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I'm studying for a qual exam. I cannot solve the following problem:

Let $f$ be holomorphic from the unit disc to itself. $f\left(\frac{1}{2}\right) = f\left(-\frac{1}{2}\right) = 0$. Show that $|f(0)| \le \frac{1}{3}$.

With an application of Schwarz lemma, I showed that $|f(0)| \le \frac{1}{2}$. I can't prove the tighter bound. Any ideas?

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1 Answer 1

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Consider the function

$$g(z) = f(z) \frac{1 - r^2 z^2}{r^2 - z^2}$$

where $r = \frac{1}{2}$, which is holomorphic on $\mathbb{D}$ since $f$ has roots at $\pm r$. When $|z| = 1$, $|f(z)| \leq 1$ and

$$|g(z)| \leq \left|\frac{1 - r^2 z^2}{r^2 - z^2} \right|$$

We have

$$16 |1 - r^2 z^2|^2 = |4 - z^2|^2 = (4 - z^2) (4 - \overline{z}^2) = 16 - 4z^2 - 4\overline{z}^2 + 1$$

and

$$16|r^2 - z^2|^2 = |1 - 4z^2|^2 = (1 - 4z^2)(1 - 4\overline{z}^2 = 1 - 4z^2 - 4\overline{z}^2 + 16$$

We conclude that $|1 - r^2 z^2| = |r^2 - z^2|$, so $|g(z)| \leq 1$ whenever $|z| = 1$.

So by the Maximum Modulus Principle, $|g(0)| \leq 1$; in particular,

$$|f(0)| \frac{1}{1/4} \leq 1 \implies |f(0)| \leq \frac{1}{4}$$

which is a stronger result.

The approach was suggested by an answer here, as indicated in Clayton's comment.

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