$\newcommand{\supp}{\operatorname{supp}}$ We are given with distributions $f,g \in D'(\Bbb R)$. If $\supp f\subset (-\infty,a)$ and $\supp(g)\subset(b,\infty)$ then prove that $f*g$ is well defined distribution, where $a$ and $b$ are real numbers.

What I know:

$\langle f*g\rangle=\langle f(x)g(x),\phi(x+y)\rangle$

[Above defined convolution will be well defined if $\supp(\phi(x+y))$ has a compact intersection with the support of $f(x)g(x)$. ]

$=\int_\Bbb R\int_\Bbb Rf(x)g(x)\phi(x+y)\,dx\,dy=\int_\Bbb Rf(x)[\int_\Bbb Rg(x)\phi(x+y)\,dy]\,dx$

Here, $\int_\Bbb Rg(x)\phi(x+y)\,dy$ is well defined since $\phi$ has a compact support. $\implies \supp(\phi)\cap \supp(g)\neq\phi$


$\supp(\phi)\cap \supp(f)\neq\supp\phi$

$\implies \supp(\phi)\cap \supp(f*g)\neq\supp\phi$, since $\supp(f*g)\subset \overline{ \supp(g)+\supp(f)}$

Pleas help me to prove it.


One has to be somewhat careful when defining the convolution of two distributions. It is not possible to proceed by computing an explicit convolution, since distributions are not necessarily functions, and so computing their integral is meaningless. When $\lambda$ is a distribution, and $h$ a smooth test function, we define $$\langle f, \lambda \ast h \rangle = \langle f \ast \tilde{h}, \lambda \rangle$$

This makes percent sense, since $\lambda$ is only being used as a distribution, and $h$ is an actual test function, so $f \ast \tilde{h}$ can be plugged into $\lambda$.

Now we want to make sense of $\lambda \ast \mu$, for $\lambda$ and $\mu$ distributions. The trick here is to use the result that test functions are dense in the space of distributions. Let $h_n$ be a sequence of test functions converging, in the distributional sense, to $\mu$. We can define $$\langle f, \lambda \ast \mu\rangle = \lim_{n \to \infty} \langle f, \lambda \ast h_n\rangle$$

Of course, this will not necessarily converge, which prevents us from convolving arbitrary $\lambda$ and $\mu$. However, in our case, since the intersection of their supports has compact closure, you should be able to demonstrate that this converges for a fixed $f$, and actually defines a distribution $\lambda \ast \mu$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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