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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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  • $\begingroup$ Haven't thought carefully, but it's undefined at $z=0$... which should matter. $\endgroup$ – abnry Mar 11 '15 at 16:23
  • $\begingroup$ So, how we can proceed? $\endgroup$ – Empty Mar 11 '15 at 16:29
  • $\begingroup$ This seems related math.stackexchange.com/questions/392682/… $\endgroup$ – kingW3 Mar 11 '15 at 16:52
  • $\begingroup$ @S717717 can you please tell from which book can I get this question? $\endgroup$ – user8795 Feb 7 '17 at 4:06
  • $\begingroup$ @user8795 Sorry..! I forget the source of this problem. $\endgroup$ – Empty Feb 7 '17 at 4:52
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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.

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    $\begingroup$ 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 $0$. $\endgroup$ – Srinivas K Mar 11 '15 at 16:46
  • $\begingroup$ @SrinivasK Yes, that's correct. $\endgroup$ – saz Mar 11 '15 at 16:47
  • $\begingroup$ @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. $\endgroup$ – user412674 Sep 29 '17 at 4:34
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    $\begingroup$ 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 $h \neq g$! $\endgroup$ – p4sch Sep 3 '18 at 16:53
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    $\begingroup$ @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. $\endgroup$ – saz Sep 4 '18 at 5:51

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