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Let $F$ be continuous on $\overline{D}(0,1)$ and holomorphic on $D(0,1)$. Suppose that $|F(z)|\leq 1$ when $|z|=1$. Prove that $|F(z)|\leq 1$ for all $z\in D(0,1)$.

Here $D(p,r)$/$\overline{D}(p,r)$ are the open/closed disk with radius $r$ centered at $p$.

I thought if I use the Cauchy integral formula $$|F(z)| = |\frac{1}{2\pi i} \oint_{D(0,1-\epsilon)} \frac{F(\zeta)}{\zeta - z} d\zeta |$$

and find a way to bound the integral on the right side with some known bounds related to the $|\int F|$ and $\int |F|$. Although I can't find any bound for that to work.

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There is a small typo: You switch from $F$ to $f$ in the middle of the displayed equation. –  Andres Caicedo May 12 '11 at 16:26
    
fixed. maximum modulus principle is proven around 100 pages later in the book. I see how this problem can be done easily if the principle is already proven. –  Chao Xu May 12 '11 at 16:35

1 Answer 1

up vote 2 down vote accepted

For $z=0$ the bound works. For $z \ne 0$, note that the fractional linear transformation $\phi(\zeta) = \frac{\zeta + z}{1 + \overline{z} \zeta}$ maps $D(0,1)$ onto itself and maps $0$ to $z$, so you can apply the previous case to $G(\zeta) = F(\phi(\zeta))$.

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Nice argument! –  Andres Caicedo May 12 '11 at 17:15

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