# 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.

• Can we first prove this for compact subsets of the unit disc ? because on compact subsets, the power series will converge uniformly.Then we can try something like this $\int \sum z^n=\sum \int z^n= z^{n+1}/{n+1}$ and use the fact that $\sum 1/n$ diverges. Feb 6, 2015 at 3:20
• It is clear that if $f(z)=\sum a_\nu z^\nu$ converges absolutely at $z=1$ then it converges absolutely at every point of the unit disk by comparison. You can prove that if $f$ is holomorphic at $z=1$ (i.e. if we can write it as a series with positive radius of convergence), then it is holomorphic at each point of the boundary of the disk.
– Pedro
Feb 6, 2015 at 20:43

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.
• @user122916: I am not satisfied with my answer. So far it is not a concrete example of a power series $\sum_n a_n z^n$ such that $\sum_n |a_n| = \infty$ while $|\sum_n a_n z^n| \le 1$ for all $|z|<1$. Still thinking about it, great question. Feb 7, 2015 at 20:51
• I was looking more for the existence of a counterexample rather than a concrete one. However, it would be nice to have a concrete formula for such $a_n$. Thank you. Feb 14, 2015 at 3:34
• @user122916: I rephrased your question and posted it again, see math.stackexchange.com/questions/1291914/…. From the answers given it seems to follow that this sequence is a concrete example: $n^ {-\frac{3}{4}} \cos ( \sqrt{8 n} + \frac{\pi}{4})$ , although I am still baffled. May 22, 2015 at 6:48