0
$\begingroup$

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 $f(-z)$?

$\endgroup$
  • 3
    $\begingroup$ What happens if you take $g(z)=f(z)f(-z)$? Especially noting that $g(0)=f(0)^2$... $\endgroup$ – Conrad May 20 at 0:25
1
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.