Uniform convergence of uniformly bounded sequence of holomorphic functions $\{f_n\}$ is a uniformly bounded sequence of holomorphic functions in $\Omega$ which pointwise converges. Prove that the convergence is uniform on every compact subset of $\Omega$. (Hint. Apply LDCT to the Cauchy formula for $f_n-f_m$.)
It is well known - I found one solution here.
However, in fact it is from Rudin RCA chapter 10, and Arzela-Ascoli is in the next chapter. So I believe that there is an relatively elementary proof. Anyone know about this?
 A: Take any $\overline{B(z_0,r)} \subset D$, then $f_n$ is uniformly bounded on $\overline{B(z_0,r)}$. Hence $f_n'$ is uniformly bounded, say $|f_n'| \le M, z \in \overline{B(z_0,r)}$.
For any $\epsilon >0$, take $\delta=\frac{\epsilon}{3M},$ for $|z_1-z_2|<\delta$ implies that $$|f(z_1)-f(z_2)| =|\int_{z_2}^{z_1}f_n'(z)dz|< \frac{\epsilon}{3}$$
Hence $f_n$ equicontinuous on $B(z_0,r)$.
Take any compace subset $K \subset \Omega$, we have a finite cover $$K \subset \bigcup_{j=1}^l B(z_j,\frac{\delta}{2})$$
So $f_n$ also equicontinuous on $K$.
Since $$\lim_{n \to \infty}f_n(z_j)$$ exists, then for any $\epsilon >0$, there exists a $N_j \in \mathbb{Z_+}$, for $n,m>N_j$ implies that $|f_n(z_j)-f_m(z_j)| < \frac{\epsilon}{3}$.
Pick $N=max(N_1,……,N_j)$, for $n,m>N$, we have $$|f_n(z)-f_m(z)|<|f_n(z)-f_n(z_k)|+|f_n(z_k)-f_m(z_k)|+|f_m(z_k)-f_m(z)|<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$$
Here $z \in B(z_k, \frac{\delta}{2})(1 \le k \le l)$, and so $f_n$ uniformly converges on any compact subset of $\Omega$.
A: you basically want to show $\forall K\subseteq \Omega$ compact, $\forall \varepsilon > 0$ $\exists n$ such that $\forall x\in K,\forall m>n$ u have:
$|f_n(x)-f_m(x)|<\varepsilon$
So what you want to do is estimate this difference using the cauchy formula. The difference is a holomorphic function:
$|f_n(x)-f_m(x)|=|\sum_{k=0}^{\infty}\frac{1}{2\pi i}\oint \frac{f_n(z)-f_m(z)}{(z-a)^{k+1}}dz\cdot (x-a)^k|$
By dominated convergence you know that the integral will converge to zero, therefore you can make it arbitrarily small by choosing n big enough. But here comes the kicker: you have to do this such that the sum also converges ...Take an open covering of K with balls of radius $<1$ (this will make the sum summable) and select a finite subcovering (K is compact) then you obtain a $n$ for every ball $B$ in your covering such that 
$|f_n(x)-f_m(x)|<\varepsilon\quad \forall x\in B$
take the biggest $n$ of your covering and you are done.
A: Take any compact set K inside the region R.  Noting that the distance between K and the complement of R is greater than some d, which is itself strictly greater than 0.  Cover K by finitely many balls of radius d/2.  Any z in K is in at least one such ball, say the one centered at a.  Represent the difference of fn and fm at z by a cauchy integral, the path of integration being a + de^it with t in [0,2pi].  Notice that, in turning the line integral (that is, the integral of a particular 1 form in the complex plain) into a lebesgue integral (that is, a reimann integral over the compact set [0,2pi]), a factor of (w-a)/(w-z) appears in the integrand.  Because the line integral is on the boundary of a circle of radius d, but z is within a concentric circle of radius d/2, this factor is at most 2.  DCT applies to the remaining factor, providing a single bound for the difference evaluated at ANY z within this d/2 ball about a.  There are only finitely many such balls to consider; the sup of a finite number of finite bounds is finite; hence the theorem.
