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Question: Suppose that $f$ is an entire function (i.e. analytic in $\mathbb{C}$), and satisfies the condition $\iint_{\mathbb{C}}|f|^p dxdy<\infty$ for some $p\in (0,1)$. I guess that $f\equiv 0$ but I do not know how to prove it.

Note. If $p\in[1,\infty)$, it is easy to prove that $f\equiv 0$. In the settting $p\in (0,1)$, one should deal with the integral of an entire function near the essential singularity point $\infty$ carefully.

EDIT. Thank Alexandre Eremenko for his answer. I also want to know the solution to the following harmonic version of question.

Question (H): Suppose that $f$ is a harmonic function ($i.e. \Delta f=0$, $f$ may be complex-harmonic in $\mathbb{C}$), and satisfies the condition $\iint_{\mathbb{C}}|f|^p dxdy<\infty$ for some $p\in (0,1).$ I believe that $f\equiv 0$.

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2 Answers 2

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Yes, this is so. And much stronger statements are available: See the beautiful survey paper

MR2567024 Rashkovskii, Alexander Classical and new loglog-theorems. Expo. Math. 27 (2009), no. 4, 271–287.

also available on the arxiv

All results in this paper are actually for subharmonic functions, so this settles the questrion for both analytic and harmonic functions.

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  • $\begingroup$ What about the answer to the harmonic version of question in my EDIT? $\endgroup$
    – woodbass
    Dec 31, 2012 at 1:01
  • $\begingroup$ Rashkovski's paper also discusses the harmonic, and more generally, subharmonic versions. $\endgroup$ Feb 8, 2013 at 14:40
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In the case of $f$ entire a much more straightforward solution is availabe. First if $f$ is entire, then $|f|^{p}$ is subharmonic. Secondly, by subharmonicity, for any $z_0$, we have \begin{equation} |f(z_0)|^{p} \leq \frac{1}{\text{meas}({D(z_0,R)})} \iint_{D(z_0,R)} |f(z)|^{p} dz \leq \frac{C}{\text{meas}(D(z_0,R))} \end{equation} where $D(z_0,R)$ is a disc of radius $R$ around $z_0$ and $C = \iint_{\mathbb{C}} |f(z)|^{p} dz$. Taking $R \rightarrow \infty$ in the above inequality we obtain $f(z_0) = 0$. Since $z_0$ was arbitrary $f \equiv 0$.

If the implications `$f$ harmonic implies $|f|^p$ subharmonic'' is correct then this would also answer the second part of the question. I haven't checked if it's true, but it seems correct at least in the case $p = 1$ (which leaves me hoping that it's correct for all $p > 0$). If anybody could check and post the answer in the comments I will be indebted.

EDIT: Here's a proof that $|f|^p$ is subharmonic if $f$ is entire and $0 < p < 1$. Jensen's formula implies that, $$ \log |f(0)| \leq \frac{1}{2\pi} \int_{C} \log |f(e^{i t})| dt $$ Multiply by $p$, exponentiate, and then use Jensen's inequality to conclude that $$ |f(0)|^p \leq \exp(\frac{1}{2\pi} \int_{C} p \log |f(e^{i t})| dt ) \leq \frac{1}{2\pi} \int_{C} |f(e^{i t})|^p dt $$ Therefore $|f|^p$ is sub-harmonic.

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  • $\begingroup$ You may want to look at the Wikipedia article on subharmonic functions. $\endgroup$
    – S. Carnahan
    May 14, 2013 at 13:58
  • $\begingroup$ @Pooper: How do you prove that $|f|^p$ is subharmonic when $p\in (0,1)$ even in the case of $f$ entire? $\endgroup$
    – woodbass
    May 22, 2013 at 15:43

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