$\{f_n\}$ 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?


3 Answers 3


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 $\Omega$.


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:


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.

  • $\begingroup$ Sorry but I cannot understand... Are $f_n$ converge uniformly in a ball? And why are we done when take the biggest $n$? $\endgroup$ Sep 16, 2015 at 6:10
  • $\begingroup$ The sum will converge in a Ball around $a$ with radius $r<1$ due to the Faktor $(x-a)^k$ and the uniform boundedness of $f_n$. For every such Ball in your covering (they have radius lass than one and different centers) you have an $n$ such that the difference is small. Therefore the difference is small for all these Balls in your covering in you take the maximal n you obtained via the above... $\endgroup$
    – Vincent.W.
    Sep 16, 2015 at 7:44
  • $\begingroup$ I just worked this problem. This post helped me to suss out my own thoughts, but there are two problems with this answer: 1) it is not clear to me that you get a single uniform bound on the integrals, because of the factor of r^n in the denominator; and 2) Rudin tends to be straight forward in his hints, if he wanted you to use the PROOF of the power series expansion theorem, rather than even its statement (which is what you did here) he would usually say so. (I've worked hundreds of Rudin problems). $\endgroup$
    – entprise
    Feb 26, 2018 at 17:44
  • $\begingroup$ Wrote up a much simpler answer that directly addresses the problems cited in my comment. $\endgroup$
    – entprise
    Feb 26, 2018 at 17:54

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.

  • $\begingroup$ Why does the factor $(w-a)/(w-z)$ appear? the differential of the path you gave is $ide^{it}$. $\endgroup$
    – Hasse1987
    Aug 17, 2019 at 12:57
  • $\begingroup$ By way of a cauchy integral formula which representats an analytic function by a line integral along a circle boundry with the integrand being partially evaluated at an interior point and the circle center (here z and a) before integration (here in w). The path I gave was a + de^it. The path you gave is just a rotated circle at the origin--we dont even know the origin is in the region. $\endgroup$
    – entprise
    Aug 18, 2019 at 13:25
  • $\begingroup$ I don't follow. You say "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]." Let $\Gamma=a+de^{it}$. I interpret your suggestion to be the integral $\int_\Gamma \frac{f_n(w)-f_m(w)}{z-w}dw = \int_{t=0}^{2\pi}\frac{f_n(a+de^{it})-f_m(a+de^{it})}{z-a-de^{it}}ide^{it}dt$, is that right? $\endgroup$
    – Hasse1987
    Aug 18, 2019 at 19:39

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