If an entire function satisfies $|f(z)| \leq C e^{M|z|}$, then $f(x)$ can't decay super-exponentially as $x\to\infty$ 
Let $f$ be a non-zero entire function.  Suppose there are positive real numbers $C$ and $M$ such that $|f| \leq C e^{M|z|}$.
Show that there is no function $g(x)$, defined on $x>0$ with $\lim_{x \to \infty} g(x)=+\infty$ such that $f(x)=O(e^{-xg(x)})$ as $x \to \infty$.

Source: I'm reading Complex Analysis by Stein.
From the assumption, I can know the order of $f$ is less than or equal to $1$. Then how can I get from there?
 A: It must follow from those factorization theorems... I don't know that stuff well enough and I'm not going to look it up.
It's enough to show that there exists a function $h(r)$ defined for $r>0$ such that $h(r)\to\infty$ and $$|f(re^{it})|\le e^{-rh(r)}\quad(r>0,0<t<\pi/4).$$
Indeed, suppose so, let $\omega=e^{2\pi i/8}$, and define $$F(z)=\prod_{k=0}^7f(\omega^kz).$$Then $F$ is an entire function which tends to zero at infinity; hence $F=0$ and it follows that $f=0$.
The existence of such a function $h$ follows from the fact that $u=\log|f|$ is subharmonic.
First let $G(r)=\inf_{r/2<t<2r}$, and note that $G(r)\to\infty$. Fix $r>0$, let $$S=\{\rho e^{it}:r/2<\rho<2r,0<t<\pi/2\}$$and $$K=\{re^{it}:0< t\le\pi/4\}.$$Think about the Dirchlet problem in $S$. It's "clear" that there exists $\alpha\in(0,1)$ such that the harmonic measure of the interval $[-r/2,2r]$ with respect to any point of $K$ is at least $\alpha$. Heh - Garnett tells us that the best intuition regarding harmonic measure is given by Brownian motion, and it's clear that if you take Brownian motion started at any point of $K$ the probability that the first point where you hit $\partial S$ is a point of this interval is at least $\alpha$. (See Complex Made Simple for a non-rigorous but totally convincing explanation of what Brownian motion has to do with harmonic measure, free from nasty details like definitions and proofs...) See Note Below
Since $u\le 2Mr$ on all of $\partial S$ while $u\le -rG(r)/2$ on the interval $[-r/2,2r]$ this shows that $$u(z)\le 2(1-\alpha)Mr-\alpha rG(r)/2\quad(z\in K);$$this gives the inequality we need, with $$h(r)=-2(1-\alpha)M+\alpha G(r)/2.$$
Note:
I'm leaving the text as it is because I think it makes the existence of $\alpha$ maximally clear. But it is a little fuzzy. Here's an actual proof with no handwaving at all:
Note first that the $S$ we get for one value of $r$ is similar to the $S$ for another value of $r$, so any $\alpha$ that works for one $r$ works for all $r$.
Now $S$ is certainly a Dirichlet domain. Say $v$ is harmonic in $S$ and continuous on the closure, with $0\le v\le 1$, such that $v$ vanishes on $\partial S\setminus[r/2,2r]$ and $v=1$ on most of $[r/2,2r]$. The harmonic measure in question is greater than or equal to $v$, and it's clear that $v$ is bounded away from $0$ on $K$ because $v$ is continuous and strictly positive on $\overline K$.
