Convolution of a Locally Integrable Function with a Compactly Supported Kernel

We work in the setting of Euclidean Space and Lebesgue Measure: Standing Notation and in the setting of Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation, both used here with a natural number $n$ satisfying $1\le n$, and in the setting of Measure Spaces and the Lebesgue Integral: Standing Notation, whose measure space $(X,\mathcal{F},\mu)$ is instantiated throughout as $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}),\lambda_{n})$; this is a measure space because $\mathcal{B}(\mathbb{R}^{n})$ is a $\sigma$-algebra on $\mathbb{R}^{n}$ and $\lambda_{n}$ is a measure on it. The three settings fix the real numbers, the natural numbers and the initial segments $[n]$ by reference to the same definitions, so their readings agree. From them we take Euclidean space $\mathbb{R}^{n}$ with its norm $\lVert\,\cdot\,\rVert$, distance, balls, topology and the notions of open, closed, bounded and compact set, measurability of maps into $\mathbb{R}$, the indicator $\mathbf{1}_{A}$ of a subset $A$, the integral and the notion of an integrable map, and the partial derivatives $\partial_{i}$ and the classes $C^{k}$ on a Euclidean open set; the set $\mathbb{R}^{n}$ is open in itself by claim 1 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous, so the calculus notions apply to maps defined on all of $\mathbb{R}^{n}$. A map $\mathbb{R}^{n}\to\mathbb{R}$ is called continuous when it is continuous from $\mathbb{R}^{n}$ with its Euclidean distance to $\mathbb{R}$ with the metric of The Absolute Value Metric on the Real Line, and smooth means smooth on $\mathbb{R}^{n}$.

Let $w:\mathbb{R}^{n}\to\mathbb{R}$ be measurable and such that $\mathbf{1}_{B}w$ is integrable for every bounded $B\in\mathcal{B}(\mathbb{R}^{n})$. Let $R\in\mathbb{R}$ with $0<R$, and let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be continuous and such that $\psi(y)=0$ for every $y\in\mathbb{R}^{n}$ with $R<\lVert y\rVert$. Then the following hold.

1. (The convolution is defined)¶ For every $x\in\mathbb{R}^{n}$ the map $y\mapsto\psi(x-y)\,w(y)$ is measurable and integrable. Consequently there is a map $\psi\star w:\mathbb{R}^{n}\to\mathbb{R}$, the convolution of $\psi$ with $w$, given by

2. (Continuity)¶ $\psi\star w$ is continuous.

3. (Linearity)¶ Let $w':\mathbb{R}^{n}\to\mathbb{R}$ satisfy the same hypotheses as $w$, let $\psi':\mathbb{R}^{n}\to\mathbb{R}$ be continuous with $\psi'(y)=0$ for every $y$ with $R<\lVert y\rVert$, and let $c\in\mathbb{R}$. Then, as maps on $\mathbb{R}^{n}$,

4. (Differentiation through the kernel)¶ Suppose in addition that $\psi$ is of class $C^{1}$ on $\mathbb{R}^{n}$, and let $i\in[n]$. Then $\partial_{i}\psi$ is continuous and satisfies $\partial_{i}\psi(y)=0$ for every $y$ with $R<\lVert y\rVert$, so claims 1 and 2 apply verbatim with the kernel $\partial_{i}\psi$ in place of $\psi$. Moreover the partial derivative of $\psi\star w$ with respect to the $i$th variable exists at every point of $\mathbb{R}^{n}$ and

Together with the continuity of $\psi\star w$ supplied by claim 2, it follows that $\psi\star w$ is of class $C^{1}$ on $\mathbb{R}^{n}$.

5. (Higher order and smoothness)¶ Let $k$ be a natural number. If $\psi$ is of class $C^{k}$ on $\mathbb{R}^{n}$, then $\psi\star w$ is of class $C^{k}$ on $\mathbb{R}^{n}$. If $\psi$ is smooth, then $\psi\star w$ is smooth.