Htykuut

Let $alpha in (0, 1)$. Then $f(z)=sum_{n=1}^{infty}frac{z^n}{(n!)^{alpha}} $ satisfies $vert f(z) vert leq A_{epsilon} e^{b_{epsilon} vert z vert^{frac{1}{alpha} + epsilon}} $ for any $epsilon>0$ for some constants $A_{epsilon}>0$ and $b_{epsilon}>0$.
I can show the result is true for $alpha geq 1$, but I don't know what to do for $alpha in (0, 1)$.

If $alpha geq 1$, then $vert f(z) vert leq sum_{n=1}^{infty} vertfrac{z^n}{(n!)^{alpha}}vert= sum_{n=1}^{infty} left(frac{ (vert z vert^{frac{1}{alpha}})^n}{n!}right)^{alpha}$ and $(frac{ (vert z vert^{frac{1}{alpha}})^n}{n!})^{alpha} leq frac{ (vert z vert^{frac{1}{alpha}})^n}{n!}$ for all except finitely many $n$.

Any hints or reference are appreciated.

For $alphain (0,1)$, I suggest Cauchy-Schwarz.

Let $g(r) = max_{|z|=r} |f(z)|$.

Fix $delta in (0,alpha)$. Let $1/p = alpha-delta$ and $1/q = 1- (alpha-delta)$. Then $p$ and $q$ are Holder conjugates.

Let's crank the CS machine:

begin{align*} g(r^{alpha-delta} ) &le sum left(frac{ r^{n}}{n!}right)^{alpha-delta}times frac{1}{n!^delta}\
& le left(sum frac{ r^{n}}{n!}right)^{alpha-delta}underset{=c_delta}{underbrace{left (sum frac{1}{n!^{frac{delta}{1-alpha+delta} }} right)^{1-alpha+delta}}}\
& = e^{(alpha-delta) r} c_delta,end{align*}

and the result follows.