# When can we conclude via the Identity theorem?

Suppose $$f:D_1\mapsto \mathbb{C}$$ and $$g:D_2\mapsto \mathbb{C}$$, where $$D_1, D_2$$ are domains in the usual complex analysis sense, are two holomorphic functions such that the following identity holds for all $$n\in\mathbb{N}$$: $$f\left(\frac{1}{n}\right) = g(n).$$ (Of course $$\{1, 1/2, \cdots\} \subset D_1$$ and $$\mathbb{N}\subset D_2$$).

We want to prove using the Identity Theorem that $$f=h$$ holds everywhere on some domain $$D'$$, where $$h(z)=g(1/z)$$ on said domain.

My attempt: For all $$z\in\{1,1/2,1/3,\cdots\}$$ we can write $$f(z) = g(1/z)$$. Let $$h(z) = g(1/z)$$. In order to conclude via the Identity Theorem that $$f=h$$, we need to restrict both $$f$$ and $$h$$ to a common domain, say $$D'$$, such that both $$f(z)$$ and $$g(1/z)$$ are defined and the point $$0$$ (limit point of the set) is in the domain. But if we take $$D'=\left\{z\;|\:z\in D_1 \cap \frac{1}{z}\in D_2\right\},$$ then surely $$z=0$$ can't be in $$D'$$ which leads me to believe that we can't conclude by the identity theorem that $$f=h$$ no matter what $$g$$ is. But if $$g(n)=0$$, then it's easy to see that $$f(z)=0$$ if $$f, g$$ have a domain of e.g. $$\mathbb{C}$$. Where am I going wrong in my thought process?

• “But if $g(n)=1/n$” – $g(z) = 1/z$ does not have $\Bbb C$ as its domain. Commented Jul 5, 2023 at 5:34
• @MartinR You are right, I will edit the post. Commented Jul 5, 2023 at 5:39
• Just to be clear, what exactly did you want to prove? That if $f$ and $g$ are holomorphic on domains $D_1$ and $D_2$, respectively, such that $f(1/n) = g(n)$ for all $n \in \mathbb{N}$, then there exists a domain $U \subseteq D_1 \cap D_2$ such that $f = g$ on $U$? Commented Jul 5, 2023 at 11:50
• If $f(1/n) = g(n)$ where $f,g$ are holomorphic on $D_1, D_2$ respectively, then $f=h$ on some domain $D' (\subseteq D_1)$ where $h(z)=g(1/z)$. I dont think $D_2$ needs to be a superset of $D'$. Commented Jul 5, 2023 at 11:59

I don't think what you have at the moment allows you to conclude what you want. The problem is that with such arbitrary domains and no limit-like conditions on $$f$$ and $$g$$, there is no reason that $$f$$ and $$g$$ need to agree on a set with an accumulation point, and hence the identity theorem can't be applied, even with that condition you have on $$f$$ and $$g$$.
As a counter-example, let $$f: \mathbb{C} \to \mathbb{C}$$ be the constant function $$z \mapsto 1$$ and $$g: \mathbb{C} \to \mathbb{C}$$ be $$z \mapsto \exp(2\pi i z)$$. Then $$f$$ and $$g$$ are entire and for all $$n \in \mathbb{N}$$,
$$f\left(\frac{1}{n}\right) = 1 = \exp(2\pi i n) = g(n).$$
But there is no open subset of $$\mathbb{C} \setminus \{0\}$$ for which $$g(1/z)$$ is constant with value $$1$$.