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**dori1123** Suppose that $\displaystyle f: A \rightarrow \mathbb{C}$ is continuous, where $\displaystyle A$ is a region in $\displaystyle \mathbb{C}$. Let $\displaystyle \gamma$ be a simple closed curve in $\displaystyle A$. Show that for each $\displaystyle n$, $\displaystyle f_n(z)=\int_\gamma \frac{f(\zeta)}{(\zeta-z)^n}d\zeta$ is continuous on $\displaystyle A-\gamma$.

Let $\displaystyle \epsilon>0$. Suppose that there is $\displaystyle \delta>0$ such that $\displaystyle |z-z_0|<\delta$. I need to show $\displaystyle |f_n(z)-f_n(z_0)|<\epsilon$.

$\displaystyle |f_n(z)-f_n(z_0)|=|\int_\gamma \frac{f(\zeta)}{(\zeta-z)^n}d\zeta-\int_\gamma \frac{f(\zeta)}{(\zeta-z_0)^n}d\zeta|=\int_\gamma f(\zeta)(\frac{1}{(\zeta-z)^n}-\frac{1}{(\zeta-z_0)^n})d\zeta$.

I got stuck here... can I get some help please?

There's also a more general proof using uniform continuity; but let's do it your way.

Remember $\displaystyle a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+a b^{n-2} + b^{n-1})$. We have

$\displaystyle \frac{1}{(\zeta-z)^n}-\frac{1}{(\zeta-z_0)^n}=\frac{(\zeta-z_0)^n-(\zeta-z)^n}{(\zeta-z)^n(\zeta-z_0)^n}$

hence, using the above formula,

$\displaystyle \left|\frac{1}{(\zeta-z)^n}-\frac{1}{(\zeta-z_0)^n}\right|\leq$ $\displaystyle |z-z_0|\frac{\sum_{k=0}^{n-1}|\zeta-z|^k |\zeta-z_0|^{n-1-k}}{|\zeta-z|^n|\zeta-z_0|^n}$.

Since the curve $\displaystyle \gamma$ is a bounded closed set (thus a compact set), you can choose $\displaystyle \delta>0$ such that the closed ball $\displaystyle \bar{B}(z_0,\delta)$ does not meet $\displaystyle \gamma$. Then we can find $\displaystyle m,M>0$ such that, for any $\displaystyle z$ in this ball, $\displaystyle m<|\zeta-z|<M$ for all $\displaystyle \zeta\in \gamma$. Using these bounds in the previous inequality enables to bound the right-hand side by $\displaystyle C|z-z_0|\leq C\delta$ for some constant $\displaystyle C$, and to conclude quickly from there.