Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.