3
$\begingroup$

I want to show that

There does not exist any holomorphic function $f$ in the open unit disc such that $$f\left(\frac{1}{n}\right)=\frac{(-1)^n}{n^2}, n=2,3,\ldots $$

My attempt:

Suppose there exists such a holomorphic function. Now $$f\left(\frac{1}{n}\right)=\begin{cases}\frac{1}{n^2}, & \text{if } n\ \text{is even.}\\ -\frac{1}{n^2}, & \text{if } n\ \text{is odd.} \end{cases}$$

Which says that $f(z)=z^2$ and $f(z)=-z^2$, which is not possible. Therefore, there does not exit any such holomorphic function.

Is my argument fine?

EDIT

I used the identity theorem to conclude that. Take $g(z)=z^2$. Since $$ f\left(\frac{1}{n}\right)=g\left(\frac{1}{n}\right)=\frac{1}{n^2} $$ As, $\frac{1}{n}\to 0\in \mathbb{D}(0,1)$ so using identity theorem I conclude that $f\equiv g$ on $\mathbb{D}$. Similar argument shows that $f(z)=-z^2$, and hence contradiction.

$\endgroup$
  • $\begingroup$ Yes, this is right. In order to be totally precise, you should say why determining $f$ on all values $\tfrac{1}{n}$ with either even or odd $n$ already determines $f$ on all complex numbers. The keyword here is the identity theorem. $\endgroup$ – Luke Sep 20 '17 at 12:15
  • $\begingroup$ In other words, $f$ coincides with $z \mapsto z^2$, on the set $\left\{\frac{1}{n} : n \in\mathbb{N}, n\text{ is even}\right\}$ and coincides with $z \mapsto -z^2$ on the set $\left\{\frac{1}{n} : n \in\mathbb{N}, n\text{ is odd}\right\}$. Both of these sets have a limit point, which is $0$, so we would have $f(z) = z^2$ and $f(z) = -z^2$. Contradiction. $\endgroup$ – mechanodroid Sep 20 '17 at 12:27
2
$\begingroup$

In an introductory complex variables course, I'd be unhappy with this solution from a student. In a discussion with an expert, I'd be fine with it, because both of us know the relevant theorems.

Why am I unhappy with this as a student answer?

Because (aside from the letter $z$), there's no mention of complex numbers here. In particular, if the solution as written were good, it'd also apply to smooth functions on the reals, but it clearly does not. (There is a smooth function from $\Bbb R$ to $\Bbb R$ that meets those requirements!)

You have to explicitly cite some property that holomorphic functions have, but which ordinary smooth functions on the reals do not have, in order for the proof to be valid. @mechandroid's comment does this, as does @Luke's, but your answer does not.

$\endgroup$

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.