Let $f$ be an entire function and define $$M(r)=\max\limits_{|z|\le r}|f(z)|=\max\limits_{|z|=r}|f(z)|.$$ Similarly, also define $$m_0(r)=\min\limits_{|z|\le r} |f(z)|,\hspace{5mm} m(r)=\min\limits_{|z|=r}|f(z)|.$$

I was thinking about what we can say about the functions $M(r), m_0(r)$ and $m(r)$ as $r\to\infty.$ For example, by the maximum modulus principle, we know that $M(r)$ is actually increasing. Therefore the limit $\lim\limits_{r\to\infty}M(r)$ exists. It is also clear that if $M(r)\to c<\infty$ then (Liouville's theorem) $f$ must be a constant.It follows that $M(r)\to \infty.$ In fact, for polynomials it is each to check that $M(r)\approx r^n$ as $r\to \infty.$ Moreover, if $\frac{M(r)}{r^n}\to c<\infty$ then using Cauchy's estimate we can show that $f$ must be a polynomial. If $f$ is non-polynomial entire function then it is clear that $M(r)/r^n\to \infty$ for every $n\ge 0.$

My question is can we strengthen it furthre? For example can we say that for a non-polynomial entire function $f,$ we must have $M(r)\approx e^{r}$ or $M(r)\geq Ce^{r^{1-\epsilon}}$ for every $\epsilon>0?$

Now coming to $m_0(r)$ and $m(r).$ We note that $m_0(r)$ is decresing and hence the limit exists, but it is not very interesting. If $f$ has any zero in the plane then $m_0(r)=0$ for all $r$ sufficiently large and therefore the limit will be zero. On the other hand, if $f$ does not have any zero and $f$ is non-constant then $f$ must go arbitrary close to $0$ by picard's theorem. It follows that $m_0(r)\to 0.$ In other words, $m_0(r)\to c<\infty,$ and $c\neq 0$ if and only if $f$ is a constant.

The most interesting one is $m(r).$ Let us start with a simple case. If $f$ is a polynomial (of degree $n\ge 1$) then $m(r)\approx r^n$ for suffiently large $r.$ Therefore, the limit $m(r)\to \infty.$ Moreover, $\frac{m(r)}{r^n}\to c\neq 0.$ If $f$ is not a polynomial and $f$ does not have a zero then it is very similar to $m_0(r)$ and $m(r)\to 0.$ In general, for a non-polynomial entire function $f,$ we know that the infinity is an essential singularity. In particular, there exists a sequence $z_n\to \infty$ such that $|f(z_n)|\to 0.$ This tells us that $m(|z_n|)\to 0.$ In particular, if $\lim m(r)$ exists, then it must be $0.$ But, I am not able to establish the existence of limit of $m(r).$

Does the limit $\lim\limits_{r\to \infty}m(r)$ always exist?

Can we make a more refined statement about the behavior of $m(r)?$ For example, if $f$ has $n$ zeroes in the complex plane can we say that $m(r)\to 0$ like $r^{-n}?$ (I am not hoping this statement to be true, it is just for the illustration of the kind of statement I want to make about $m(r).$)

  • $\begingroup$ I would recommend checking out a book on entire functions (Levin's Lectures on Entire functions for a masterful concise exposition or Boas Entire Functions for a more leisurely approach) as your questions are answered there since they are part of this theory; for example $\cos \sqrt z=\sum {\frac{(-z)^k}{(2k)!}}$ is an entire function of order $\frac{1}{2}$ which means it satisfies $M(r)$~$e^{\sqrt r}$ (in the logarithmic asymptotic sense that the ratio of the logarithms of the two goes to $1$) so things are subtler than you think $\endgroup$
    – Conrad
    Apr 3, 2020 at 12:25
  • $\begingroup$ Maybe first start here: en.wikipedia.org/wiki/Entire_function $\endgroup$
    – Conrad
    Apr 3, 2020 at 12:32


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.