\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 $\SchwartzFunctions{\RR^n}$ 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}{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 $u * f \in \TemperedDistributions{\RR^n}$ defined by
  \begin{equation*}
    (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
\end{equation*}
emphasizing its character of a smoothing operation.
The purpose of this appendix is to state and prove
a vector-valued version of this formula.

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 $\alpha \in \NN^n$.
Recall that the space $\SchwartzFunctions{\RR^n}$ of \emph{Schwartz functions} is defined to be the vector space
\begin{equation*}
  \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}} \abs{\partial^{\beta} f(x)}.
\end{equation*}
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*}
  \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}{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 \SchwartzFunctions{\RR^n}.
  \end{equation*}
\end{proposition}

\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}
