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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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    @rlgordonma The brackets is the action of a distribution on a test function.2012-12-23

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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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    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?2012-12-23