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Texas A&M U has on their August 2012 qualifying exam this problem:

let $f$ be analytic on $\mathbb{D}$ with $|f(z)|\leq 1$. Then we have

$$\frac{|f(0)|-|z|}{1-|f(0)||z|}\leq |f(z)| \leq \frac{|f(0)|+|z|}{1+|f(0)||z|}$$

However, I am only getting this result:

$$\frac{|f(0)|-|z|}{1+|f(0)||z|}\leq |f(z)| \leq \frac{|f(0)|+|z|}{1-|f(0)||z|}$$

How are these related? Is there a typo in Texas' problem? I find that unlikely.

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  • $\begingroup$ not a typo. the first inequality is true. $\endgroup$ – ts375_zk26 May 2 '16 at 20:57
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Suppose that $|f(z)|<1$ for $z$ with $|z|<1$ and consider $$ g(z)=\frac{f(z)-c}{1-\bar{c}f(z)},$$ where $c=f(0).$ Then $g(z)$ is analytic on $\mathbb{D}$ and satisfies $|g(z)|<1, g(0)=0.$ By Schwarz lemma we get $|g(z)|\le |z|$, that is, $$ \left|\frac{f(z)-c}{1-\bar{c}f(z)}\right|\le |z|. $$ Write $w=f(z), r=|z|$ for simplicity. Then we have \begin{align} \left|\frac{w-c}{1-\bar{c}w}\right|&\le r,\\ 1-\left|\frac{w-c}{1-\bar{c}w}\right|^2&\ge 1-r^2,\\ |1-\bar{c}w|^2-|w-c|^2&\ge (1-r^2)|1-\bar{c}w|^2. \end{align} Since \begin{align} LHS&=1-|c|^2-|w|^2-|c|^2|w|^2,\\ RHS&\ge (1-r^2)(1-|c||w|)^2\\ &=1-2|c||w|+|c|^2|w|^2-r^2(1-|c||w|)^2, \end{align} we have \begin{align} 0&\ge |w|^2-2|w||c|+|c|^2- r^2(1-|c||w|)^2\\ &=(|w|-|c|)^2-r^2(1-|c||w|)^2\\ &=\left(|w|-|c|+r(1-|c||w|)\right)\left(|w|-|c|-r(1-|c||w|)\right)\\ &=\left((1-|c|r)|w|-(|c|-r)\right)\left((1+|c|r)|w|-(|c|+r)\right) \end{align} Therefore we have $$ \frac{|c|-r}{1-|c|r}\le |w|\le \frac{|c|+r}{1+|c|r}.$$ The proof is complete.

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