I am working on the part (a) of Stein Complex Analysis, Chapter 5 Problem 4, which states as follows:
Let $F(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be an entire function of finite order. Suppose $|F(z)|\leq Ae^{a|z|^{\rho}}$. Then show that $$\limsup_{n\rightarrow\infty}|a_{n}|^{\frac{1}{n}}n^{\frac{1}{\rho}}<\infty.$$
Stein also gave hints, so I followed the hint.
First Step: Use Cauchy inequality to bound the coefficient.
Okay this one is easy: Let $R>0$, as $F(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ is entire, it must be holomorphic in $|z|<R$, so for any $0<r<R$ and $n\in\mathbb{N}$, we have $$|a_{n}|\leq \dfrac{\sup_{|z|=r}|F(z)|}{r^{n}}\leq \dfrac{Ae^{ar^{\rho}}}{r^{n}},$$ where the second inequality was obtained because the hypothesis that $F|(z)|\leq Ae^{a|z|^{\rho}}$ for any $z$ implies that the $\sup$ over all $|z|=r$ is the numerator of the quotient on the RHS.
Second Step: Prove the fact that the function $u^{-n}e^{u^{\rho}}$, $0<u<\rho$ attains its minimum value $e^{n/\rho}(\rho/n)^{n/\rho}$ at $u=n^{1/\rho}/\rho^{1/\rho}$. Then choose $r$ in terms of $n$ to achieves this minimum.
I got stuck in this step. Firstly, Consider the function $f(x)$ defined by $f(x):=x^{-n}e^{x^{\rho}}$ for $0<x<\rho$. Its derivative is $$f'(x)=e^{x^{\rho}}x^{-n-1}(\rho x^{\rho}-n),$$ which means that the root for $f'(x)=0$ is $x_{0}=\rho^{-\frac{1}{\rho}}n^{\frac{1}{\rho}}.$ But how do you make sure $x_{0}\in (0,\rho)$? It is varying with $n$ right? Think about if we compare $x_{0}^{\rho}$ with $\rho^{\rho}$, we then compare $\rho^{-1}n$ with $\rho^{\rho}$, so we actually compare $n$ with $\rho^{\rho+1}$. Yes, at first $n$ may be small, but when $n$ goes higher it will eventually bigger than $\rho^{\rho+1}$. How can I conclude that this one is even in the interval $0<x_{0}<\rho$?
Also, for similar reason I do not know how to argue $f(x_{0})$ is the minimum in this interval since I don't know how to compare $f(x_{0})$ to $f(\rho)$. $f(x_{0})$ is varying with $n$.
Then I want to just skip this step for now and to see what's gonna happen. But Stein said it attains minimum right? I think we need a maximum to further the bound right?
I am really confused about this hint.. Any idea?
Best Answer
While the question is about the hint and is addressed in the comments, I would suggest doing this in a different way by taking logarithms.
The problem reduces to showing $\frac{\log |a_n|}{n}+\frac{\log n}{\rho} <M$ for some positive $M$ uniformly in $n$ where the inequality is assumed satisfied automatically for $a_n=0$ as then $\log |a_n| =-\infty$
By Cauchy one knows that $\frac{\log |a_n|}{n} \leq C/n+\frac{ar^{\rho}}{n}-\log r$ for any positive $r$ so one needs to find some $r$ (depending on $n$) for which $\frac{ar^{\rho}}{n}-\log r+\frac{\log n}{\rho} <M$ uniformly in $n$
But now if we take $\log r =\frac{\log n}{\rho}$ (as that term is clearly going to infinity with $n$ so needs getting rid of) we notice that $r^{\rho}=e^{\rho \log r}=n$ so $\frac{ar^{\rho}}{n}=a$ constant and we are done!