Convolution of distributions property

Question

I'm trying to prove the following equation:

For distributions $S$ and $T$ on the space of test functions:

$e^{\lambda x} (T \circledast S) = e^{\lambda x} T \circledast e^{\lambda x} S$

What I have is:

$<e^{\lambda x}(T \circledast S), \phi (x)> \\ = <T_{x}, <S_{y}, e^{\lambda (x+y)} \phi (x+y)>> \\ =<e^{\lambda x} T_{x}, <e^{\lambda y}S_{y}, \phi (x+y)>> \\ =<e^{\lambda x} T \circledast e^{\lambda x} S, \phi (x)> \\$

But I'm not sure wether this is a legit proof...

score 1 · Accepted Answer · answered Jan 20 '18 at 15:46

1

You need more hypotheses! In general the convolution of two distributions does not even exist.

The second line in the display in your post is, or at least may be, meaningless. The problem is that the function $$x\mapsto<S_{y}, e^{\lambda (x+y)} \phi (x+y)>$$need not be a test function (for example there's no reason to think it has compact support).

answered Jan 20 '18 at 15:46

David C. Ullrich

89,985

This is true, indeed. But supposing that the convolution has a compact support, would you say my proof is then true? – xzeo Jan 20 '18 at 19:14
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Supposing that $S$ has compact support I guess it's ok. – David C. Ullrich Jan 20 '18 at 20:19

Convolution of distributions property

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