The Hardy space $H^1(\mathbf{D})$ is the vector space of holomorphic functions $f$ on the open unit disk that satisfy: $$ \sup_{0\leq r<1}\left(\frac{1}{2\pi} \int_0^{2\pi}\left|f \left (re^{i\theta}\right )\right| \; \mathrm{d}\theta\right)<\infty.$$ The radial limit $$ f^*(e^{i\theta})= \lim_{r \to 1} f(re^{i\theta}) $$ Question Prove that $$\sup_{0\leq r<1}\left(\frac{1}{2\pi} \int_0^{2\pi}\left|f \left (re^{i\theta}\right )\right| \; \mathrm{d}\theta\right)=\frac{1}{2\pi}\int_{0}^{2\pi}|f^*(e^ {i\theta})|d\theta $$
Attempt For $H^1$ functions the radial limit exists a.e. on the circle (Boundary). But how to prove the above equality.? Also $H^1$ forms a Banach Space.
Fatou's Lemma shows that $$||f^*||_1\le||f||_{H^1}:=\sup_{0<r<1}\frac1{2\pi} \int_0^{2\pi}|f(re^{it}|\,dt.$$
For the other inequality, note first that $f$ is the Poisson integral of $f^*$. If we define $$f_r(t)=f(re^{it})$$this says that $$f_r=f^**P_r,$$where the $*$ denotes convolution and $P_r$ is the Poisson kernel. All that's relevant here is that $||P_r||_1=1$; now Young's inequality (or someone's well-known inequality) says that$$||f_r||_1\le||f^*||_1||P_r||_1=||f^*||_1.$$Which is qed, since $||f||_{H^1}=\sup_{0<r<1}||f_r||_1$.