Hi I'm confused by a proof on page 53 in Strichartz book on Fourier Transforms. Specifically, in the first equation on page 53, why is it valid to interchange the action of the distribution with the integral? I know that distributions are linear, but integrals are in general infinite sums. I have tried looking earlier in the book to find a justification for this, but I could not find anything. I would really appreciate some help with understanding this.


welcome to SE.

In the above, the author is not justifying his calculation. In fact, he says `formally.' Here is one sketch of proof: Let $O_\phi = \{y : supp(\phi_y) \subset \Omega\}$. We can take $\psi^\delta = \eta_\delta * \psi$, where $\eta_\delta$ is a mollifier. In particular, we have that $\psi^\delta * \phi \rightarrow \psi * \phi$ uniformly as $\delta \rightarrow 0$, and $\int_{O_\phi} |\psi^\delta - \psi| < \delta$. By continuity of $T$, the former bound gives $T(\psi^{\delta} * \phi) \rightarrow T(\psi * \phi)$. These two facts yield

\begin{align*} \left|T(\psi*\phi) - \int_{O_\phi} \psi(y) T(\phi_y) dy\right| &\leq |T(\psi*\phi)-T(\psi^\delta*\phi)| + \left|\int_{O_\phi}(\psi(y)-\psi^{\delta}(y))T(\phi_y) dy\right| \\ &\leq \epsilon + \sup|T(\phi_y)|\int_{O_\phi} |\psi(y)-\psi^{\delta}(y)|dy \end{align*} Hence, we prove the claim for $\psi \in C^\infty$. Since $\phi \in C^\infty$ (and so $\phi_y \in C^\infty$), we can approximate the integral by Riemann sums. It follows that \begin{align*} \int_{O_\phi} \psi(x)T(\phi_y) dx &\leq \sum_{i=1}^m \Delta_i \psi(x_i)T(\phi_y) + \epsilon \\ &= T\left(\sum_{i=1}^n \Delta_i \psi(x_i)\phi(x_i-y)\right) + \epsilon \end{align*} where we approximated with lower Riemann sums. We can also approximate with upper Riemann sums to get the reverse inequality and find that in the limit the two expressions must be equal.

  • $\begingroup$ Thanks for the answer, but I have a couple of questions. Is the uniform convergence of $\psi^\delta \ast \phi \to \psi \ast \phi$ sufficient here? The topology on the Schwartz space require that it converges in every single norm in the countable family, and the uniform norm is only one of these. Also, it is possible that the integral is taken over all of $\mathbb{R}^n$ since Schwartz functions are not always compactly supported. I was under the impression that Riemann sums do not work for infinite domains of integration. $\endgroup$ – coffee_equals_donut Nov 1 '13 at 0:26
  • $\begingroup$ Another thing: I am only familiar with convergence of sequences in Schwartz space. Does the proof work if I take an arbitrary sequence $\delta_n \to 0$ rather than a continuous $\delta \to 0$? $\endgroup$ – coffee_equals_donut Nov 1 '13 at 0:29

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.