# Inequality

• Mar 11th 2009, 11:37 PM
manjohn12
Inequality
Lemma. Let $f(z)$ be analytic inside and on the boundary of the disk $|z-z_0| < \rho_0$ and satisfy $|f(z)| \leq M_0$ for $|z-z_0| \leq \rho_0$. If $|f(z_0)| = M_0$ then $|f(z)| = M_0$ for all $z$ in the disk.

Proof. Consider any circle $|z-z_0| = \rho \leq \rho_0$. By the Cauchy Integral Formula we know that $f(z_0) = \frac{1}{2 \pi i} \oint_{|\xi-z_0| = \rho} \frac{f(\xi) \ d \xi}{\xi - z_0}$. For each $\xi$ on this circle with $|f(\xi)| < M_0$ or $|f(\xi)| = M_0$. Assume that there is a $\xi_1$ on this circle for which $|f(\xi_1)| < M_0$. Then since $|f(\xi)|$ is continuous there must be a whole arc $C_1$ of the circle on which $|f(\xi)| < M_0$ for $\xi$ on $C_1$. Also $|f(\xi)| \leq M_0$ on the rest of the circle. Write $f(z_0) = \frac{1}{2 \pi i} \left[ \int_{C_1} \frac{f(\xi) \ d \xi}{\xi- z_0}+ \int_{\text{rest}} \frac{f(\xi)}{\xi-z_0} \ d \xi \right]$.

From here how do we establish that $|f(z_0)| < M_0$, a contradiction? From the M-L Inequality, we know that $|f(z_0)| < \frac{1}{2 \pi i} \left(2 \pi \rho M_0+ 2 \pi \rho M_0 \right) = \frac{2 \rho M_0}{i}$. How do we get this to $M_0$?
• Mar 12th 2009, 09:33 AM
ThePerfectHacker
Quote:

Originally Posted by manjohn12
Lemma. Let $f(z)$ be analytic inside and on the boundary of the disk $|z-z_0| < \rho_0$ and satisfy $|f(z)| \leq M_0$ for $|z-z_0| \leq \rho_0$. If $|f(z_0)| = M_0$ then $|f(z)| = M_0$ for all $z$ in the disk.

This is similar to the maximum-modulos theorem. By Cauchy's theorem we know that:
$f(z_0) = \frac{1}{2\pi i} \oint_{|z-z_0|=r} \frac{f(z)}{z-z_0}dz \text{ where }0
Therefore, (by definition of contour integral)
$f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) d\theta$
Thus,
$M_0 = |f(z_0)| \leq \frac{1}{2\pi } \int_0^{2\pi} |f(z_0 + re^{i\theta}) |d\theta \leq \max\{ |f(z_0+re^{i\theta})| : \theta\in [0,2\pi]\}$

We have shown that on maximum value of $|f|$ on the disk $|z-z_0| = r$ is at least $M_0$. However, $|f|\leq M_0$ and so the maximum value of $|f|$ must be equal to $M_0$. Therefore, it follows that,
$\frac{1}{2\pi} \int_0^{2\pi} |f(z_0 + re^{i\theta})| d\theta = M_0$
The only way that integral can equal to $M_0$ is it at each $z$ on $|z-z_0|=r$ we have $|f|=M_0$.
Thus, $|f| = M_0$ for $|z-z_0| = r$.
But $r$ was arbitrary and by shrinking and expanding $r$ we get the whole disk $|z-z_0|\leq \rho$ with $|f| = M_0$.