Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Generally,when taking convolution of two distributions,at least one of which is supposed to be of compact support.

But when u,$v\in S'(\mathbb{R})$ ( temperate distributions) have suports on the positive half axis,then $u\ast v \in S'(\mathbb{R})$

how to prove this and generalize to high dimensions?

share|cite|improve this question
It has been a long time since I thought about this kind of stuff, but isn't the idea that the support of "t\mapsto u(x-t)v(t)" is compactly supported (in t) for all x? Of course this only makes sense if the distributions are integrations over functions $u$,$v$. This should lead to generalizations – Thomas Rot Aug 15 '12 at 7:42
I think this is also related to the product of two distributions( some perticular information about it's wavefront set are needed to make the definition work.),since we have $\widehat{u\ast v}=\hat{u}\hat{v}$transfer the convolution to the product of the fourier transformation of the distributions. – sun Aug 15 '12 at 8:21
up vote 2 down vote accepted

Since this is homework, I probably shouldn't write down a complete solution. But let's at least write down a definition for the convolution general enough for the situation described above (taken from my lecture notes of the course "Distribution et équations aux derivées partiélles" by André Cérezo):

Théorême Soient $S,T \in \mathcal{D}'(\mathbb{R}^n)$, $F=(\operatorname{supp} S_x)\times(\operatorname{supp} T_y)\subset \mathbb{R}^{2n}$, et $\Delta=\{ (x,-x)|x\in \mathbb{R}^n\}\subset \mathbb{R}^{2n}$. Supposons que, pour tout $K\Subset\mathbb{R}^n$, le fermé $(K\times\{0\}+\Delta)\cap F$ soit un compact de $\mathbb{R}^{2n}$. Alors la formule $$(*)\qquad\forall \varphi\in \mathcal{D}(\mathbb{R}^n)\qquad <S*T,\varphi>=<S_x\otimes T_y,\varphi(x+y)>$$ définit une distribution sur $\mathbb{R}^n$, appelée "produit de convolution" de $S$ et $T$.

Here $K\Subset\mathbb{R}^n$ means that $K$ is compact. We have $\mathcal{S}'(\mathbb R)\subset\mathcal{D}'(\mathbb R)$, so the first step is to verify the additional condition. This gives us $u*v\in\mathcal{D}'(\mathbb R)$. Now all that is left to show is $u*v\in\mathcal{S}'(\mathbb R)$.

Edit (the requested translation of the cited theorem)

Theorem Let $S,T \in \mathcal{D}'(\mathbb{R}^n)$, $F=(\operatorname{supp} S_x)\times(\operatorname{supp} T_y)\subset \mathbb{R}^{2n}$, and $\Delta=\{ (x,-x)|x\in \mathbb{R}^n\}\subset \mathbb{R}^{2n}$. Assume that for all $K\Subset\mathbb{R}^n$, the closed set $(K\times\{0\}+\Delta)\cap F$ is always compact. Then the formula $$(*)\qquad\forall \varphi\in \mathcal{D}(\mathbb{R}^n)\qquad <S*T,\varphi>=<S_x\otimes T_y,\varphi(x+y)>$$ defines a distribution on $\mathbb{R}^n$. It is called the "convolution" of $S$ and $T$.

share|cite|improve this answer
I added a translation into english. I just found the mentioned lecture notes online: The cited theorem is from section II.4. However, it won't evoke the same emotions and memories for somebody who hasn't met André Cérezo... – Thomas Klimpel Aug 15 '12 at 10:53
I think the condition is equivalent to that the map:$suppS \times suppT\ni (x,y) \to x+y\in \mathbb{R}^{n}$ is proper,that is the inverse image of each compact set is compact. – sun Aug 20 '12 at 9:01
@ShanLinHuang Yes, this is the intention of the condition. By the way, did you notice that I didn't really indicate how to show $u*v \in \mathcal{S}'(\mathbb R)$? – Thomas Klimpel Aug 20 '12 at 9:24
@ Thomas Klimpel right,that's what I'm still confused about(I think the obstacal is choosing some proper cut-off functions,maybe i'm wrong).Since in this special case,u,v are temperate distributions,it should say something more about the convolution – sun Aug 20 '12 at 13:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.