# Convolution between two distributions

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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What are the brackets? Inner product? –  Ron Gordon Dec 23 '12 at 20:42
@rlgordonma The brackets is the action of a distribution on a test function. –  mrf Dec 23 '12 at 23:27
The problem with your approach is that $T*\phi$ is not necessarily a test function.
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? –  Saltimbanco Dec 23 '12 at 21:07
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? –  mrf Dec 23 '12 at 23:25