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Math Help - Proof question

  1. #1
    Junior Member enjam's Avatar
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    Proof question

    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.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by enjam View Post
    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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