For $\alpha\in (0,\ 1)$, $f(z)=\sum_{n=1}^{\infty}\frac{z^n}{(n!)^{\alpha}}$ satisfies $|f(z)|\leq A_{ε}e^{b_{ε}|z|^{\frac{1}{\alpha}+\epsilon}}$

Question

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 $\vert f(z) \vert \leq \sum_{n=1}^{\infty} \vert\frac{z^n}{(n!)^{\alpha}}\vert= \sum_{n=1}^{\infty} \left(\frac{ (\vert z \vert^{\frac{1}{\alpha}})^n}{n!}\right)^{\alpha}$ 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.

Answer 1

For $\alpha\in (0,1)$, 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.