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Let $ f : \mathbb{C} \to \mathbb{C} ,$ entire and $|f(z)|\le \log|z|,\ |z|>1. $

Show that $f$ is constant.

What first comes to mind is Louville's theorem, but log 's problems with analyticity confuse me.

kimtahe6
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helplessKirk
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    What first comes to mind are the Cauchy estimates. – Daniel Fischer Aug 19 '14 at 13:39
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    You can use Louville's theorem. Study the function $g(z) = \frac{f(z) - f(0)}{z}$ with $g(0) = f'(0)$. Since $f$ is entire, $g$ is entire and bounded by a constant $\to$ $f(z) = az+b$. Now use $|f(z)| = |az+b| > \frac{|a|}{2}|z|$ for large enough $|z|$ to show that we must have $a=0$. – Winther Aug 19 '14 at 15:04
  • Look at my answer. It is very easy. You only need Cauchy Integral. – xpaul Aug 20 '14 at 03:05
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    @@ Winther) How $f(0)$ is defined ? – Empty Oct 24 '15 at 20:16

3 Answers3

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If you use the Cauchy Integral Formula, the proof is very straight forward. Note, for $\forall n\ge 1$, $$ |f^{(n)}(0)|=\frac{n!}{2\pi}\bigg|\int_{|z|=r}\frac{f(z)}{z^{n+1}}dz\bigg|\le\frac{n!}{2\pi}\frac{\ln r}{r^{n+1}}2\pi r=\frac{n!\ln r}{2\pi r^n}\to0\text{ as }r\to\infty $$ and hene $f\equiv C$ is a constant. Note $|f(z)|\le \ln |z|$ for $|z|>1$, namely, $$ |C|\le \ln|z| \text{ for }\forall |z|>1. $$ Letting $|z|\to1$ gives $C=0$, namely, $$ f(z)\equiv0.$$ You can use the same way to show a more general result: If there is a constant $\alpha>0$ such that $$ |f(z)|\le C|z|^\alpha, \forall z\in\mathbb{C}, $$ then $f(z)$ is a polynomial.

xpaul
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  • Thanks! I did indeed use this way to prove that if $|f(z)|\ne C|z|^a$ then f(z) is a polynomial of degree at most a. – helplessKirk Aug 20 '14 at 07:14
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Pick $\epsilon>0$ and let $r=\exp(\epsilon)>1$. Then $|f(z)|\le\epsilon$ for all $z$ with $|z|=r$ by hypothesis. By the maximum modulus principle, $|f(z)|\le \epsilon$ for all $z$ with $|z|<r$, especially for all $z\in\mathbb D$ . Since $\epsilon$ was arbitrary, $f|_{\mathbb D}=0$, hence $f=0$.

Hagen von Eitzen
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  • Can you please explain how are we using the Maximum modulus principle here. I know the following version of maximum modulus principle "The absolute value of a nonconstant analytic function on a connected open set G ⊂ C cannot have a local maximum point in G" – Shweta Aggrawal Jun 11 '19 at 10:21
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Let $z\in \mathbb{C}$, with $|z|=1$. Put $a_n=1+1/n$ and $z_n=a_nz$. By your hypothesis, we have $0\leq |f(z_n)|\leq \log a_n$. Hence $|f(z_n|\to 0$ as $n\to \infty$. But $|f(z_n)|\to |f(z)|$. Hence $f(z)=0$ for all $z$ on the unit circle, and $f=0$.

Kelenner
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