I am studying Banach Spaces of Analytic Function by Hoffman. After giving a description of nonzero $H^1$ functions as a factorization of Blaschke, Singular and Outer function, the author makes a remark that if $f$ is holomorphic in the closed unit disc then "singular part" of $f$ is zero and the Blaschke part of $f$ is a finite Blaschke product.
I wonder what the author means by the "singular part" of $f$ is zero because the factorization theorem would tell us that $f\equiv 0$.
I do not see how this remark follows. Here's my attempt, however to show that Blaschke product is finite: Let $f$ be analytic in the closed disc. Then $f$ must be analytic in a slightly bigger open connected set containing the closed unit disc. Now if there were infinitely many zeroes of $f$ in the open unit disc then they must accumulate somewhere in the closed unit disc (by Bolzano Weierstrass) and hence must be a zero function which would contradict our assumption. Thus the Blaschke part of $f$ is indeed a finite Blaschke product.
