Using the identity theorem: can there be an analytic function $f$ with $f\left(\frac{1}{n^2}\right) = \frac{1}{n}$
As Conway states it the theorem is as follows:
Let $G$ be an open connected set and let $f:G\rightarrow \mathbb{C}$ is analytic on $G$. Then the TFAE:
- $f\equiv0$
- $\{z\in G: f(z)=0 \}$ has a limit point in $G$.
I get confused when I have to use this to solve problems. My understanding on this is sort of like "if $f$ goes to zero along a certain sequence" then the function must be identically zero. Is this thinking correct? Even if I am I want some good explanation on this. Also, if any of you could give a good reference to this that would help too.
For example what can we say about an analytic function $f:\mathbb C\rightarrow \mathbb C$, such that $f\left(\frac{1}{n^2}\right) = \frac{1}{n}$. Can such a function exist? Can we even use the identity theorem to answer this question? Thanks so much for your time.
What you said about $f$ going to zero along a sequence isn't quite correct. Consider $a_n = \frac{1}{n}$ and $f(z) = z$. We have $f\left(\frac{1}{n}\right) \to 0$ as $n \to \infty$, but $f$ is not identically zero. However, if $f : G \to \mathbb{C}$ is holomorphic (analytic), and there is a sequence ${a_n}$ in $G$ with $a_n \to a \in G$, and $f(a_n) = 0$ for all $n$, then $f(a) = 0$ (by continuity), and hence the set $\{z \in G : f(z) = 0\}$ contains the limit point $a$. Therefore $f$ is identically zero.
As for your second question, suppose $f$ exists and consider the holomorphic function $z \mapsto f(z^2) - z$. What are its values at $a_n = \frac{1}{n}$? What does that tell us about $f$?