This looks simple, but for some reason I am stuck:
Let $A=\{\alpha>0|\exists C_{\alpha}>0 \ \ \forall z \in \mathbb{C} \ \ \ |f(z)| \leq C_{\alpha} \cdot e^{|z|^{\alpha}} \}$.
I have shown that $infA \leq \frac{1}{2} (1)$ and that $\forall z \in \mathbb{R}_{-} \ \ |f(z)| \geq \frac{1}{2} e^{|z|^{\frac{1}{2}}} (2)$ where $\mathbb{R}_{-}= \{r \in \mathbb{R}|r<0 \}$
I want to conclude that $infA=\frac{1}{2}$.
Does $(2)$ imply that $infA \geq \frac{1}{2}$? If yes , can someone prove it in detail?
PS: Here $f: \mathbb{C} \rightarrow \mathbb{C}$ is the entire function $cos(\sqrt{z})$, but I don't think that matters. The goal was to show that the order of growth of f ($=infA$) is exactly $\frac{1}{2}$.
We have $\frac 1 2 e^{\sqrt x} \leq |f(-x)|\leq C_{\alpha} e^{|-x|^{\alpha}}$ for all $x >0$. This implies that $\sqrt x -x^{\alpha}$ is bounded above so $x^{\alpha} (x^{\frac 1 2 -\alpha}-1) $ is bounded above. Clearly this $\to \infty$ as $x \to \infty$ if $\frac 1 2 -\alpha >0$ so we must have $\frac 1 2 -\alpha \leq 0$ for every $\alpha \in A$ which implies $\inf A \geq \frac 1 2 $.