Suppose \(f: \DD\to \DD\) with \(f(0) = 0\) and \(\abs{f(z)} \leq \abs{e^z}\) when \(\abs{z} = 1\). Find an upper bound for \(f\qty{1+i\over 2}\).
Solution
Assume \(f\) extends continuously to \(\overline\DD\), so that the boundary condition makes sense. Consider \(g(z) \da f(z)e^{-z}\), holomorphic on \(\DD\) and continuous on \(\overline\DD\) with \(g(0) = 0\). On \(\abs z=1\), \(\abs{g(z)}\le1\), so by the maximum modulus principle \(\abs g\le1\) on \(\DD\); as \(g(0)=0\), \(g\) is not a unimodular constant, so \(g(\DD)\subseteq\DD\). Schwarz applies and \[ \abs{g(z)}\leq \abs{z} \implies \abs{f(z)} \leq \abs{z}\abs{e^z} = \abs z e^{\Re z} .\] So \[ \abs{f\qty{1+i\over 2}} \leq \abs{1+i\over 2}e^{1/2}= {\sqrt{2e} \over 2} .\]