Is there an analytic function $f$ on $B(0,1)\subset\mathbb{C}$ such that $f(1/n)=e^{-n}$ for $n=2,3,4,...$? I know the following doesn't work:

Let $g(z)=\exp(-1/z)$. Then, $f=g$ on a sequence with a limit point in $B(0,1)$ and so $f=g$ on $B(0,1)$. Since $g$ is not $\mathbb{C}$-differentiable at $0$, neither is $f$ and so such a function cannot exist.

This is not the solution because you cannot use the identity principle with a non analytic function like $\exp(-1/z)$ is not analytic at $z=0$. Any help using the identity principle another way?


1 Answer 1


Here's another way to do it. Assume that such an $f$ exists. Then we can write $$ f(z) = z^k g(z) $$ for some (positive due to the assumptions) integer $k$ and some holomorphic function $g$ on $B(0,1)$ with $g(0) \neq 0$. Plug in $z=1/n$: $$ e^{-n} = f(1/n) = \frac{1}{n^k} g(1/n) $$ i.e. $$ g(1/n) = n^k e^{-n}. $$ Let $n \to \infty$. This gives (by continuity of $g$) that $g(0)=0$, which is a contradiction.

  • $\begingroup$ Why can we represent $f$ this way? $\endgroup$ Dec 7, 2015 at 22:30
  • $\begingroup$ Is it because $f(0)=0$ and $f$ is analytic so it has taylor series in a neighborhood of 0 $f=\sum\limits_{n=1}^{\infty}a_nz^n=z^kg(z)$? $\endgroup$ Dec 7, 2015 at 22:34
  • 1
    $\begingroup$ @TonyS.F. Yes, that's right. $\endgroup$
    – mrf
    Dec 7, 2015 at 22:45

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