Analytic and bounded implies uniform continuity

Question

Let $f$ be analytic and bounded in $\{z\in\mathbb{C}\mid Re(z)>0\}$. Prove that $f$ is uniformly continuous in $\{z\in\mathbb{C}\mid Re(z)>C\}=:D$ for every $C>0.$

For uniform continuity, I have to show that for every $\varepsilon>0$ there exists $\delta>0$ such that for all $x,y\in D$ with $\|x-y\|<\delta$ we have $\|f(x)-f(y)\|<\varepsilon$. How can I find such a $\delta$? I don't know how to use the assumptions that $f$ is analytic (i.e. can be written as a power series) and bounded.

Answer 1

Hint: Suppose $|f|\le M$ in the right half plane. Cauchy's estimates give $|f'(x+iy)|\le M/x$ everywhere in the right half plane.