Analytic + bounded --> Uniformly continoius

• Feb 5th 2011, 02:43 PM
skyking
Analytic + bounded --> Uniformly continoius
Here is a problem I got stuck on.

Let $\displaystyle f$ be analytic and bounded in the half plane $\displaystyle \{z:Rez>0\}$. Show that for every $\displaystyle c>0$, $\displaystyle f$ is uniformly continous in the half plane $\displaystyle \{ z:Rez>c\}$.

I tried the following but got stuck.

for every $\displaystyle \epsilon >0$ choose $\displaystyle \delta = min(\frac{\epsilon}{M},c)$, where $\displaystyle M$ is the bound of $\displaystyle f$. So for every $\displaystyle |z_2-z_1|<\delta$ we get by Cauchys Integral formula $\displaystyle |f(z_2)-f(z_1)|=\frac{1}{2\pi}|\int_{C_{\delta}} \frac{f(w)}{w-z_2} - \frac{f(w)}{w-z_1} dw|=\frac{1}{2\pi}|\int_{C_{\delta}} f(w)\frac{z_2-z_1}{(w-z_2)(w-z_1)} dw|\leq \frac{4*2\pi \delta}{2\pi \delta^2}M|z_1-z_2|\leq 2M$. Which gets me nowhere.

I'd appreciate some direction.
• Feb 5th 2011, 05:09 PM
Abu-Khalil
What is $\displaystyle C_\delta$?
• Feb 6th 2011, 02:08 AM
skyking
Quote:

Originally Posted by Abu-Khalil
What is $\displaystyle C_\delta$?

Sorry I should have explained.

It is a circle with radius $\displaystyle \delta$ that containes both $\displaystyle z_1,z_2$.

SK