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I want to define the convolution $*$ between two distributions $S$ and $T$. For a test function $\varphi$, can I say:

$$\langle S * T, \varphi \rangle \doteqdot \langle S, T*\varphi \rangle $$

where the convolution between a distribution and a test function is a function that I define as:

$$ T*\varphi \doteqdot x \mapsto \langle T,\tau_x \varphi \rangle $$

With $\tau$ the translation operator, i.e., $\tau_x (t \mapsto \varphi(t))\doteqdot t \mapsto \varphi(t-x) $ .

Does this make any sense? I'm trying to follow what my textbook says but the author is not exactly clear.

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  • $\begingroup$ What are the brackets? Inner product? $\endgroup$
    – Ron Gordon
    Dec 23, 2012 at 20:42
  • $\begingroup$ @rlgordonma The brackets is the action of a distribution on a test function. $\endgroup$
    – mrf
    Dec 23, 2012 at 23:27

1 Answer 1

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In general, convolutions of distributions cannot be defined. (It's possible with some extra conditions, for example that at least one of the distributions has compact support.)

The problem with your approach is that $T*\phi$ is not necessarily a test function.

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  • $\begingroup$ What if $T$ is tempered? Then I can write $ (T * \varphi) (x) = \int_{-\infty}^{+\infty} f(t)\varphi(x-t) \mathrm{d}t$ for some Schwartz function $f$ (i.e., $T=[f]$); wouldn't this be a test function in this case since $f$ is infinitely differentiable? $\endgroup$
    – Saltimbanco
    Dec 23, 2012 at 21:07
  • $\begingroup$ If $S$ is a tempered distribution and $T$ is a Schwarz function, your're ok, but the convolution of two tempered distributions is in general not defined: what would $1*1$ be? $\endgroup$
    – mrf
    Dec 23, 2012 at 23:25

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