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Prove that there does not exist an analytic function $f$ in an unit disc containing $0$ such that $$f\left(\frac{1}{n}\right)=2^{-n}.$$

I tried by using Identity theorem.

Suppose that $f$ is analytic in the unit disc.

Consider the function $g(z)=f(z)-2^{-\frac{1}{z}}$.

Then the zeros of the function $g(z)$ are $\{\frac{1}{n}:n\in \mathbb N\}$ which has a limit point $0$ in the disc. So, $g$ is identically zero. Then, $f(z)=2^{-1/z}$. But, I am unable to find a point such that we arrive at a contradiction.

Please help to find it OR any other technique to prove the question.

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1 Answers1

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Note that

$$f(z) = 2^{-1/z}$$

is not continuous at $z=0$. Indeed: For $(z_n)_{n \in \mathbb{N}} \subseteq (0,\infty)$ such that $z_n \to 0$, we have

$$f(z_n) \to 0.$$

On the other hand, for $w_n := -z_n \to 0$,

$$f(w_n) \to \infty.$$

Since an analytic function has to be continuous, this contradicts our assumption.

saz
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    Since $f(z)=e^{log(1/2)\over z}$ can we actually prove that $0$ is an essential singularity of $f$ ? The function $g(z)=e^{1\over z}$ has an essential singularity at $0$. – Srinivas K Mar 11 '15 at 16:46
  • @SrinivasK Yes, that's correct. – saz Mar 11 '15 at 16:47
  • @saz I do not see how this solves the problem. By the identity theorem, if both $f$ and $g$ are analytic in a domain $D$, and agree on some sequence with limit point in the interior of $D$, then $f \equiv g$. You have shown that the function $g(z) = 2^{-1/z}$ is not analytic in $D$. I do not see how this shows that $f$ cannot be analytic. –  Sep 29 '17 at 04:34
  • @saz The identity theorem does not give us the existence of an analytic function $g$, it requires it. –  Sep 29 '17 at 04:35
  • @user412674 No, that's not correct. The idea is to prove the statement by contradiction. If $f$ was analytic, then $g$ would be analytic. By the identity theorem this would imply $g=0$, and so $f(z) = g(z)-2^{-1/z} = 2^{-1/z}$ would be analytic. As explained in my answer, $z \mapsto 2^{-1/z}$ is not continuous at $0$ (hence not analytic), and so we arrive at a contradiction. Consequently, there cannot exist an analatic function $f$ with the desired property. – saz Sep 29 '17 at 04:52
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    That's just wrong: How you show that $g$ is also analytic? We have $g(1/n) =f(1/n)$, but the identity theorem requires that both functions are analytic on the whole unit disc. Thus, we cannot conclude that $f=g$ on the punctured unit disc. For example $h(z) = 2^{-1/z}(1+ \sin(2\pi/z))$ satisfies also $h(1/n) = 2^{-n}$, but $h \neq g$! – p4sch Sep 03 '18 at 16:53
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    @p4sch There is nothing wrong about my answer itself (because I'm just saying that $2^{-1/z}$ is not continuous at $z=0$) but you are right that the reasoning via the identity theorem doesn't work that way... which I failed to realize. Sorry about that. – saz Sep 04 '18 at 05:51