How to show that functions are equicontinuous

Question

Let $g\in A(\mathbb{D})$ (Dirichlet algebra), i.e., $g(z)$ is analytic in $\mathbb{D}=(|z|<1)$ and continuous on $\overline {\mathbb{D}}=(|z|\leq 1)$.

How is it possible to show that:

1) The family of functions $G_r(t)=|g(re^{it})|^2,\;0<r\leq1$ is equicontinuous;

2) $G_r(t)\to G_1(t)$ uniformly when $r\to1$.

What I understand now:

First part (equicontinuity)

It seems that it's necessary to show that $\left|g(re^{it_2})|^2-|g(re^{it_1})|^2\right|<C|t_2-t_1|,\; \forall r$ or get a similar bound.

Answer 1

This doesn't have to do with analyticity. The function $h(z) = |g(z)|^2$ is continuous on the compact metric space $\overline D$, hence uniformly continuous.

You can combine this with some elementary estimates:

$|re^{i s} - r e^{it}| = r | e^{i(s -t)} - 1| < |s -t|$, and

$|r e^{it} - e^{it}| = (1 -r)$.

These observations together give both uniform equicontinuity of your functions $G_r$, and the required uniform convergence.

Explicitly: Given $\varepsilon > 0$, there exists a $\delta > 0$ such that for all $z, w \in \overline D$, if $|z-w| < \delta$ then $|h(z) - h(w)| < \varepsilon$. For uniform equicontinuity, let $s, t \in \mathbb R$ with $|s - t| < \delta$. Then for all $r$ with $0 \le r \le 1$, $|r e^{it} - r e^{i s}| < |s - t| < \delta$. Hence $|G_r(t) - G_r(s)| = |h(r e^{it}) - h(re^{i s})| < \varepsilon$. This shows uniform equicontinuity of the functions $G_r$. For uniform convergence, let $r > 1 - \delta$. Then for all $t \in \mathbb R$, $|G_r(t) - G_1(t)| = |h(re^{it}) - h(e^{it})| < \varepsilon$.

Answer 2

The function $g$ is continuous on the closed unit disk $\overline{\mathbb{D}}$, which is a compact set. So $g$ is uniformly continuous on the closed unit disk. That means, for every $\epsilon > 0$ there exists $\delta > 0$ such that $$ |g(z)-g(w)| < \epsilon \mbox{ whenever } |z-w| < \delta, \; z,w\in\overline{\mathbb{D}} $$ Therefore, for every $\epsilon > 0$ the above $\delta$ gives $$ |g(re^{i\theta})-g(re^{i\theta'})| < \epsilon \tag{*} $$ whenever $|re^{i\theta}-re^{i\theta'}| < \delta$, which is certainly satisfied if $$ |re^{i\theta}-re^{i\theta'}|=|ir\int_{\theta'}^{\theta}e^{i\theta''}d\theta''| \le r|\theta-\theta'| < \delta. $$ If $r < \delta$, then no restriction is needed to obtain $(*)$ for all $\theta,\theta'$. And if $r \ge \delta$, then $|\theta-\theta'| < \delta/r$ gives $(*)$.