Prove that a square-integrable entire function is identically zero

Suppose $f$ is entire and $$\iint_\mathbb{C}|f(z)|^2dxdy < \infty$$Prove that $f\equiv 0.$

So far I have:

Suppose $f$ is bounded. Then $f$ is constant by virtue of Liouville and so the conclusion is obvious. Thus, assume $f$ non-bounded. Then $$\Big|\iint_\mathbb{C}f^2(z)dA\Big| \leq \iint_\mathbb{C}|f(z)|^2dxdy < \infty$$ We can parameterize \begin{align*} \iint_\mathbb{C}f^2(z)dA &= \int_0^{\infty}\int_0^{2\pi}f^2(re^{i\theta})\;d\theta\;rdr \\ &= \int_0^{\infty}\int_0^{2\pi}f^2(re^{i\theta})\;ire^{i\theta}[-i\frac{1}{r}e^{-i\theta}]d\theta\;rdr \\ &= \int_0^{\infty}\oint_{C_R}f^2(z)\;[-i\frac{1}{z}]dz\;rdr \\ &= \int_0^{\infty}2\pi r dr\frac{1}{2\pi i}\oint_{C_R}\frac{f^2(z)}{z}dz \\ \\ &= 2\pi f^2(0)\int_0^\infty rdr \end{align*}

Whence f(0) = 0. And I have no idea where to go from there.

The "entire" bit seems to suggest Liouville but I have already dealt with the bounded case. Please advise ...

Solution 1:

Fix $z_0\in \Bbb C$. For all $r > 0$, $$f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta})\, d\theta$$ By the Cauchy-Schwarz inequality, $$\lvert f(z_0)\rvert^2 \le \frac{1}{2\pi}\int_0^{2\pi} \lvert f(z_0 + re^{i\theta})\rvert^2\, d\theta$$ for all $r > 0$. Hence, for every $R > 0$,

$$\int_0^{R} \lvert f(z_0)\rvert^2 r\, dr \le \frac{1}{2\pi}\int_0^R \int_0^{2\pi} \lvert f(z_0 + re^{i\theta})\rvert^2 r\, d\theta\, dr $$ or $$\lvert f(z_0)\rvert^2 \frac{R^2}{2} \le \frac{1}{2\pi}\iint_{D(z_0;R)} \lvert f(z)\rvert^2\, dx\, dy$$ Hence, $$\lvert f(z_0)\rvert^2 \le \frac{1}{\pi R^2}\iint_{D(z_0;R)} \lvert f(z)\rvert^2\, dx\, dy\le \frac{1}{\pi R^2}\iint_{\Bbb C} \lvert f(z)\rvert^2\, dx\, dy \to 0\quad \text{as} \quad R\to \infty$$

So $f(z_0) = 0$. Since $z_0$ was arbitrary, $f \equiv 0$.