If $\lim_{r \to 1} \frac{1}{2\pi}\int_0^{2\pi} \left| \log \left| f(re^{it}) \right| \right| dt = 0$ then $\left| f(z) \right| \leq 1$

Question

My aim is to prove that Blaschke products are the only holomorphic funcions that verify the property $$ \lim_{r \to 1} \frac{1}{2\pi} \int_0^{2\pi} \left| \log \left| f(re^{it}) \right| \right| dt = 0. $$ For this, I need to show that if $f$ is a holomorphic function defined in the open unit disc $\mathbb{D}$ such that $$ \lim_{r \to 1} \frac{1}{2\pi} \int_0^{2\pi} \left| \log \left| f(re^{it}) \right| \right| dt = 0 $$ then $\left| f(z) \right| \leq 1$ for all $z \in \mathbb{D}$. I know that $z \in \mathbb{D} \mapsto \log \left| f(z) \right| $ is a subharmonic function and hence $$ t \in (0,1) \mapsto \frac{1}{2\pi} \int_0^{2\pi} \log \left| f(re^{it}) \right| dt $$ is an increasing function. How can I conclude?

Answer 1

$|\log |f||=\log^+ |f| +\log^- |f|$ so $\log^+ |f| \le |\log |f||$, hence:

$\limsup_{r \to 1} \int_0^{2\pi} \log^+ \left| f(re^{it}) \right| dt \le \ \lim_{r \to 1} \int_0^{2\pi} \left| \log \left| f(re^{it}) \right| \right| dt = 0$ which means that $\lim_{r \to 1}\int_0^{2\pi} \log^+ \left| f(re^{it}) \right| dt =0$ by the non-negativity of the integrand.

But $\log^+|f|$ is also subharmonic so $ r \to \int_0^{2\pi} \log^+ \left| f(re^{it}) \right| dt$ is increasing and non-negative, so since it has limit zero, it must be identically zero for all $r<1$; hence $\log^+|f|(z)=0, |f(z)| \le 1, |z| <1$ and we are done!