# Bound $|f(1/4)|$, given $f(1)=0$, $f$ analytic, $|f|\le2$ for $|z|\lt2$

How to show that problem

Given $f$ analytic in $|z| < 2,$ bounded there by 2, and such that $f(1) = 0,$ find the best possible bound for $|f(1/4)|$

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Let $D$ the open unit disk, and $g(z):=\frac{f\left(2z\right)}2$, which maps $D$ to itself. Consider the transformation $\phi\colon D\to D$ defined by $\phi(z):=\frac{2z-1}{2-z}$. It's a bijective map, whose inverse is given by $\phi^{-1}(z)=\frac{2z+1}{2+z}$. Let $h(z):=\phi(g(\phi^{—1}(z)))$. It's a holomorphic function from $D$ to $D$, and $h(0)=0$ hence $|h(z)|\leq |z|$, by Schwarz lemma. This gives that $$\left|\frac{f\left(\frac 14\right)-1}{2-\frac 12f\left(\frac 14\right)}\right|\leq \frac{10}{17}.$$ Assuming WLOG that $f\left(\frac 14\right)$ is real, we can find an upper bound.

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We don't need f(1/4) to be real, because we're finding a bound for its modulus – Bey Aug 11 '12 at 16:03
@Bey You are right, but I think it's easier to see what could be the maximum value of $x$ such that $\left|\frac{x-1}{2-x/2}\right|\leq 10/17$. – Davide Giraudo Aug 11 '12 at 17:45