Let $f$ be an entire function that is not a polynomial. Denote $$ M(r)=\max_{|z|=r}|f(z)|. $$ Show that $$ \lim_{r\to\infty}\frac{M(r/2)}{M(r)}=0. $$

I believe I have a proof for this, but I'm concerned because I don't think I'm using the assumption that $f$ is not a polynomial. My proof is the following:

First, notice that \begin{align*} \log\left(\frac{M(r/2)}{M(r)}\right)&=\log M(r/2)-\log M(r)\\ &=-\frac{r}{2}\frac{d}{dr}\left(\log M(r)\right)\big|_{r=r_0} \end{align*} for some $r/2\leq r_0<r$ by the mean value theorem. Let $\log r=x$. By Hadamard's three-circles theorem, $\log M(r)$ is a convex function of $\log r$, i.e., $x$. Hence $\frac{d}{dx}\left(\log M(r)\right)>0$ for sufficiently large $x$. The chain rule shows $$ \frac{d}{dx}\left(\log M(r)\right)=e^x\cdot\frac{d}{dr}\left(\log M(r)\right). $$ Since $\frac{d}{dx}(\log M(r)),e^x>0$ for large $x$, $\frac{d}{dr}(\log M(r))>0$ for large $x$ as well. This shows that $$ \lim_{r\to\infty}\log\left(\frac{M(r/2)}{M(r)}\right)=-\infty $$ and so $\lim_{r\to\infty}\frac{M(r/2)}{M(r)}=0$.

Could someone please enlighten me on where my proof goes wrong or where I may subtlety use the assumption? Alternative proofs are welcome. Thank you in advance for your time.

  • $\begingroup$ Why does $\frac{d}{dr}(\log M(r))>0$ for large $r$ imply $\lim_{r\to\infty}\log\left(\frac{M(r/2)}{M(r)}\right)=-\infty$? Take $f(z)=z+1$; then $M(r)=r+1$, $\log M(r)=\log(r+1)$, and $\frac{d}{dr}(\log M(r))>0$ but the limit is $-\log 2$. $\endgroup$
    – Michael M
    Jun 20, 2015 at 18:30

1 Answer 1


Michael M. correctly identified the mistake: you have $\frac{d}{dr}(\log M(r))>0$ for large $r$, but in order to conclude $$ \log M(r)-\log M(r/2) \to \infty $$ from the Mean value theorem, you would need to know this derivative tends to infinity. So far you just have $\log M(r)-\log M(r/2) >0$.

To fix this, apply the same argument to $g(z)=f(z)/z^n$ where $n$ is a large integer. This function is holomorphic on $0<|z|<\infty$, so the three-circle theorem still applies. Since $f$ is not a polynomial, $\log M_g(r)\to\infty$ as $r\to\infty$. Being convex, $\log M_g$ must be eventually increasing. So, $\log M_g(r) > \log M_g(r/2)$ for large $r$. But this means $\log M_f(r) > 2^n \log M_f(r/2)$ and since $n$ was arbitrary, the conclusion follows.


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