Let \(f\) be analytic in the open unit disk and suppose \[ |f(z)|\le\frac{C}{1-|z|} \] for all \(z\) in the disk, where \(C>0\). Prove that \[ |f'(z)|\le\frac{4C}{(1-|z|)^2}. \]
Solution
Fix \(z\) in the open unit disk and set \[ r\coloneqq\frac{1-\abs{z}}{2}. \]
1. If \(\abs{w-z}=r\), then\[ 1-\abs{w}\geq r. \]
By the triangle inequality, \[ \abs{w} \leq \abs{z}+\abs{w-z} = \abs{z}+r. \] Therefore \[ 1-\abs{w} \geq 1-\abs{z}-r = \frac{1-\abs{z}}2 =r. \] In particular, the circle \(\abs{w-z}=r\) lies inside the open unit disk.
2. On the circle \(\abs{w-z}=r\),\[ \abs{f(w)} \leq \frac{2C}{1-\abs{z}}. \]
By the hypothesis and step 1, \[ \abs{f(w)} \leq \frac{C}{1-\abs{w}} \leq \frac{C}{r} = \frac{2C}{1-\abs{z}}. \]
3. One has\[ \boxed{ \abs{f'(z)} \leq \frac{4C}{(1-\abs{z})^2} }. \]
The Cauchy derivative estimate on the circle \(\abs{w-z}=r\) gives \[ \abs{f'(z)} \leq \frac{1}{r} \max_{\abs{w-z}=r}\abs{f(w)}. \] Using step 2 and the definition of \(r\), \[ \abs{f'(z)} \leq \frac{2}{1-\abs{z}} \cdot \frac{2C}{1-\abs{z}} = \frac{4C}{(1-\abs{z})^2}. \]
4. Q.E.D.
The point \(z\) was arbitrary, so step 3 proves the required estimate throughout the unit disk.