly decreasing #-derivatives. The space of tempered distributions (with its natural topology) is denoted $S^{\#}(*\mathbb{R}_{c}^{\#n})$. Every #-compactly supported distribution is a tempered distribution, yielding an inclusion $\mathcal{E}^{\#}(*\mathbb{R}_{c}^{\#n}) \hookrightarrow S^{\#}(*\mathbb{R}_{c}^{\#n})$.

## § 3. The Fourier transform on $S^{\#}(*\mathbb{R}_{c}^{\#n})$, $S_{fin}^{\#}(*\mathbb{R}_{c}^{\#n})$

We begin by defining the Fourier transform, and the inverse transform, on $S^{\#}(*\mathbb{R}_{c}^{\#n})$,

$n \in \mathbb{N}^{\#}$, the Schwartz space of $C^{\infty}^{\#}$ functions of rapid decrease.

**Definition 3.1.** Suppose $f \in S^{\#}(*\mathbb{R}_{c}^{\#n})$. The Fourier transform of $f(x)$ is the function $\hat{f}(\lambda)$

given by

$$\hat{f}(\lambda) = \frac{1}{(2\pi_{\#})^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_{c}^{\#n}} f(x) [\mathrm{Ext} - \exp(-ix \cdot \lambda)] d^{\#n}x \right), \qquad (3.1)$$

where $x \cdot \lambda = \mathrm{Ext} - \sum_{i=1}^{n} x_i \lambda_i$. The inverse Fourier transform of $f$, denoted by $\check{f}$, is the function

$$\check{f}(\lambda) = \frac{1}{(2\pi_{\#})^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_{c}^{\#n}} f(x) [\mathrm{Ext} - \exp(ix \cdot \lambda)] d^{\#n}x \right). \qquad (3.2)$$

We will usually write $\hat{f} = \mathcal{F}[f]$ and $\check{f} = \mathcal{F}^{-1}[f]$.

Since every function in Schwartz space is in $\mathcal{L}_{1}^{\#}(*\mathbb{R}_{c}^{\#n})$, the above integrals (1.1) and (1.2) make sense.

We will use the standard multi-index notation. A multi-index $\alpha = \langle \alpha_1, \dots, \alpha_n \rangle, n \in \mathbb{N}^{\#}$ is an $n$-tuple of nonnegative integers. The collection of all multi-indices will be denoted by $I_{+}^{n}$. The symbols $|\alpha|, x^{\alpha}, D^{\#\alpha}$, and $x^2$ are defined as follows:
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 $P(x) = \text{Ext}-\sum_{n=1}^{N} c_n x^n$, $N \in *\mathbb{N}$. 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

$$
(P(A) - \mu)^{-1} = a^{-1} \left( \operatorname{Ext} - \prod_{i=1}^{n} (A - \lambda_i)^{-1} \right) \quad (4.1.6)
$$

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

$$
\begin{align}
\|P(A)\|_{\#}^2 &= \|P(A)^* P(A)\|_{\#} = \|(\bar{P}P)(A)\|_{\#} = \nonumber \\
&= \sup_{\lambda \in \sigma((\bar{P}P)(A))} |\lambda| = \sup_{\lambda \in \sigma(A)} |\bar{P}P(\lambda)| = \left( \sup_{\lambda \in \sigma(A)} |P(\lambda)| \right)^2. \tag{4.1.8}
\end{align}
$$

**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 $\mathbf{I}$, 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 $g$ $*\mathbb{R}_c^\#$-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 an 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.
$$|\alpha| = \mathrm{Ext} - \sum_{i=1}^{n} \alpha_i$$

$$x^{\alpha} = \mathrm{Ext} - \prod_{i=1}^{n} x_i^{\alpha_i} \quad \text{or} \quad \mathrm{Ext} - (x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_n^{\alpha_n}) \quad \text{or} \quad \text{simbolically} \quad x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_n^{\alpha_n}$$

$$D^{\#\alpha} f(x) = \mathrm{Ext} - \prod_{i=1}^{n} \frac{\partial^{\#\alpha_i} f(x)}{\partial^{\#} x^{\alpha_i}} \quad \text{or simbolically} \quad D^{\#\alpha} f(x) = \frac{\partial^{\#|\alpha|} f(x)}{\partial^{\#} x^{\alpha_1} \partial^{\#} x^{\alpha_2} \cdots \partial^{\#} x^{\alpha_n}} \quad (3.3)$$

$$x^2 = \mathrm{Ext} - \sum_{i=1}^{n} x_i^2.$$

**Lemma 1.1.** The maps $f \mapsto \hat{f}$ and $f \mapsto \check{f}$ are #-continuous linear transformations of $S^{\#}(\mathbb{R}_c^{\#n})$ into $S^{\#}(\mathbb{R}_c^{\#n})$. Furthermore, if $\alpha$ and $\beta$ are multi-indices, then

$$((i\lambda)^\alpha D^{\#\beta} \hat{f})(\lambda) = \overline{D^{\#\alpha} ((-ix)^\beta f(x))}(\lambda). \quad (3.4)$$

**Proof** The map $f \mapsto \hat{f}$ is clearly linear. Since

$$((i\lambda)^\alpha D^{\#\beta} \hat{f})(\lambda) =$$

$$\frac{1}{(2\pi_\#)^{n/2}} \left( \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} (\lambda^\alpha)(-ix)^\beta f(x) [\mathrm{Ext}\text{-exp}(-ix\cdot\lambda)] f(x) d^{\#n}x \right) =$$

$$\frac{1}{(2\pi_\#)^{n/2}} \left( \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} \frac{1}{(-i)^\alpha} (D_x^{\#\alpha} [\mathrm{Ext}\text{-exp}(-ix\cdot\lambda)]) (-ix)^\beta f(x) d^{\#n}x \right) = \quad (3.5)$$

$$\frac{(-i)^\alpha}{(2\pi_\#)^{n/2}} \left( \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n} x}^{\#\beta} [\mathrm{Ext}\text{-exp}(-ix\cdot\lambda)] D_x^{\#\alpha} ((-ix)^\beta f(x)) d^{\#n}x \right).$$

We conclude that

$$\|\hat{f}\|_{\alpha,\beta} = \sup_{\lambda \in \mathbb{R}_c^{\#n}} |\lambda^\alpha (D^{\#\beta} \hat{f})(\lambda)| \le \frac{1}{(2\pi_\#)^{n/2}} \left( \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} |D_x^{\#\alpha} (x^\beta f(x))| d^{\#n}x \right) < \infty^\# \quad (3.6)$$

so $f \mapsto \hat{f}$ takes $S^{\#}(\mathbb{R}_c^{\#n})$ into $S^{\#}(\mathbb{R}_c^{\#n})$, and we have also proven (1.4). Furthermore, if $k$ is large enough, $\int (1+x^2)^{-k} d^{\#n}x < \infty^\#$ so that

$$\|\hat{f}\|_{\alpha,\beta} \le \frac{1}{(2\pi_\#)^{n/2}} \left( \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} \frac{(1+x^2)^{-k}}{(1+x^2)^{-k}} |D_x^{\#\alpha} ((-ix)^\beta f(x))| d^{\#n}x \right) \le \quad (3.7)$$

$$\frac{1}{(2\pi_\#)^{n/2}} \left( \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} (1+x^2)^{-k} d^{\#n}x \right) \sup_{x \in \mathbb{R}_c^{\#n}} \left\{ (1+x^2)^{+k} |D_x^{\#\alpha} ((-ix)^\beta f(x))| \right\}.$$

Using generalized Leibnitz's rule we easily conclude that there exist multi-indices $\alpha_j, \beta_j$ and constants $c_j$ so that

$$\|\hat{f}\|_{\alpha,\beta} \le \sum_{j=1}^{M} c_j \|f\|_{\alpha_j,\beta_j}. \quad (1.8)$$
**Proposition 4.1.1.** Suppose that $\phi: C^{\#}(X) \to \mathcal{L}(H^{\#})$ is an algebraic $*$-homomorphism, $X$ a $\#$-compact metric space. Then

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

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

**Definition 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

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

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

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

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

$$ B_n^{\#}(x, f) = \mathrm{Ext}-\sum_{k=0}^{n} f\left(\frac{k}{n}\right) \binom{n}{k} x^k (1-x)^{n-k}. \qquad (4.1.9) $$
Thus the map $f \mapsto \hat{f}$ is bounded and therefore #-continuous. The proof for $f \mapsto \tilde{f}$ is the same.

**Theorem 1.1.** (Generalized Fourier inversion theorem) The Fourier transform (3.1) is a linear bicontinuous bijection from $S^{#}(*\mathbb{R}_c^{\#n})$ onto $S^{#}(*\mathbb{R}_c^{\#n})$. Its inverse map is the inverse Fourier transform, i.e., $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$ and $\mathcal{F}(\mathcal{F}^{-1}[f]) = f$.

**Proof.** We will prove that $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$. The proof that $\mathcal{F}(\mathcal{F}^{-1}[f]) = f$ is similar. $\mathcal{F}(\mathcal{F}^{-1}[f]) = f$ implies that $\mathcal{F}[f]$ is surjective and $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$ implies that $\mathcal{F}[f]$ is injective. Since $\mathcal{F}[f]$ and $\mathcal{F}^{-1}[f]$ are #-continuous maps of $S^{#}(*\mathbb{R}_c^{\#n})$ onto $S^{#}(*\mathbb{R}_c^{\#n})$, it is sufficient to prove that $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$ for $f$ contained in the dense set $C_0^{\infty}(*\mathbb{R}_c^{\#n})$. Choose $\varepsilon \approx 0$ infinite small enough so that the support of $f$ is contained in $C_\varepsilon$. Let $\mathbf{K}_\varepsilon = \{\mathbf{k} \in *\mathbb{R}_c^{\#n} | \text{ each } k_i/\varepsilon\pi_\# \in \mathbf{k} \text{ is an integer } \}$

$$f(x) = \mathrm{Ext} - \sum_{\mathbf{k} \in \mathbf{K}_{\varepsilon}} \left( \left( \frac{1}{2} \varepsilon \right)^{n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)], f \right) \left( \frac{1}{2} \varepsilon \right)^{n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)] \quad (3.9)$$

is just the hyper infinite Fourier series of $f$ which #-converges uniformly in $C_\varepsilon$ to $f$ since $f$ is #-continuously #-differentiable. Thus

$$f(x) = \mathrm{Ext} - \sum_{\mathbf{k} \in \mathbf{K}_{\varepsilon}} \frac{\hat{f}(k)[\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)]}{(2\pi\#)^{n/2}} (\varepsilon\pi_{\#})^n. \quad (3.10)$$

Since $*\mathbb{R}_c^{\#n}$ is the disjoint union of the cubes of volume $(\varepsilon\pi_\#)^n$ centered about the points in $\mathbf{K}_\varepsilon$, the right-hand side of (1.10) is just a hyper finite Riemann sum for the integral of the function $\hat{f}(k)[\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)]$. By the lemma 3.1, $\hat{f}(k)[\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)] \in S^{#}(*\mathbb{R}_c^{\#n})$, so the hyperfinite Riemann sums (1.10) #-converge to the integral. Thus $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$.

**Corollary 3.1.** Suppose $f \in S^{#}(*\mathbb{R}_c^{\#n})$. Then

$$\mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} |f(x)|^2 d^{\#n}x = \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} |f(k)|^2 d^{\#n}k. \quad (3.11)$$

**Proof.** This is really a corollary of the proof rather than the statement of Theorem 1.1. If $f$ has #-compact support, then for $\varepsilon \approx 0$ small enough,

$$f(x) = \mathrm{Ext} - \sum_{\mathbf{k} \in \mathbf{K}_\varepsilon} \left( \left( \frac{1}{2} \varepsilon \right)^{n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)], f \right) \left( \frac{1}{2} \varepsilon \right)^{n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)] \quad (3.12)$$

Since $\left\{ \left( \frac{1}{2} \varepsilon \right)^{n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)] \right\}_{\mathbf{k} \in \mathbf{K}_\varepsilon}$ is an orthonormal basis for $\mathcal{L}_2^\#(C_\varepsilon)$,

$$\mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} |f(x)|^2 d^{\#n}x = \mathrm{Ext} - \int_{C_\varepsilon} |f(x)|^2 d^{\#n}x = \sum_{\mathbf{k} \in \mathbf{K}_\varepsilon} \left| \left( \frac{1}{2} \varepsilon \right)^{n/2} ([\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)], f(x)) \right|^2 = \\ \sum_{\mathbf{k} \in \mathbf{K}_\varepsilon} \left| \hat{f}(k) \right|^2 (\varepsilon\pi_\#)^n \underset{\varepsilon \to \#}{\to} \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} |f(k)|^2 d^{\#n}k. \quad (3.13)$$

This proves the corollary for $f \in C_0^{\infty}(*\mathbb{R}_c^{\#n})$. Since $f \mapsto \hat{f}$ and $\|\cdot\|_2$ are #-continuous on $S^{#}(*\mathbb{R}_c^{\#n})$ and $C_0^{\infty}(*\mathbb{R}_c^{\#n})$ is #-dense, the result holds for all of $S^{#}(*\mathbb{R}_c^{\#n})$.

**Definition 3.2.** Let $T \in S^{\#}(*\tilde{\mathbb{R}}_c^{\#n})$ the Fourier transform of $T$, denoted by $\hat{T}$ or $\mathcal{F}[T]$, is the tempered distribution defined by $\hat{T}(\varphi) = T(\hat{\varphi})$.

Suppose that $h, \varphi \in S^{#}(*\mathbb{R}_c^{\#n})$, then by the polarization identity and the corollary to
**Lemma 4.1.3.** For any $n \in *\mathbb{N}$ :

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

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

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

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

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

By the generalized binomial theorem we have

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

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

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

Thus

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

and therefore

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

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

$$ \mathrm{Ext}-\sum_{k=0}^{n} \binom{n}{k} k x^k (1-x)^{n-k} = nx. \qquad (4.1.18) $$

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

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

Thus

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

and therefore

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

Replacing y by x and z by $1-x$ in the expression (4.1.22) we have identity
**Theorem 1.1** we have $(h, \varphi) = (\hat{h}, \hat{\varphi})$. Substituting $\overline{\mathcal{F}[g]} = \mathcal{F}^{-1}[\overline{g}]$ for $h$, we obtain
$$T_{\hat{g}}(\varphi) = \mathrm{Ext}-\int_{*\tilde{\mathbb{R}}_c^{\#n}} \hat{g}(x)\varphi(x)d^{\#n}x = \mathrm{Ext}-\int_{*\tilde{\mathbb{R}}_c^{\#n}} g(x)\hat{\varphi}(x)d^{\#n}x = T_g(\hat{\varphi}) = \hat{T}_g(\varphi).$$

where $T_{\hat{g}}$ and $T_g$ are the distributions corresponding to the functions $\hat{g}$ and $g$
respectively. This shows that the Fourier transform on $S^{\#'}(*\tilde{\mathbb{R}}_c^{\#n})$ extends the
transform we previously defined on $S^{\#}(*\tilde{\mathbb{R}}_c^{\#n})$.

**Theorem 3.2.** The Fourier transform is a one-to-one linear bijection from $S^{\#'}(*\tilde{\mathbb{R}}_c^{\#n})$ to $S^{\#'}(*\tilde{\mathbb{R}}_c^{\#n})$ which is the unique weakly #-continuous extension of the Fourier transform on $S^{\#}(*\tilde{\mathbb{R}}_c^{\#n})$.

**Proof.** If hyper infinite sequence {$\varphi_n$}$_{n \in \mathbb{N}^\#}$ #-convergence to $\varphi \in S^\#$, then by
Theorem 1.1, hyper infinite sequence {$\hat{\varphi}_n$}$_{n \in \mathbb{N}^\#}$ #-convergence to $\hat{\varphi} \in S^\#$, so
$T(\hat{\varphi}_n) \to_\# T(\hat{\varphi})$ for each $T \in S^\#$. Thus #-lim$_{n \to \infty^\#}$ $T(\hat{\varphi}_n) = T(\hat{\varphi})$, which shows that
$T$ is a #-continuous linear functional on $S^\#$. Furthermore, if $T_n \xrightarrow{w} T$, then $\hat{T}_n \xrightarrow{w} \hat{T}$
because $T(\hat{\varphi}_n) \to_\# T(\hat{\varphi})$ implies $\hat{T}(\varphi_n) \to_\# \hat{T}(\varphi)$. Thus $T \mapsto \hat{T}$ is weakly #-continuous.

**Definition 3.3.** Suppose that $f,g \in S^{\#}(*\tilde{\mathbb{R}}_c^{\#n})$. Then the convolution of $f$ and $g$, denoted by $f * g$, is the function

$$
(f * g)(y) = \mathrm{Ext}-\int_{*\tilde{\mathbb{R}}_c^{\#n}} f(y-x)g(x)d^{\#n}x. \quad (3.14)
$$

Convolutions frequently occur when one uses the Fourier transform because the
Fourier transform takes products into convolutions.

**Theorem 3.3.** (a) For each $f \in S^{\#}(*\mathbb{R}_c^{\#n})$, $g \mapsto f * g$ is a #-continuous map of $S^{\#}(*\mathbb{R}_c^{\#n})$ into $S^{\#}(*\mathbb{R}_c^{\#n})$.

(b) $\hat{f}g = (2\pi\#)^{-n/2}\hat{f} * \hat{g}$ and $\hat{f} * \hat{g} = (2\pi\#)^{n/2}\hat{f}\hat{g}$.

(c) For $f,g,h \in S^{\#}(*\mathbb{R}_c^{\#n})$, $f * g = g * f$ and $f * (g * h) = (f * g) * h$.

**Definition 3.4.** Suppose that $f \in S^{\#}(*\mathbb{R}_c^{\#n})$, $T \in S^{\#'}(*\mathbb{R}_c^{\#n})$ and let $\tilde{f}(x)$ denote the function, $f(-x)$. Then, the convolution of $T$ and $f$ denoted $T * f$ is the distribution in, $S^{\#'}(*\tilde{\mathbb{R}}_c^{\#n})$ given by $(T * f)(\varphi) = T(\tilde{f} * \varphi)$ for all $\varphi \in S^{\#}(*\mathbb{R}_c^{\#n})$.

The fact that $g \to \tilde{f} * g$ is a #-continuous transformation guarantees that
$T * f \in S^{\#'}(*\tilde{\mathbb{R}}_c^{\#n})$.

**Abbreviation 3.1.** Let $f_y$ denote the function $f_y(x) = f(x-y)$ and $\tilde{f}_y$ the function $f(y-x)$. When $f$ is given by a longe expression $(\cdots)$, we will sometimes write $(\cdots)^{\sim}$ rather than $(\cdot \succ \cdot)$.

**Theorem 3.4.** For each $f \in S^{\#}(*\tilde{\mathbb{R}}_c^{\#n})$ the map $T \to T * f$ is a weakly #-continuous map of $S^{\#'}(*\tilde{\mathbb{R}}_c^{\#n})$ into $S^{\#'}(*\tilde{\mathbb{R}}_c^{\#n})$ which extends the convolution on $S^{\#}(*\tilde{\mathbb{R}}_c^{\#n})$.
Furthermore,

(a) $T * f$ is a polynomially bounded $C^{\infty}$ function. In fact, $(T * f)(y) = T(\tilde{f}_y)$ and
$D^{\#\beta}(T * f) = (D^{\#\beta}T) * f = T * D^{\#\beta}f$;

(b) $(T * f) * g = T * (f * g)$;

(c) $\hat{T} * \hat{f} = (2\pi\#)^{n/2}\hat{f}\hat{T}$.

**Theorem 3.5.** Let $T \in S^{\#}(*\tilde{\mathbb{R}}_c^{\#n})$ and $f \in S^{\#}(*\tilde{\mathbb{R}}_c^{\#n})$. Then $\hat{f}T \in O_M^n$ and
$\hat{f}T(k) = (2\pi\#)^{n/2}T(f[\mathrm{Ext}-\exp(-ik \cdot x)])$. In particular, if $T$ has #-compact support and
$$ \mathrm{Ext} - \sum_{k=0}^{n} \binom{n}{k} k^2 x^k (1-x)^{n-k} = nx + n(n-1)x^2. \quad (4.1.23) $$

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

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

From the identity (4.1.24) one obtains that

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

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

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

and

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

From (4.1.9) we obtain

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

From (4.1.9) we obtain

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

Finally we obtain

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

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

**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
$\psi \in S^{\#}(\tilde{\mathbb{R}}_{c}^{\#n})$ is identically one on a \#-neighborhood of the support of $T$, then

$$ \hat{T}(k) = (2\pi_{\#})^{n/2} T(\psi[\mathrm{Ext}\text{-exp}(-ik \cdot x)]). \qquad (3.15) $$

**Proof** By Theorem 3.4.c and the Fourier inversion formula we have

$$ \hat{fT} = (2\pi_{\#})^{n/2} \hat{f} * \hat{T}. \text{ Thus } \hat{fT} \in O_{M}^{n} \text{ and } \hat{fT}(k) = (2\pi_{\#})^{n/2} \hat{T}(\tilde{f}_{k}) = (2\pi_{\#})^{n/2} T(f[\mathrm{Ext}\text{-exp}(-ik \cdot x)]). $$

**Remark 3.1.** We remark that one can also define the convolution of a distribution $T \in \mathcal{D}^{\#}(\mathbb{R}_{c}^{\#n})$ with an $f \in \mathcal{D}^{\#}(\mathbb{R}_{c}^{\#n})$ by $(T * f)(y) = T(\tilde{f}_{y})$.

**Definition 3.5.** Let $j(x)$ be a positive $C^{\infty}$ function whose support lies in the sphere of radius one about the origin in $\mathbb{R}_{c}^{\#n}$ and which satisfies $\mathrm{Ext}-\int_{\mathbb{R}_{c}^{\#n}} j(x) d^{\#n}x = 1$. The function $j_{\varepsilon}(x) = \varepsilon^{-n}j(x/\varepsilon)$, $\varepsilon \approx 0$ is called an approximate identity.

**Proposition 3.1.** Suppose $T \in S^{\#}(\mathbb{R}_{c}^{\#n})$ and let $j_{\varepsilon}(x)$ be an approximate identity. Then $T * j_{\varepsilon}(x) \to_{\#} T$ weakly as $\varepsilon \to_{\#} 0$.

**Proof.** If $\varphi \in S^{\#}(\mathbb{R}_{c}^{\#n})$, then $(T * j_{\varepsilon})(\varphi) = T(\tilde{j}_{\varepsilon} * \varphi)$, so it is sufficient to show that $\tilde{j}_{\varepsilon} * \varphi \to_{\#} \varphi$ in $S^{\#}(\mathbb{R}_{c}^{\#n})$. To do this it is sufficient to show that $(2\pi_{\#})^{n/2} \tilde{j}_{\varepsilon} \hat{\varphi} \to_{\#} \hat{\varphi}$ in $S^{\#}(\mathbb{R}_{c}^{\#n})$. Since $\tilde{j}_{\varepsilon}(\lambda) = j(\varepsilon\lambda)$ and $j(0) = (2\pi_{\#})^{n/2}$, it follows that $(2\pi_{\#})^{n/2} \tilde{j}_{\varepsilon}(x)$ \#-converges to 1 uniformly on \#-compact sets and is uniformly bounded. Similarly, $D^{\#\alpha} \tilde{j}_{\varepsilon}$ \#-converges uniformly to zero. We conclude that $(2\pi_{\#})^{n/2} \tilde{j}_{\varepsilon} \hat{\varphi} \to_{\#} \hat{\varphi}$.

**Theorem 3.6** (The generalized Plancherel theorem) The Fourier transform extends uniquely to a unitary map of $L_{2}^{\#}(\mathbb{R}_{c}^{\#n})$ onto $L_{2}^{\#}(\mathbb{R}_{c}^{\#n})$. The inverse transform extends uniquely to its adjoint.

**Proof** The corollary to Theorem 3.1 states that if $f \in S^{\#}(\mathbb{R}_{c}^{\#n})$, then $\|f\|_{2} = \|\hat{f}\|_{2}$. Since $\mathcal{F}[S^{\#}] = S^{\#}$ is a surjective isometry on $L_{2}^{\#}(\mathbb{R}_{c}^{\#n})$.

**Theorem 3.7** (The generalized Riemann-Lebesgue lemma) The Fourier transform extends uniquely to a bounded map from $L_{1}^{\#}(\mathbb{R}_{c}^{\#n})$ into $C^{\infty}(\mathbb{R}_{c}^{\#n})$, the \#-continuous functions vanishing at $\infty$.

**Proof** For $f \in S^{\#}(\mathbb{R}_{c}^{\#n})$, we know that $\hat{f} \in S^{\#}(\mathbb{R}_{c}^{\#n})$ and thus $\hat{f} \in C^{\infty}(\mathbb{R}_{c}^{\#n})$. The estimate is trivial. The Fourier transform is thus a bounded linear map from a \#-dense set of $L_{1}^{\#}(\mathbb{R}_{c}^{\#n})$ into $C^{\infty}(\mathbb{R}_{c}^{\#n})$. By the generalized B.L.T. theorem, extends uniquely to a bounded linear transformation of $C^{\infty}(\mathbb{R}_{c}^{\#n})$ into $C^{\infty}(\mathbb{R}_{c}^{\#n})$.

**Remark 3.2.** We remark that the Fourier transform takes $L_{1}^{\#}(\mathbb{R}_{c}^{\#n})$ into, but not onto $C^{\infty}(\mathbb{R}_{c}^{\#n})$.

A simple argument with test functions shows that the extended transform on $L_{1}^{\#}(\mathbb{R}_{c}^{\#n})$ and $L_{2}^{\#}(\mathbb{R}_{c}^{\#n})$ is the restriction of the transform on $S^{\#}(\mathbb{R}_{c}^{\#n})$, but it is useful to have an explicit integral representation. For $f \in L_{1}^{\#}(\mathbb{R}_{c}^{\#n})$, this is easy since we can find $f_{m} \in S^{\#}(\mathbb{R}_{c}^{\#n})$ so that $\#\lim_{m \to \infty} \|f - f_{m}\|_{1} = 0$. Then, for each $\lambda$,
$\|Tz\|_{\#} \le C \|z\|_{\#}$ 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 $\psi \in \mathbf{H}^\#$. 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

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

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

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

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

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

**Theorem 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,
$$
\begin{equation}
\begin{split}
f(\lambda) = \#\text{-lim}_{m\to\infty} & (\hat{f}_m(\lambda)) = \\
& \#\text{-lim}_{m\to\infty} \left\{ \frac{1}{(2\pi\#)^{n/2}} \left( \text{Ext-} \int_{\mathbb{R}_c^{\#n}} [\text{Ext-exp}(-ik \cdot x)] f_m(x) d^\#x \right) \right\} = \\
& \qquad \frac{1}{(2\pi\#)^{n/2}} \left( \text{Ext-} \int_{\mathbb{R}_c^{\#n}} [\text{Ext-exp}(-ik \cdot x)] f(x) d^\#x \right).
\end{split}
\tag{3.16}
\end{equation}
$$

So, the Fourier transform of a function in $\mathcal{L}_1^{\#}(\ast\mathbb{R}_c^{\#n})$ is given by the usual formula.
Next, suppose $f \in \mathcal{L}_2^{\#}(\ast\mathbb{R}_c^{\#n})$ and let

$$
\chi_R(x) = \begin{cases} 1 & \text{if } |x| \le R \\ 0 & \text{if } |x| > R \end{cases} \tag{3.17}
$$

Then $\chi_R f \in \mathcal{L}_1^{\#}(\ast\mathbb{R}_c^{\#n})$ and $\#\text{-lim}_{R\to\infty} \chi_R f = f$ in $\mathcal{L}_2^{\#}$, so by the generalized Plancherel theorem $\widehat{\#\text{-lim}_{R\to\infty} \chi_R f} = \hat{f}$ in $\mathcal{L}_2^{\#}$. Thus

$$
f(\lambda) = \#\text{-lim}_{R \to \infty} \frac{1}{(2\pi\#)^{n/2}} \left( \text{Ext} - \int_{|x| \le R} [\text{Ext-exp}(-ik \cdot x)] f(x) d^\#x \right) \quad (3.18)
$$

where by $\#\text{-lim}_{R\to\infty}$ we mean the $\#\text{-limit}$ in the $\mathcal{L}_2^\#$-norm. Sometimes we will dispense
with $|x| \le R$ and just write

$$
f(\lambda) = \#\text{-lim}_{R \to \infty} \frac{1}{(2\pi\#)^{n/2}} \left( \text{Ext} - \int [\text{Ext-exp}(-ik \cdot x)] f(x) d^\#x \right) \quad (3.19)
$$

for functions $f \in \mathcal{L}_2^\#(\ast\mathbb{R}_c^{\#n})$.

We have proven above that $\mathcal{F} : \mathcal{L}_2^{\#}(\ast\mathbb{R}_c^{\#n}) \to \mathcal{L}_2^{\#}(\ast\mathbb{R}_c^{\#n})$ and $\mathcal{F} : \mathcal{L}_1^{\#}(\ast\mathbb{R}_c^{\#n}) \to \mathcal{L}_{\infty}^{\#}(\ast\mathbb{R}_c^{\#n})$
and in both cases is a bounded operator.

**Theorem 3.8 (Generalized Hausdorff-Young inequality)** Suppose $1 \le q \le 2$, and $p^{-1} + q^{-1} = 1$. Then the Fourier transform is a bounded map of $\mathcal{L}_p^{\#}(\ast\mathbb{R}_c^{\#n})$ to $\mathcal{L}_q^{\#}(\ast\mathbb{R}_c^{\#n})$ and its norm is less than or equal to $(2\pi\#)^{n(1/2-1/q)}$. distribution in $\mathcal{D}^{\#}(\ast\mathbb{R}_c^{\#n})$.

§ 4.Schwartz space $S^\#(\tilde{\mathbb{R}}_c^{\#n})$.

**Definition 4.1.** A function $f : X \to \tilde{\mathbb{R}}_c^\#$ defined on some set $X$ is called *finitely bounded* (or *bounded*) if the set of its values is finitely bounded, i.e., $f(X) \subset [a, b]$ where $a, b \in \tilde{\mathbb{R}}_{c, \text{fin}}^\#$. In other words, there exists a finite hyperreal number $M \in \tilde{\mathbb{R}}_{c, \text{fin}}^\#$ such that

$$
|f(X)| \leq M. \tag{4.1}
$$

**Definition 4.2.** A function $f : X \to \tilde{\mathbb{R}}_c^\#$ defined on some set $X$ is called *hyper finitely bounded* (or *hyper bounded*) if the set of its values is hyper finitely bounded, i.e., $f(X) \subset [a, b]$ where $a, b \in \tilde{\mathbb{R}}_c^\# \setminus \tilde{\mathbb{R}}_{c, \text{fin}}^\#$. In other words, there exists a hyperfinite hyperreal number $M \in \tilde{\mathbb{R}}_c^\# \setminus \tilde{\mathbb{R}}_{c, \text{fin}}^\#$ such that $|f(X)| \le M$.

**Definition 4.3.** For $n \in \mathbb{N}^{\#}$, an $\#\text{-integrable function } \phi : \tilde{\mathbb{R}}_c^{\#n} \to \tilde{\mathbb{R}}_c^{\#}$ is called $\#\text{-rapidly}$ decreasing if for all $\alpha \in \mathbb{N}^{\#n}$ the product function $x \mapsto x^{\alpha}\phi(x)$ is a finitely bounded 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 $B^{#}(\ast\mathbb{R}_c^{#})$ is the smallest family closed under #-limits of form (d) containing all of $C^{#}(\ast\mathbb{R}_c^{#})$, we know that any $\hat{\phi}(f)$ is in the Smallest non Archimedean $C^*$-algebra containing $A$ which is also strongly #-closed; such an algebra is called a von Neumann #-algebra or non Archimedean $W^*$-algebra. When we study von Neumann #-algebras we will see that this follows from (g).

(iii) The #-norm equality of Theorem 4.2.1 carries over if we define $\|f\|_{\infty}^{\prime}$ to be the $L_{\infty}^{\#}$ #-norm with respect to a suitable notion of "#-almost everywhere." Namely, pick an orthonormal basis $\{\psi_n\}_{n=1}^{\infty}$ and say that a property is true #-a.e. if it is true #-a.e. with respect to each $\mu_{\psi_n}^{\#}$ Then $\|\hat{\phi}(f)\|_{L(\mathbb{H}^{\#})} = \|f\|_{\infty}^{\prime}$.

**Definition 4.2.2.** A vector $\psi \in \mathbb{H}^\#$ is called a cyclic vector for $A$ if hyperfinite linear combinations of the elements $\{A^n\psi\}_{n=0}^\infty$ are #-dense in $\mathbb{H}^\#$. Not all operators have cyclic vectors, but if they do.

**Lemma 4.2.1.** Let $A$ be a bounded in $\ast\mathbb{R}_c^\#$ self-#-adjoint operator with cyclic vector $\psi$. Then, there is a unitary operator $U : \mathbb{H}^\# \to L_2^\#(\sigma(A), d^\#\mu_\psi^\#)$, 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
$$ \|\phi(f)\|_{\#}^2 = \langle\psi, \phi^*(f)\phi(f)\rangle_{\#} = \langle\psi, \phi(\tilde{f} \times f)\rangle_{\#} = \text{Ext}-\int|f(\lambda)|^2 d^\#\mu_\psi^\# $$

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 $\#-\{\phi(f)\psi|f \in C^\#(\sigma(A))\} = \mathbb{H}^\#$ so by the generalized B.L.T. theorem $U$ extends to an #-isometric map of $\mathbb{H}^\#$ into $L_2^\#(\sigma(A), d^\#\mu_\psi^\#)$. Since $C^\#(\sigma(A))$ is #-dense in $L_2^\#$, $\text{Ran } U = L_2^\#(\sigma(A), d^\#\mu_\psi^\#)$. 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 $\mathbb{H}^\#$ 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 $\mathbb{H}^\#$.

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

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

(a) $A$ leaves each $\mathbb{H}_n^\#$ invariant, that is, $\psi \in \mathbb{H}_n^\#$ implies $A\psi \in \mathbb{H}_n^\# = \text{Ext}-\bigoplus_{n=1}^\infty \mathbb{H}_n^\#$ so that:

(b) For each $n \in \aleph$, there is a $\phi_n \in \mathbb{H}_n^\#$ which is cyclic for $A \upharpoonright \mathbb{H}_n^\#$, i.e. $\mathbb{H}_n^\# = \#-\{f(A)\phi_n|f \in C^\#(\sigma(A))\}$

**Theorem 4.2.3 (spectral theorem-multiplication operator form)** Let $A$ be a bounded in $\ast\mathbb{R}_c^\#$ self-#-adjoint operator on $\mathbb{H}^\#$, a $\ast$-separable Hilbert space. Then, there exist #-measures $\{\mu_n^\#\}_{n=1}^N$ with $N \in \aleph$ or $\{\mu_n^\#\}_{n=1}^\infty$ on $\sigma(A)$ and a
hyper finitely bounded function.

**Remark 4.1.** If $\phi$ is a #-rapidly decreasing function, then its integral exists

$$ \operatorname{Ext-} \int_{*\tilde{\mathbb{R}}_{c}^{\#n}} \phi(x) d^{\#n}x < \infty^{\#} \qquad (4.2) $$

In fact for all $\alpha \in \mathbb{N}^{\#n}$ the integral of $x \mapsto x^{\alpha}\phi(x)$ exists

$$ \operatorname{Ext-} \int_{*\tilde{\mathbb{R}}_{c}^{\#n}} x^{\alpha}\phi(x) d^{\#n}x < \infty^{\#}. \qquad (4.3) $$

**Definition 4.4.** The Schwartz space, $S^#(*\tilde{\mathbb{R}}_{c}^{\#n})$, is the space of all #-smooth functions in $C^{\#\infty#}(*\tilde{\mathbb{R}}_{c}^{\#n})$ that are rapidly decreasing at #-infinity along with all partial #-derivatives.

Thus

$\phi : *\tilde{\mathbb{R}}_{c}^{\#n} \to *\tilde{\mathbb{R}}_{c}^{\#} $ is in the Schwartz space provided that any #-derivative of $\phi$, multiplied with any power of $|x|$, #-converges to 0 as $|x| \to \infty$. These functions form a #-complete TVS with a suitably defined family of seminorms. More precisely, for any multi-indices $\alpha$ and $\beta$ define:

$$ p_{\alpha,\beta}(\phi) = \sup_{x \in *\tilde{\mathbb{R}}_{c}^{\#n}} |x^{\alpha} \partial^{\#\beta} \phi(x)|. \qquad (4.4) $$

Then $\phi$ is in the Schwartz space $S^#(*\tilde{\mathbb{R}}_{c}^{\#n})$ if all the values satisfy: $p_{\alpha,\beta}(\phi) < \infty$.

Thus

$$ S^#(*\tilde{\mathbb{R}}_{c}^{\#n}, *\tilde{\mathbb{R}}_{c}^{\#}) \triangleq \{\phi \in C^{\infty#}(*\tilde{\mathbb{R}}_{c}^{\#n}, *\tilde{\mathbb{R}}_{c}^{\#}) | \forall \alpha, \beta \in \mathbb{N}^{\#n} (p_{\alpha,\beta}(\phi) < \infty)\}. \qquad (4.5) $$

Similarly

$$ S^#(*\tilde{\mathbb{R}}_{c}^{\#n}, *\tilde{\mathbb{C}}_{c}^{\#}) \triangleq \{\phi \in C^{\infty#}(*\tilde{\mathbb{R}}_{c}^{\#n}, *\tilde{\mathbb{C}}_{c}^{\#}) | \forall \alpha, \beta \in \mathbb{N}^{\#n} (p_{\alpha,\beta}(\phi) < \infty)\}. \qquad (4.6) $$

The family of seminorms $p_{\alpha,\beta}(\cdot)$ defines a locally convex topology on the Schwartz space $S^#(*\tilde{\mathbb{R}}_{c}^{\#n})$.

For $n=1$, the seminorms are norms on the Schwartz space $S^#(*\tilde{\mathbb{R}}_{c}^{\#n})$. One can also use the following family of seminorms to define the topology:

$$ |f|_{m,k} = \sup_{|p| \le m} \left( \sup_{x \in *\tilde{\mathbb{R}}_{c}^{\#n}} \left\{ (1 + |x|)^k \left| (\partial^{\#\alpha} f)(x) \right| \right\} \right), k, m \in \mathbb{N}^{\#n}. \qquad (4.7) $$

Otherwise, one can define a norm on $S^#(*\tilde{\mathbb{R}}_{c}^{\#n})$ by

$$ \|\phi\|_k = \max_{|\alpha|+|\beta| \le k} \sup_{x \in *\tilde{\mathbb{R}}_{c}^{\#n}} |x^{\alpha} \partial^{\beta} \phi(x)|, k \ge 1. \qquad (4.8) $$

The Schwartz space $S^#(*\tilde{\mathbb{R}}_{c}^{\#n})$ is a Fréchet space (that is, a #-complete metrizable locally convex space). Because the Fourier transform changes $\partial^{\#\alpha}$ into multiplication by $x^\alpha$ and vice versa, this symmetry implies that the Fourier transform of a Schwartz function is also a Schwartz function.

**Definition 4.5.** A sequence $\{f_i\}_{i=1}^{\infty}$ #-converges to 0 in $S^#(*\tilde{\mathbb{R}}_{c}^{\#n})$ if and only if the functions $(1+|x|)^k(\partial^{\#p}f_i)(x)$ #-converge to 0 uniformly in the whole of $*\tilde{\mathbb{R}}_{c}^{\#n}$, which implies that such a sequence must converge to zero in $C^{\#\infty#}(*\tilde{\mathbb{R}}_{c}^{\#n})$.

The subset of all #-analytic Schwartz functions is #-dense in $S^#(*\tilde{\mathbb{R}}_{c}^{\#n})$

The Schwartz space is nuclear and the tensor product of two maps induces a canonical surjective TVS-isomorphisms $S^#(*\tilde{\mathbb{R}}_{c}^{\#m}) \hat{\otimes} S^#(*\tilde{\mathbb{R}}_{c}^{\#n}) \to S^#(*\tilde{\mathbb{R}}_{c}^{\#m+n})$, where $\hat{\otimes}$ represents the #-completion of the injective tensor product
unitary operator $U : \mathbf{H}^{\#} \to \bigoplus_{n=1}^{N} L_{2}^{\#}(\ast\mathbb{R}_{c}^{\#}, d^{\#}\mu_{n}^{\#})$ or $U : \mathbf{H}^{\#} \to \bigoplus_{n=1}^{\infty} L_{2}^{\#}(\ast\mathbb{R}_{c}^{\#}, d^{\#}\mu_{n}^{\#})$

so that $(UAU^{-1}\psi)_{n}(\lambda) = \lambda\psi_{n}(\lambda)$

where we write an element $\psi \in \bigoplus_{n=1}^{N} L_{2}^{\#}(\ast\mathbb{R}_{c}^{\#}, d^{\#}\mu_{n}^{\#})$ as an $N$-tuple $\langle\psi_{1}(\lambda), \dots, \psi_{N}(\lambda)\rangle$ or $*$-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 $\ast\mathbb{R}_{c}^{\#}$ self-adjoint operator on a $*$-separable Hilbert space $\mathbf{H}^{\#}$. Then there exists a finite in $\ast\mathbb{R}_{c}^{\#}$ measure space $\langle M, \mu^{\#} \rangle$, a bounded in $\ast\mathbb{R}_{c}^{\#}$ 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 $\ast\mathbb{R}_{c}^{\#}$. Define $\mu$ by requiring that its restriction to the $n$-th copy of $\ast\mathbb{R}_{c}^{\#}$ be $\mu_{n}$. Since $\mu(M) = \text{Ext}-\sum_{n=1}^{N} \mu_{n}^{\#}(\ast\mathbb{R}_{c}^{\#}) < \infty$, $\mu_{n}$ is finite in $\ast\mathbb{R}_{c}^{\#}$. 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

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

**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

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

**Definition 4.2.4.** Let $F$ be a $\ast\mathbb{R}_{c}^{\#}$-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 $\ast\mathbb{R}_{c}^{\#}$ $\ast\mathbb{R}_{c}^{\#}$-valued function on a $*$-measure space $\langle M, \mu^{\#} \rangle$. Let $T_{f}$ be the operator on $L_{2}^{\#}(M, d^{\#}\mu^{\#})$ given by $(T_{f}g)(m) = F(m)g(m)$
## § 4.2. Schwartz space $S_{\text{fin}}^{\#} (*\tilde{\mathbb{R}}_{c,\text{fin}}^{\#n})$

**Definition 4.6.** For $n \in \mathbb{N}$, an $(*\tilde{\mathbb{R}}_{c,\text{fin}}^{\#n})$-valued and $\#$-integrable function $\phi: (*\tilde{\mathbb{R}}_{c}^{\#n} \to (*\tilde{\mathbb{R}}_{c,\text{fin}}^{\#n})$ is called $\#$-rapidly decreasing if for all $\alpha \in \mathbb{N}^n$ the product function $x \mapsto x^\alpha \phi(x)$ is a finitely bounded function.

**Remark 4.2.** If $\phi$ is a $\#$-rapidly decreasing $(*\tilde{\mathbb{R}}_{c,\text{fin}}^{\#n})$-valued function, then its integral exists and finite or hyperfinite, i.e.,

$$ \operatorname{Ext-} \int_{(*\tilde{\mathbb{R}}_{c}^{\#n})} \phi(x) d^{\#n} x \in (*\tilde{\mathbb{R}}_{c,\text{fin}}^{\#n}). \qquad (4.9) $$

In fact for all $\alpha \in \mathbb{N}^n$ the integral of $x \mapsto x^\alpha \phi(x)$ exists and finite, i.e.,

$$ \operatorname{Ext-} \int_{(*\tilde{\mathbb{R}}_{c}^{\#n})} x^{\alpha} \phi(x) d^{\#n} x \in (*\tilde{\mathbb{R}}_{c,\text{fin}}^{\#n}). \qquad (4.10) $$

It follows that for all $\alpha \in \mathbb{N}^n$ and for any $R \in (*\tilde{\mathbb{R}}_{c}^{\#n} \setminus (*\tilde{\mathbb{R}}_{c,\text{fin}}^{\#n})$

$$ \operatorname{Ext-} \int_{(*\tilde{\mathbb{R}}_{c}^{\#n} \setminus B(R)} x^{\alpha} \phi(x) d^{\#n} x \approx 0 \qquad (4.11) $$

where $B(R) \triangleq \{x \in (*\tilde{\mathbb{R}}_{c}^{\#n} \mid |x| \le R)\}$

**Definition 4.7.** The Schwartz space, $S_{\text{fin}}^{\#} (*\tilde{\mathbb{R}}_{c,\text{fin}}^{\#n})$, is the space of all $(*\tilde{\mathbb{R}}_{c,\text{fin}}^{\#n})$-valued $\#$-smooth functions that are rapidly decreasing at $\#$-infinity along with all partial $\#$-derivatives any finite order $1 \le m < \infty$.

Thus

$\phi: (*\tilde{\mathbb{R}}_{c}^{\#n} \to \mathbb{R}^{\#})$ is in the Schwartz space provided that any $\#$-derivative of $\phi$, multiplied with any power of $|x|$, $\#$-converges to 0 as $|x| \to \infty$. These functions form a $\#$-complete TVS with a suitably defined family of seminorms. More precisely, for any multi-indices $\alpha$ and $\beta$ define:

$$ p_{\alpha, \beta}(\phi) = \sup_{x \in (*\tilde{\mathbb{R}}_{c}^{\#n})} |x^{\alpha} \partial^{\#\beta} \phi(x)|. \qquad (4.12) $$

## § 5. Non-Archimedean tempered distributions $S^{\#} (*\tilde{\mathbb{R}}_{c}^{\#n})$.

A non-Archimedean tempered distribution is a distribution $u \in D' (*\tilde{\mathbb{R}}_{c}^{\#n})$ that does not "grow too fast" – at most polynomial (or tempered) growth – at $\#$-infinity in all directions; in particular it is only defined on $(*\tilde{\mathbb{R}}_{c}^{\#n})$, not on any $\#$-open subset.

Formally, a tempered distribution is a $\#$-continuous linear functional on the Schwartz space $S^{\#} (*\tilde{\mathbb{R}}_{c}^{\#n})$ of smooth functions with $\#$-rapidly decreasing $\#$-derivatives. The space of tempered distributions (with its natural topology) is denoted $S^{\#} (*\tilde{\mathbb{R}}_{c}^{\#n})$.

Every $\#$-compactly supported distribution is a tempered distribution, yielding an inclusion $\mathcal{E}^{\#} (*\tilde{\mathbb{R}}_{c}^{\#n}) \hookrightarrow S^{\#} (*\tilde{\mathbb{R}}_{c}^{\#n})$.

### § 5.1 The Fourier transform on $S^{\#} (*\tilde{\mathbb{R}}_{c}^{\#n})$, $S_{\text{fin}}^{\#} (*\tilde{\mathbb{R}}_{c}^{\#n})$

We begin by defining the Fourier transform, and the inverse transform, on $S^{\#} (*\tilde{\mathbb{R}}_{c}^{\#n})$, $n \in \mathbb{N}^{\#}$, the Schwartz space of $C^{\infty}^{\#}$ functions of rapid decrease.

**Definition 5.1.** Suppose $f \in S^{\#} (*\tilde{\mathbb{R}}_{c}^{\#n})$. The Fourier transform of $f(x)$ is the function
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 $H^\#$

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

We have thus proven.

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

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

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

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

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

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

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

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

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

The sets need not be disjoint, however. The reader should be warned that $\sigma_{\text{sing}}(A)$ 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 unitarily equivalent to multiplication by $x$ on $L_2^*(\mathbb{R}_c^\#, d^\#\mu^\#)$, that is, when only one spectral #-measure is needed. An simple 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 $*\mathbb{R}_c^\#$ self-#-adjoint operator $A$ is called **multiplicity free** if and only if $A$ is unitarily equivalent to multiplication by $A$ on $L_2^*(\mathbb{R}_c^\#, d^\#\mu^\#)$ for some #-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
$\hat{f}(\lambda)$
given by

$$
\hat{f}(\lambda) = \frac{1}{(2\pi\#)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} f(x)[\mathrm{Ext}\text{-exp}(-ix \cdot \lambda)]d^{\#n}x \right), \quad (5.1)
$$

where $x \cdot \lambda = \mathrm{Ext}-\sum_{i=1}^{n} x_i \lambda_i$. The inverse Fourier transform of $f$, denoted by $\check{f}$, is the function

$$
\check{f}(\lambda) = \frac{1}{(2\pi\#)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} f(x)[\mathrm{Ext}\text{-exp}(ix \cdot \lambda)]d^{\#n}x \right). \quad (5.2)
$$

We will usually write $\hat{f} = \mathcal{F}[f]$ and $\check{f} = \mathcal{F}^{-1}[f]$.

Since every function in Schwartz space is in $L_1^\#(*\mathbb{R}_c^{\#n})$, the above integrals (1.1) and (1.2) make sense.

We will use the standard multi-index notation. A multi-index $\alpha = \langle \alpha_1, \dots, \alpha_n \rangle, n \in \mathbb{N}^\#$ is an $n$-tuple of nonnegative integers. The collection of all multi-indices will be denoted by $I_+^\alpha$. The symbols $|\alpha|, x^\alpha, D^{\#\alpha}$, and $x^2$ are defined as follows:

$$
|\alpha| = \mathrm{Ext} - \sum_{i=1}^{n} \alpha_i
$$

$$
x^\alpha = \mathrm{Ext} - \prod_{i=1}^{n} x_i^{\alpha_i} \quad \text{or} \quad \mathrm{Ext} - (x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_n^{\alpha_n}) \quad \text{or} \quad \text{simbolically} \quad x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_n^{\alpha_n}
$$

$$
D^{\#\alpha} f(x) = \mathrm{Ext} - \prod_{i=1}^{n} \frac{\partial^{\#\alpha_i} f(x)}{\partial^{\#} x^{\alpha_i}} \quad \text{or simbolically} \quad D^{\#\alpha} f(x) = \frac{\partial^{\#|\alpha|} f(x)}{\partial^{\#} x^{\alpha_1} \partial^{\#} x^{\alpha_2} \cdots \partial^{\#} x^{\alpha_n}} \quad (5.3)
$$

$$
x^2 = \mathrm{Ext} - \sum_{i=1}^{n} x_i^2.
$$

**Lemma 5.1.** The maps $f \mapsto \hat{f}$ and $f \mapsto \check{f}$ are \#-continuous linear transformations of $S^\#(*\mathbb{R}_c^{\#n})$ into $S^\#(*\mathbb{R}_c^{\#n})$. Furthermore, if $\alpha$ and $\beta$ are multi-indices, then

$$
((i\lambda)^\alpha D^{\#\beta} \hat{f})(\lambda) = \overline{D^{\#\alpha} ((-ix)^\beta f(x))}(\lambda). \quad (5.4)
$$

**Proof** The map $f \mapsto \hat{f}$ is clearly linear. Since

$$
((i\lambda)^\alpha D^{\#\beta} \hat{f})(\lambda) =
$$

$$
\frac{1}{(2\pi\#)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} (\lambda^\alpha)(-ix)^\beta f(x) [\mathrm{Ext}\text{-exp}(-ix\cdot\lambda)] f(x) d^{\#n}x \right) =
$$

$$
\frac{1}{(2\pi\#)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} \frac{1}{(-i)^\alpha} (D_x^{\#\alpha} [\mathrm{Ext}\text{-exp}(-ix\cdot\lambda)]) (-ix)^\beta f(x) d^{\#n}x \right) = \tag{5.5}
$$

$$
\frac{(-i)^\alpha}{(2\pi\#)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n} x}^{\#\beta} [\mathrm{Ext}\text{-exp}(-ix\cdot\lambda)] D_x^{\#\alpha} ((-ix)^\beta f(x)) d^{\#n}x \right).
$$

We conclude that
#-measure and the map, $U : L_1^{\#}(\ast\mathbb{R}_c^{\#}, d^{\#}v) \to L_1^{\#}(\ast\mathbb{R}_c^{\#}, d^{\#}\mu^{\#})$ 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 $v$. By the generalized Radon-Nikodym theorem, $d^{\#}v = Fd^{\#}\mu^{\#}$ with $F$ #-a.e. nonzero if and only if $v^{\#}$ and $\mu^{\#}$ have the same sets of #-measure zero. This suggests the definition:

**Definition 4.2.8.** Two #-Borel #-measures $\mu^{\#}$ and $v^{\#}$ 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 $v^{\#}$ be #-Borel #-measures on $\ast\mathbb{R}_c^{\#}$ with bounded in $\ast\mathbb{R}_c^{\#}$ support. Let $A_{\mu^{\#}}$ be the operator on $L_2^{\#}(\ast\mathbb{R}_c^{\#}, d^{\#}\mu^{\#})$ given by $(A_{\mu^{\#}}f)(\lambda) = \lambda f(\lambda)$ and similarly for $A_{v^{\#}}$ on $L_2^{\#}(\ast\mathbb{R}_c^{\#}, d^{\#}v^{\#})$. Then $A_{\mu^{\#}}$ and $A_{v^{\#}}$ are unitarily equivalent if and only if $\mu^{\#}$ and $v^{\#}$ 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 $m \in \mathbb{N}$ if $A$ is unitarily equivalent to multiplication by $\lambda$ on $\text{Ext} \oplus L_2^{\#}(\ast\mathbb{R}_c^{\#}, d^{\#}\mu^{\#})$ where there are $m$ terms in the external sum and $\mu^{\#}$ 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 $\lambda$ on $\text{Ext} \oplus L_2^{\#}(\ast\mathbb{R}_c^{\#}, d^{\#}\mu^{\#})$ ($m$ times) and on $\text{Ext} \oplus L_2^{\#}(\ast\mathbb{R}_c^{\#}, d^{\#}v)$ ($n$ 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

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

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 $\ast\mathbb{R}_c^{\#}$ self-#-adjoint operator on a Hilbert space $H^{\#}$. Then there is a decomposition $\text{Ext} \oplus H_m^{\#}$ so that
$$ \|\hat{f}\|_{\alpha, \beta} = \sup_{\lambda \in *\mathbb{R}_c^{\#n}} |\lambda^\alpha (D^{\#\beta}\hat{f})(\lambda)| \le \frac{1}{(2\pi^\#)^{n/2}} \left( \text{Ext-} \int_{*\mathbb{R}_c^{\#n}} |D_x^{\#\alpha}(x^\beta f(x))| d^{\#n}x \right) < \infty^\# \quad (5.6) $$

so $f \mapsto \hat{f}$ takes $S^#(*\tilde{\mathbb{R}}_c^{\#n})$ into $S^#(*\tilde{\mathbb{R}}_c^{\#n})$, and we have also proven (5.4). Furthermore, if $k$ is large enough, $\int (1+x^2)^{-k} d^{\#n}x < \infty^\#$ so that

$$ \begin{aligned} \|\hat{f}\|_{\alpha, \beta} &\le \frac{1}{(2\pi^\#)^{n/2}} \left( \text{Ext-} \int_{*\mathbb{R}_c^{\#n}} \frac{(1+x^2)^{-k}}{(1+x^2)^{-k}} |D_x^{\#\alpha}((-ix)^\beta f(x))| d^{\#n}x \right) \le \\ &\frac{1}{(2\pi^\#)^{n/2}} \left( \text{Ext-} \int_{*\mathbb{R}_c^{\#n}} (1+x^2)^{-k} d^{\#n}x \right) \sup_{x \in *\mathbb{R}_c^{\#n}} \left\{ (1+x^2)^{+k} |D_x^{\#\alpha}((-ix)^\beta f(x))| \right\}. \end{aligned} \quad (5.7) $$

Using generalized Leibnitz's rule we easily conclude that there exist multi-indices $\alpha_j, \beta_j$ and constants $c_j$ so that

$$ \|\hat{f}\|_{\alpha, \beta} \le \sum_{j=1}^{M} c_j \|f\|_{\alpha_j, \beta_j}. \quad (5.8) $$

Thus the map $f \mapsto \hat{f}$ is bounded and therefore #-continuous. The proof for $f \mapsto \check{f}$ is the same.

**Theorem 5.1.** (Generalized Fourier inversion theorem) The Fourier transform (3.1) is a linear bicontinuous bijection from $S^#(*\tilde{\mathbb{R}}_c^{\#n})$ onto $S^#(*\tilde{\mathbb{R}}_c^{\#n})$. Its inverse map is the inverse Fourier transform, i.e., $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$ and $\mathcal{F}(\mathcal{F}^{-1}[f]) = f$.

**Proof.** We will prove that $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$. The proof that $\mathcal{F}(\mathcal{F}^{-1}[f]) = f$ is similar. $\mathcal{F}(\mathcal{F}^{-1}[f]) = f$ implies that $\mathcal{F}[f]$ is surjective and $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$ implies that $\mathcal{F}[f]$ is injective. Since $\mathcal{F}[f]$ and $\mathcal{F}^{-1}[f]$ are #-continuous maps of $S^#(*\tilde{\mathbb{R}}_c^{\#n})$ onto $S^#(*\tilde{\mathbb{R}}_c^{\#n})$, it is sufficient to prove that $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$ for $f$ contained in the dense set $C_0^{\infty}(*\tilde{\mathbb{R}}_c^{\#n})$. Let $C_\varepsilon, \varepsilon \approx 0$ be the cube of volume $(2/\varepsilon)^n$ centered at the origin in $*\tilde{\mathbb{R}}_c^{\#n}$. Choose $\varepsilon \approx 0$ infinite small enough so that the support of $f$ is contained in $C_\varepsilon$. Let $\mathbf{K}_\varepsilon = \{\mathbf{k} \in *\tilde{\mathbb{R}}_c^{\#n} | \text{ each } k_i/\varepsilon\pi_\# \in \mathbf{k} \text{ is an integer}\}$

$$ f(x) = \text{Ext-} \sum_{\mathbf{k} \in \mathbf{K}_\varepsilon} \left( \left( \frac{1}{2} \varepsilon \right)^{n/2} [\text{Ext-} \exp(i\mathbf{k} \cdot x)], f \right) \left( \frac{1}{2} \varepsilon \right)^{n/2} [\text{Ext-} \exp(i\mathbf{k} \cdot x)] \quad (5.9) $$

is just the hyper infinite Fourier series of $f$ which #-converges uniformly in $C_\varepsilon$ to $f$ since $f$ is #-continuously #-differentiable. Thus

$$ f(x) = \text{Ext-} \sum_{\mathbf{k} \in \mathbf{K}_\varepsilon} \frac{\hat{f}(k) [\text{Ext-} \exp(i\mathbf{k} \cdot x)]}{(2\pi^\#)^{n/2}} (\varepsilon\pi^\#)^n. \quad (5.10) $$

Since $*\tilde{\mathbb{R}}_c^{\#n}$ is the disjoint union of the cubes of volume $(\varepsilon\pi^\#)^n$ centered about the points in $\mathbf{K}_\varepsilon$, the right-hand side of (1.10) is just a hyper finite Riemann sum for the integral of the function $\hat{f}(k)[\text{Ext-} \exp(i\mathbf{k} \cdot x)]$. By the lemma 3.1, $\hat{f}(k)[\text{Ext-} \exp(i\mathbf{k} \cdot x)] \in S^#(*\tilde{\mathbb{R}}_c^{\#n})$, so the hyperfinite Riemann sums (1.10) #-converge to the integral. Thus $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$.

**Corollary 5.1.** Suppose $f \in S^#(*\tilde{\mathbb{R}}_c^{\#n})$. Then
(a) A leaves each $H_m^\#^\#$ invariant.

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

(c) The \#-measure classes $\langle \mu_m^\# \rangle$ associated with the spectral representation of $A \uparrow H_m^\#$ 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 $\{P_\Omega|\Omega \text{ an arbitrary } \#\text{-Borel set}\}$ 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

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

(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

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

**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
$$ \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} |f(x)|^2 d^{\#n}x = \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} |f(k)|^2 d^{\#n}k. \qquad (5.11) $$

**Proof.** This is really a corollary of the proof rather than the statement of Theorem 1.1. If $f$ has #-compact support, then for $\varepsilon \approx 0$ small enough,

$$ f(x) = \mathrm{Ext} - \sum_{\mathbf{k} \in \mathbb{K}_{\varepsilon}} \left( \left( \frac{1}{2} \varepsilon \right)^{n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)], f \right) \left( \frac{1}{2} \varepsilon \right)^{n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)] \quad (5.12) $$

Since $\{ (\frac{1}{2}\varepsilon)^{n/2n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)] \}_{\mathbf{k} \in \mathbb{K}_{\varepsilon}}$ is an orthonormal basis for $\mathcal{L}_2^{\#}(C_{\varepsilon})$,

$$ \begin{aligned} \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} |f(x)|^2 d^{\#n}x &= \mathrm{Ext} - \int_{C_{\varepsilon}} |f(x)|^2 d^{\#n}x = \sum_{\mathbf{k} \in \mathbb{K}_{\varepsilon}} \left| \left(\frac{1}{2}\varepsilon\right)^{n/2} ([\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)], f(x)) \right|^2 = \\ & \qquad \sum_{\mathbf{k} \in \mathbb{K}_{\varepsilon}} |\hat{f}(k)|^2 (\varepsilon\pi_{\#})^n \xrightarrow{\varepsilon \to 0} \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} |f(k)|^2 d^{\#n}k. \end{aligned} \qquad (5.13) $$

This proves the corollary for $f \in C_0^{\infty}(*\mathbb{R}_c^{\#n})$. Since $f \mapsto \hat{f}$ and $\|\cdot\|_2$ are #-continuous on $S^{\#}(*\mathbb{R}_c^{\#n})$ and $C_0^{\infty}(*\mathbb{R}_c^{\#n})$ is #-dense, the result holds for all of $S^{\#}(*\mathbb{R}_c^{\#n})$.

**Definition 5.2.** Let $T \in S^{\#}(*\mathbb{R}_c^{\#n})$ the Fourier transform of $T$, denoted by $\hat{T}$ or $\mathcal{F}[T]$, is the tempered distribution defined by $\hat{T}(\varphi) = T(\hat{\varphi})$.

Suppose that $h, \varphi \in S^{\#}(*\mathbb{R}_c^{\#n})$, then by the polarization identity and the corollary to Theorem 5.1 we have $(h, \varphi) = (\hat{h}, \hat{\varphi})$. Substituting $\mathcal{F}[\hat{g}] = \mathcal{F}^{-1}[\hat{g}]$ for $h$, we obtain

$$ T_{\hat{g}}(\varphi) = \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} \hat{g}(x)\varphi(x)d^{\#n}x = \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} g(x)\hat{\varphi}(x)d^{\#n}x = T_g(\hat{\varphi}) = \hat{T}_g(\varphi). $$

where $T_{\hat{g}}$ and $T_g$ are the distributions corresponding to the functions $\hat{g}$ and $g$ respectively. This shows that the Fourier transform on $S^{\#}(*\mathbb{R}_c^{\#n})$ extends the transform we previously defined on $S^{\#}(*\mathbb{R}_c^{\#n})$.

**Theorem 5.2.** The Fourier transform is a one-to-one linear bijection from $S^{\#}(*\mathbb{R}_c^{\#n})$ to $S^{\#}(*\mathbb{R}_c^{\#n})$ which is the unique weakly #-continuous extension of the Fourier transform on $S^{\#}(*\mathbb{R}_c^{\#n})$.

**Proof.** If hyper infinite sequence $\{\varphi_n\}_{n \in \mathbb{N}^\#}$ #-convergence to $\varphi \in S^\#$, then by Theorem 1.1, hyper infinite sequence $\{\hat{\varphi}_n\}_{n \in \mathbb{N}^\#}$ #-convergence to $\hat{\varphi} \in S^\#$, so

$T(\hat{\varphi}_n) \xrightarrow{\#} T(\hat{\varphi})$ for each $T \in S^{\#}$. Thus $\#\mathrm{lim}_{n \to \infty} T(\hat{\varphi}_n) = T(\hat{\varphi})$, which shows that $T$ is a #-continuous linear functional on $S^\#$. Furthermore, if $T_n \xrightarrow{w} T$, then $\hat{T}_n \xrightarrow{w} \hat{T}$ because $T(\hat{\varphi}_n) \xrightarrow{\#} T(\hat{\varphi})$ implies $\hat{T}(\varphi_n) \xrightarrow{\#} \hat{T}(\varphi)$. Thus $T \mapsto \hat{T}$ is weakly #-continuous.

**Definition 5.3.** Suppose that $f, g \in S^{\#}(*\mathbb{R}_c^{\#n})$. Then the convolution of $f$ and $g$, denoted by $f * g$, is the function

$$ (f * g)(y) = \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} f(y-x)g(x)d^{\#n}x. \qquad (5.14) $$

Convolutions frequently occur when one uses the Fourier transform because the Fourier transform takes products into convolutions.

**Theorem 5.3.** (a) For each $f \in S^{\#}(*\mathbb{R}_c^{\#n})$, $g \mapsto f * g$ is a #-continuous map of $S^{\#}(*\mathbb{R}_c^{\#n})$ into $S^{\#}(*\mathbb{R}_c^{\#n})$.

$$ (b) \quad \hat{f}\hat{g} = (2\pi\#)^{-n/2} \hat{f} * \hat{g} \quad \text{and} \quad \hat{f} * \hat{g} = (2\pi\#)^{n/2} \hat{f}\hat{g}. $$
$$A \mapsto \{P_{\Omega}\} = \{\chi_{\Omega}(A)\} \qquad (4.3.3)$$

and

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

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 $\|(A - \lambda)^{-1}\|_{\#} = [\mathrm{dist}(\lambda, \sigma(A))]^{-1}$.

This suggests that we distinguish between two types of spectrum.

**Definition 4.3.3.** We say that (i) $\lambda \in \sigma_{\mathrm{ess}}(A)$, 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 $\lambda \in \sigma_{\mathrm{disc}}(A)$, the discrete spectrum of A. P is hyper infinite dimensional means $\mathrm{Ran}(P)$ 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 $\sigma_{\mathrm{disc}}$ is not necessarily #-closed, but notice that.

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

**Proof** Let $\lambda_n \to \# \lambda$ with each $\lambda_n \in \sigma_{\mathrm{ess}}(A)$. 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 $\sigma_{\mathrm{disc}}$ and $\sigma_{\mathrm{ess}}$;

**Theorem 4.3.4** $\lambda \in \sigma_{\mathrm{disc}}$ 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** $\lambda \in \sigma_{\mathrm{ess}}$ if and only if one or more of the following holds:

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

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

(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) $\lambda \in \sigma_{\mathrm{ess}}(A)$ 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 $\sigma_{\mathrm{ess}}(A) = \sigma_{\mathrm{ess}}(B)$ 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
(c) For $f,g,h \in S^{#}(*\mathbb{R}_c^{\#n})$, $f \ast g = g \ast f$ and $f \ast (g \ast h) = (f \ast g) \ast h$.

**Definition 5.4.** Suppose that $f \in S^{#}(*\mathbb{R}_c^{\#n})$, $T \in S^{#}(*\mathbb{R}_c^{\#n})$ and let $\tilde{f}(x)$ denote the function, $f(-x)$. Then, the convolution of $T$ and $f$ denoted $T \ast f$ is the distribution in $S^{#}(*\mathbb{R}_c^{\#n})$ given by $(T \ast f)(\varphi) = T(\tilde{f} \ast \varphi)$ for all $\varphi \in S^{#}(*\mathbb{R}_c^{\#n})$.

The fact that $g \to \tilde{f} \ast g$ is a #-continuous transformation guarantees that $T \ast f \in S^{#}(*\mathbb{R}_c^{\#n})$.

**Abbreviation 5.1.** Let $f_y$ denote the function $f_y(x) = f(x - y)$ and $\tilde{f}_y$ the function $f(y - x)$. When $f$ is given by a longe expression $(\cdots)$, we will sometimes write $(\cdots)^{\sim}$ rather than $(\cdot \cdot \cdot)$.

**Theorem 5.4.** For each $f \in S^{#}(*\mathbb{R}_c^{\#n})$ the map $T \to T \ast f$ is a weakly #-continuous map of $S^{#}(*\mathbb{R}_c^{\#n})$ into $S^{#}(*\mathbb{R}_c^{\#n})$ which extends the convolution on $S^{#}(*\mathbb{R}_c^{\#n})$. Furthermore,

(a) $T \ast f$ is a polynomially bounded $C^{\infty}$ function. In fact, $(T \ast f)(y) = T(\tilde{f}_y)$ and
$D^{\#\beta}(T \ast f) = (D^{\#\beta}T) \ast f = T \ast D^{\#\beta}f$;

(b) $(T \ast f) \ast g = T \ast (f \ast g)$;

(c) $\hat{T} \ast \hat{f} = (2\pi_{\#})^{n/2} \hat{f} \hat{T}$.

**Theorem 5.5.** Let $T \in S^{#}(*\mathbb{R}_c^{\#n})$ and $f \in S^{#}(*\mathbb{R}_c^{\#n})$. Then $\hat{fT} \in O_M^n$ and

$\hat{fT}(k) = (2\pi_{\#})^{n/2}T(f[\mathrm{Ext}-\exp(-ik \cdot x)])$. In particular, if $T$ has #-compact support and $\psi \in S^{#}(*\mathbb{R}_c^{\#n})$ is identically one on a #-neighborhood of the support of $T$, then

$$ \hat{T}(k) = (2\pi_{\#})^{n/2} T(\psi[\mathrm{Ext}-\exp(-ik \cdot x)]). \quad (5.15) $$

**Proof** By Theorem 3.4.c and the Fourier inversion formula we have

$$ \hat{fT} = (2\pi_{\#})^{n/2} \hat{f} \ast \hat{T}. \text{ Thus } \hat{fT} \in O_M^n \text{ and } \hat{fT}(k) = (2\pi_{\#})^{n/2} \hat{T}(\tilde{f}_k) = (2\pi_{\#})^{n/2} T(f[\mathrm{Ext}-\exp(-ik \cdot x)]) $$

**Remark 5.1.** We remark that one can also define the convolution of a distribution $T \in D^{#}(*\mathbb{R}_c^{\#n})$ with an $f \in D^{#}(*\mathbb{R}_c^{\#n})$ by $(T \ast f)(y) = T(\tilde{f}_y)$.

**Definition 5.5.** Let $j(x)$ be a positive $C^{\infty}$ function whose support lies in the sphere of radius one about the origin in $*\mathbb{R}_c^{\#n}$ and which satisfies $\mathrm{Ext}-\int_{*\mathbb{R}_c^{\#n}} j(x) d^{\#n}x = 1$. The function $j_{\varepsilon}(x) = \varepsilon^{-n}j(x/\varepsilon)$, $\varepsilon \approx 0$ is called an approximate identity.

**Proposition 5.1.** Suppose $T \in S^{#}(*\mathbb{R}_c^{\#n})$ and let $j_{\varepsilon}(x)$ be an approximate identity. Then $T \ast j_{\varepsilon}(x) \to_{\#} T$ weakly as $\varepsilon \to_{\#} 0$.

**Proof.** If $\varphi \in S^{#}(*\mathbb{R}_c^{\#n})$, then $(T \ast j_{\varepsilon})(\varphi) = T(\tilde{j}_{\varepsilon} \ast \varphi)$, so it is sufficient to show that $\tilde{j}_{\varepsilon} \ast \varphi \to_{\#} \varphi$ in $S^{#}(*\mathbb{R}_c^{\#n})$. To do this it is sufficient to show that $(2\pi_{\#})^{n/2} \tilde{j}_{\varepsilon} \hat{\varphi} \to_{\#} \hat{\varphi}$ in $S^{#}(*\mathbb{R}_c^{\#n})$. Since $\tilde{j}_{\varepsilon}(\lambda) = j(\varepsilon\lambda)$ and $j(0) = (2\pi_{\#})^{n/2}$, it follows that $(2\pi_{\#})^{n/2} \tilde{j}_{\varepsilon}(x)$ #-converges to 1 uniformly on #-compact sets and is uniformly bounded. Similarly, $D^{\#\alpha} \tilde{j}_{\varepsilon}$ #-converges uniformly to zero. We conclude that $(2\pi_{\#})^{n/2} \tilde{j}_{\varepsilon} \hat{\varphi} \to_{\#} \hat{\varphi}$.

**Theorem 5.6 (The generalized Plancherel theorem)** The Fourier transform extends uniquely to a unitary map of $L_2^{\#}(*\mathbb{R}_c^{\#n})$ onto $L_2^{\#}(*\mathbb{R}_c^{\#n})$. The inverse transform extends uniquely to its adjoint.

**Proof** The corollary to Theorem 5.1 states that if $f \in S^{#}(*\mathbb{R}_c^{\#n})$, then $\|f\|_2 = \|\hat{f}\|_2$. Since $\mathcal{F}[S^{#}] = S^{#}$ is a surjective isometry on $L_2^{\#}(*\mathbb{R}_c^{\#n})$.

**Theorem 5.7 (The generalized Riemann-Lebesgue lemma)** The Fourier transform
$$f_{\varepsilon}(x) = (2\pi\#i)^{-1} \left( \mathrm{Ext} - \int_a^b [(x-\lambda - i\varepsilon)^{-1} - (x-\lambda + i\varepsilon)^{-1}] d^\#\lambda \right). \quad (4.3.5)$$

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

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

## § 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
extends uniquely to a bounded map from $\mathcal{L}_1^\#(*\mathbb{R}_c^{\#n})$ into $C^\infty(*\mathbb{R}_c^{\#n})$, the #-continuous functions vanishing at $\infty^\#$.

**Proof** For $f \in S^\#(*\mathbb{R}_c^{\#n})$, we know that $\hat{f} \in S^\#(*\mathbb{R}_c^{\#n})$ and thus $\hat{f} \in C^\infty(*\mathbb{R}_c^{\#n})$. The estimate is trivial. The Fourier transform is thus a bounded linear map from a #-dense set of $\mathcal{L}_1^\#(*\mathbb{R}_c^{\#n})$ into $C^\infty(*\mathbb{R}_c^{\#n})$. By the generalized B.L.T. theorem, extends uniquely to a bounded linear transformation of $C^\infty(*\mathbb{R}_c^{\#n})$ into $C^\infty(*\mathbb{R}_c^{\#n})$.

**Remark 5.2.** We remark that the Fourier transform takes $\mathcal{L}_1^\#(*\mathbb{R}_c^{\#n})$ into, but not onto $C^\infty(*\mathbb{R}_c^{\#n})$.

A simple argument with test functions shows that the extended transform on $\mathcal{L}_1^\#(*\mathbb{R}_c^{\#n})$ and $\mathcal{L}_2^\#(*\mathbb{R}_c^{\#n})$ is the restriction of the transform on $S^\#(*\mathbb{R}_c^{\#n})$, but it is useful to have an explicit integral representation. For $f \in \mathcal{L}_1^\#(*\mathbb{R}_c^{\#n})$, this is easy since we can find $f_m \in S^\#(*\mathbb{R}_c^{\#n})$ so that $\#\lim_{m \to \infty^\#} \|f - f_m\|_1 = 0$. Then, for each $\lambda$,

$$f(\lambda) = \#\lim_{m \to \infty^\#} (\hat{f}_m(\lambda)) = \\ \#\lim_{m \to \infty^\#} \left\{ \frac{1}{(2\pi)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} [\mathrm{Ext}\text{-exp}(-ik \cdot x)]f_m(x)d^\#x \right) \right\} = \\ \frac{1}{(2\pi)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} [\mathrm{Ext}\text{-exp}(-ik \cdot x)]f(x)d^\#x \right). \quad (5.16)$$

So, the Fourier transform of a function in $\mathcal{L}_1^\#(*\mathbb{R}_c^{\#n})$ is given by the usual formula. Next, suppose $f \in \mathcal{L}_2^\#(*\mathbb{R}_c^{\#n})$ and let

$$\chi_R(x) = \begin{cases} 1 & \text{if } |x| \le R \\ 0 & \text{if } |x| > R \end{cases} \quad (5.17)$$

Then $\chi_R f \in \mathcal{L}_1^\#(*\mathbb{R}_c^{\#n})$ and $\#\lim_{R \to \infty^\#} \chi_R f = f$ in $\mathcal{L}_2^\#$, so by the generalized Plancherel theorem $\#\lim_{R \to \infty^\#} \chi_R f = \hat{f}$ in $\mathcal{L}_2^\#$. Thus

$$f(\lambda) = \#\lim_{R \to \infty^\#} \frac{1}{(2\pi)^{n/2}} \left( \mathrm{Ext} - \int_{|x| \le R} [\mathrm{Ext}\text{-exp}(-ik \cdot x)]f(x)d^\#x \right) \quad (5.18)$$

where by $\#\lim_{R \to \infty^\#}$ we mean the $\#\text{-limit}$ in the $\mathcal{L}_2^\#$-norm. Sometimes we will dispense with $|x| \le R$ and just write

$$f(\lambda) = \#\lim_{R \to \infty^\#} \frac{1}{(2\pi)^{n/2}} \left( \mathrm{Ext} - \int [\mathrm{Ext}\text{-exp}(-ik \cdot x)]f(x)d^\#x \right) \quad (5.19)$$

for functions $f \in \mathcal{L}_2^\#(*\tilde{\mathbb{R}}_c^{\#n})$.

We have proven above that $\mathcal{F}: \mathcal{L}_2^\#(*\tilde{\mathbb{R}}_c^{\#n}) \to \mathcal{L}_2^\#(*\tilde{\mathbb{R}}_c^{\#n})$ and $\mathcal{F}: \mathcal{L}_1^\#(*\tilde{\mathbb{R}}_c^{\#n}) \to \mathcal{L}_\infty^\#(*\tilde{\mathbb{R}}_c^{\#n})$ and in both cases is a bounded operator.

**Theorem 5.8** (Generalized Hausdorff-Young inequality) Suppose $1 \le q \le 2$, and $p^{-1} + q^{-1} = 1$. Then the Fourier transform is a bounded map of $\mathcal{L}_p^\#(*\tilde{\mathbb{R}}_c^{\#n})$ to $\mathcal{L}_q^\#(*\tilde{\mathbb{R}}_c^{\#n})$ and its norm is less than or equal to $(2\pi)^{n(1/2-1/q)}$. distribution in $\mathcal{D}^\#(*\tilde{\mathbb{R}}_c^{\#n})$.

# Chapter VI. Non-Archiedean Hilbert Spaces over field
$|g(m)| > n$ and equal to $g$ otherwise. By the generalized dominated convergence theorem, $g_n \to_g g$ and $fg_n \to_g fg$ 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 $g_n \to_g g$ in $L_q^\#$. Since $\|g_n - g\|_{\#2} \le \|1\|_{\#p} \cdot \|g_n - g\|_{\#q}$ and $\|T_f(g_n - g)\|_{\#2} \le \|f\|_{\#p} \cdot \|g_n - g\|_{\#q}$, $g \in \text{\#- } D(T_f \uparrow D)$.

Thus $L_q^\# \subset D(T_f \uparrow D)$ 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) $\psi \in D(A)$ if and only if $f(\cdot)(U\psi)(\cdot) \in L_2^\#(M, d^\#\mu^\#)$.

(b) If $\varphi \in U[D(A)]$, 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 $U : H^\# \to L_2^\#(M, d^\#\mu^\#)$, 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, $g(m) \ne 0$ 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 $\varphi \in H^\#$ 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 $\varphi \in H^\#$ 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 $\varphi \in H^\#$ 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

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

where $T_{h(f)}$ is the operator on $L_2^\#(M, d^\#\mu^\#)$ which acts by multiplication by the function
*$\tilde{\mathbb{C}}_c^\#$.

§ 1. Non-Archiedean Hilbert Spaces over field *$\tilde{\mathbb{C}}_c^\#$*
Basics.

**Definition 1.1.** (i) Let $H$ be external hyper infinite dimensional vector space over field $\mathbb{C}^\# = \mathbb{R}_c + i\mathbb{R}_c$. An inner #-product(or non-Archiedean inner product) on $H$ is a $\mathbb{C}^\#$-valued function, $\langle \cdot, \cdot \rangle_\# : H \times H \to \mathbb{C}_c$, such that

(1) $\langle ax + by, z \rangle_\# = \alpha \langle x, z \rangle + \beta \langle y, z \rangle_\#,$ i.e. $x \to \langle x, z \rangle_\#$ is linear.

(2) $\langle x, y \rangle_\# = \langle y, x \rangle_\#.$

(3) $\|x\|_\#^2 = \langle x, x \rangle_\# \ge 0$ with equality $\|x\|_\#^2 = 0$ iff $x = 0$.

Notice that combining properties (1) and (2) that $x \to \langle z, x \rangle$ is anti-linear for fixed $z \in H$, i.e. $\langle z, ax + by \rangle_\# = \bar{a}\langle z, x \rangle_\# + \bar{b}\langle z, y \rangle_\#.$

(ii) Let $\{a_n\}_{n=0}^k$ be finite or hyperfinite sequence in $H$, $\{a_n\}_{n=0}^k \subset H$.

We define external hyper infinite sequence $\overbrace{\{a_n\}_{n=0}^k} \subset H$ by

$$
\begin{align}
\{A_n; k\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^k}^{k} = \nonumber \\
&= (a_0, a_1, \dots, a_m, \dots, a_{k-1}, a_k, \widehat{a_k}). \tag{0.1}
\end{align}
\[
\widehat{a} = (a, a, \dots), a \in H.
\]
$$

(iii) Let {$a_n$}$_{n=0}^\infty$ be countable sequence in $H : \{a_n\}_{n=0}^\infty \subset H$.

We define hyper infinite sequence $\{A_n\}_{n=0}^\infty = \overbrace{\{a_n\}_{n=0}^\infty} \subset H$ by

$$
\begin{align}
\{A'_n; \infty\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^{\infty}}^{k} = \nonumber \\
&= \left( \widehat{a}_0, a_1, \dots, a_k, \dots, \{a_n\}_{n=0}^{\infty}, \overbrace{\{a_n\}_{n=0}^{\infty}} \right). \tag{0.2}
\end{align}
$$

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

We define hyper infinite sequence $\overbrace{\{a_n\}_{n=0}^N} \subset H$ by

$$
\begin{align}
\{A_n; N\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^{N}}^{N} = \tag{0.3} \\
&= (a_0, a_1, \dots, a_m, \dots, a_{N-1}a_N, \widehat{a_N}). \nonumber
\end{align}
$$

(v) Let {$a_n$}$_{n=0}^k$ be finite sequence in $H$, {$a_n$}$_{n=0}^N \subset H$.

We define external finite sum $\text{Ext}-\sum_{n=0}^{n=k} a_n$ by

$$
\mathrm{Ext}-\sum_{n=0}^{n=k} a_n = \overbrace{\{c_n\}_{n=0}^k}^{k} = (c_0, c_1, \dots, c_k, \dots, \hat{c}_k) \in [[\hat{c}_k]] \quad (0.4)
$$

where $c_0 = a_0, c_j = \text{Ext}-\sum_{n=0}^{n=j} a_n, 0 \le j \le k$.

(vi) Let {$a_n$}$_{n=0}^\infty$ be countable sequence in $H : \{a_n\}_{n=0}^\infty \subset H$. We define external countable sum $\text{Ext}-\sum_{n=0}^{n=\infty} a_n$ by