I have some problems proving that if $f$ is analytic in the open disc, then
$$\int_{\mathbb{T}}\mid\log(|f(rz)|)\mid \,< \infty.$$
I seen some proofs when $f$ is non zero since then we have no problem with the logarithm and we get that the integrand is the real part of an analytic function. But I can't seem to find a way around this even with subharmonic functions.
Suggestions?
It is possible that this statement is false and that the person who wrote it forgot to assume that $f$ is in $H^p$.
A direct proof : Let $f(z)$ be analytic on $|z| < 1$ and not identically zero.
for any $|a| < 1$ and $|z|$ small enough, $f(a+z) = \sum_{n=0}^\infty c_n z^n$ so if $f(a) = 0$ then $f(a+z) \sim c_k z^k$, so $a$ is a zero of order $k$.
take a compact in $|z| < 1$, and suppose $f(z)$ has infinitely many zeros $a_1,a_2, \ldots$. Then $(a_n)$ has at least one convergent subsequence $a_{n_1},a_{n_2}, \ldots$. Let $A = \lim_k a_{n_k}$. By continuity $f(A) = \lim_k f(a_{n_k}) = \lim_k 0 = 0$. So $A$ is a non isolated zero of $f$, a contradiction with $f(A+z) \sim c_k z^k$ for some $k, c_k \ne 0$.
hence on $|z| = r < 1$, $f$ is $C^\infty$ and has finitely many zeros of finite order, so that $\int_0^{2\pi} |\log f(r e^{it})| dt$ converges.