# Given $f^2\left(\frac{1}{n}\right)g\left(\frac{1}{n}\right)= 0$ prove $f\equiv0$ or $g\equiv0$

Given the entire functions $$f$$,$$g:\mathbb{C} \rightarrow \mathbb{C}$$ for which : $$f^2\left(\frac{1}{n}\right)g\left(\frac{1}{n}\right)= 0 \ \ \ \ \ \ \ \forall\ n \in \mathbb{N}$$

I want to prove that $$f\equiv0$$ or $$g\equiv0$$

Here is my attempt.

Suppose that both $$f$$ and $$g$$ have a countable number of roots inside the disk $$D(0,1)$$. Then $$f^2g$$ would also have a countable number of roots which leads to a contradiction because of the above relationship.

Therefore, either $$f$$ or $$g$$ has an uncountable number of roots in $$D(0,1)$$.

Let $$g$$ be that function.

This is where I'm stuck. From other examples I've seen, I think I need to make use of "accumulation points" somehow.

Is this approach correct? How should I procced or what other ways are there to solve this kind of problem?

• Poles and zeroes of entire functions are isolated. Commented Jan 23, 2023 at 17:40
• Try looking into this theorem: en.wikipedia.org/wiki/Identity_theorem Commented Jan 23, 2023 at 18:42

The Identity theorem (https://en.wikipedia.org/wiki/Identity_theorem) states that if we have two holomorphic functions $$f$$ and $$g$$ on some set $$D$$ and $$f=g$$ on $$S \subseteq D$$ where $$S$$ has an accumulation point, then $$f = g$$ on $$D$$.

Let $$h(z) = f^2(z)g(z)$$ and $$p(z) = 0$$. Since $$f,g$$ are entire, $$h$$ is entire, and so it's also holomorphic.

If we form the following sequence $$(1/n)_{n\in\mathbb{N}}$$, we can show that indeed $$0$$ is an accumulation point. Use the fact that an accumulation point of a set is the limit of some sequence of distinct elements of that set.

In this case our set $$S = \{1/n : n \in \mathbb{N}\}$$

Hence, the only point which could satisfy this is $$0$$ for our set $$S$$.

We also know that $$h(1/n) = p(1/n)$$ on our set $$S$$, but $$S$$ is just a subset of $$D = \mathbb{C}$$ which $$h$$ is analytic on.

Since all conditions are satisfied from the theorem, $$h(z) = p(z) = 0$$ on $$D$$.

But $$h(z) = f^2(z)g(z) = 0$$ on $$D = \mathbb{C}$$ which can only happen if $$f \equiv 0$$ or $$g \equiv 0$$ on $$\mathbb{C}$$.

There is a countably infinite subset of $$\{\frac{1}{n}\}_{n\in \Bbb N}$$ on which either $$f$$ or $$g$$ is zero. By continuity, that function vanishes at zero.

Can you take it from here?

• I'll admit I have some blanks here. I had to look at mathinsight.org/definition/countably_infinite to understand what "countably infinite" means. But "vanishes" essentially means the function is 0 according to math.stackexchange.com/questions/208022/…. So there is nothing else to prove, is there? Commented Jan 23, 2023 at 17:50
• Using the comment of @jp boucheron, you can say that since zeroes are isolated, one of these functions will be identically zero. Commented Jan 23, 2023 at 17:56