> [!definition] > > Let $U \subset \real^d$ be [[Open Set|open]], $\psi \in \cd(U)$ be a [[Space of Test Functions|test function]], $F \in \cd'(U)$ be a [[Distribution|distribution]], and define > $ > V = \bracs{x \in \real^d:x - \supp{\psi} \subset U} > $ > then there exists a function $F * \psi \in C^\infty(V)$ such that > 1. $(F * \psi)(x) = \anglesn{F, \tau_x \td \psi}$. > 2. $\angles{F * \psi, \phi} = \anglesn{F, \phi * \td \psi}$. > 3. $\partial^\alpha (F * \psi) = (\partial^\alpha F) * \psi = F * (\partial^\alpha \psi)$. > > known as the **convolution** of $F$ and $\psi$. > > *Proof*. Since for any $g \in \loci(U)$, > $ > \begin{align*} > g * \psi(x) &= \int_{\supp{\psi}} g(x - y)\psi(y) \\ > &= \int_{x - \supp{\psi}} g(y) \psi(x - y) = \langle g, \tau_x\td \psi \rangle > \end{align*} > $ > is well-defined for any $x \in V$. Similarly, define $(F * \psi)(x) = \anglesn{F, \tau_x \td \psi}$. Let $\phi \in \cd(U)$, then > $ > \phi * \td \psi = \int\phi(y)\tau_y\td\psi(x)dy = \limv{n}\Delta_n\sum_{j = 1}^n \phi(y_{n, j})\tau_{y_{n, j}}\td\psi(x) > $ > as a limit of Riemann sums. Since the sum, and all of its derivatives converge uniformly to $\phi * \td \psi$, the sum converges to $\phi * \td \psi$ in $\cd$. This allows writing > $ > \begin{align*} > \anglesn{F, \phi * \td \psi} &= \limv{n}\Delta_n \sum_{j = 1}^n \phi(y_{n, j})\anglesn{F, \tau_{y_{n, j}}\td \psi} \\ > &= \int \phi(y)\anglesn{F, \tau_y \td\psi}dy = \angles{F * \psi, \phi} > \end{align*} > $ > To show smoothness, since the difference quotient converges to the derivative in $\mathcal D$, $\partial^\alpha f = F * (\partial^\alpha \psi)$. Lastly, > $ > \begin{align*} > (\partial^\alpha F) * \psi(x) &= \anglesn{\partial^\alpha F, \tau_x \td \psi} = (-1)^{\abs{\alpha}}\anglesn{F, \partial^\alpha\tau_x \td{\psi}} \\ > &= \anglesn{F, \tau_x \td{\partial^\alpha \psi}} = F * (\partial^\alpha \psi)(x) > \end{align*} > $ > since $\partial^\alpha \td \psi = (-1)^{\abs{\alpha}}\td{\partial^\alpha \psi}$. Let $F \in \cd'$ be a distribution and $G \in \cd'$ be a [[Compactly Supported|compactly supported]] distribution. For any $\phi \in \cd$, define $ \angles{F * G, \phi}_{\cd} = \anglesn{F, \td G * \phi}_{\cd} \quad \angles{G * F, \phi}_{\cd} = \anglesn{G, \td F * \phi}_{C^\infty} $ then 1. $F * G$ is a distribution. 2. $F * G = G * F$. *Proof*. For $(1)$, it is sufficient to show that for any $\seq{\phi_n} \subset \cd$ with $\phi_n \to \phi$ in $\cd$, $\td G * \phi_n \to \td G * \phi$ in $\cd$. Let $K$ be a [[Compactness|compact]] set such that $\seq{\phi_n}$ are commonly supported in $K$, then $\{\td G * \phi_n: n \in \nat\}$ are commonly supported in $S = K + \text{supp}(\td G)$. In particular, $\{\tau_x \phi_n: n \in \nat, x \in S\}$ are also commonly supported in a compact set. As $\td G \in C^\infty(U)^*$, there exists $N \in \nat$ and $C \ge 0$ such that $ |\anglesn{\td G, \tau_x(\phi_n - \phi)}| \le C\sum_{\abs{\alpha} \le N}\norm{\partial^\alpha(\phi_n - \phi)}_{u} \to 0 $ as $n \to \infty$, independent of the choice of $x$. Therefore $\td G * \phi_n \to \td G * \phi$ uniformly. Similarly, since the derivative commutes with convolutions, $ \partial^\alpha(\td G * \phi_n) = \td G * (\partial^\alpha\phi_n) \to \td G * (\partial^\alpha \phi) = \partial^\alpha (\td G * \phi) $ For $(2)$, $ \begin{align*} \angles{F * G, \phi} = \anglesn{F, \td G * \phi} \end{align*} $