Qual-type problem

Problem: Let f be entire and suppose that for all z\in{\mathbb C}:

\displaystyle |f(z)|\leq \frac{1}{|{\mathcal Re} z|}.

Show that f(z)\equiv 0.

Solution: For k=0,1,2,..., let

\displaystyle I(r,k)=\int_{|z|=r}\left(1+\frac{z^2}{|z|^2}\right)\frac{f(z)}{z^{k+1}}\frac{dz}{2\pi i}.

On one hand, using the fact that

\displaystyle \left|1+\frac{z^2}{|z|^2}\right|=\frac{2|{\mathcal Re} z|}{|z|},

we get that |I(r,k)|\rightarrow 0 as r\rightarrow\infty, for each k.

And on the other hand, direct integration shows that I(r,k)\rightarrow a_k where f(z)=\sum_{n=0}^\infty a_n z^n. \Box.

Questions: Is there a different proof, say, using the maximum principle? What if f is simply harmonic?

Update: The same power series approach seems to give the harmonic case.

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3 Responses to Qual-type problem

  1. Leonid says:

    Here’s another proof (for entire functions): u=log|f| is subharmonic, and its averages over circles |z-z_0|=R tend to -infinity as R->infinity. (Because the singularity at Re z=0 is now negligible). Not as slick, of course.

  2. Ah yes! I was looking at the harmonic case and didn’t see this simple argument.

    As an aside, I didn’t know before that 1/|{\mathcal Re} z| is actually subharmonic. In fact it seems that \Delta |{\mathcal Re} z|^\alpha=\alpha(\alpha-1)|{\mathcal Re} z|^{\alpha-2}, which is positive unless 0<\alpha<1.

  3. Oops I guess there are some integrability issues with 1/|{\mathcal Re} z|….

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