Let \(f\) be analytic in \(|z|<1\) and suppose \[ |f(z)|\le\frac1{1-|z|}. \] Show that \[ |f'(0)|\le4. \]
Solution
1. For every \(0<r<1\),\[ |f'(0)| \leq \frac{1}{r(1-r)}. \]
On the circle \(|z|=r\), the hypothesis gives \[ |f(z)| \leq \frac1{1-r}. \] By Cauchy’s estimate for the first derivative at the center, \[ |f'(0)| \leq \frac{1}{r}\max_{|z|=r}|f(z)| \leq \frac{1}{r(1-r)}. \]
2. The quantity \(r(1-r)\) is maximized on \((0,1)\) at\[ r=\frac12, \]with maximum value \(1/4\).
Completing the square gives \[ r(1-r) = \frac14-\left(r-\frac12\right)^2 \leq \frac14. \] Equality holds at \(r=1/2\).
3. Therefore\[ \boxed{|f'(0)|\leq4}. \]
Substitute \(r=1/2\) into step 1: \[ |f'(0)| \leq \frac{1}{(1/2)(1/2)} = 4. \]
4. Q.E.D.
Step 3 is the required bound.