Convolution and Differentiation of Distributions

calculuscomplex analysis

Why you can freely pass derivatives through convolutions in distribution theory.

The purpose of this note is to provide a proof of the following theorem

TheoremConvolving with a Distribution

Let FF be a distribution, gg a smooth compactly supported function, and ∇\nabla some differential operator. ∇(F⋆g)=(∇F)⋆g\nabla (F\star g)=(\nabla F)\star g

I’ll review (extremely rapidly!) to fix notation. We equip the vector space Cc∞C_c^\infty of smooth compactly supported real valued functions on Rn\mathbb{R}^n with the following topology: a sequence ϕn\phi_n converges to ϕ\phi if the ϕn\phi_n and all their derivatives converge to ϕ\phi and all of its derivatives with respect to the norm ∥ϕ∥=∫Rn∣ϕ∣dvol\|\phi\|=\int_{\mathbb{R}^n}|\phi|d\mathrm{vol}. The topological dual D\mathcal{D} of Cc∞C_c^\infty is the set of distributions on Rn\mathbb{R}^n, or continuous linear functionals Cc∞→RC_c^\infty\to \mathbb{R}.

Every smooth function f∈Cc∞f\in C_c^\infty naturally gives rise to a distribution F∈DF\in\mathcal{D} via integration, f↦FF(ϕ)=∫Rnfϕdvol,f\mapsto F\hspace{1cm} F(\phi)=\int_{\mathbb{R}^n}f\phi d\mathrm{vol}, but not all distributions are of this form. Important examples are given by the delta distributions: for any p∈Rnp\in \mathbb{R}^n we define δp∈D\delta_p\in \mathcal{D} by δp(ϕ):=ϕ(p)\delta_p(\phi):=\phi(p), and we write δ\delta for δ0\delta_0.

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

Distributions can be multiplied by smooth functions: if ψ ⁣:Rn→R\psi\colon\mathbb{R}^n\to\mathbb{R} is smooth and F∈DF\in \mathcal{D}, we define ψF\psi F to be the distribution such that (ψF)(ϕ)=F(ψϕ)(\psi F)(\phi)=F(\psi\phi) for all ϕ∈Cc∞\phi\in C_c^\infty. Distributions can also be convolved with functions in Cc∞C_c^\infty, but this operation now yields smooth functions rather than distributions: if F∈DF\in\mathcal{D} and g∈Cc∞g\in C_c^\infty, their convolutional product F⋆gF\star g is defined below, where g(p−⋅)g(p-\cdot) is the function x↦g(p−x)x\mapsto g(p-x). F⋆g ⁣:p↦F(g(p−⋅))F\star g\colon\hspace{3mm} p\mapsto F\left(g(p-\cdot)\right)

Having defined everything in the theorem, our goal is clear: we wish to show that differentiating the function resulting from F⋆gF\star g is the same as the function that arises from convolving the distribution ∇F\nabla F with gg.

Why We Care

One use of this is to justify a rather clever approach to finding general solutions to PDEs. The goal is to calculate (hopefully easier to find) distributional solutions to a differential equation to actual, real-valued function solutions through convolution. More precisely, if ∇\nabla is a differential operator we say a distribution FF is a fundamental solution for ∇\nabla if ∇F=δ\nabla F=\delta.

The delta distribution is important here because of its particular relationship to convolution. For any ϕ∈Cc∞\phi\in C_c^\infty, we compute the value of δ⋆ϕ\delta\star\phi at p∈Rnp\in\mathbb{R}^n as (δ⋆ϕ)(p):=δ(ϕ(p−⋅))=ϕ(p−0)=ϕ(p)(\delta\star\phi)(p):=\delta\left(\phi(p-\cdot)\right)=\phi(p-0)=\phi(p) Thus, δ⋆ϕ=ϕ\delta\star\phi=\phi, and convolution with δ\delta realizes the identity operator on Cc∞C_c^\infty.

Knowing this, a fundamental solution lets us find a real solution to the differential equation ∇u=g\nabla u=g, as follows. Since g=δ⋆gg=\delta\star g and δ=∇F\delta = \nabla F we see that g=(∇F)⋆gg=(\nabla F)\star g. But, using the claimed theorem this can be rewritten as g=∇(F⋆g)g=\nabla(F\star g). But this is precisely the statement that the function u=F⋆gu=F\star g solves the equation ∇u=g\nabla u = g, as desired! Thus, the general solution to our PDE is simply the convolution of the fundamental solution with the initial condition.

Of course, this (crucially!) relies on the main theorem of this note 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 the Theorem

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

On R\mathbb{R}, suppose F∈DF\in \mathcal{D} and g∈Cc∞g\in C_c^\infty. Then F⋆gF\star g is differentiable and (F⋆g)′=(F′)⋆g(F\star g)^\prime=(F^\prime)\star g.

Proof

Here’s Daniel’s argument: Let x∈Rx\in \mathbb{R}, h≠0h\neq 0 and consider the difference quotient ψh(x):=(F⋆g)(x+h)−(F⋆g)(x)h\psi_h(x):=\frac{(F\star g)(x+h)-(F\star g)(x)}{h} If the limit lim⁡h→0ψh(x)\lim_{h\to 0}\psi_h(x) exists, then F⋆gF\star g is differentiable at xx. Using the definition of F⋆gF\star g and the linearity of FF, we may evaluate this as

ψh(x)=F(g(x+h−⋅))−F(g(x−⋅))h=F(g(x+h−⋅)−g(x−⋅)h)\begin{aligned} \psi_h(x)&=\frac{F\left(g(x+h-\cdot)\right)-F\left(g(x-\cdot)\right)}{h}\\[4pt] &=F\left(\frac{g(x+h-\cdot)-g(x-\cdot)}{h}\right) \end{aligned}

Using the continuity of FF, we may take the limit inside, and so lim⁡h→0ψh(x)=F(lim⁡h→0g(x+h−⋅)−g(x−⋅)h).\lim_{h\to 0}\psi_h(x)=F\left(\lim_{h\to 0}\frac{g(x+h-\cdot)-g(x-\cdot)}{h}\right).

The quantity inside of FF attempts to assign to each p∈Rp\in \mathbb{R} the value p↦lim⁡h→0g(x+h−p)−g(x−p)h=g′(x−p)p\mapsto \lim_{h\to 0}\frac{g(x+h-p)-g(x-p)}{h}=g^\prime(x-p) so in our notation, this is the function g′(x−⋅)g^\prime(x-\cdot), which itself is in Cc∞C_c^\infty as gg was. Thus, (F⋆g)′(x)(F\star g)^\prime(x) exists, and (F⋆g)′(x)=F(g′(x−⋅))(F\star g)^\prime(x)=F(g^\prime(x-\cdot)). But this new term is exactly the definition of FF convolved with g′g^\prime, when evaluated at xx! Thus as functions, we have shown (F⋆g)′=F⋆g′(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′F^\prime is the linear functional such that F′(ϕ)=−F(ϕ′)F^\prime(\phi)=-F(\phi^\prime) for all ϕ∈Cc∞\phi\in C_c^\infty, so computing (F′)⋆g(F^\prime)\star g, we see for x∈Rx\in\mathbb{R} (F′⋆g)(x)=F′(g(x−⋅)):=−F(g(x−⋅)′)(F^\prime\star g)(x)=F^\prime\left(g(x-\cdot)\right):=-F(g(x-\cdot)^\prime)

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

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

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

Corollary

Let DkD^k be the kthk^{th} derivative operator on Cc∞(R)C_c^\infty(\mathbb{R}). Then for any F∈DF\in\mathcal{D} and g∈Cc∞g\in C_c^\infty, the convolution F⋆gF\star g is kk times differentiable and Dk(F⋆g)=(DkF)⋆g=F⋆(Dkg)D^k(F\star g)=(D^k F)\star g=F\star (D^k g)

Proof

We can proceed inductively using Theorem 1, as Dk=D∘Dk−1D^k=D\circ D^{k-1} is the kk-fold composition of the first derivative operator.

Lemma

On Rn\mathbb{R}^n, let ∂x\partial_x denote the directional derivative with respect to the first coordinate. Then for any F∈DF\in\mathcal{D} and g∈Cc∞g\in C_c^\infty, ∂x(F⋆g)\partial_x(F\star g) is smooth, and ∂x(F⋆g)=(∂xF)⋆g=F⋆(∂xg)\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∈Rnp\in\mathbb{R}^n, where ∂x(F⋆g)(p)=lim⁡h→0(F⋆g)(p+he1)−(F⋆g)(p)h\partial_x(F\star g)(p)=\lim_{h\to 0}\frac{(F\star g)(p+he_1)-(F\star g)(p)}{h} Evaluating the convolutions and using the linearity and continuity of FF shows this to be F(∂xg(p−⋅))F\left(\partial_x g(p-\cdot)\right), which is the convolution of FF with ∂xg\partial_xg evaluated at pp. Thus, ∂x(F⋆g)=F⋆(∂xg).\partial_x(F\star g)=F\star(\partial_x g). The second equality again follows simply by stating the definition of ∂xF\partial_xF to compute (∂xF)⋆g(\partial_xF)\star g at pp, resulting in (∂xF)⋆g=F⋆(∂xg).(\partial_xF)\star g=F\star(\partial_x g).

Corollary

If L=∂xa∂yb∂zc⋯L=\partial_x^a\partial_y^b\partial_z^c\cdots is any monomial in the coordinate partial derivative operators on Rn\mathbb{R}^n, then for any F∈DF\in\mathcal{D} and g∈Cc∞g\in C_c^\infty, L(F⋆g)=(LF)⋆g=F⋆(Lg)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 LL.

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

Lemma

If F,ΦF,\Phi are distributions and g,γg,\gamma are smooth compactly supported functions, then (F+Φ)⋆g=F⋆g+Φ⋆g(F+\Phi)\star g=F\star g+\Phi\star g and F⋆(g+γ)=F⋆g+F⋆γF\star(g+\gamma)=F\star g+F\star\gamma.

Proof

Let x∈Rnx\in\mathbb{R}^n. First consider F⋆(g+γ)F\star(g+\gamma) evaluated at xx. This is by definition F((g+γ)(x−⋅))F((g+\gamma)(x-\cdot)), that is, F(g(x−⋅)+γ(x−⋅))F(g(x-\cdot)+\gamma(x-\cdot)). Using the linearity of FF, we see this to be F(g(x−⋅))+F(γ(x−⋅))F(g(x-\cdot))+F(\gamma(x-\cdot)), which is by definition (F⋆g)(x)+(F⋆γ)(x)(F\star g)(x)+(F\star\gamma)(x). Thus, F⋆(g+γ)=F⋆g+F⋆γ.F\star(g+\gamma)=F\star g+F\star \gamma.

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

Lemma

Let L1,L2L_1,L_2 be a differential operator on Rn\mathbb{R}^n such that Li(F⋆g)=(LiF)⋆g=F⋆(Lig)L_i(F\star g)=(L_iF)\star g=F\star(L_ig) for i∈{1,2}i\in\{1,2\} and any F∈DF\in\mathcal{D}, g∈Cc∞g\in C_c^\infty. Then L=L1+L2L=L_1+L_2 also satisfies L(F⋆g)=(LF)⋆g=F⋆LgL(F\star g)=(LF)\star g=F\star Lg for all F,gF,g.

Proof

The differential operator L=L1+L2L=L_1+L_2 acts on functions by Lϕ=L1ϕ+L2ϕL\phi=L_1\phi+L_2\phi. Thus if F∈DF\in\mathcal{D}, g∈Cc∞g\in C_c^\infty, (L1+L2)(F⋆g)=L1(F⋆g)+L2(F⋆g)(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 LiL_i to re-write this as (L1F)⋆g+(L2F)⋆g(L_1F)\star g+(L_2 F)\star g, and then use Lemma 5 to factor out the convolution giving (L1+L2)(F⋆g)=(L1F+L2F)⋆g(L_1+L_2)(F\star g)=(L_1F+L_2F)\star g. Factoring out the FF gives what we wanted: (L1+L2)(F⋆g)=((L1+L2)F)⋆g(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 LiL_i to rewrite L1(F⋆g)+L2(F⋆g)L_1(F\star g)+L_2(F\star g) as F⋆(L1g)+F⋆(L2g)F\star(L_1g)+F\star(L_2g). We use Lemma 5 to factor out the distribution from this convolution, followed by the further factoring L1g+L2g=(L1+L2)gL_1g+L_2g=(L_1+L_2)g. All together, this gives what we wanted: (L1+L2)(F⋆g)=F⋆((L1+L2)g)(L_1+L_2)(F\star g)=F\star((L_1+L_2)g)

Lemma

Let LL be a differential operator on Rn\mathbb{R}^n such that L(F⋆g)=(LF)⋆g=F⋆(Lg)L(F\star g)=(LF)\star g=F\star(Lg) for any F∈DF\in\mathcal{D} and g∈Cc∞g\in C_c^\infty. Then if ψ ⁣:Rn→R\psi\colon\mathbb{R}^n\to\mathbb{R} is any smooth function, the differential operator K=ψLK=\psi L defined by Kϕ=ψL(ϕ)K\phi=\psi L(\phi) also satisfies K(F⋆g)=(KF)⋆g=F⋆(Kg)K(F\star g)=(KF)\star g=F\star (Kg) for all F,gF,g.

Proof

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

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

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

Fact

Any linear differential operator on Rn\mathbb{R}^n is a multinomial in the partial derivative operators, with coefficients in C∞(Rn)C^\infty(\mathbb{R}^n).

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

Theorem

Let ∇\nabla be any linear differential operator on Rn\mathbb{R}^n. Then ∇(F⋆g)=(∇F)⋆g=F⋆(∇g)\nabla(F\star g)=(\nabla F)\star g=F\star(\nabla g) for all F∈DF\in \mathcal{D}, g∈Cc∞g\in C_c^\infty.

Proof

We write ∇\nabla as a multinomial in the partial derivatives, ∇=∑[α]ψ[α]∂[α]\nabla = \sum_{[\alpha]} \psi_{[\alpha]} \partial^{[\alpha]} where [α]=[a,b,c,⋯ ][\alpha]=[a,b,c,\cdots] ranges over some finite subset of all multi-indices, ∂[α]=∂xa∂yb∂zc⋯\partial^{[\alpha]}=\partial_x^a\partial_y^b\partial_z^c\cdots, and for each index ψ[α]\psi_{[\alpha]} is some smooth function Rn→R\mathbb{R}^n\to \mathbb{R}. But as each ∂[α]\partial^{[\alpha]} satisfies the desired property by Corollary 4, we can apply lemmas 6 and 7 finitely many times to conclude that ∇\nabla does as well.

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