Does a bounded holomorphic function on the unit disc have summable Taylor coefficients? Given a bounded holomorphic function $\phi (z)=\sum_{n=0}^\infty a_nz^n$ on the open unit disc, is it true that $\sum_{n=0}^\infty |a_n|<\infty$? I was thinking that this question could possibly be answered using some theory on analytic continuation, but unfortunately I do not know this theory very well.
Any hint or counterexample would be most appreciated. Thank you.
 A: No, not really. If $\phi(z)= \sum a_n z^n$ has $\sum_n |a_n|< \infty$ then 
$$z \mapsto \sum_n a_n z^n$$
defines a continuous extension of $\phi$ to the closed disk $\{ |z|\le 1\}$. 
Consider now $\phi(z)$ a  Blaschke product. $\phi(z)$ will be a function on the open unit disk with zeroes exactly the set $\{\lambda_n\}$. Moreover, $|\phi(z)| < 1$ for all $z$ in the unit open disk. Now pick the sequence $\lambda_n$ so that any point on the circle $\{ |z|=1\}$ is a limit point of it. To do that, just ensure that the arguments of $\lambda_n$ are dense in  $\mathbb{R}/2 \pi \mathbb{Z}$. Claim: $\phi(z)$ does not extend continuously  to the closed disk. Indeed, if it did then $\phi(z) = 0$ on the circle, and by the maximum principle $\phi(z) \equiv 0$, which is not the case. 
It follows that for the series of this $\phi(z) = \sum _n a_n z^n$ we have $\sum_n |a_n| = \infty$.
Still, the sequence  $(a_n)$ itself will be bounded if $\phi$ is bounded on the disk. ( this is not always true for functions on the open unit disk, for instance $ \frac{1}{(1-z)^2}= \sum_{n\ge 1} n z^{n-1} )$ See: Hardy spaces.
