The problem states that $f_n$ is a sequence of functions which are continuous on the closure of $\Omega$ and holomorphic on $\Omega$ where $\Omega$ is a bounded region and were asked to show that if $f_n$ converges uniformly on the boundary $\partial \Omega$ then it converges uniformly on the closure.
So I was hoping someone could check my work. This is what I have so far:
Let $z$ be in the closure of $\Omega$. If $z\in \partial \Omega$ then there is nothing to prove. Otherwise, we can write $$f_n(z)=\frac{1}{2\pi i}\int_{\partial \Omega} \frac{f_n(\zeta)}{\zeta-z}d\zeta$$ and now pass to the limit $$f(x)=\lim \frac{1}{2\pi i}\int_{\partial \Omega} \frac{f_n(\zeta)}{\zeta-z}d\zeta$$ then, by the uniform convergence of $f_n$ on the boundary, the limit can be taken inside the integral. Is this all that needs to be said?
Since $\Omega$ is bounded, we can use the maximum modulus principle. For each $m,n$ we have that $f_n-f_m$ is holomorphic on $\Omega$ and is continuous on the boundary. Therefore, $$\sup_{\overline{\Omega}} |f_n-f_m|=\sup_{\partial{\Omega}} |f_n-f_m|.$$ Since the $f_n$ converge uniformly on $\partial{\Omega}$, the right hand side is small for $m,n$ large. Therefore, the $f_n$ are uniformly Cauchy on $\overline{\Omega}$ and hence converge uniformly there.