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    Schwarz lemma and $g(\{|z|<r\})\subseteq f(\{|z|<r\})$ for holomorphic $f,g:\DD\to\Omega$ with $f$ injective and $f(0)=g(0)$

    P-6YHN7

    Area
    Complex Analysis
    Topics
    Biholomorphisms, Conformal Maps, Schwarz Lemma
    Status
    draft

    Suppose \(f, g: \DD\to \Omega\) are holomorphic with \(f\) injective and \(f(0) = g(0)\).

    Show that \[ \Forall 0 < r < 1,\qquad g\qty{\theset{\abs{z} < r}} \subseteq f\qty{\theset{\abs{z} < r}} .\]

    The first part of this problem asks for a statement of the Schwarz lemma.

    Solution

    The statement is false as written. Take \(\Omega=\DD\), \[ f(z)=\frac z2, \qquad g(z)=z. \] Then \(f\) is injective and \(f(0)=g(0)=0\), but for every \(0<r<1\), \[ g(\{|z|<r\})=\{|w|<r\}\not\subseteq\{|w|<r/2\} =f(\{|z|<r\}). \]

    The natural Schwarz-lemma version is true if \(f\) is a biholomorphism \(\DD\to\Omega\). Then \[ h=f^{-1}\circ g:\DD\to\DD \] is holomorphic and fixes \(0\). By Schwarz’s lemma, \(|h(z)|\le |z|\), hence \(h(\{|z|<r\})\subseteq\{|z|<r\}\) for every \(0<r<1\). Applying \(f\) gives \[ g(\{|z|<r\}) =f(h(\{|z|<r\})) \subseteq f(\{|z|<r\}). \]