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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 ...

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    $\begingroup$ Try to consider the translation $g(z)=f(z-z_0)$. $\endgroup$ – Xiang Yu Jun 12 '16 at 2:05
  • $\begingroup$ How do you get the expression in the second line following the words "we can parametrize"? $\endgroup$ – DanielWainfleet Jun 12 '16 at 6:33
  • $\begingroup$ @user254665 I made an error with parentheses, now corrected. $\endgroup$ – matty_k_walrus Jun 13 '16 at 16:37
  • $\begingroup$ @XiangYu But of course. Thank you! $\endgroup$ – matty_k_walrus Jun 13 '16 at 16:39
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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, $f \equiv 0$.

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  • $\begingroup$ Okay. How is Cauchy-Schwarz applicable? What is the inner product? Is it $$\langle u, v\rangle = \int_0^{2\pi} uv$$ Since the functions are complex-valued it is not necessarily true that $$\langle v,u\rangle = \overline{\langle u, v\rangle}$$. If $$\langle u, v\rangle = \int_0^{2\pi} |uv|$$ then we don't have linearity in the first argument. $\endgroup$ – matty_k_walrus Jun 13 '16 at 16:33
  • $\begingroup$ Think simpler than that. We have $4\pi^2 \lvert f(z_0)\rvert^2 \le \left(\int_0^{2\pi} \lvert f(z_0 + re^{i\theta})\rvert^2\, d\theta\right)^2$, and by the Cauchy-Schwarz inequality, $$\left(\int_0^{2\pi} \lvert f(z_0 + re^{i\theta})\rvert^2\, d\theta\right)^2 \le \int_0^{2\pi} 1^2\, d\theta\cdot \int_0^{2\pi} \lvert f(z_0 + re^{i\theta})\rvert^2\, d\theta = 2\pi \int_0^{2\pi} \lvert f(z_0 + re^{i\theta})\rvert^2\, d\theta$$ Dividing through by $4\pi^2$ yields the second line in the above answer. $\endgroup$ – kobe Jun 13 '16 at 17:03
  • $\begingroup$ @kobe It follows directly from $|f(z_0)|\le\frac{1}{2\pi}\int_0^{2\pi}|f(z_0+re^{i\theta})|d\theta$ and by the Cauchy-Schwarz inequality we have the seond line in your answer. $\endgroup$ – Bach May 21 at 1:14
  • $\begingroup$ @user549397 that's how I did it originally, but I explained it to the OP in a different way. $\endgroup$ – kobe May 21 at 5:12
  • $\begingroup$ @kobe But I can't see any reason why you have $4\pi^2|f(z_0)|^2\le \left( \int_0^{2\pi}|f(z_0+re^{i\theta})|^2d\theta \right)^2$. $\endgroup$ – Bach May 21 at 5:16

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