# A question about convolution of two distributions

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?

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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

## 1 Answer

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 $T$.

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I added a translation into english. I just found the mentioned lecture notes online: math.unice.fr/~frou/ACdistributions.html 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