From e1a26e4528eb7b9c2f462562c8265cf963f34dfb Mon Sep 17 00:00:00 2001
From: Justin Gassner <justin.gassner@mailbox.org>
Date: Thu, 20 Jun 2024 03:56:45 +0200
Subject: weiter

---
 convolution.tex | 286 +++++++++++++++++++++++++++++++++++++++++++++++++++-----
 1 file changed, 263 insertions(+), 23 deletions(-)

(limited to 'convolution.tex')

diff --git a/convolution.tex b/convolution.tex
index d3fb8bc..eee6e16 100644
--- a/convolution.tex
+++ b/convolution.tex
@@ -1,27 +1,32 @@
 \chapter{A Convolution Formula for Vector-Valued Tempered Distributions}
 \label{chapter:convolution}
-
 \blockcquote{Bisognano1975}{%
   The extension to vector-valued tempered distributions is trivial.
 }
-Recall that the class $\schwartz{\RR^n}$ of complex-valued Schwartz functions on $\RR^n$
+Recall that the class $\SchwartzFunctions{\RR^n}$ of complex-valued Schwartz functions on $\RR^n$
 is closed under convolution, a operation that assigns to functions $f$ and $g$ a third one, $f * g$,
 given by
 \begin{equation*}
   (f*g)(x) = \int f(x-y) g(y) \, dy
   \qquad x \in \RR^n.
 \end{equation*}
-
-\begin{definition}{Convolution of a Distribution with a Test Function}{}
-  Let $u \in \tempdistrib{\RR^n}$ be tempered distribution and
-  let $f \in \schwartz{\RR^n}$ be a Schwartz test function.
+\begin{definition}{Convolution of a Distribution with a Test Function}{convolution-distribution-test-function}
+  Let $u \in \TemperedDistributions{\RR^n}$ be a tempered distribution and
+  let $f \in \SchwartzFunctions{\RR^n}$ be a Schwartz test function.
   Then the \emph{convolution} of $u$ with $f$ is
-  the tempered distribution $u * f \in \tempdistrib{\RR^n}$ defined by
+  the tempered distribution $u * f \in \TemperedDistributions{\RR^n}$ defined by
   \begin{equation*}
-    (u * f)(g) \defequal u(\tilde{f} * g) \qquad g \in \schwartz{\RR^n},
+    (u * f)(g) \defequal u(\tilde{f} * g) \qquad g \in \SchwartzFunctions{\RR^n},
   \end{equation*}
   where $\tilde{f}(x) = f(-x)$ for all $x \in \RR^n$.
 \end{definition}
+The motivation and justification for this definition is provided by the adjoint identity
+\begin{equation*}
+  \int (h * f)(x) \, g(x) \, dx =
+  \int h(x) \, (\tilde{f} * g)(x) \, dx
+\end{equation*}
+holding for all $f,g,h \in \SchwartzFunctions{\RR^n}$.
+
 It is well-known that the convolution can be expressed by the integral
 \begin{equation*}
   (u * f)(g) = \int u(\tau_x \tilde{f}@@) g(x) \, dx
@@ -30,34 +35,269 @@ emphasizing its character of a smoothing operation.
 The purpose of this appendix is to state and prove
 a vector-valued version of this formula.
 
-Let $X$ be a complex Banach space.
-Denote by $C^{\infty}(\RR^n,X)$  the vector space of all functions $f : \RR^n \to X$
+We proceed to develop a generalization of the Bochner integral
+for functions valued in a separable Fréchet space,
+as this will facilitate our proof of the convolution formula.
+
+We consider a $\sigma$-finite measure space $(X,\SigmaAlgebra{A},\mu)$,
+a separable Fréchet space $Y$ (over $\CC$) and the task is
+to define the integral of functions $f \vcentcolon X \to Y$.
+Recall that a measure space is said to be \emph{$\sigma$-finite}
+if it can be exhausted by a countable number of measurable subsets of finite measure.
+By \emph{Fréchet space} we mean a complete Hausdorff locally convex (topological vector) space
+which possesses countable neighborhood bases.
+We will make use of a countable family $P@@$ of seminorms that generates the topology of $@@Y$.
+A topological space is called \emph{separable} if it contains a countable dense subset.
+
+A function $f \vcentcolon X \to Y$ will be called \emph{simple}
+if it is of the form $\sum_{i=1}^n \chi_{A_i} y_i$
+where $n \in \NN$, $A_i \in \SigmaAlgebra{A}$ with $\mu(A_i) < \infty$, and $y_i \in Y$.
+Naturally, the \emph{integral} of $f$ is defined to be the vector $\int f = \sum_{i=1}^n \mu(A_i) y_i \in Y$.
+We say that a function $f \vcentcolon X \to Y$ is \emph{strongly measurable}
+if it is the $\mu$-almost everywhere pointwise limit of simple functions.
+
+\begin{definition}{Generalized Bochner Integral}{}
+  Suppose $(X,\SigmaAlgebra{A},\mu)$ is a $\sigma$-finite measure space,
+  and $Y@@$ is a separable Fréchet space
+  whose topology is generated by a family $P@@$ of seminorms.
+A strongly measurable function $f \vcentcolon X \to Y$ is called \emph{(generalized Bochner) integrable}
+if there exists a sequence $(f_n)$ of simple functions such that
+\begin{equation}
+  \label{equation:bochner-integrable}
+  \lim_{n \to \infty} \int_X p \circ (f_n - f) \, d\mu = 0
+  \qquad \forall p \in P.
+\end{equation}
+In this case, the \emph{(generalized Bochner) integral} of $f$ is defined by
+\begin{equation}
+  \label{equation:bochner-integral}
+  \int_X f \ d\mu \defequal
+  \lim_{n \to \infty} \int_X f_n \, d\mu.
+\end{equation}
+\end{definition}
+This definition needs justification.
+First, for the integral in~\eqref{equation:bochner-integrable} to be meaningful,
+the functions $p \circ (f_n - f)$ must be $\mu$-measurable.
+Since $f$ is strongly measurable, there exists simple functions $s_k$ such that $f (x) = \lim_{k \to \infty} s_k(x)$ for almost all $x \in X$.
+The continuity of $p$ implies that $p \circ (f_n - f)$ is the almost everywhere limit simple scalar functions, namely $p \circ (f_n - s_k)$,
+and as such must be measurable.
+%We will defer this question for now.
+Second, we have to verify that the limit in~\eqref{equation:bochner-integral} exists
+and is independent of the particular sequence $(f_n)$.
+Remember that the sets $U_{F,\epsilon} = \braces{y \in Y \vcentcolon p(y) < \epsilon \forall p \in F}$,
+where $F \subset P$ is finite and $\epsilon > 0$,
+form a neighborhood basis for $0 \in Y$.
+Consider any such $U_{F,\epsilon}$.
+Then, for all $p \in F$ and $m,n \in \NN$
+\begin{equation*}
+  p \parens*{\smallint f_n - \smallint f_m}
+  %= p \parens*{\smallint (f_n - f_m)}
+  \le \smallint p \circ (f_n - f_m)
+  \le \smallint p \circ (f - f_n) +  \smallint p \circ (f - f_m).
+\end{equation*}
+By~\eqref{equation:bochner-integrable} there exists $N_p \in \NN$ such that
+$p \parens*{\int f_n - \int f_m} < \epsilon$ for all $m,n \ge N_p$.
+If we set $N = \max \braces{N_p \vcentcolon p \in F}$, then $\int f_n - \int f_m \in U_{F,\epsilon}$ for all $m,n \ge N$.
+This shows that $(\int f_n)$ is a Cauchy sequence in the topological vector space $Y$.
+Now the existence of a limit point follows from the completeness of $Y$.
+It is unique because the topology is Hausdorff.
+
+
+\begin{theorem}{Generalized Bochner Integrability Criterion}{generalized-bochner}
+  Suppose $X$ is a $\sigma$-finite measure space,
+  and $Y@@$ is a separable Fréchet space
+  whose topology is generated by a countable family $P@@$ of seminorms.
+  A function $f \vcentcolon X \to Y@@$ is generalized Bochner integrable if and only if it is strongly measurable and
+  \begin{equation*}
+    \int_X p \circ f \ d\mu < \infty
+    \qquad \forall p \in P.
+  \end{equation*}
+\end{theorem}
+
+\begin{proof}
+  Since $X$ is $\sigma$-finite,
+  $X = \bigcup_{m=1}^{\infty} X_m$ with $\mu(X_m) < \infty$ and $X_m \subset X_{m+1}$.
+  Clearly, $f$ is the pointwise limit of
+  the functions $f_m = f \chi_{X_m}$, as $m \to \infty$.
+  Let $(p_i)_{i \in \NN}$ be an enumeration of the countable family $P$ of seminorms
+  generating the locally convex topology on $Y$.
+  Since $Y$ is separable,
+  there is a dense sequence $(y_j)_{j \in \NN}$ of vectors in $Y$.
+  For $n,j \in \NN$ let
+  \begin{gather*}
+    C_{nj} = y_j + U_{\braces{p_1, \ldots, p_n},1/n}
+    = \braces[\big]{y \in Y \vcentcolon p_i(y - y_j) \le \tfrac{1}{n} \forall i=1,\ldots,n} \\
+    B_{nj} = f^{-1} C_{nj} \qquad
+    A_{nj} = B_{nj} \setminus \bigcup_{k=1}^{j-1} B_{nk}
+  \end{gather*}
+  Observe that for each fixed $n$ the sets $C_{nj}$ cover $Y$,
+  the sets $B_{nj}$ cover $X$ and
+  the sets $A_{nj}$ partition $X$.
+  Moreover, the sets $B_{nj}$, and consequently $A_{nj}$, are $\mu$-measurable
+  because the functions $x \mapsto p_i \parens[\big]{f(x) - y_j}$ are $\mu$-measurable.
+  Then, the functions
+  \begin{equation*}
+    f_{mn} = \sum_{j=1}^{\infty} \chi_{X_m \cap A_{nj}} y_j
+  \end{equation*}
+  satisfy $p_i(f(x) - f_{mn}(x)) \le \frac{1}{n}$ for all $x \in X$ when $i \le n$.
+  Hence, $p_i \circ f_{mn} \le p_i \circ f + \frac{1}{n}$.
+  Since $f_{mn}$ is supported in $X_m$, a set of finite measure, and $\int p_i \circ f < \infty$,
+  we conclude $\int p_i \circ f_{mn} < \infty$ for all $i \le n$.
+  For each $(m,n) \in \NN^2$ choose $J(m,n)$ so large that
+  \begin{equation*}
+    \int_{\bigcup_{j=J(m,n)+1}^{\infty} X_m \cap A_{nj}} p_i \circ f_{mn} < \frac{\mu(X_m)}{n}
+    \qquad \forall i=1,\ldots,n.
+  \end{equation*}
+  The functions
+  \begin{equation*}
+    s_{mn} = \sum_{j=1}^{J(m,n)} \chi_{X_m \cap A_{nj}} y_j
+  \end{equation*}
+  are simple and satisfy
+  \begin{equation*}
+    \int p_i \circ (f_m - s_{mn})
+    \le \int p_i \circ (f_m - f_{mn}) + \int p_i \circ (f_{mn} - s_{mn})
+    < \frac{2\mu(X_m)}{n}
+  \end{equation*}
+  for $n \ge i$.
+  It follows that
+  \begin{equation*}
+    \lim_{n \to \infty} \int p_i \circ (f_m - s_{mn}) = 0
+    \qquad \forall i \in \NN.
+  \end{equation*}
+  For each $m \in \NN$ choose $N(m)$ so large that
+  \begin{equation*}
+    \int p_i \circ (f_m - s_{mN(m)}) < \frac{1}{m} 
+    \qquad \forall i=1,\ldots,m.
+  \end{equation*}
+  and therefore
+  \begin{equation*}
+    \int p_i \circ (f - s_{m N(m)})
+    \le \frac{1}{m} + \int p_i \circ (f - f_m) 
+  \end{equation*}
+  by the triangle inequality.
+  %This implies
+  %\begin{equation*}
+    %\lim_{n \to \infty} \int p_i \circ (f_m - s_{mN(m)}) = 0
+    %\qquad \forall i \in \NN.
+  %\end{equation*}
+  For each $i \in \NN$ the increasing sequence $(p_i \circ f_{m})_m$ of positive real-valued measurable functions
+  converges pointwise to the function $p_i \circ f$,
+  which is by hypothesis is integrable.
+  By Dominated Convergence, $\int p_i \circ (f-f_m) \to  0$, as $m \to \infty$.
+  \begin{equation*}
+    \lim_{m \to \infty} \int p_i \circ (f - s_{m N(m)}) = 0
+    \qquad \forall i \in \NN.
+  \end{equation*}
+  This proves that $f$ is generalized Bochner integrable.
+\end{proof}
+
+\begin{theorem}{}{integral-commutes-with-operator}
+  Suppose $X$ is a $\sigma$-finite measure space.
+  Let $Y$ and $Z$ be separable Fréchet spaces,
+  and let $T \vcentcolon Y \to Z$ be a continuous linear operator.
+  If $f \vcentcolon X \to Y$ is generalized Bochner integrable,
+  then $T \circ f \vcentcolon X \to Z$ is generalized Bochner integrable, and
+  \begin{equation*}
+    \int T \circ f =
+    T \! \int \! f.
+  \end{equation*}
+\end{theorem}
+
+\begin{proof}
+  Clearly, the composition $T \circ f$ is strongly measurable
+  because $T$ is continuous and $f$ is strongly measurable.
+  Suppose that the locally convex topologies on $Y$ and $Z$
+  are generated by the seminorm families $P$ and $Q$, respectively.
+  If $q \in Q$, then the fact that $T$ is continuous and linear implies that
+  there exists a finite subset $F \subset P$ and a constant $M \ge 0$
+  such that $q \circ T \le M \max_{p \in F} p$.
+  If $(f_n)$ is a sequence of simple functions such that $\int p \circ (f - f_n) \to 0$, 
+  then $\int q \circ T \circ (f-f_n) \to 0$.
+  This shows that $T \circ f$ is generalized Bochner integrable, and
+  \begin{equation*}
+    \int T \circ f = \lim_{n \to \infty} \int T \circ f_n
+    = T \lim_{n \to \infty} \int f_n = T \int f.\qedhere
+  \end{equation*}
+  %By \cref{theorem:generalized-bochner},
+  %it follows that $\int q \circ T \circ f < \infty$,
+\end{proof}
+
+We now return to tempered distributions.
+Denote by $\TestFunctions{\RR^n}$  the vector space of all functions $f \vcentcolon \RR^n \to \CC$
 such that the derivatives $\partial^{\alpha} f$ exist and are continuous for all multi-indices $\alpha \in \NN^n$.
-We define the space $\schwartz{\RR^n,X}$ of \emph{$X$-valued Schwartz functions} to be the vector space
+Recall that the space $\SchwartzFunctions{\RR^n}$ of \emph{Schwartz functions} is defined to be the vector space
 \begin{equation*}
-  \schwartz{\RR^n,X} \defequal \braces{f \in C^{\infty}(\RR^n,X) \vcentcolon \norm{f}_{\alpha,\beta} < \infty \forall \alpha,\beta \in \NN^n}
+  \SchwartzFunctions{\RR^n,X} \defequal \braces{f \in \TestFunctions{\RR^n} \vcentcolon \norm{f}_{\alpha,\beta} < \infty \ \forall \alpha,\beta \in \NN^n}
 \end{equation*}
 equipped with the locally convex topology induced by the family of seminorms
 \begin{equation*}
-  \norm{f}_{\alpha,\beta} = \sup_{x \in \RR^n} \abs{x^{\alpha}} \norm{\partial^{\beta} f(x)}_X.
+  \norm{f}_{\alpha,\beta} = \sup_{x \in \RR^n} \abs{x^{\alpha}} \abs{\partial^{\beta} f(x)}.
 \end{equation*}
-We define the space $\tempdistrib{\RR^n,X}$ of \emph{$X$-valued tempered distributions} to be the vector space
+It is well known that the Schwartz space is a separable Fréchet space.
+Now let $X$ be any separable Fréchet space.
+We define the space $\TemperedDistributions{\RR^n\!,X}$ of \emph{$X$-valued tempered distributions} to be the vector space
 \begin{equation*}
-  \tempdistrib{\RR^n,X} \defequal \BoundedLinearOperators[\big]{\schwartz{\RR^n},X}.
+  \TemperedDistributions{\RR^n\!,X} \defequal \ContinousLinearOperators[\big]{\SchwartzFunctions{\RR^n},X}.
 \end{equation*}
+of all continuous linear operators $\SchwartzFunctions{\RR^n} \to X$
 equipped with the bounded convergence topology.
+The convolution of a $X$-valued tempered distribution $v$ with a Schwartz function $f$
+is defined in the same way as in \cref{definition:convolution-distribution-test-function}, that is by
+  \begin{equation*}
+    (v * f)(g) \defequal v(\tilde{f} * g) \qquad g \in \SchwartzFunctions{\RR^n}.
+  \end{equation*}
 
-\begin{proposition}{Vector-Valued Convolution Formula}{}
-  Let $v \in \tempdistrib{\RR^n,X}$ be tempered distribution with values in a Banach space $X$, and
-  let $f \in \schwartz{\RR^n}$ be a Schwartz test function. Then one has
+\begin{proposition}{Vector-Valued Convolution Formula}{vector-valued-convolution-formula}
+  Let $v \in \TemperedDistributions{\RR^n\!,X}$ be a tempered distribution with values in a separable Fréchet space $X$, and
+  let $f \in \SchwartzFunctions{\RR^n}$ be a Schwartz test function. Then one has
   \begin{equation*}
-    (v * f)(g) = \int v(\tau_x \tilde{f}@@) g(x) \, dx \qquad g \in \schwartz{\RR^n}.
+    (v * f)(g) = \int v(\tau_x \tilde{f}@@) g(x) \, dx \qquad g \in \SchwartzFunctions{\RR^n}.
   \end{equation*}
 \end{proposition}
 
-Der Beweis ist in Arbeit ;)
+\begin{proof}
+  We fix a Schwartz function $g$, and consider the finite measure $\mu = \abs{g} \lambda$ on $\RR^n$,
+  where $\lambda(x) = dx$ is the Lebesgue measure.
+  We show that the mapping $x \mapsto \tau_x \tilde{f}$ is a generalized Bochner $\mu$-integrable function $\RR^n \to \SchwartzFunctions{\RR^n}$
+  using \cref{theorem:generalized-bochner}.
+  For all $\alpha,\beta \in \NN^n$ we see by substituting $x+y$ for $y$ that
+  \begin{equation*}
+    \norm{\tau_x \tilde{f}}_{\alpha,\beta} =
+    \sup_{y} \abs{y^{\alpha} \partial^{\beta} (\tau_x \tilde{f})(y)} =
+    \sup_{y} \abs{(x+y)^{\alpha} \partial^{\beta} \tilde{f}(y)}.
+  \end{equation*}
+  There exists constants $c_{\gamma \delta}$ with
+  $\abs{(x+y)^{\alpha}} \le \sum_{\gamma + \delta = \alpha} c_{\gamma \delta} \abs{x^{\gamma} y^{\delta}}$,
+  and it follows that
+  \begin{equation*}
+    \int \norm{\tau_x \tilde{f}}_{\alpha,\beta} \, d \mu(x) 
+    \le \sum_{\gamma + \delta = \alpha} c_{\gamma \delta} \norm{\tilde{f}}_{\delta,\beta} \int \abs{x^{\gamma}} g(x) \, dx < \infty 
+  \end{equation*}
+  because $g$ is Schwartz class.
+  Hence, $x \mapsto \tau_x \tilde{f}$ defines an integrable function.
+  
+  The mapping $v \vcentcolon \SchwartzFunctions{\RR^n} \to X$ is linear and continuous by definition.
+  By \cref{theorem:integral-commutes-with-operator},
+  the composite mapping $x \mapsto v(\tau_x \tilde{f})$ is a $\mu$-integrable function $\RR^n \to X$, and
+  \begin{equation}
+    \label{equation:general-bochner-appears}
+    \int v(\tau_x \tilde{f}) \, d\mu(x) = v \parens[\bigg]{\int \tau_x \tilde{f} \, d\mu(x)}
+  \end{equation}
+  For every fixed $y \in \RR^4$ the evaluation mapping $\ev_{\! @@y} \vcentcolon \SchwartzFunctions{\RR^4} \to \CC$, $h \mapsto h(y)$, clearly is continuous.
+  A second invocation of \cref{theorem:integral-commutes-with-operator} delivers
+  \begin{equation*}
+    \ev_{\! @@y} \parens[\bigg]{\int \tau_x \tilde{f} \, d\mu(x)} =
+    \int \ev_{\! @@y}(\tau_x \tilde{f}) \, d\mu(x) =
+    \int \tilde{f}(y-x) g(x) \, dx =
+    (\tilde{f} * g)(y) 
+  \end{equation*}
+  and the proof is complete.
+\end{proof}
+
+Let us point out that even in the special case that $X$ is a Banach space
+the integral on the right hand side of~\eqref{equation:general-bochner-appears}
+only has meaning as a generalized Bochner integral,
+since the integrand takes values in $\SchwartzFunctions{\RR^n}$,
+which is not a Banach space.
+We could not have performed this step with the ordinary Bochner integral.
 
 %\nomenclature[B]{$\BoundedLinearOperators{X,Y}$}{bounded linear operators from $X$ to $Y$\nomnorefpage}
-
-\chapterbib
-\cleardoublepage
-- 
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