Exercise 8.4.14

Question

Finish the proof of Theorem 8.4.6.

Ulisse Mini · Accepted Answer

Note that
    $$\frac{F(z) - F(x)}{z-x} = \int_c^d \frac{f(z,t) - f(x,t)}{z-x} dt$$
so choosing $\delta$ so that $0 < |z-x| < \delta$ implies
$$\left|f_x(x,t) - \frac{f(z,t) - f(x,t)}{z-x}\right| < \frac{\epsilon}{d-c}$$
we have
$$\left|\frac{F(z) - F(x)}{z-x} - \int_c^d f_x(x,t)dt\right| = \left|\int_c^d \frac{f(z,t) - f(x,t)}{z-x} - f_x(x,t)\right| < \int_c^d \left|\frac{\epsilon}{d-c}\right| = \epsilon$$
as desired.

Exercise 8.4.14

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Exercise 8.4.14

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