# Growth of holomorphic functions

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 $$m(r).$$)

• 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 Apr 3, 2020 at 12:25
• Maybe first start here: en.wikipedia.org/wiki/Entire_function Apr 3, 2020 at 12:32