Let \(f:\Delta\to\mathbb C\) be analytic and suppose \[ \operatorname{Re}f(z)\ge0. \] Show that \[ |f(z)|\le |f(0)|\frac{1+|z|}{1-|z|}. \]
Solution
Set \(a=f(0)\).
1. If \(\Re a=0\), then \(f\) is constant and the estimate holds.
The harmonic function \(u=\Re f\) satisfies \(u\ge0\) on \(\Delta\) and \(u(0)=0\). Thus \(u\) attains its minimum at an interior point, so the minimum principle gives \(u\equiv0\). A holomorphic function with identically zero real part is constant, hence \(f\equiv a\). Therefore \[ |f(z)|=|a| \le |a|\frac{1+|z|}{1-|z|}. \]
2. Suppose \(\Re a>0\). Then\[ \Phi(w)=\frac{w-a}{w+\overline a} \]maps the right half-plane holomorphically into \(\Delta\) and satisfies\(\Phi(a)=0\).
For \(\Re w>0\), \[ |w+\overline a|^2-|w-a|^2 =4(\Re w)(\Re a)>0. \] Hence \(|\Phi(w)|<1\). The identity \(\Phi(a)=0\) is immediate.
3. For every \(z\in\Delta\),\[ \left|\frac{f(z)-a}{f(z)+\overline a}\right|\le |z|. \]
By 2, the map \[ g=\Phi\circ f:\Delta\to\Delta \] is holomorphic and satisfies \(g(0)=0\). Schwarz’s lemma therefore gives \(|g(z)|\le|z|\).
4. The desired estimate follows.
Let \(r=|z|\). From 3, \[ |f(z)-a| \le r\,|f(z)+\overline a| \le r\bigl(|f(z)|+|a|\bigr). \] The triangle inequality also gives \[ |f(z)|\le |f(z)-a|+|a|. \] Combining these inequalities yields \[ |f(z)| \le r|f(z)|+(1+r)|a|, \] so \[ (1-r)|f(z)|\le(1+r)|a|. \] Since \(a=f(0)\), \[ |f(z)|\le |f(0)|\frac{1+|z|}{1-|z|}. \]