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Prove the following extension of the Maximum modulus principle. Let $f$ be holomorphic and bounded on $ |z|<1$ and continuous on $ |z|\le 1$ except maybe at $z=1$. If $ |f(e^{i\theta}| \le A $ $ \forall$ $ \theta$ such that $ 0<\theta <2\pi$, then $ |f(z)|\le A$ for all $ |z| <1 $.

Well... I was trying to use maximum modulus principle, fixing the domain, i was searching for a curve that joins two distinct points of the boundary (not equal 1 ), lives in the interior of the disk, such that all the norms of the values of the curve are bounded by A, and then aplying maximum modulus in this new region. Well I'm a little lost... please help me :(

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Given $|w|<1$, to show $|f(w)|\le A$, consider the function $g(z)=f(\frac{w-z}{1-\bar{w}z})$. Then $g(0)=f(w)$, $g$ is holomorphic and bounded on $|z|<1$, and $g$ is continuous on $|z|\le 1$ except for at most one point $\frac{w-1}{1-\bar{w}}$. Then for each $0\le r<1$, $$|g(0)|=|\frac{1}{2\pi}\int_{0}^{2\pi}g(re^{i\theta})d\theta|\le\frac{1}{2\pi}\int_{0}^{2\pi}|g(re^{i\theta})|d\theta.$$

Noting that $|g(z)|$ is bounded on $|z|<1$ and $\lim_{r\to 1}|g(re^{i\theta})|=|g(e^{i\theta})|\le A$ except for at most one $0\le\theta<2\pi$, by dominated convergence theorem,

$$|f(w)|=|g(0)|\le\lim_{r\to 1}\frac{1}{2\pi}\int_{0}^{2\pi}|g(re^{i\theta})|d\theta\le A.$$

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I don't know that theorem , please could you tell me what is? –  Joseph Nov 11 '12 at 6:10
    
Well it's a very powerfull theorem, I can't use it :/ –  Joseph Nov 11 '12 at 6:17
    
But thanks! Your answer was very usefull!! –  Joseph Nov 11 '12 at 6:26
    
@Joseph: Do you mean dominated conve.rgence theorem? I hope you have got it. –  23rd Nov 11 '12 at 6:39
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