Prove there does not exist a holomorphic $g(z)$ defined on a neighborhood $U$ of $z=1$ equal to $f(z)=\sum_{n\ge0}z^{2^n}$

Asked May 22 '21 at 11:37

Active May 22 '21 at 11:47

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Let $D = \{z\in\mathbb C \vert\vert {z}\vert \lt 1\}$ be the open unit disc in the complex plane, and let $f(z)$ be the following power series of the complex variable $z$: $$f(z)=z+z^2+z^4+\cdots+z^{2^n}+\cdots.$$ By considering the limit $\lim_{x\to 1^{-}} f(x) $ as $0\lt x \lt 1 $ approaches $1$ from the left.

How to show that there does not ecist a holomorphic function $g(z)$ defined on a neighborhood $U$ of $z=1$ such that $f(z)=g(z)$ for all $z\in U\cap D$

And how to deal with the limit $\lim_{x\to 1^{-}} f(x) $ as $0\lt x \lt 1 $

edited May 22 '21 at 11:47

J.G.

115,835

asked May 22 '21 at 11:37

yumcwy

Said limit is infinite, so you can't even extend $f$ continuously to 1, let alone complex differentiably. – Vercassivelaunos May 22 '21 at 11:40
Check this: https://math.stackexchange.com/q/264413/42969, and this: https://math.stackexchange.com/q/2264935/42969 – Martin R May 22 '21 at 11:42
Thank you! @Martin R – yumcwy May 22 '21 at 11:48

Prove there does not exist a holomorphic $g(z)$ defined on a neighborhood $U$ of $z=1$ equal to $f(z)=\sum_{n\ge0}z^{2^n}$

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