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I'm having trouble with proving the following:

Let $f(z)$ be an entire function such that for all $z$, $|f(z)| \leq \sqrt{|z|}$. Show that $f(z) = 0$ for all $z$.

I know you need Cauchy's formula and take the first derivative for $f$ and then prove if we take $R$ sufficiently large then we have a function that is constant or equal to $0$.

But I'm sure how to get to that stage with the above entire function.

Michael Albanese
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DJ_
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  • where do you get stuck? did you write down Cauchy's formula for the derivative in this case? You have the right recipe to solve this, now you need to at least try. Tell us where you got stuck and you'll get a hint. – Ittay Weiss Apr 21 '13 at 06:30
  • Well for one, how do I find f' if I'm only given |f(z)|, sorry if this is a basic question. – DJ_ Apr 21 '13 at 06:31
  • @EuYu: The questions are slightly different, right? That one has an extra "$1+$" in the RHS of the inequality and only requires (or enables) proving that $f$ is constant. – ShreevatsaR Apr 21 '13 at 09:42
  • @ShreevatsaR They're essentially the same. Letting $g = f + 1$ we have $$|g|\le 1 + |f| \le 1 + |z|^{1/2}$$ That question implies $g = f+1 = c$ and hence $f$ is constant. Evaluating at $0$ gives the constant as $0$. – EuYu Apr 21 '13 at 09:50

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Using Cauchy's integral formula, write $$f^{(n)}(a)=\frac1{2\pi i}\int_{|z|=R}\frac{f(z)}{(z-a)^{n+1}}dz$$ Apply absolute value on each side to get: $$\begin{align*}|f^{(n)}(a)|&=\left|\frac1{2\pi i}\int_{|z|=R}\frac{f(z)}{(z-a)^{n+1}}dz\right|\leq\frac1{2\pi}\int_0^{2\pi}\left|\frac{f(Re^{it})}{(Re^{it}-a)^{n+1}}iRe^{it}\right|dt\\ &=\frac R{2\pi}\int_0^{2\pi}\frac{|f(Re^{it})|}{|Re^{it}-a|^{n+1}}dt \leq \frac R{2\pi}\int_0^{2\pi}\frac{\sqrt{|Re^{it}|}}{(R-|a|)^{n+1}}dt=\frac{R^{3/2}}{(R-|a|)^{n+1}} \end{align*}$$ Now observe that for all $n\geq1$, $\lim_{R\to\infty} \frac{R^{3/2}}{(R-|a|)^{n+1}}=0$.
From here you can easily conclude the proof.

Dennis Gulko
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