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Suppose $f$ is continuous on $\overline{\mathbb{D}}$ and holomorphic on $\mathbb{D}$. If $|f(e^{i\theta})| \leq a$ for $\theta \in [0, \pi)$ and $|f(e^{i\theta})| \leq b$ for $\theta \in [\pi, 2\pi)$ with $a, b \neq 0$ then the problem I am working on is to show that $|f(0)| \leq \sqrt{ab}$.

I can prove this if $f$ is also holomorphic on $\overline{\mathbb{D}}$ as follows: Since $\overline{\mathbb{D}}$ is compact and $f$ is holomorphic, $f$ must have only finitely many zeros in the closed unit disc since zeros of $f$ are isolated. Denote the zeros by $z_{1}, \ldots, z_{n}$. Consider $g(z) = f(z)\prod_{k = 1}^{n}\frac{1 - \overline{z_{k}}z}{z - z_{k}}$. This function has no zeros in $\overline{\mathbb{D}}$ and satisfies the same bounds as $f$. Applying (essentially) the Mean Value Property to $\log |g(z)|$ at $z = 0$ proves the result.

However, if $f$ is just holomorphic on $\mathbb{D}$, I am not sure what to do, does the same argument still work? All I need for the above argument to work is that $f$ has finitely many zeros in $\mathbb{D}$, but I am not sure how to get that.

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Consider the function $g(z)=f(z)f(-z)$ and use the maximum modulus principle.

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  • $\begingroup$ Wait, do you instead mean $f(z)\overline{f(\overline{z})}$? $\endgroup$ – jyt123 Nov 10 '13 at 18:28

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