Convolution and Differentiation of Distributions

I’ll review (extremely rapidly!) to fix notation. We equip the vector space $C_c^{\mathrm{\infty }}$ of smooth compactly supported real valued functions on ${\mathbb{R}}^n$ with the following topology: a sequence ${\varphi }_n$ converges to $\varphi$ if the ${\varphi }_n$ and all their derivatives converge to $\varphi$ and all of its derivatives with respect to the norm $||\varphi ||={\int }_{{\mathbb{R}}^n}|\varphi |d\mathrm{vol}$. The topological dual of $\mathcal{D}$ of $C_c^{\mathrm{\infty }}$ is the set of distributions on ${\mathbb{R}}^n$, or continuous linear functionals $C_c^{\mathrm{\infty }}\to \mathbb{R}$.

Every smooth function $f\in C_c^{\mathrm{\infty }}$ naturally gives rise to a distribution $F\in \mathcal{D}$ via integration, $$f\mapsto FF(\varphi )={\int }_{{\mathbb{R}}^n}f\varphi d\mathrm{vol},$$ but not all distributions are of this form. Important examples are given by the delta distributions: for any $p\in {\mathbb{R}}^n$ we define ${\delta }_p\in \mathcal{D}$ by ${\delta }_p(\varphi ):=\varphi (p)$, and we write $\delta$ for ${\delta }_0$.

The set of distributions is closed under differentiation, where the derivative of a distribution $F$ is defined by its action on a function $\varphi$ in analogy to integration by parts. When $M=\mathbb{R}$ this lets us define the first derivative $F^{\prime }$ of $F$ by $F^{\prime }(\varphi ):=-F({\varphi }^{\prime })$. More generally, if $\mathrm{\nabla }$ is some differential operator on $C_c^{\mathrm{\infty }}$ we define $\mathrm{\nabla }$ on $\mathcal{D}$ by $$\mathrm{\nabla }F(\varphi ):=-F\left(\mathrm{\nabla }\varphi \right)$$

Distributions can be multiplied by smooth functions: if $\psi :{\mathbb{R}}^n\to \mathbb{R}$ is smooth and $F\in \mathcal{D}$, we define $\psi F$ to be the distribution such that $(\psi F)(\varphi )=F(\psi \varphi )$ for all $\varphi \in C_c^{\mathrm{\infty }}$. Distributions can also be convolved with functions in $C_c^{\mathrm{\infty }}$, but this operation now yields smooth functions rather than distributions: if $F\in \mathcal{D}$ and $g\in C_c^{\mathrm{\infty }}$, their convolutional product $F\star g$ is defined below, where $g(p-\cdot )$ is the function $x\mapsto g(p-x)$. $$F\star g:p\mapsto F(g(p-\cdot ))$$

As an example, if $\varphi \in C_c^{\mathrm{\infty }}$, we compute the value of $\delta \star \varphi$ at $p\in {\mathbb{R}}^n$ aa $$(\delta \star \varphi )(p):=\delta (\varphi (p-\cdot ))=\varphi (p-0)=\varphi (p)$$ Thus, $\delta \star \varphi =\varphi$, and colvolution with $\delta$ realizes the identity operator on $C_c^{\mathrm{\infty }}$.

Differentiation and convolution interact in a particularly simple way, if $\mathrm{\nabla }$ is a linear differential operator and $F\in \mathcal{D}$, $g\in C_c^{\mathrm{\infty }}$, then we may differentiate the function $F\star g$ by $$\mathrm{\nabla }(F\star g)=(\mathrm{\nabla }F)\star g=F\star (\mathrm{\nabla }g)$$

Once you know this line of equalities, the motivation for introducing distributions to solve PDE’s becomes clear! The general goal is to calculate (hopefully easier to find) distributional solutions to a differential equation to actual, real-valued function solutions through convolution.If $\mathrm{\nabla }$ is a differential operator we say a distribution $F$ is a fundamental solution for $\mathrm{\nabla }$ if $\mathrm{\nabla }F=\delta$ Such a fundamental solution lets us find a real solution to the differential equation $\mathrm{\nabla }u=g$, with $g\in C_c^{\mathrm{\infty }}$ by simply taking $u=F\star g$, as we easily confirm:

$$\mathrm{\nabla }(F\star g)=(\mathrm{\nabla }F)\star g=\delta \star g=g$$

Of course, this (crucially!) relies on the fact that $\mathrm{\nabla }(F\star g)=(\mathrm{\nabla }F)\star g$, and recently I realized I had completely forgotten how to prove this fact! Luckily, shortly after Daniel O’Connor showed me how it works, and so I want to write it down for the next time that I forget.

Proving $\mathrm{\nabla }(F\star g)=(\mathrm{\nabla }F)\star g=F\star (\mathrm{\nabla }g)$

To prove this, we start small and build up to the general case. The interesting part is actually this small beginning however, the rest is just packaging.

Theorem 1: On $\mathbb{R}$, suppose $F\in \mathcal{D}$ and $g\in C_c^{\mathrm{\infty }}$. Then $F\star g$ is differentiable and $(F\star g{)}^{\prime }=(F^{\prime })\star g$.

Proof: Here’s Daniel’s argument: Let $x\in \mathbb{R}$, $h\ne 0$ and consider the difference quotient $${\psi }_h(x):=\frac{(F\star g)(x+h)-(F\star g)(x)}{h}$$ If the limit $\underset{h\to 0}{lim}{\psi }_h(x)$ exists, then $F\star g$ is differentiable at $x$. Using the definition of $F\star g$ and the linearity of $F$, we may evaluate this as $${\psi }_h(x)=\frac{F(g(x+h-\cdot ))-F(g(x-\cdot ))}{h}$$ $$=F\left(\frac{g(x+h-\cdot )-g(x-\cdot )}{h}\right)$$

Using the continuity of $F$, we may take the limit inside, and so $$\underset{h\to 0}{lim}{\psi }_h(x)=F\left(\underset{h\to 0}{lim}\frac{g(x+h-\cdot )-g(x-\cdot )}{h}\right).$$

The quantity inside of $F$ attempts to assign to each $p\in \mathbb{R}$ the value $$p\mapsto \underset{h\to 0}{lim}\frac{g(x+h-p)-g(x-p)}{h}=g^{\prime }(x-p)$$ so in our notation, this is the function $g^{\prime }(x-\cdot )$, which itself is in $C_c^{\mathrm{\infty }}$ as $g$ was. Thus, $(F\star g{)}^{\prime }(x)$ exists, and $(F\star g{)}^{\prime }(x)=F(g^{\prime }(x-\cdot ))$. But this new term is exactly the definition of $F$ convolved with $g^{\prime }$, when evaluated at $x$! Thus as functions, we have shown $$(F\star g{)}^{\prime }=F\star g^{\prime }$$

This is half of what we want, but the rest is just a straightforward application of the definition of the distributional derivative. By definition, $F^{\prime }$ is the linear functional such that $F^{\prime }(\varphi )=-F({\varphi }^{\prime })$ for all $\varphi \in C_c^{\mathrm{\infty }}$, so computing $(F^{\prime })\star g$, we see for $x\in \mathbb{R}$ $$(F^{\prime }\star g)(x)=F^{\prime }(g(x-\cdot )):=-F(g(x-\cdot {)}^{\prime })$$

Where $g(x-\cdot {)}^{\prime }$ is the function sending $p\mapsto \frac{d}{dp}g(x-p)$. Computing this derivative with the chain rule shows $g(x-\cdot {)}^{\prime }=-g^{\prime }(x-\cdot )$, and so $$-F(g(x-\cdot {)}^{\prime })=-F(-g^{\prime }(x-\cdot ))=F(g^{\prime }(x-\cdot ))$$

Stringing all this together, we see $(F^{\prime }\star g)(x)=F(g^{\prime }(x-\cdot ))$, where we recognize this second term as defining the convolution $F\star g^{\prime }(x)$. As this equality holds for all $x\in \mathbb{R}$ we have equality between functions: $$F^{\prime }\star g=F\star g^{\prime }$$

Combining with our earlier result proves the theorem, as we have shown both $(F\star g{)}^{\prime }$ and $F^{\prime }\star g$ are equal to $F\star g^{\prime }$.

Corollary 2: Let $D^k$ be the $k^{th}$ derivative operator on $C_c^{\mathrm{\infty }}(\mathbb{R})$. Then for any $F\in \mathcal{D}$ and $g\in C_c^{\mathrm{\infty }}$, the convolution $F\star g$ is $k$ times differentiable and $D^k(F\star g)=(D^{k}F)\star g=F\star (D^{k}g)$

Proof: We can proceed inductively using Theorem 1, as $D^k=D\circ D^{k-1}$ is the $k$-fold composition of the first derivative operator.

Lemma 3: On ${\mathbb{R}}^n$, let ${\partial }_x$ denote the directional derivative with respect to the first coordinate. Then for any $F\in \mathcal{D}$ and $g\in C_c^{\mathrm{\infty }}$, ${\partial }_x(F\star g)$ is smooth, and ${\partial }_x(F\star g)=({\partial }_{x}F)\star g=F\star ({\partial }_{x}g)$.

Proof: The proof is exactly analogous to the one dimensional case in Theorem 1, so we can proceed rather quickly. It suffices to check this equality holds at an arbitrary fixed $p\in {\mathbb{R}}^n$, where $${\partial }_x(F\star g)(p)=\underset{h\to 0}{lim}\frac{(F\star g)(p+h{e}_1)-(F\star g)(p)}{h}$$ Evaluatiting the convolutions and using the linearity and continuity of $F$ shows this to be $F({\partial }_{x}g(p-\cdot ))$, which is the convolution of $F$ with ${\partial }_{x}g$ evaluated at $p$. Thus, ${\partial }_x(F\star g)=F\star ({\partial }_{x}g).$ The second equality again follows simply by starting the definition of ${\partial }_{x}F$ to compute $({\partial }_{x}F)\star g$ at $p$, resulting in $({\partial }_{x}F)\star g=F\star ({\partial }_{x}g).$

Corollary 4: If $L={\partial }_x^a{\partial }_y^b{\partial }_z^c\cdots$ is any monomial in the coordinate partial derivative operators on ${\mathbb{R}}^n$, then for any $F\in \mathcal{D}$ and $g\in C_c^{\mathrm{\infty }}$, $$L(F\star g)=(LF)\star g=F\star (Lg)$$

Proof: As in corollary 2, we inductively apply Lemma 3 to each partial derivative operator which shows up in $L$.

To build upwards from this, its useful to stop for a second and factorize out a little argument about convolution:

Lemma 5: If $F,\mathrm{\Phi }$ are distributions and $g,\gamma$ a smooth compactly supported functions, then $(F+\mathrm{\Phi })\star g=F\star g+\mathrm{\Phi }\star g$ and $F\star (g+\gamma )=F\star g+F\star \gamma$.

Proof: Let $x\in {\mathbb{R}}^n$. First consider $F\star (g+\gamma )$ evaluated at $x$. This is by definition $F((g+\gamma )(x-\cdot ))$, that is, $F(g(x-\cdot )+\gamma (x-\cdot ))$. Using the linearity of $F$, we see this to be $F(g(x-\cdot ))+F(\gamma (x-\cdot ))$, which is by definition $(F\star g)(x)+(F\star \gamma )(x)$. Thus, $$F\star (g+\gamma )=F\star g+F\star \gamma .$$

Next, consider$(F+\mathrm{\Phi })\star g$ evaluated at $x$. By the definition of convolution, $((F+\mathrm{\Phi })\star g)(x)=$$(F+\mathrm{\Phi })(g(x-\cdot ))$. Using the definition of $+$ in $\mathcal{D}$, we distribute as $(F+\mathrm{\Phi })(g(x-\cdot ))=$$F(g(x-\cdot ))+\mathrm{\Phi }(g(x-\cdot ))$, where the last terms are each by definition equal to $(F\star g)(x)$ and $(\mathrm{\Phi }\star g)(x)$ respectively. Thus, $$(F+\mathrm{\Phi })\star g=F\star g+\mathrm{\Phi }\star g.$$

Lemma 6: Let $L_1,L_2$ be a differential operator on ${\mathbb{R}}^n$ such that $L_i(F\star g)=(L_{i}F)\star g=F\star (L_{i}g)$ for $i\in 1,2$ and any $F\in \mathcal{D}$, $g\in C_c^{\mathrm{\infty }}$. Then $L=L_1+L_2$ also satisfies $L(F\star g)=(LF)\star g=F\star Lg$ for all $F,g$.

Proof: The differential operator $L=L_1+L_2$ acts on functions by $L\varphi =L_1\varphi +L_2\varphi$. Thus if $F\in \mathcal{D}$, $g\in C_c^{\mathrm{\infty }}$, $(L_1+L_2)(F\star g)=$$L_1(F\star g)+L_2(F\star g)$.

To get the first of the two claimed equalities, we can use half our hypothesis on the $L_i$ to re-write this as $(L_{1}F)\star g+(L_{2}F)\star g$, and then use Lemma 5 to factor out the convolution giving $(L_1+L_2)(F\star g)=$$(L_{1}F+L_{2}F)\star g$. Factoring out the $F$ gives what we wanted: $$(L_1+L_2)(F\star g)=((L_1+L_2)F)\star g$$

To get the second equality, we use the other half of our assumption on the $L_i$ to rewrite $L_1(F\star g)+L_2(F\star g)$ as $F\star (L_{1}g)+F\star (L_{2}g)$. We use Lemma 5 to factor out the distribution from this convolution, followed by the further factoring $L_{1}g+L_{2}g=(L_1+L_2)g$. All together, this gives what we wanted: $$(L_1+L_2)(F\star g)=F\star ((L_1+L_2)g)$$

Lemma 7: Let $L$ be a differential operator on ${\mathbb{R}}^n$ such that $L(F\star g)=(LF)\star g=F\star (Lg)$ for any $F\in \mathcal{D}$ and $g\in C_c^{\mathrm{\infty }}$. Then if $\psi :{\mathbb{R}}^n\to R$ is any smooth function, the differential operator $K=\psi L$ defined by $K\varphi =\psi L(\varphi )$ also satisfies $K(F\star g)=(KF)\star g=F\star (Kg)$ for all $F,g$.

Proof: Evaluating $K(F\star g)=\psi \cdot L(F\star g)$,we can use that $L$ satisfies our hypothesis to conclude this is equal to $\psi \cdot ((LF)\star g)$ and $\psi \cdot (F\star (Lg))$. Taking the former and evaluating at $x\in {\mathbb{R}}^n$, we see it to equal $$\psi (x)((LF)\star g)(x)=\psi (x)(LF)(g(x-\cdot ))$$ At this fixed $x$, $\psi (x)$ is a constant, and so this is the same as evaluating the distribution $(\psi (x)LF)$ on $g(x-\cdot )$. But this is the definition of the convolution of $\psi (x)LF$ with the function $g$, evaluated at $x$! Thus, all together $\psi \cdot ((LF)\star g)$ is the function $x\mapsto ((\psi (x)LF)\star g)(x)$ which is the first half of what we want: $$K(F\star g)=\psi \cdot ((LF)\star g)=(\psi LF)\star g=(KF)\star g$$

The other case is similar, considering $\psi \cdot (F\star (Lg))$ evaluated at $x$. This yields $\psi (x)F((Lg)(x-\cdot ))$, and as $\psi (x)$ is a constant at this $x$, we may pull it inside to get $F(\psi (x)(Lg)(x-\cdot ))$. That is, $\psi \cdot (F\star (Lg))$ sends $x$ to the result of convolving the function $\psi (x)Lg$ with $F$, so $$K(F\star g)=\psi \cdot (F\star (Lg))=F\star (\psi Lg)=F\star (Kg)$$

Finally, we need a description of the class of linear differential operators on ${\mathbb{R}}^n$ which is amenable to our start-small-and-build-upwards approach:

Lemma 8: Any linear differential operrator on ${\mathbb{R}}^n$ is a multinomial in the partial derivative operators, with coefficients in $C^{\mathrm{\infty }}({\mathbb{R}}^n)$.

All the hard work is done, now its just putting the pieces together to state the main result:

Theorem 9: Let $\mathrm{\nabla }$ be any linear differential operator on ${\mathbb{R}}^n$. Then $\mathrm{\nabla }(F\star g)=(\mathrm{\nabla }F)\star g=F\star (\mathrm{\nabla }g)$ for all $F\in \mathcal{D}$, $g\in C_c^{\mathrm{\infty }}$.

Proof: We write $\mathrm{\nabla }$ as a multinomial in the partial derivatives, $$\mathrm{\nabla }=\sum _{[\alpha ]}{\psi }_{[\alpha ]}{\partial }^{[\alpha ]}$$ where $[\alpha ]=[a,b,c,\cdots ]$ ranges over some finite subset of all multi-indices, ${\partial }^{[\alpha ]}={\partial }_x^a{\partial }_y^b{\partial }_z^c\cdots$, and for each index ${\psi }_{[\alpha ]}$ is some smooth function ${\mathbb{R}}^n\to \mathbb{R}$. But as each ${\partial }^{[\alpha ]}$ satisfies the desired property by Corollary 4, we can apply Lemmas 6 and 7 finitely many times to conclude that $\mathrm{\nabla }$ does as well.