$f$ entire, $g(x, y) := |f(x+iy)|$ integrable on $\mathbb{R}^{2}$ $\implies$ $f \equiv 0$.

Question

Let $f \in H(\mathbb{C})$ and $g(x, y) := |f(x+iy)|$, and assume $g$ is integrable on $\mathbb{R}^{2}$. Show then that $f \equiv 0$.

Things I Know: $f$ is bounded on each $D(0, R)$ for $R > 0$, but the bounds must increase as $R \rightarrow \infty$. Intuitively, this should mean $\int_{\partial D(0, R)} |f(z)| dz \rightarrow \infty$ as $R \rightarrow \infty$ (assuming $f \not\equiv 0$), which could say something about $g$. But here I'm talking about upper bounds, which don't help. I could consider the infimum $f$ takes on $\partial D(0, r)$ to bound below, but these inf's don't necessarily grow larger as $R$ does.

My intuition tells me Liouville will come into play, but I'm not sure... Any ideas? Thank you!

Edit: Though this is the same as this question essentially, I want to include proofs that use any complex analytic tools, while the linked question wants a solution based only in elementary calculus.

Answer 1

By the mean-value property of harmonic functions, $f(z)$ is the average of $f$ over any circle centred at $z$, and therefore over any disk centred at $z$. A disk $D$ of radius $r$ centred at $z$ has area $\pi r^2$, so $$\int_D |f(x+iy)|\; dx\; dy \ge |f(z)| \pi r^2$$ and as $r \to \infty$ the only way to avoid an infinite limit is $f(z)=0$.