A previous complex analysis qualifying exam problem:
Let $\{f_n\}_{n=1}^\infty$ be a sequence of functions holomorphic on the open unit disc $D = \{z:|z|<1\}$ and continuous on the closed unit disc $\bar{D} = \{z:|z|\leq 1\}$. Show that if the sequence $\{f_n\}_{n=1}^\infty$ converges uniformly on the unit circle $\partial D = \{z:|z|=1\}$, then there is a function $f$ such that
(i) $f_n\to f$ on $D$, i.e. $\lim_{n\to\infty} f_n(z) = f(z)$ for every $z\in D$, and
(ii) $f$ is holomorphic on $D$.
Based on comments, I think it goes something like this:
(i) If $f_n\to f$ on the unit circle, then there exists an $\epsilon > 0$ and an $N$ such that for $n,m>N$, $|f_n - f|<\epsilon$. Since $f_n - f_m$ is holomorphic, $|f_n-f_m|$ has its maximum on the boundary $|z|=1$. Thus, $|f_n-f_m|<\epsilon$ on the boundary implies $|f_n-f_m|<\epsilon$ on all of $D$ and we have that $f_n\to f$ on all of $D$ uniformly.
(ii) $f_n\to f \implies \oint f_n dz \to \oint f dz = 0$ for all simple closed curves in $D$, thus $f$ is holomorphic.
However, I struggle a good deal with this type of problem. A clear and fully detailed answer would still be incredibly helpful.