2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $\Bbb C$.

$\endgroup$
6
  • $\begingroup$ 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

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.