# There is no holomorphic function $f$ on the open unit disk such that $f(1/n)=2^{-n}$

Prove that there is no holomorphic function $f$ on the open unit disk such that $f(1/n)=2^{-n}$ for $n=2,3,..$

I know a similar question was asked on this website before but this is different.

I define $g(z)=2^{-\frac{1}{z}}$, Can I use this function and identity theorem to show that $f=g$ on the disk.

But $g$ is not analytic at $z=0$, so we get a contradiction?

Is there a loophole in my argument.

Edited the definition of function $g$

• How is this question different from yours...? (Note that your function $g$ doesn't do the job since $g(1/n) \neq 2^{-n}$.) – saz Sep 3 '18 at 16:27
• I saw a different question. Thank you. Should I delete this question? – StammeringMathematician Sep 3 '18 at 16:35
• user412674 is right that the identity theorem cannot be used. In fact, the answer in the other thread is wrong. – p4sch Sep 3 '18 at 16:47
• To complete my comment: For example $h(z) = 2^{-1/z}(1+ \sin(2\pi/z))$ has also the property that $h(1/n) = 2^{-n}$, but $h \neq g$ on the punctured unit disk! Thus, assuming that $f$ is holmorphic on the punctured unit disk, $f$ dont have to be equal to your $g$. – p4sch Sep 3 '18 at 16:54

You can't just apply the identity theorem, because the accumulation point must be in the region of holomorphy. $g$ is not holomorphic in the unit-disk, only in the punctured-unit disk $D^* = \{z \in \mathbb{C} \colon 0 < |z| < 1\}$. Thus $(2^{-n})_{n \in \mathbb{N}}$ has no accumulation points in $D^*$.
Therefore, we cannot apply the identity theorem! Assume that there exists an analytic function $f$ with these property. By continuity we must have $f(0)=0$. Using that $f$ has a representation as a Taylor series with convergence radius at least $1$, we can write $f(z) = z^r g(z)$ with $r \geq 1$ and a function $g$ which is holomorphic on the unit disc and $g(0) \neq 0$ (because $f$ cannot be identically zero). But $f(1/n) = n^{-r} g(1/n)$ has only decay rate $n^{-r}$ and not $2^{-n}$. A contradiction!