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    Bound on $\abs{f(2i)}$ for $f:\HH\to\DD$ with $f(i)=0$

    E-GPFKM

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

    Suppose \(f:\HH\to \CC\) with \(\abs{f(z)}< 1\) and \(f(i) = 0\). Find an upper bound for \(\abs{f(2i)}\).

    Solution

    Compose with the inverse Cayley map \(g(z) \da i{1+z\over 1-z}\) so \(g: \DD\to \HH\) to get \(F\da f\circ g:\DD\to \DD\), where \(F(0) =f(g(0))=f(i) = 0\). So Schwarz applies and \(\abs{F(z)}\leq \abs{z}\). The inverse of \(g\) is \[ g^{-1}(w)={w-i\over w+i}, \qquad g^{-1}(2i)={1\over3}. \] Therefore \[ \abs{f(2i)} =\abs{F(g^{-1}(2i))} =\abs{F(1/3)} \le {1\over3}. \]

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