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I have trouble with the following small exercise in complex analysis.
Let $R=Ann(0;0,1)=\{z\in\mathbb{C}\,|\,0<|z|<1\}$, and $f$ is an analytic function on $R$. For a fixed
$p>0$, if $f$ satisfies
$$\iint\limits_R |f(x+iy)|^p dxdy < \infty$$, what can we say about the type of singularity $f$ has at $0$?

This exercise is taken from Conway,John B. Functions of One Complex Variable. v1 (GTM 11) p.112.
I have no trouble to deal with the case when $p\ge 2$. Since when $p=2$, one can use Parseval's identity
to derive that the singular part of the Laurent series of $f$ is $0$. But when $0<p<2$, it seems not easy
to reduce to the case when $p=2$ simply by considering $z^n f(z)$ for a suitable $n$ as I had expected to
work. But it doesn't work for me.

I am guessing the answer is that $0$ is in fact a pole of order $m=\lceil\frac{2}{p}-1\rceil$ by considering the function $f(z)=z^{-m}$. Am I right ? Can anyone give me some hints ? Any help is appreciated.

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migrated from Jan 28 '14 at 0:27

This question came from our site for professional mathematicians.

up vote 1 down vote accepted

Function $|f|^p$ is subharmonic for each $p>0$. For small $r>0$ solve the Dirichlet problem for the exterior of the disc $|z|<r$ with boundary values $|f(re^{i\theta})|^p$. This can be done explicitly by a version of the Poisson formula. Let $u(z)$ be this solution. Using subharmonicity, from the explicit formula, you obtain an estimate of the form $$|f(2re^{i\theta})|\leq u(re^{i\theta})+C_1\leq C\int_0^{2\pi}|f(re^{i\theta})|^pd\theta+C_1=:Cm_p(r)+C_1,$$ where $C$ is an absolute constant and $C_1$ depends on $f$ only, and where $m_p$ is the $L^p$ "norm". This does the business, by passing from $L_p$ to the uniform estimate. So under your condition, the function has at most a pole, and the order of the pole can be easily determined, as you did.

Remark. I know from experience that some problems in Conway can be much harder than expected in an undergraduate course. So a teacher making home assignment should be careful. In one extreme case, he himself could not provide a solution, when asked:-)

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