a small argument about proving holomorphic function as constant

Question

Let $f$ be holomorphic on $|z|>R$, for some large $R>0$. If there exist $r_2>r_1>R$ with $max_{|z|=r_1}|f(z)| = max_{|z|=r_2}|f(z)|$, then $f$ is constant.

NB: $\lim_{z\to\infty}f(z)<\infty$

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Previous Note: That is my argument. I am not sure whether it is true or false. Also, if necessary, you may add that $\lim_{z\to\infty}f(z)$ exists to consider the argument above.

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UPDATE: The argument is false for some $f$ with $\lim_{z\to\infty}f(z)=\infty$. Now if we add that $\lim_{z\to\infty}f(z)<\infty$, is it true?

Thanks a lot!

Answer 1

No. Counterexample $f(z)=z+1/z$ with $r_1=2$ and $r_2=1/2$.

Answer 2

Thank you, @MartinR for answering my previous question Application of Max Module Theorem! I think that it is also OK to write it here directly to prove that $f$ is constant!

(Weak Version: do not need bounded $D$)

Max Modulus Principle: Let $f$ be holomorphic in a district $D$. Then if $f$ is not constant, then $|f(z)|$ cannot take maximum in $D$.

Now that $max_{|z|=r_1}|f(z)|=max_{|z|=r_2}|f(z)|$. Then take $D=\{z||z|>r_1\}$. Then there exists some point $w\in \partial B(\infty,r_2)$, so that $|f|$ takes maximum at $w$ in $D$, which may contradict with the principle. So the only possibility is that $f$ is constant in $D$, which implys that $f=const$ everywhere.

I hope this solution is explicit.