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Let a function $f \in \mathcal{O(\mathbb{\bar{D}})}$ where $\mathcal{O(\mathbb{\bar{D}})}$ is the ring of holomorphic functions on the closed unit disk $\bar{\mathbb{D}}$.

Assuming

$\lvert{f(e^{i \theta})}\rvert \leq e^{\sin{\theta}},\qquad 0\leq\theta\leq 2\pi$

how to show that

$\lvert{f(0)}\rvert \leq 1$

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Hint: Put $g(z)=f(z)\exp(iz)$, and compute $|g(\exp(i\theta))|$.

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Since $f$ is a bounded holomorphic function in the disk it has a factorization $$f=Be^{u+iv}$$where $B$ is a Blaschke product. Now $B$ has boundary values of modulus $1$ almost everywhere, hence $$e^{u(e^{i\theta})}\le e^{\sin\theta}$$at almost every boundary point. Which says that $u(e^{i\theta})\le\sin\theta$ almost everywhere. Hence $$u(0)\le\frac1{2\pi}\int_0^{2\pi}\sin\theta\,d\theta=0.$$Since $|B(0)|\le1$ this shows that $|f(0)|\le1$.


Or if you wanted to do it without the machinery of Blaschke products, use the fact that $\log|f|$ is subharmonic.

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  • $\begingroup$ Thanks! I was looking for a simpler way to do it, without involving Blaschke products or subharmonic functions. $\endgroup$ – marcg Jan 14 '16 at 15:22
  • $\begingroup$ I predict that any "simpler" argument you find is going to look, at least to me, like it's using one of those two arguments implicitly. Since after all the subharmonicity of $\log|f|$ hits it right on the nose... or you come up with some explicit function $g$ and look at $f/g$; that's going to be "equivalent" to the Blachke product thing, just with $u$ made explicit. $\endgroup$ – David C. Ullrich Jan 14 '16 at 15:34
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Let $g(z)=f(z)f(-z)$.

Then $|g(e^{i\theta})|\leq|f(e^{i\theta}).f(e^{i(\theta+\pi)})|\leq e^{\sin\theta-\sin\theta}=1$

By the maximum principle,

$|f(0)|^2=|g(0)|\leq\max_{0\leq\theta\leq2\pi}|g(e^{i\theta})|\leq1$

Therefore $|f(0)|\leq1$

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