Let \(f\) be a holomorphic function on the unit disc \(\{z : |z| < 1\}\) satisfying \(f(0) = 0\) and \(\mathrm{Re}\, f(z) \leq A\) for some positive number \(A > 0\). Prove:
\[|f(z)| \leq \frac{2A|z|}{1 - |z|}.\]
Solution
1. A fractional transformation of \(f\) satisfies the Schwarz estimate.
Put \(D=\{z:|z|<1\}\) and \[ h(z)=\frac{f(z)}{2A-f(z)}. \] The denominator cannot vanish: \(f(z)=2A\) would contradict \(\operatorname{Re}f(z)\leq A\) and \(A>0\). Thus \(h\) is holomorphic on \(D\) and \(h(0)=0\). Moreover, \[ |2A-f(z)|^2-|f(z)|^2=4A\bigl(A-\operatorname{Re}f(z)\bigr)\geq0, \] so \(|h(z)|\leq1\).
The Schwarz bound \(|h(z)|\leq|z|\) holds under the non-strict hypothesis \(|h|\leq1\). The quotient \(H(z)=h(z)/z\) extends holomorphically to zero by the Taylor expansion of \(h\). For \(0<\rho<1\), it satisfies \(|H|\leq1/\rho\) on \(|z|=\rho\). The maximum modulus principle gives this bound throughout that closed disk [@SS03]. For fixed \(z\in D\), let \(\rho\uparrow1\) with \(\rho>|z|\) to obtain \(|H(z)|\leq1\). Therefore \(|h(z)|\leq|z|\) on \(D\).
2. Inverting the transformation yields the required modulus bound.
The identity defining \(h\) rearranges to \(f(1+h)=2Ah\). Step 1 gives \(|h(z)|\leq|z|<1\), so \(1+h(z)\) is nonzero and \[ |f(z)|=\frac{2A|h(z)|}{|1+h(z)|} \leq\frac{2A|z|}{1-|z|}. \] Here the denominator estimate is \(|1+h(z)|\geq1-|h(z)|\geq1-|z|>0\). This also covers \(z=0\), where both sides vanish.