3
$\begingroup$

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 $g$

$\endgroup$
  • 1
    $\begingroup$ How is this question different from yours...? (Note that your function $g$ doesn't do the job since $g(1/n) \neq 2^{-n}$.) $\endgroup$ – saz Sep 3 '18 at 16:27
  • $\begingroup$ I saw a different question. Thank you. Should I delete this question? $\endgroup$ – StammeringMathematician Sep 3 '18 at 16:35
  • $\begingroup$ user412674 is right that the identity theorem cannot be used. In fact, the answer in the other thread is wrong. $\endgroup$ – p4sch Sep 3 '18 at 16:47
  • 1
    $\begingroup$ 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 $g$. $\endgroup$ – p4sch Sep 3 '18 at 16:54
6
$\begingroup$

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

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.