Let $\alpha$ be a simple closed curve. $\Omega$ is the interior of $\alpha$. For any continuous function $g:\alpha\to\mathbb{C}$, define $f_g:\Omega\to\mathbb{C}$ by $$f_g(z)=\int_\alpha\frac{g(\xi)}{\xi-z}d\xi$$ Show that $\{f_g:|g(z)|\le1,z\in\alpha\}$ is a normal family in $\Omega$
My attempt:$|f_g(z)|\le\int_\alpha\frac{|g(\xi)|}{|\xi-z|}d|\xi|\le\int_\alpha\frac{1}{|\xi-z|}d|\xi|$
I am not sure if $\int_\alpha\frac{1}{|\xi-z|}d|\xi|=2\pi$ for the $\alpha$ which is not in the form of $z+Re^{i\theta},\theta\in[0,2\pi)$.
If so then it's the functions are bounded, thus by Montel's theorem, the family is normal.
Normality is equivalent to uniform boundedness on compact subsets. Let $K \subset \Omega$ be compact. Then there is a positive distance $\delta$ between the disjoint compact sets $\alpha$ and $K$. Hence $|\xi -z| \geq \delta$ for all $\xi$ on $\alpha$. This gives $|f_g(z)| \leq \frac 1 {\delta} L$ where $L$ is the length of $\alpha$ and the proof is complete.