The function : $\sum _{n=0}^{\infty} \frac{z^{2^n}} {2^n}$ convergs to holomorphic function $f$ on $D_1(0)$ and is continious on $\overline{D_1(0)}$. I need to prove that f can't be exteneded to any domain $\Omega$ such that $\overline{D_1(0)}\subseteq\Omega$.

Any ideas?


If it has an analytical continuation on some domain $\Omega \supset \overline{D}(0,1)$, then $\Omega$ contains some $\overline{D}(0,r)$ for some $r >1$. Using the common properties of analytical functions, this requires the series $\sum_n{2^{-n}r^{2^n}}$ to converge. Well, this series does not.

  • $\begingroup$ Could you please elaborate on the part of "Using the common properties..."? Obviously, I know that the series is a taylor expansion around 0. But maybe, around some $z$ where $|z|>1$, it has another expansion that converges. $\endgroup$ – user3708158 Jan 22 at 14:45
  • $\begingroup$ The Taylor expansion of any holomorphic function converges normally on any closed disc that is a subset of the domain. $\endgroup$ – Mindlack Jan 22 at 14:46
  • $\begingroup$ I am not sure, for example you have the function $f(z)=\frac {1}{1-z} = \sum _n z^n$ that does not converge on the domain where $|z|>1$. I mean, you could have another expansions around different points in the domain. $\endgroup$ – user3708158 Jan 22 at 14:48
  • $\begingroup$ That is why I wrote: converges normally on any closed disc that is a subset of the domain. $\overline{D}(0,1)$ (or anything greater than $1$) is not a subset of the domain of $z \longmapsto (1-z)^{-1}$. $\endgroup$ – Mindlack Jan 22 at 14:50
  • $\begingroup$ Ok, I will change my question: why for points s.t $|z|>1$, I need that series to converge? Why can't I have another expansions, around s.t that do converge? $\endgroup$ – user3708158 Jan 22 at 14:52

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