Let $H^\infty$ denote space of holomorphic and bounded functions on the unit disk $U$ in $\mathbb{C}.$ For $f \in H^\infty$ we let $||f||_\infty$ be the supremum of the values of $|f(z)|,$ where $z \in U.$
One can show that any $f \in H^\infty$ has radial limits almost everywhere; denote the radial limit function by $f^*.$ On the unit circle, we have the $L^\infty$-norm. If we let $$||f^*||_\infty$$ be the $L^\infty$-norm on the unit circle, then why do we have that $$||f||_\infty = ||f^*||_\infty?$$
It's clear that $||f^*||_\infty\le||f||_\infty$, since the values of $f^*$ are limits of values of $f$ in the disk.
The other direction follows from the fact that $$f=P[f^*],$$where $P[.]$ denotes the Poisson integral.
(How do you prove that, you ask? For $0<r<1$ let $f_r(z)=f(rz)$. Then $f_r$ is continuous on the closed disk, so $f_r=P[f_r^*]$. Let $r\to1$ and apply dominated convergence.)