How to show that functions are equicontinuous 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.
 A: 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 $t \in \mathbb R$,  $|G_r(t) - G_1(t)| = |h(re^{it}) - h(e^{it})| < \varepsilon$.
A: 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 $(*)$.
