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

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.