# Proof question

• Aug 3rd 2009, 02:26 AM
enjam
Proof question
Hey guys, I've been staring at this problem for quite some time and not really sure where to start.
http://img26.imageshack.us/img26/6676/question13.jpg
Any assistance would be greatly appreciated. Thanks.
• Aug 3rd 2009, 05:19 AM
Opalg
Quote:

Originally Posted by enjam
Hey guys, I've been staring at this problem for quite some time and not really sure where to start.
http://img26.imageshack.us/img26/6676/question13.jpg
Any assistance would be greatly appreciated. Thanks.

Start by guessing what $\displaystyle H'(z)$ ought to be. Differentiating under the integral sign would suggest that $\displaystyle H'(z) = \int_0^1\frac{h(t)}{(t-z)^2}\,dt$.

Now see if you can justify that guess by doing as the hint suggests and using the definition of the derivative. First, note that $\displaystyle H(z)-H(z_0) = \int_0^1\frac{h(t)(z-z_0)}{(t-z)(t-z_0)}\,dt$. Then $\displaystyle \frac{H(z)-H(z_0)}{z-z_0} - \int_0^1\frac{h(t)}{(t-z)^2}\,dt = \int_0^1\frac{h(t)(z-z_0)}{(t-z)(t-z_0)^2}\,dt$, so that $\displaystyle \left|\frac{H(z)-H(z_0)}{z-z_0} - \int_0^1\frac{h(t)}{(t-z)^2}\,dt\right| \leqslant |z-z_0|\int_0^1\frac{|h(t)|}{|t-z||t-z_0|^2}\,dt$.

The idea now is to find a bound for that last integral. The other factor $\displaystyle |z-z_0|$ will then ensure that the expression on the left side of the inequality goes to 0 as $\displaystyle z\to z_0$, which will show that $\displaystyle H$ is differentiable at $\displaystyle z_0$.

To get a bound on that integral, let d be the distance from $\displaystyle z_0$ to the interval [0,1] (you are told that $\displaystyle z_0\notin[0,1]$, so you know that d>0), and choose z close enough to $\displaystyle z_0$ to ensure that the distance from z to [0,1] is greater than d/2. That will give you a lower bound on the denominator of the integrand; and the fact that h is continuous means that it is bounded on [0,1], so that gives you an upper bound for the numerator.