# Consequence of maximum modulus principle [duplicate]

Let $$f$$ be continuous on the closed unit disc $$\overline D=\{z\in \mathbb C~:~|z|\leq 1\}$$ and analytic in the open unit disc $$D=\{z\in \mathbb C~:~|z|<1\}$$. Suppose that $$|f(z)|\leq a$$ on the set $$\{z\in \mathbb C~:~|z|=1, Im(z)\geq0\}$$ and $$|f(z)|\leq b$$ on the set $$\{z\in \mathbb C~:~|z|=1, Im(z)\leq 0\}$$. Then prove that $$|f(0)|\leq \sqrt{ab}$$.

How maximum modulus principle plays here to get an estimate for $$|f(0)|$$, Is there any role for $$f(-z)$$?

• What happens if you take $g(z)=f(z)f(-z)$? Especially noting that $g(0)=f(0)^2$... May 20, 2020 at 0:25

## 1 Answer

Define $$h(z) = f(z)f(-z)$$. One can quickly note that $$h(z)$$ is holomorphic on the open unit disk and continuous on $$\overline{D}$$ as it is a product of holomorphic/continuous functions. By the maximum modulus principle applied to $$h$$, we have that there exists $$z_0 \in \partial D$$ such that $$|h(z)| \leq |h(z_0)|$$ for all $$z \in \overline{D}$$. Without loss of generality, one can assume that $$z_0 \in \{z\in \mathbb C~:~|z|=1, \space \text{Im}(z)\geq0\}$$ Then $$-z_0 \in \{z\in \mathbb C~:~|z|=1, \space \text{Im}(z)\leq 0\}$$. Can you finish?