Confused by a proof in Strichartz' book on Fourier Transforms

Question

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.

Answer 1

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.