Let $T\in\mathcal S'(\mathbb{R})$ be a Schwartz distribution and $\phi\in\mathcal D(\mathbb{R})$ smooth and compactely supported. Does the convolution $T\ast\phi$ have compact support? If not, what are sufficient (and weakest possible) assumptions on $T$ and/or $\phi$ in order to ensure that?
Support of convolutions
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real-analysis
distribution-theory
1 Answers
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For the first question the answer is no. Take $T= \exp(-x^2)$ and a non-negative test function $\phi$. The convolution $T\ast\phi$ is strictly positive (hence the support is $\Bbb R$).
For the second question, if $T$ is compactly supported, then for all compactly supported $\phi$ the convolution $T\ast\phi$ is also compactly supported. This should be easy to show by definition. Obviously, this is sufficient, but I do not think that it is necessary.