For an exercise I'm being asked to prove that, given two functions $f,g \in\mathcal{L}^2(\mathbb{R}^d)$, their convolution $ f \star g$ is "almost everywhere defined, and in $\mathcal{L}^\infty$ once we fill in the undefined points. But it looks like to me like $ f \star g $ should be defined everywhere. My "proof" goes by noting that, if $x \in \mathbb{R}^d$, the function given by $t \mapsto g(x-t) $ is also squared integrable, since its square norm is the same one as $g$'s by substitution. So we should have:

$$ f \star g (x) = \int_{\mathbb{R}^d} f(t)g(x-t) \, dt \leq \lVert f \rVert_2 \lVert g \rVert_2 $$

by the Hölder inequality, which also gives us that the function is in $\mathcal{L}^\infty$.

Can anyone tell me why this doesn't work? Or is the convolution truly defined anywhere?

  • $\begingroup$ I edited my answer, I'm quite sure your teacher meant Riemann vs Lebesgue integral $\endgroup$ – reuns Mar 20 '17 at 18:12
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    $\begingroup$ You are correct. It is defined everywhere. If the functions are in $L^1$, then it is defined a.e., and that requires Fubini's Theorem. $\endgroup$ – DisintegratingByParts Mar 21 '17 at 15:08

Your teacher probably meant $\int_{\mathbb{R}^d} f(t)g(x-t) \, dt$ as a Riemann Stjejes integral. If you interpret it as a Lebesgue integral :

By Cauchy-Schwarz $$| f \ast g(x)| = |\langle f(.) , g(x-.) \rangle| \le \|f\|_{L^2} \|g\|_{L^2}$$ is well-defined.

Also letting $\varphi_\epsilon(x) = \epsilon e^{-|x|^2/\epsilon^2}$, for every $\epsilon > 0$ we have $$\lim_{h \to 0}\|\varphi_\epsilon \ast (g(.+h)-g(.))\|_{L^2} = \lim_{h \to 0}\|g \ast (\varphi_\epsilon(.+h)-\varphi_\epsilon(.))\|_{L^2}=0$$ so that $$\lim_{h \to 0}\|g(.+h)-g(.)\|_{L^2} = 0$$ hence $$| f \ast g(x+h)- f \ast g(x)| \le \|f\|_{L^2} \|g(.+h)-g(.)\|_{L^2} $$ shows that $f \ast g$ is uniformly continuous

  • $\begingroup$ No, I'm positive it was meant to be a Lebesgue integral. $\endgroup$ – DR6 Mar 20 '17 at 18:17
  • $\begingroup$ @DR6 the Lebesgue integral is made exactly for that : making $\int F(t)dt$ well-defined whenever $F \in L^1$. And if $f,g \in L^2$ then $f g \in L^1$ $\endgroup$ – reuns Mar 20 '17 at 18:27
  • $\begingroup$ It's probably just a mistake then. Thank you. $\endgroup$ – DR6 Mar 21 '17 at 15:23

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