harmonic function is zero on complex plane

Question

Let $u(z)$ be harmonic on all of $\mathbb{C}$ such that

$$\int \int_{\mathbb{C}} \vert u(z) \vert^2 dx dy < \infty$$

Prove then that $u(z)=0$ for all $z$.

So a few ideas I had, since we know $u(z)$ is harmonic, this means $u_{xx}+u_{yy}=0$ but I don't know how to use that with my given information.

or do I use Fubini's Theorem?

Answer 1

Pick $w$ arbitrary in the plane and use the mean value property and Cauchy-Schwarz to get: $$|\pi R^2u(w)|^2=|\int \int_{B(u,R)} u(z) dx dy|^2 \le \int_{B(u,R)} |u(z)|^2 dx dy\int_{B(u,R)} 1 dx dy$$

But letting $M=\int \int_{\mathbb{C}} \vert u(z) \vert^2 dx dy$, we get that:

$$|\pi R^2u(w)|^2 \le M^2\pi R^2$$

Let $R \to \infty$ and conclude $u(w)=0$ so $u$ is identically zero as $w$ was arbitrary

Answer 2

Firstly, we know for harmonic function, there is mean value property. Then fixed $x_0\in \mathbb R^2$, and for any ball $B_R(x_0)$ we have $$ \begin{aligned} |u(x_0)|&=\frac{1}{\pi R^2}\left|\int_{B_R(x_0)}u(y)\,dy\right|\\[5pt] &\leq \frac{1}{\pi R^2}\left(\int_{B_R(x_0)}u^2(y)\,dy\right)^\frac12 \left(\int_{B_R(x_0)}1\,dy\right)^\frac12\\[5pt] &= \frac{1}{\sqrt\pi R} \left(\int_{B_R(x_0)}u^2(y)\,dy\right)^\frac12\\[5pt] &\leq \frac{1}{\sqrt\pi R} \left(\int_{\mathbb R^2}u^2(y)\,dy\right)^\frac12. \end{aligned}$$ Hence, letting $R\to+\infty$, we have $u(x_0)=0$. Since $x_0$ is arbitrary, we know $u= 0$ in $\mathbb R^2$.