0

Show that if $ |f(z)| < M $ on $ |z| = 1 $ and $ f(z) = z^2 g(z) $, then \begin{equation} |f(z)| \le \frac{M}{4} \ for \ |z| \le \frac{1}{2}. \end{equation} Here $ f(z) $ and $ g(z) $ are analytic in $ |z| \le 1 $.

Using Cauchy's estimate it is easy to prove that \begin{equation} |f(z)| \le \frac{M}{2}\ \ for\ |z| \le \frac{1}{2}, \end{equation}

since

\begin{equation} |g^{(n)}(0)| \le Mn! \\ \end{equation}

and therefore

\begin{equation} |f(z)| = |z^2\sum\limits_{n=0}^{\infty}\frac{g^{(n)}(0)}{n!}z^n| \le \frac{1}{4}\sum\limits_{n=0}^{\infty}M|z^n| \le \frac{1}{4}\frac{M}{1 - |z|} \le \frac{M}{2} \end{equation}

But how to prove that $ |f(z)| \le \frac{M}{4} $?

Community
  • 1

1 Answers1

1

Got it, thanks. Set $ B = \{z:|z| \le 1\} $ is a compact, then exists $ z_0 \in B : |g(z_0)| \ge |g(z)| $ on $ B $. $ |z_0| $ can't be strictly less than $ 1 $ by maximum modulus principle, then $ |z_0| = 1$ and therefore $ |g(z_0)| < M $ and $ |f(z)| \le |z^2|M \le \frac{M}{4} $ for $ |z| \le \frac{1}{2} $