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I'll post the full problem before I'll show my (rather limited) progress:

i) Find all $z \in \mathbb{C}$ so that the following power series converge around $0$: a) $\sum_{k=0}^\infty z^k$, b) $\sum_{k=0}^\infty z^{(2^k)}$

ii) Is there a domain $G \supset B_1(0)$ and a holomorphic function $f\colon G\to \mathbb{C}$ that is identic to the resp. power series in i) on $B_1(0)$?

iii) Construct a function that is holomorphic in $B_1(0)$, continuous in $\overline{B_1(0)}$ and that cannot be continued holomorphically over $B_1(0)$.

i) is easy (?). Both a) and b) converge on $|z|<1$, because the geometric series is a convergent majorant. Both diverge on $|z|\geq 1$, because neither $|z|^k$ nor $|z|^{(2^k)}$ is a zero sequence.

ii) and iii) I have no idea how to approach. One idea is to try and write $\frac{d}{dz}\sum_{k=0}^\infty z^k$ and $\frac{d}{dz}\sum_{k=0}^\infty z^{(2^k)}$ as the derivative of a holomorphic function, but somehow I'm still missing the central idea, I guess. Any hints?

So (thanks froggie) for iia) I have $\sum z^k = \frac{1}{1-z}=:f(z)$. Then $f(z)$ is holomorphic on $B_1(0)$ (when $B_1(0)$ denotes the open unit disk).

For iib) we look at $\sum z^{2^k}$ for $z=r\zeta$ with $0<r<1$ and $\zeta$ a $2^n$th root of unity. Then $\sum z^{2^k} = \sum (r\zeta)^{2^k} = \sum r^{2^k}$ which obviously diverges for $r\to 1$. So every "holomorphic extension" $f(z)$ would be automatically singular in every root of unity. Since these are dense on the unit circle, there cannot be a holomorphic function that is identic to $\sum z^{2^k}$ on $B_1(0)$.

The last argument seems to be a bit "fishy" to me. Any comments?

For iii) we look at $\sum \frac{1}{2^k} z^{2^k}$. This converges on $B_1(0)$ since $\sum \frac{1}{2^k} z^k$ is a convergent majorant. On the other hand we have $\frac{d}{dz} \sum \frac{1}{2^k} z^{2^k} = \sum_{k=1} z^{(2^k)-1} = \frac{1}{z}\sum\frac{1}{2^k} z^{2^k}$ which diverges because of ii).

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Your reasoning for (i) seems fine to me. For (iia), you know what such a holomorphic function $f$ would have to be, since you know how to sum the geometric series $\sum z^k$. – froggie May 17 '12 at 16:03
Also, for (iib), I suggest looking at the sum $\sum z^{2^k}$ when $z = r\zeta$, with $\zeta$ a $2^n$th root of unity, and $0<r<1$. What happens when $r\to 1^-$? – froggie May 17 '12 at 16:14
For (iii) I would suggest looking at $\sum z^{n!}/n^2$. It is easy to see this converges on the closed unit disk, so you just have to show that it can't be extending. To do this, I suggest looking at the derivative of this function near the unit circle. – froggie May 17 '12 at 17:03
For part (ia), I believe the series converges for all $|z|\leq1$ except $z=1$; i.e. the only singular point on the boundary is $z=1$. On the other hand, every boundary point with argument $2\pi im/2^n$ is a singular point of the function defined in (ib). Thus, the set of singular points here is dense on the unit circle, and so we say the unit circle is a natural boundary. The series converges for $|z|<1$ only. – John Adamski May 17 '12 at 18:19
Thanks guys, I've added my thoughts to the original question. What do you think? – gnirx May 18 '12 at 17:57

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