Show that no entire function $f: \mathbb{C} \to \mathbb{C}$ exists with $$f\bigg(\frac{1}{n}\bigg)= \frac{n}{2n-1}$$ for $n \in \mathbb{N}$. What is the domain of a holomorphic function with such values?

I was thinking of putting $g(n) = \frac{1}{2-n}$ with $n \in \mathbb{N}$ which is holomorphic on $\mathbb{C}\setminus \{2\}$ and then arguing that because both functions "meet" at $0$, according to identity theorem, $g(n) = f\big(\frac{1}{n}\big)$ and because $g(n)$ is not an entire function per definition, there is no entire function $f\big(\frac{1}{n}\big)$ on $\mathbb{C}$. The domain for which a function with such values is defined would be defined for values in $(0, 1]$, right?


Let $g(z)=\frac1{2-z}$, for each $z\in\Bbb C\setminus\{2\}$. Suppose that such an entire function $f$ exists. Then$$(\forall n\in\Bbb N):f\left(\frac1n\right)=\frac n{2n-1}=g\left(\frac1n\right).$$On the other hand, since $f$ is continuous,\begin{align}f(0)&=\lim_{n\to\infty}f\left(\frac1n\right)\\&=\lim_{n\to\infty}\frac n{2n-1}\\&=\frac12\\&=g(0).\end{align}So,$$\{z\in\Bbb C\setminus\{2\}\mid f(z)=g(z)\}\supset\{0\}\cup\left\{\frac1n\,\middle|\,n\in\Bbb N\right\}.$$In particular, the set $\{z\in\Bbb C\setminus\{2\}\mid f(z)=g(z)\}$ contains an accumulation point ($0$) and therefore, by the identity theorem and because $\Bbb C\setminus\{2\}$ is connected,$$(\forall z\in\Bbb C\setminus\{2\}):f(z)=g(z).$$But this is impossible, since the limit $\lim_{z\to 2}f(z)$ exists in $\Bbb C$ (it is $f(2)$), whereas the limit $\lim_{z\to 2}g(z)$ does not exist in $\Bbb C$.

  • $\begingroup$ Thanks. And what would be the domain of definition of a function $f$ with such values? $\endgroup$
    – MJP
    Jun 8 '20 at 21:29
  • $\begingroup$ Since $f$ is an entire function, its domain would be $\Bbb C$. $\endgroup$ Jun 8 '20 at 21:31
  • $\begingroup$ Wait, but I thought I was showing that there is no entire function $f$? $\endgroup$
    – MJP
    Jun 8 '20 at 21:32
  • $\begingroup$ Yes, what I did was to prove that there is no entire function $f$ such that $(\forall n\in\Bbb N):f\left(\frac1n\right)=\frac n{2n-1}$. $\endgroup$ Jun 8 '20 at 21:33
  • $\begingroup$ Yes, that was my intent as well, but I was also trying to answer the part of the question where it is asking what the maximum domain of definition of such a holomorphic function is. Maybe I'm not understanding that second part of the question right... $\endgroup$
    – MJP
    Jun 8 '20 at 21:37

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