About the growth of entire functions

Define $M(r)=\sup_{|z|=r}|f(z)|$. Given an increasing function $\phi(r)$ as $r\to\infty$, how to construct an entire function $f(z)$ to satisfy the inequality $M(r)>1+\phi(r)$?

Attempt to solve this problem:

If $M(r)>1+\phi(r)$, there exists $\theta_0$ such that $|f(re^{i\theta_0})|>1+\phi(r)$. Therefore, I need to find an entire function $f(z)$ such that $$\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(re^{i\theta})|^2d\theta>[1+\phi(r)]^2$$.

If $f(z)=\sum_{n=0}^\infty a_nz^n$, we have $$\sum_{n=0}^\infty|a_n|^2r^{2n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(re^{i\theta})|^2d\theta$$

Therefore, I need to find a sequence of $\{a_n\}$ such that $$\sum_{n=0}^\infty|a_n|^2r^{2n}>[1+\phi(r)]^2\qquad\forall r>0$$

and $$\lim_{n\to\infty} |a_n|^{1/n}=0$$

Alternatively, I need to find a sequence of $\{a_n\}$ such that $$\sup_n |a_n|r^n>1+\phi(r)\qquad\forall r>0$$

and $$\lim_{n\to\infty} |a_n|^{1/n}=0$$

However, since I know little information about the growth of $\phi(r)$, I can hardly pin down the $a_n$'s. Any hints will be appreciated.

Thanks.

As a hint, for any increasing sequence of positive integers $(a_n)$, and any constant $C$, the function $$f(z) = C + \sum_{n=1}^\infty \left(\frac{z}{n}\right)^{a_n}$$ is entire and satisfies $M(r) \ge C$ for $0 \le r < 2$, and $M(n+1) \ge \left(\frac{n+1}{n}\right)^{a_n}$ for all integers $n \ge 1$. By choosing $C$ and $a_n$ sufficiently large, you can find the desired function.
• Do you suggest that I should choose a sufficient large $C$ such that $C>1+\phi(2)$ and a sequence $a_n$ such that $M(n+1)>1+\phi(n)$. However, with this choice, $M$ will be larger than $\phi$ only for integers not less than 1. I need to construct an entire function such that $M(r)>1+\phi(r)$ for any $r>0$, not just integers. I know this comment is definitely stupid, but I don't get your point... – YYF Jan 4 '13 at 23:27
• You almost got it, then you just have to throw in the fact that $M(r)$ is increasing (by the maximum principle), and then you get the estimates for non-integer $r$ as well. – Lukas Geyer Jan 5 '13 at 3:47