Can there exist a holomorphic function defined on the unit disk $D = \{ z \mid |z| <1\}$ such that $f(\frac1n) = \frac{1}{2^n}$ where $n \in \Bbb Z$?
I thought of using CR equations to find a contradiction but unable to do.
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1So an obvious candidate would be “$2^{-1/z}$”, that is: $D → D, z ↦ \exp (-{1/z} \log 2)$ – what about it? (Thanks @b00nheT.) – k.stm Jan 31 '17 at 08:52
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1@b00nheT Yeah, right. Thanks. Corrected it. – k.stm Jan 31 '17 at 08:54
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1@user8795 there is no way to use CR equations. You must use the identity theorem or something in that direction – b00n heT Jan 31 '17 at 08:56
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Possible duplicate of There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$? See also: http://math.stackexchange.com/questions/2031222/how-do-i-show-that-a-holomorphic-function-which-satisfies-this-bound-on-reciproc ; http://math.stackexchange.com/questions/1175315/is-that-function-must-be-constant-under-the-following-conditions – Winther Jan 31 '17 at 09:13