# Showing a Holomorphic Function is Constant

1. Jul 26, 2010

### misnomered

1. The problem statement, all variables and given/known data
Let $$f$$ be a holomorphic function defined for $$\vert z \vert < 1$$. If there is a $$t \in (0, 1)$$ such that

$$\left|f\left(\frac{1}{n}\right)\right| \leq t^n$$

for all $$n \in \mathbb{N}$$ show that $$f$$ is constant on $$\vert z \vert < 1$$

2. Relevant equations
Hearing the word constant, I immediately think of the Identity Theorem, Open Mapping Theorem, and Maximum Modulus Principle

3. The attempt at a solution
The function $$f$$ has a power series representation in the unit circle about $$0$$; that is

$$f(z) = \sum_{k = 0}^{\infty} a_{k}z^{k}$$.

From this we have that

$$f\left(\frac{1}{n}\right) = \sum_{k = 0}^{\infty} \frac{a_{k}}{n^{k}}$$

and by the condition that

$$\left|f\left(\frac{1}{n}\right)\right| \leq t^n$$

we see that $$f(0) = 0$$. Suppose now that $$a_{0} = a_{1} = ... = a_{m - 1} = 0$$ holds for some $$m$$. Then we have that

$$\frac{1}{n^{m}}\left|a_{m} + \sum_{k = m + 1}^{\infty} \frac{a_{k}}{n^{m - k}}\right| = \left|\sum_{k = 0}^{\infty} \frac{a_{k}}{n^{k}}\right| = \left|f\left(\frac{1}{n}\right) \right| \leq t^{n}$$

so that

$$\left|a_{m} + \sum_{k = m + 1}^{\infty} \frac{a_{k}}{n^{m - k}}\right| \leq n^{m}t^{n}$$.

Letting $$n \rightarrow \infty$$ we see that $$a_{m} = 0$$ (note that exponentials overtake polynomials). The result follows.

Last edited: Jul 26, 2010