5
$\begingroup$

I have the following question in which it is easy to use Fourier transform to get the answer if the function is nice enough, for example $u\in C_{0}^{\infty}(\mathbb{R}^{n})$, however here $u$ is only in $L^{1}$. How can I get through this?

The question: Let $u\in L^{1}\left(\mathbb{R}^{n}\right)$ so that $\Delta u=3u$ in the distribution sense. Prove that $u\equiv0$.

Thanks.

$\endgroup$
  • 1
    $\begingroup$ Why do you think you cannot use the Fourier transform in this case? $\endgroup$ – MaoWao Mar 18 at 10:57
4
$\begingroup$

JustDroppedIn's answer works fine for $u\in L^1$. Here is why: Since $\Delta u=3u\in L^1$, it is a tempered distribution. Hence we can test against functions $\phi\in\mathscr{S}(\mathbb{R}^n)$. If we do so, we get $$ \langle \phi,\widehat{\Delta u}\rangle=\langle \Delta \hat \phi,u\rangle=-4\pi^2\langle \widehat{|\cdot|^2\phi},u\rangle=-4\pi^2\langle \phi,|\cdot|^2\hat u\rangle. $$ Thus $\widehat{\Delta u}(\xi)=-4\pi^2|\xi|^2\hat u(\xi)$ without any additional differentiabiliyt assumption on $u$. From here on you can proceed exactly as suggest by JustDroppedIn.

$\endgroup$
  • $\begingroup$ Thanks for good explanations. $\endgroup$ – Hahn Mar 24 at 0:16
3
$\begingroup$

You can use the Fourier transform for any $L^1$ function, since it is defined as an operator from $L^1$ to $L^\infty$ (or $C_0$ to be more precise); moreover, it is a continuous operator.

Let's use the Fourier transform: We have $\hat{\Delta u}(\xi)=\displaystyle{\sum_{j=1}^{n}\int_{\mathbb{R}^n}e^{-2\pi ix\cdot\xi}\frac{\partial^2 u}{\partial x_j^2}(x)dx=\sum_{j=1}^{n}\int_{\mathbb{R}^n}(-2\pi i\xi_j)^2e^{-2\pi ix\cdot\xi}u(x)dx=-4\pi^2\|\xi\|^2\hat{u}(\xi)}$, hence $-4\pi^2\|\xi\|^2\hat{u}(\xi)=3\hat{u}(\xi)$ and that holds for all $\xi\in\mathbb{R}^n$. This forces $\hat{u}$ to be identically $0$, hence $u\equiv0$, since the Fourier transform is injective.

$\endgroup$
  • 1
    $\begingroup$ Well, I figured since the Laplacian was defined, u would be at least twice differentiable. The case that $u$ is not differentiable (or a.e. differentiable) is beyond my knowledge. I would advise you to specify it in your question that you are talking about a more general case. $\endgroup$ – JustDroppedIn Mar 18 at 13:40
  • $\begingroup$ As I mentioned the difficult part is $u$ only $L^{1}$, not $C^{2}$. $\endgroup$ – Hahn Mar 18 at 13:52
  • 1
    $\begingroup$ You mentioned that the difficult part is that u is not $C^\infty$ $\endgroup$ – JustDroppedIn Mar 18 at 13:54

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.