Let $f\in C(T)$, and define $Hf$ on the closed unit disk $\overline U,$ by
$\begin{equation} Hf(re^{i\theta})= \begin{cases} f(e^{i\theta}), & \text{if}\ r=1 \\ P[f](re^{i\theta}), & \text{if}\ 0\leq r<1 \end{cases} \end{equation}$
where $P[f](re^{i\theta})=\frac{1}{2\pi}\int_TP_r(\theta-t)f(t)dt.$
Then, $Hf$ is continuous on $\overline U$.
Rudin proves this by noting that $\|Hf\|_{\overline U}=\|f\|_{T}$ and then using an approximation argument by trigonometric polynomials.
Later in the section, he defines $u_{r}(e^{i\theta})=u(re^{i\theta})$ for any $u:U\to \mathbb C$ and claims, appealing to the above result, that if $f\in C(T)$ and $F=P[f]$, that $F_r\to f$ uniformly as $r\to 1.$
I do not see how this follows $immediately$ from Rudn's proof, which he seems to be claiming.
I ended up doing it by direct calculation, by considering $\|F_r-f\|_{\infty}$ and splitting the integral up into two parts, controlling the first using the continuity of $f$ and the second by noting that $P(re^{i\theta},e^{it})\to 0$ as $r\to 1$, and the estimate does not depend on $\theta$ so the convergence is uniform.
I would still like to know how this follows from Rudin's proof, which I paste below:
$\phi \equiv Hf \in C(\bar U)$. So $\phi$ is uniformly continuous on $\bar U$. This implies that $\phi (re^{i\theta}) \to \phi (e^{i\theta})$ uniformly as $r \to 1$. This I think is what Rudin is claiming.