I'm trying to prove the following:
Suppose that $\gamma$ is a rectifiable curve in $\mathbb{C}$ and $\varphi$ is defined and continuous on the trace of $\gamma$. Show that $$g(z)=\int_{\gamma}\frac{\varphi(w)}{w-z}\:dw$$ is analytic on $\mathbb{C}-\{\gamma\}$ and that $$g^{(n)}(z)=n!\int_{\gamma}\frac{\varphi(w)}{(w-z)^{n+1}}\:dw.$$
I know that to show that the function is continuous at some arbitrary $z_{0}\in\mathbb{C}-\{\gamma\}$, we must show that $$\lim_{h\to0}\frac{g(z_{0}+h)-g(z_{0})}{h}=\lim_{h\to0}\frac{1}{h}\left(\int_{\gamma}\frac{\varphi(w)}{w-(z_{0}+h)}\:dw-\int_{\gamma}\frac{\varphi(w)}{w-z_{0}}\:dw\right)$$ exists. I'm stuck trying to show this fact, and I have no idea how to prove that the expression for $g^{(n)}(z)$ is as claimed.
Perhaps since the trace of $\gamma$ is a compact set and $\varphi$ is a continuous function on said set, I can use some sort of uniform continuity property?
Thanks for any help!
Let $\delta =d(z_0, \gamma)$ i.e., $\delta = inf \{|z_o-w|:w \in \gamma\}$. Since trace of $\gamma$ is a compact set $\delta >0$. For $w$ on $\gamma$ $|w-z_0| \geq \delta$. If $|h|<\delta /2$ then $|w-z_o -h| \geq |w-z_0| -|h| > \delta -\delta /2=\delta /2$. Also $\phi$ is bounded . Noting that $\frac 1 h \{\frac 1 {w-z_0 -h} -\frac 1 {w-z_0}\}=\frac 1 {(w-z_o-h)(w-z_0)}$ it follows that you can apply Dominated Convergence Theorem to get the formula for $g'$. For higher order derivatives you just have to induction. In each step DCT can applied using the same argument.