Continuous function on simple closed contour

Question

Let $f$ denote a function that is continuous on a simple closed contour $C$. Using the Cauchy Integral formula, prove that the function $g(z)=\frac{1}{2\pi i}$ $\int_C$ $\frac{f(s)ds}{s-z}$ is analytic at each point z interior to $C$ and that $g'(z)=\frac{1}{2\pi i}$ $\int_C$ $\frac{f(s)ds}{(s-z)^2}$ at such a point.

Answer 1

You can do it directly using the definition of the derivative.

$$g'(z) = \lim_{h \to 0} \frac{1}{2\pi i} \frac{1}{h} \int_C{ \left( \frac{1}{w-(z+h)} - \frac{1}{w-z} \right) f(w) dw}$$

$$ = \lim_{h \to 0} \frac{1}{2\pi i} \frac{1}{h} \int_C{ \left( \frac{h}{(w-(z+h)) (w-z)} \right) f(w) dw} $$

$$ = \lim_{h \to 0} \frac{1}{2\pi i} \int_C{ \left( \frac{1}{(w-(z+h)) (w-z)} \right) f(w) dw} $$

$$ = \frac{1}{2\pi i} \int_C{ \left( \frac{1}{(w-z)^2} \right) f(w) dw} $$

where the last step involves some $\delta - \epsilon$ details and the $ML$ inequality for complex line integrals.