1
$\begingroup$

This is related to verifying that the Fourier transform method of yields a solution that one can make sense of appropriately:

Let $$u(t,x)=\frac{e^{Mt}}{2\pi} \int\limits_{\mathbb{R}}e^{-i\sigma t} R(\sigma+iM) \varphi\, d\sigma,$$ where $M>0$ is large, $\varphi\in L^2(\mathbb{R}^3),$ and $R(\cdot)$ denotes the resolvent, which satisfies the estimate $$\|R(\lambda)\|_{L^2(\mathbb{R}^3)\rightarrow L^2(\mathbb{R}^3)}\leq \frac{C}{|\lambda|\text{Im }\lambda},\qquad \lambda\in \mathbb{C},\ \text{Im }\lambda>0.$$ The claim is that $u\in L^2_{loc}(\mathbb{R}_t, L^2_x(\mathbb{R}^3)).$ The reason that I've been given is that when we write $\lambda=\sigma+iM$, then the above result is $L^2$ in $\sigma,$ so it follows from Plancherel. This doesn't make sense to me since (1) I'm doing the spatial norm first, and bringing it into the integral to actually use the resolvent estimate wipes out the complex exponential, (2) the time norm is local. If I want to use Plancherel, then I need to apply it over all of time. If I insert a cut-off, then I need to use convolution when doing Plancherel. There's also an exponential in $t$ outside of the $\sigma$ integral, so that needs to be accounted for similarly.

I tried to show it directly, but to no avail:

For fixed $t\in\mathbb{R},$ the resolvent estimate and Minkowski's inequality give that $$\|u(t,\cdot)\|_{L^2(\mathbb{R}^3)}\leq Ce^{Mt}\int_{\mathbb{R}}\frac{1}{M\sqrt{M^2+\sigma^2}}\, d\sigma, $$ and the $\sigma$ integral does not converge regardless of $t$.

Clearly, the issue is wiping out the complex exponential. How do I properly apply Plancherel?

This comes from page 17 of this: https://math.mit.edu/~phintz/snap19/notes.pdf (Proposition 4.2).

EDIT: Actually, I think that (1) is not an issue since I can swap the order of integration. Still not sure about (2).

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.