Proof question

**enjam** · August 3rd 2009, 02:26 AM

Hey guys, I've been staring at this problem for quite some time and not really sure where to start.

Any assistance would be greatly appreciated. Thanks.

**Opalg** · August 3rd 2009, 05:19 AM

Originally Posted by enjam

Hey guys, I've been staring at this problem for quite some time and not really sure where to start.

Any assistance would be greatly appreciated. Thanks.

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

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

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

To get a bound on that integral, let d be the distance from $z_0$ to the interval [0,1] (you are told that $z_0\notin[0,1]$ , so you know that d>0), and choose z close enough to $z_0$ 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.

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