$\Delta u=3u$ then $u\equiv0$

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.

• Why do you think you cannot use the Fourier transform in this case? – MaoWao Mar 18 at 10:57

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.
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.
• 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. – JustDroppedIn Mar 18 at 13:40
• As I mentioned the difficult part is $u$ only $L^{1}$, not $C^{2}$. – Hahn Mar 18 at 13:52
• You mentioned that the difficult part is that u is not $C^\infty$ – JustDroppedIn Mar 18 at 13:54