1
$\begingroup$

I am preparing for my qualifying exam in complex analysis, and I'm working on the following problem.

Let $f_n = u_n + iv_n$ be a family of functions that are analytic on $D_1(0)$ and continuous up to $\overline{D_1(0)}$. Furthermore, assume that, for each $n$, we have $\int_0^{2\pi} |f_n(e^{i\theta})|d\theta < 1.$ Prove that $\{f_n\}$ has a subsequence that converges uniformly on compact subsets of $D_1(0)$.

I'm pretty sure that I need to apply Montel's Theorem, which says that if a family of analytic functions is uniformly bounded on compact subsets of the domain, then that family has a uniformly convergent subsequence.

Here's my attempt at a solution. I'd like to get some feeback about whether I'm on the right track. Let $K \subset D_1(0)$ be compact, and set $D = \operatorname{dist}(K, \{z : |z| =1 \}) > 0$. Since $f_n$ is analytic on $D_1(0)$ and continuous up to $\overline{D_1(0)}$, both $u_n$ and $v_n$ are harmonic on $D_1(0)$ and continuous up to $\overline{D_1(0)}$. Thus $u_n$ and $v_n$ satisfy the Poisson integral formula. That is, for $a \in K$, we have

$$u_n(a) = \int_0^{2\pi} P(a, \theta)u_n(e^{i\theta})d\theta,$$

$$v_n(a) = \int_0^{2\pi} P(a, \theta)v_n(e^{i\theta})d\theta, $$

where $P(a,\theta)$ is the Poisson kernel given by $P(a, \theta) = \frac{1}{2\pi} \frac{1- |a|^2}{|e^{i\theta} - a|^2}$.

Noting that, $|P(a,\theta)| \le \frac{1}{\pi D^2}$, we get

$$|f_n(a)| \le |u_n(a)| + |v_n(a)| \le \frac{1}{\pi D^2}(\int_0^{2\pi} |u_n(e^{i\theta})|d\theta + \int_0^{2\pi} |v_n(e^{i\theta})|d\theta) \le \frac{2}{\pi D^2}\int_0^{2\pi} |f_n(e^{i\theta})|d\theta < \frac{2}{\pi D^2}, $$

for all $n$ and $a \in K$. Thus the $f_n$ are uniformly bounded on $K$, and Montel's Theorem gives us what we want.

Am I on the right track here with exploiting the fact that $u_n$, $v_n$ are harmonic?

$\endgroup$

1 Answer 1

1
$\begingroup$

Yes, your approach is fine. Notice however that you can simplify it by using Cauchy's integral formula (which is more or less equivalent to what you did):

$$ f(z)=\frac{1}{2\pi i} \int_{\partial D_1} \frac{f(\zeta)}{\zeta-z} d \zeta $$ from which you obtain $|f(z)| < \text{dist}(z, \partial D_1)^{-1}$. In fact taking derivatives on both sides you get $|f'(z)|<\text{dist}(z, \partial D_1)^{-2}$ from which Arzela-Ascoli gives convergence in compact subsets.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .