Show $f(z)=\int_{c} \frac{g(w)}{w-z}$ $dw$ is analytic on any domain $D$ not containing points of $C$

Question

I know how to show that $f(z)$ is continuous but I'm stuck at showing that $f(z)$ is analytic.

$f(z)$ is continuous:

$$|f(z)-f(a)|=\left|\int_{c} \frac{g(w)}{w-z}-\frac{g(w)}{w-a} dw\right|=\left|\int_{c} \frac{g(w)(z-a)}{(w-z)(w-a)}dw\right|$$

By ML-Formula:

$$\left|\int_{c} \frac{g(w)(z-a)}{(w-z)(w-a)}dw\right|\leq L(C)M|z-a|$$

where $L(c)$ is the length of the curve and M is bound of $\frac{g(w)}{(w-z)(w-a)}$ over the curve $C$ which exists since it is continuous on $C$.

Hence, we can let $\delta = \dfrac{\epsilon}{L(C)(M+1)}$

Any hints or solutions as to show $f(z)$ is analytic?

Answer 1

One way to do this would be to use Morera's theorem: https://en.wikipedia.org/wiki/Morera%27s_theorem

Since proving analyticity is local we can assume that $D = B(0,R)$ for some $R > 0$. Let $\gamma: [0,1] \rightarrow D$, then $\int_{\gamma}\int_{C}\frac{f(z)}{w-z}\,dw\,dz = \int_{C}g(w)\int_{\gamma}\frac{1}{w-z}\,dz\,dw = 0 $, since the inner integral is $0$.

Answer 2

Define $$f_1(z)=\int_C\frac{g(w)\,dz}{(w-z)^2}.$$ This is what $f'(z)$ "ought" to be. Now prove that $$\frac{f(z)-f(a)}{z-a}-f_1(a)\to0\tag{*}$$ as $z\to a$. One can write the LHS of $(*)$ as an integral, and it should not be too hard to get an estimate on the integrand sufficient to prove $(*)$.

Indeed $$\frac{f(z)-f(a)}{z-a}-f_1(a) =\int_C g(w)\left(\frac{a-z}{(w-z)^2(w-a)}\right)\,dw$$ etc.