Analysis Problem of the Day 94

Today’s problem appeared as Problem 11 on the UCLA Spring 2016 Analysis Qual:

Problem 11. Assume $f$ is holomorphic on a domain $U$ containing the closed unit disc. Show that $\max_{|z|=1}\left|\frac{1}{z}-f(z)\right| \geq 1.$

Solution: Suppose not. Then, $\left|\frac{1}{z}-f(z)\right| \leq c <1$ on $|z|=1,$ and by the triangle inequality and residue theorem,

$2\pi=|2\pi i - 0|=\left|\int_{|z|=1} \frac{1}{z}-f(z)dz\right|\leq \int_{|z|=1} \left|\frac{1}{z}-f(z)\right| dz \leq 2\pi c,$

which is a contradiction. Thus, the claim holds.

Leave a Reply Cancel reply