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In Stein's complex analysis text, the growth order of entire function is defined as follows.

Def 1) Let $f$ be an entire map on $\mathbb{C}$. We say that $f$ has a growth order $\le \rho$ if and only if there are positive constants $A,B$ and the positive number $\rho$ such that $|f(z)|\le Ae^{B|z|^{\rho}}$ on $\mathbb{C}$. The growth order $Ord_g (f)$ is defined as an infimum of all above $\rho$'s.

However, another definition of the growth order is introduced in wiki.

Def 2) Let $f$ be an entire map on $\mathbb{C}$. The growth order $Ord_g (f)$ is defined by $$\limsup_{r\to\infty } \frac{{\rm{ln}(ln}(||f||_{\infty,B_r}))}{{\rm{ln}}r}.$$

I proved the second part $\le$ $Ord_g (f)$ by the limit computations. How can I show the opposite direction?

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Let $$\lambda =\limsup_{r\to\infty } \frac{{\rm{ln}(ln}(||f||_{\infty,B_r}))}{{\rm{ln}}r}.$$ Then by the property of $\limsup$ $$\forall \varepsilon >0,\quad \exists R\,;\quad r>R\implies \frac{{\rm{ln}(ln}(||f||_{\infty,B_r}))}{{\rm{ln}}r}<\lambda +\varepsilon .$$ This implies $$ ||f||_{\infty,B_r}\le e^{r^{\lambda +\varepsilon }} \quad (r>R), $$

in other words $$ |f(z)|\le e^{|z|^{\lambda +\varepsilon }}\quad (|z|>R)\tag{1}. $$ Define $$ A=1+\max_{|z|\le R} |f(z)|.$$

Then from $(1)$ we have $$ |f(z)|\le Ae^{|z|^{\lambda +\varepsilon }}\quad \text{for all }z\in \mathbb{C}. $$ Thus we conclude that $f(z)$ has a growth order $\le \lambda +\varepsilon $. Since $\varepsilon $ can be arbitrary small, we conclude $$Ord_g (f)\le \lambda .$$

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  • $\begingroup$ Your approach is so elegant. Thank you! :) $\endgroup$ – Chris kim Feb 23 '17 at 5:49

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