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    $\abs{f(z)}\le\abs{\frac{z-i}{z+i}}$ for holomorphic $f:\HH\to\DD$ with $f(i)=0$

    E-FJHDQ

    Area
    Complex Analysis
    Topics
    Conformal Maps, Fractional Linear Transformations, Schwarz Lemma
    Status
    draft

    Show that if \(f:\HH\to \DD\) is holomorphic and \(f(i) = 0\) then \(\abs{f(z)} \leq \abs{z-i\over z+i}\).

    Solution

    Let \(g(z) \da {z-i\over z+i}: \HH\to \DD\) be the Cayley map, a biholomorphism with inverse \(\inverseof{g}(w) = i{1+w\over 1-w}: \DD\to \HH\), and let \(F \da f\circ \inverseof{g}:\DD\to\DD\). Then \(F(0) = f(\inverseof{g}(0)) = f(i) = 0\) by assumption, so the Schwarz lemma gives \(\abs{F(w)} \leq \abs{w}\) for \(w\in\DD\). For \(z\in\HH\), take \(w=g(z)\): \[ \abs{f(z)}=\abs{F(g(z))}\le\abs{g(z)}=\abs{z-i\over z+i} .\]

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