Proof of Convolution of a Locally Integrable Function with a Compactly Supported Kernel

Each result cited below is universally quantified over the data appearing in its own statement, and is applied here to the data named in the statement of this lemma: the natural number $n$, the map $w$, the real number $R$, the kernel $\psi$, and the measure space $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}),\lambda_{n})$. We write $\bar{B}(z,r)$ for the closed ball of centre $z$ and radius $r$ in $\mathbb{R}^{n}$ with its Euclidean distance $d_{E}$, and we use throughout that $d_{E}(a,b)=\lVert a-b\rVert$, by claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square.

Step 1. The kernel is compactly supported, bounded, integrable and uniformly continuous.

Put $S=\{y\in\mathbb{R}^{n}:\psi(y)\ne0\}$. If $y\notin\bar{B}(0,R)$ then $R<\lVert y\rVert$, so $\psi(y)=0$ by hypothesis and $y\notin S$; hence $S\subseteq\bar{B}(0,R)$. The support of $\psi$ is the closure of $S$, that is, the intersection of all closed sets containing $S$. By Elementary Properties of the Closed Ball in a Metric Space the set $\bar{B}(0,R)$ is closed and bounded, so the support of $\psi$ is contained in $\bar{B}(0,R)$ and is therefore bounded, and it is closed. By The Heine-Borel Theorem in $\mathbb{R}^n$, and Compactness of Closed Balls and the Unit Sphere it is compact, so $\psi$ is compactly supported.

Consequently $\psi$ is bounded by claim 1 of A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Bounded and Integrable and integrable with respect to $\lambda_{n}$ by claim 2 of that lemma. Fix a real number $M$ with $0\le M$ and $|\psi(y)|\le M$ for every $y\in\mathbb{R}^{n}$. Finally, $\psi$ is uniformly continuous on $\mathbb{R}^{n}$ by A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Uniformly Continuous.

Step 2. Two facts used repeatedly.

(a) For $x\in\mathbb{R}^{n}$ let $T_{x}:\mathbb{R}^{n}\to\mathbb{R}^{n}$ be given by $T_{x}(y)=x-y$. Then $T_{x}$ is continuous: for $y,y'\in\mathbb{R}^{n}$ we have $(x-y)-(x-y')=y'-y$, hence, using claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square with the scalar $-1$ together with claim 2 of Properties of the Absolute Value in an Ordered Field,

$$d_{E}(T_{x}(y),T_{x}(y'))=\lVert y'-y\rVert=\lVert y-y'\rVert=d_{E}(y,y'),$$

so the definition of continuity at each point is met with $\delta=\varepsilon$.

(b) For $x\in\mathbb{R}^{n}$ and a real number $r$ with $0<r$, the set $\bar{B}(x,r)$ is closed and bounded by Elementary Properties of the Closed Ball in a Metric Space, hence belongs to $\mathcal{B}(\mathbb{R}^{n})$ by claims 4 and 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets. It is therefore a bounded Borel set, so $\mathbf{1}_{\bar{B}(x,r)}w$ is integrable by the hypothesis on $w$.

We also record the following, used several times. If $g:\mathbb{R}^{n}\to\mathbb{R}$ is integrable, then $-|g|\le g\le|g|$ pointwise by claim 3 of Properties of the Absolute Value in an Ordered Field, so by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, which supplies both linearity and monotonicity of the integral of integrable maps,

$$-\int_{\mathbb{R}^{n}}|g|\,d\lambda_{n}\le\int_{\mathbb{R}^{n}}g\,d\lambda_{n}\le\int_{\mathbb{R}^{n}}|g|\,d\lambda_{n},$$

whence by claim 6 of Properties of the Absolute Value in an Ordered Field,

$$\Bigl|\int_{\mathbb{R}^{n}}g\,d\lambda_{n}\Bigr|\le\int_{\mathbb{R}^{n}}|g|\,d\lambda_{n}. \tag{$\ast$}$$

Step 3. Proof of claim 1.

Fix $x\in\mathbb{R}^{n}$ and put $f_{x}(y)=\psi(x-y)\,w(y)$ for $y\in\mathbb{R}^{n}$, so that $f_{x}=(\psi\circ T_{x})\,w$. By step 2(a) and Composition of Continuous Euclidean Maps the map $\psi\circ T_{x}$ is continuous, hence measurable with respect to $\mathcal{B}(\mathbb{R}^{n})$ by claims 3(a) and 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets; since $w$ is measurable, $f_{x}$ is measurable by Arithmetic of Measurable Real- and [0,∞]-Valued Maps.

Put $B=\bar{B}(x,R)$. If $y\notin B$ then $R<d_{E}(x,y)=\lVert x-y\rVert$, so $\psi(x-y)=0$ and $f_{x}(y)=0$. Hence for every $y\in\mathbb{R}^{n}$,

$$|f_{x}(y)|\le M\,\mathbf{1}_{B}(y)\,|w(y)|,$$

since for $y\in B$ we have $|f_{x}(y)|=|\psi(x-y)|\,|w(y)|\le M|w(y)|$ by claim 4 of Properties of the Absolute Value in an Ordered Field and claim 5 of Elementary Arithmetic in an Ordered Field, while for $y\notin B$ both sides are $0$. The map $\mathbf{1}_{B}w$ is integrable by step 2(b); its absolute value $\mathbf{1}_{B}|w|$ is then integrable, because by Integrable Functions and Their Lebesgue Integral a measurable map is integrable exactly when the integral of its absolute value is finite, and $|\mathbf{1}_{B}w|=\mathbf{1}_{B}|w|$ by claims 1 and 4 of Properties of the Absolute Value in an Ordered Field, the first giving $|\mathbf{1}_{B}|=\mathbf{1}_{B}$ from nonnegativity of the indicator. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion the map $M\,\mathbf{1}_{B}|w|$ is integrable. Since $|f_{x}|\le M\,\mathbf{1}_{B}|w|$ pointwise and both sides are measurable maps into $[0,\infty)$, claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives

$$\int_{\mathbb{R}^{n}}|f_{x}|\,d\lambda_{n}\le\int_{\mathbb{R}^{n}}M\,\mathbf{1}_{B}|w|\,d\lambda_{n}<\infty,$$

so $f_{x}$ is integrable by Integrable Functions and Their Lebesgue Integral, a measurable map being integrable exactly when the integral of its absolute value is finite. We refer to this argument below as the domination criterion. This proves claim 1.

Step 4. Proof of claim 2.

Fix $x\in\mathbb{R}^{n}$ and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Put $B'=\bar{B}(x,R+1)$ and

$$I=\int_{\mathbb{R}^{n}}\mathbf{1}_{B'}|w|\,d\lambda_{n},$$

a real number with $0\le I$ by step 2(b) and the argument of step 3. The number $1+I$ is positive, so $\theta=\varepsilon/(2(1+I))$ is a positive real number by claims 5 and 7 of Elementary Order Arithmetic in an Ordered Field. By step 1 there is a real number $\eta$ with $0<\eta$ such that $|\psi(a)-\psi(b)|<\theta$ whenever $d_{E}(a,b)<\eta$. Put $\eta'=\min\{\eta,1\}$, a positive real by claim 9 of Elementary Order Arithmetic in an Ordered Field.

Let $x'\in\mathbb{R}^{n}$ with $d_{E}(x,x')<\eta'$, and let $y\in\mathbb{R}^{n}$. First, $\lVert(x'-y)-(x-y)\rVert=\lVert x'-x\rVert<\eta'\le\eta$, so $|\psi(x'-y)-\psi(x-y)|<\theta$. Second, if $y\notin B'$ then $R+1<\lVert x-y\rVert$, and by the triangle inequality of claim 6 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square in the form $\lVert x-y\rVert\le\lVert x'-y\rVert+\lVert x-x'\rVert$ we get $\lVert x'-y\rVert\ge\lVert x-y\rVert-\lVert x-x'\rVert>(R+1)-1=R$, so $\psi(x'-y)=\psi(x-y)=0$. Combining,

$$|f_{x'}(y)-f_{x}(y)|=|\psi(x'-y)-\psi(x-y)|\,|w(y)|\le\theta\,\mathbf{1}_{B'}(y)\,|w(y)|$$

for every $y$. Both $f_{x'}$ and $f_{x}$ are integrable by claim 1, so $f_{x'}-f_{x}$ is integrable by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, and by $(\ast)$, monotonicity, and linearity,

$$\bigl|(\psi\star w)(x')-(\psi\star w)(x)\bigr|\le\int_{\mathbb{R}^{n}}|f_{x'}-f_{x}|\,d\lambda_{n}\le\theta I\le\theta(1+I)=\varepsilon/2<\varepsilon .$$

Since $\varepsilon$ was arbitrary, $\psi\star w$ is continuous at $x$; since $x$ was arbitrary, claim 2 follows.

Step 5. Proof of claim 3.

The map $w+c\,w'$ is measurable by Arithmetic of Measurable Real- and [0,∞]-Valued Maps, and for bounded $B\in\mathcal{B}(\mathbb{R}^{n})$ the map $\mathbf{1}_{B}(w+c\,w')=\mathbf{1}_{B}w+c\,\mathbf{1}_{B}w'$ is integrable by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion; so $w+c\,w'$ satisfies the hypotheses placed on $w$. The map $\psi+c\,\psi'$ is continuous by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space and vanishes at every $y$ with $R<\lVert y\rVert$, since both $\psi$ and $\psi'$ do; so it satisfies the hypotheses placed on $\psi$. For each fixed $x$, the pointwise identities

$$\psi(x-y)\bigl(w(y)+c\,w'(y)\bigr)=\psi(x-y)w(y)+c\,\psi(x-y)w'(y),$$ $$\bigl(\psi+c\,\psi'\bigr)(x-y)\,w(y)=\psi(x-y)w(y)+c\,\psi'(x-y)w(y)$$

hold for every $y$, all the maps involved being integrable by claim 1; integrating and applying claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives both identities of claim 3.

Step 6. Proof of claim 4.

Assume $\psi$ is of class $C^{1}$ on $\mathbb{R}^{n}$ and let $i\in[n]$. By Functions of Class Cᵏ on an Open Subset of ℝⁿ the partial derivative $\partial_{i}\psi$ exists at every point of $\mathbb{R}^{n}$ and is continuous.

The differentiated kernel vanishes outside the ball. Let $y$ satisfy $R<\lVert y\rVert$. The set $V=\{z\in\mathbb{R}^{n}:R<\lVert z\rVert\}$ is the complement of $\bar{B}(0,R)$, which is closed by Elementary Properties of the Closed Ball in a Metric Space, so $V$ is open and there is a real $r$ with $0<r$ such that every $z$ with $d_{E}(y,z)<r$ lies in $V$. Let $e_{i}\in\mathbb{R}^{n}$ be the point whose $i$th coordinate is $1$ and whose other coordinates are $0$, so that $\lVert e_{i}\rVert=1$ by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and hence $\lVert t\,e_{i}\rVert=|t|$ for $t\in\mathbb{R}$ by claim 5 of the same lemma. For $0<|t|<r$ we then have $y+t\,e_{i}\in V$, hence $\psi(y+t\,e_{i})=\psi(y)=0$ and the difference quotient $\bigl(\psi(y+t\,e_{i})-\psi(y)\bigr)/t$ equals $0$. By Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$ the partial derivative $\partial_{i}\psi(y)$ is the limit of these quotients as $t$ tends to $0$, and that limit is $0$: given $\varepsilon>0$, the choice $\delta=r$ makes the quotient equal to $0$ for $0<|t|<\delta$. Thus $\partial_{i}\psi(y)=0$ whenever $R<\lVert y\rVert$, and claims 1 and 2 apply verbatim with the kernel $\partial_{i}\psi$ in place of $\psi$; in particular $(\partial_{i}\psi)\star w$ is defined and continuous, and by step 1 applied to $\partial_{i}\psi$ there is a real $M_{i}$ with $0\le M_{i}$ and $|\partial_{i}\psi(z)|\le M_{i}$ for every $z$.

Differentiation. Fix $x\in\mathbb{R}^{n}$, and let $U$ be the open interval with endpoints $-1$ and $1$. Define $F:U\times\mathbb{R}^{n}\to\mathbb{R}$ by $F(t,y)=\psi(x+t\,e_{i}-y)\,w(y)$. We verify the three hypotheses of Differentiation under the Integral Sign for the measure space $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}),\lambda_{n})$.

(i) For each $t\in U$ the map $y\mapsto F(t,y)$ is $f_{x+t\,e_{i}}$, integrable by claim 1.

(ii) For each $y\in\mathbb{R}^{n}$ and each $t\in U$, the map $s\mapsto\psi(x+s\,e_{i}-y)$ has at $s=t$ the difference quotients $\bigl(\psi\bigl((x+t\,e_{i}-y)+h\,e_{i}\bigr)-\psi(x+t\,e_{i}-y)\bigr)/h$, whose limit as $h$ tends to $0$ is $\partial_{i}\psi(x+t\,e_{i}-y)$ by Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$; multiplying by the constant $w(y)$, the map $s\mapsto F(s,y)$ is differentiable at $t$ with $D_{1}F(t,y)=\partial_{i}\psi(x+t\,e_{i}-y)\,w(y)$, by the constant-multiple rule, claim 2 of Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers.

(iii) Put $B''=\bar{B}(x,R+1)$. For $t\in U$ we have $\lVert t\,e_{i}\rVert=|t|<1$. If $y\notin B''$ then $R+1<\lVert x-y\rVert$, and as in step 4, $\lVert x+t\,e_{i}-y\rVert\ge\lVert x-y\rVert-\lVert t\,e_{i}\rVert>R$, so $\partial_{i}\psi(x+t\,e_{i}-y)=0$. Hence $|D_{1}F(t,y)|\le M_{i}\,\mathbf{1}_{B''}(y)\,|w(y)|$ for every $t\in U$ and every $y$, and the right-hand side is integrable exactly as in step 3.

By Differentiation under the Integral Sign the map $G:U\to\mathbb{R}$ with $G(t)=\int_{\mathbb{R}^{n}}F(t,y)\,d\lambda_{n}(y)=(\psi\star w)(x+t\,e_{i})$ is differentiable at every $t\in U$, with

$$G'(t)=\int_{\mathbb{R}^{n}}\partial_{i}\psi(x+t\,e_{i}-y)\,w(y)\,d\lambda_{n}(y).$$

Taking $t=0$, the difference quotients of $G$ at $0$ are exactly the difference quotients defining $\partial_{i}(\psi\star w)(x)$ in Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$, so that partial derivative exists and

$$\partial_{i}(\psi\star w)(x)=\int_{\mathbb{R}^{n}}\partial_{i}\psi(x-y)\,w(y)\,d\lambda_{n}(y)=\bigl((\partial_{i}\psi)\star w\bigr)(x).$$

As $x$ and $i$ were arbitrary, the displayed identity of claim 4 holds. Each $\partial_{i}(\psi\star w)=(\partial_{i}\psi)\star w$ is continuous by claim 2 applied to the kernel $\partial_{i}\psi$, and $\psi\star w$ is itself continuous by claim 2; by Functions of Class Cᵏ on an Open Subset of ℝⁿ this makes $\psi\star w$ of class $C^{1}$ on $\mathbb{R}^{n}$.

Step 7. Proof of claim 5.

We argue by induction on the natural number $k$, the statement being: for every continuous $\psi$ vanishing at every $y$ with $R<\lVert y\rVert$ and of class $C^{k}$ on $\mathbb{R}^{n}$, the map $\psi\star w$ is of class $C^{k}$ on $\mathbb{R}^{n}$.

The case $k=1$ is the last assertion of claim 4. Assume the statement for $k$ and let $\psi$ be of class $C^{k+1}$. By Functions of Class Cᵏ on an Open Subset of ℝⁿ the map $\psi$ is of class $C^{1}$ and each $\partial_{i}\psi$ is of class $C^{k}$. By claim 4, $\psi\star w$ is of class $C^{1}$ with $\partial_{i}(\psi\star w)=(\partial_{i}\psi)\star w$; by step 6 each $\partial_{i}\psi$ is continuous and vanishes at every $y$ with $R<\lVert y\rVert$, so the induction hypothesis applies to the kernel $\partial_{i}\psi$ and gives that $(\partial_{i}\psi)\star w$ is of class $C^{k}$. Hence $\psi\star w$ is of class $C^{1}$ with all its first partial derivatives of class $C^{k}$, that is, of class $C^{k+1}$ by Functions of Class Cᵏ on an Open Subset of ℝⁿ. This completes the induction.

Finally, by Smooth Functions on an Open Subset of ℝⁿ a map is smooth on $\mathbb{R}^{n}$ exactly when it is of class $C^{k}$ on $\mathbb{R}^{n}$ for every natural number $k$. If $\psi$ is smooth then $\psi$ is of class $C^{k}$ for every $k$, so $\psi\star w$ is of class $C^{k}$ for every $k$ by what has just been proved, and therefore smooth. $\blacksquare$