0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.