From e1a26e4528eb7b9c2f462562c8265cf963f34dfb Mon Sep 17 00:00:00 2001 From: Justin Gassner 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 -- cgit v1.2.3-70-g09d2