In Rudin's book Real and Complex Analysis, in the eleventh chapter there is a theorem that states:
To every $f\in H^\infty$ corresponds a function $f^*\in L^\infty (T),$ defined almost everywhere by $$f^*(e^{i\theta})=\lim_{r\to 1^-}f(re^{i\theta}).$$ The equality $\left\Vert f\right\Vert_{\infty}=\left\Vert f^*\right\Vert_{\infty}$ holds.
In the proof of the theorem for the equality $\left\Vert f\right\Vert_{\infty}=\left\Vert f^*\right\Vert_{\infty}$ there is stated that $\left\Vert f\right\Vert_{\infty}\geq\left\Vert f^*\right\Vert_{\infty}$ is obvious. My question is how to prove that $\left\Vert f\right\Vert_{\infty}\geq\left\Vert f^*\right\Vert_{\infty}$ holds?
In this question $H^\infty$ is the space of all holomorphic functions in the open unit ball $U$ equipped with the supremum norm, while $L^\infty$ is the space of all essentially bounded functions on $T=\partial U$ normed by the essential supremum norm, relative to the Lebesgue measure, so $$\left\Vert f^*\right\Vert_\infty=\inf\left\{ \alpha\vert m\left( \left( f^* \right)^{-1} \left( \left[ \alpha,\infty \right)\right) \right)=0 \right\}$$ for $f^*:T\to \left[0, \infty \right).$
My attempt: Let suppose that $\left\Vert f\right\Vert_{\infty}<\left\Vert f^*\right\Vert_{\infty}.$ Then there exists an $\alpha '=\inf\left\{ \alpha\vert m\left( \left( f^* \right)^{-1} \left( \left[ \alpha,\infty \right)\right) \right)=0 \right\}$ s.t. $\alpha'>\Vert f\Vert_\infty$, so $m\left( \left( f^* \right)^{-1} \left( \left[ \Vert f\Vert_\infty,\infty \right)\right)\right)>0.$
How should I continue until a contradiction?
Let $C \subset T$ be the set where the radial limit $$\lim_{r \to 1^-} f(re^{i\theta})$$ exists. By theorems 11.30 and 11.23, $m(T\setminus C) = 0$, and $f$ is the Poisson integral of $f^{\ast}$. On $T\setminus C$, we can define $f^{\ast}$ arbitrarily if we want an everywhere defined function $f^{\ast}$.
Since $\lvert f(re^{i\theta})\rvert \leqslant \lVert f\rVert_{\infty}$ for $r \in [0,1)$, we have $$\lvert f^{\ast}(e^{i\theta})\rvert = \lim_{r \to 1^-}\: \lvert f(re^{i\theta})\rvert \leqslant \lVert f\rVert_{\infty}$$ for all $e^{i\theta} \in C$.
Let $\alpha > \lVert f\rVert_{\infty}$. Then $$\bigl\{ e^{i\theta} : \lvert f^{\ast}(e^{i\theta})\rvert \geqslant \alpha \bigr\} \subset T\setminus C\,,$$ so this is a null set. Hence $$(\lVert f\rVert_{\infty}, +\infty) \subset \bigl\{ \alpha \in \mathbb{R} : m\bigl(\bigl(\lvert f^{\ast}\rvert\bigr)^{-1}([\alpha, +\infty))\bigr) = 0\bigr\}$$ and consequently $$\lVert f^{\ast}\rVert_{\infty} = \inf\: \bigl\{ \alpha \in \mathbb{R} : m\bigl(\bigl(\lvert f^{\ast}\rvert\bigr)^{-1}([\alpha, +\infty))\bigr) = 0\bigr\} \leqslant \inf\: (\lVert f\rVert_{\infty}, +\infty) = \lVert f\rVert_{\infty}\,.$$
[Note that $f^{\ast}$ is complex-valued, so we must consider $\bigl(\lvert f^{\ast}\rvert\bigr)^{-1}([\alpha,+\infty))$ and not $\bigl(f^{\ast}\bigr)^{-1}([\alpha,+\infty))$.]