Assume $f$ is entire and $\int \int |f(x+iy)|^2 dx dy < \infty$ , then show that $f \equiv 0$.
My approach was to notice that $|f|^2$ is a subharmonic function and then we can use the mean value inequality. Note that $$\int \int |f(x+iy)|^2 dxdy = \int_0^{\infty} \int_{0} ^{2\pi} |f(re^{i\theta})|^2 rd\theta dr$$Now we know from subharmonicity $|f(0)|^2 \leq \frac{1}{2\pi} \int_0^{2\pi} |f(Re^{i\theta})|^2d\theta = \frac {1}{\pi R^2}\int_0^{R} \int_{0} ^{2\pi} |f(re^{i\theta})|^2 rd\theta dr$
As $R \rightarrow \infty$ , the rightmost term tends to $0$ as we know $\int_0^{\infty} \int_{0} ^{2\pi} |f(re^{i\theta})|^2 rd\theta dr < \infty$, thus we have |f(0)| = 0. We could have chosen any $z$ in place of 0 and the argument would follow the same and thus we see $f \equiv 0$.
I was wondering if there is any way to get this result without the subharmonicity, I was trying to apply Liouville's but couldn't, if there is a simpler solution please let me know.