Proof continuity of following function: $g(z)=\int_{\gamma} \frac{f(z)-f(w)}{z-w}dw$ Suppose $f(z)$ is a analytic in G and $\gamma$ is a curve in G. Show that 
$g(z)=\int_{\gamma} \frac{f(z)-f(w)}{z-w}dw$ is continuous in G
I attempted to bound the integral but can’t seem to remove a $|z-z_0|$ term.
 A: Use my answer in the following along with the fact that continuous functions on a compact set are uniformly continuous: Prove continuity of the following Function (Question in Conway)
A: I guess you have to use only elementary methods, so here is an approach: We will show that $g$ is continuous on $G\setminus\gamma$ (otherwise the integral doesn't make sense). Let $z_0\in G\setminus\gamma$ and let $\delta>0$ be such that the circle centered at $z_0$ and with radius $\delta$ doesn't intersect $\gamma$ (for example, $\delta=\frac12dist(z_0,\gamma)$). It follows that $|z_0-w|>\delta$, whenever $w\in\gamma$. Now let $|z-z_0|<\delta/2$. Then for any such $z$ and for any $w\in\gamma$, we have that $|z-w|>\delta/2$. After some basic algebra we see that
$$g(z)-g(z_0)=(z-z_0)\int_{\gamma}\frac{f(w)}{(z_0-w)(z-w)}dw+(f(z)-f(z_0))\int_{\gamma}\frac{1}{(z-w)}dw-f(z_0)(z-z_0)\int_{\gamma}\frac1{(z_0-w)(z-w)}dw.$$  
The first integral can be bounded by $\frac2{\delta^2}ML$, where $M$ is an upper bound for $|f|$ and $L$ is the length of $\gamma$. The second and third integral are similarly bounded by $\frac2{\delta}L$ and $\frac2{\delta^2}L$ respectively. For the second term you should also use the fact that $f$ is continuous, since it is analytic, in order to make the difference $f(z)-f(z_0)$ small. You should be able the figure the $\epsilon-\delta$ details on your own.
