Example that differentiable functions do not form an integral domain?

Question

Let $G$ be an open connected subset of $\mathbb{C}$ and $f,g$ be holomorphic functions on $G$ such that $fg=0$.

If neither $f$ nor $g$ is zero over $G$, since $f$ and $g$ have countably many zeros respectively and $G$ is uncountable, there must exist a point $z\in G$ which is neither a zero of $f$ nor $g$, which is a contradiction.

Hence, the space of holomorphic functions $H(G)$ forms an integral domain under natural binary operations.

However, I'm not sure whether this is also true for real case.

That is, let $U$ be an open connected subset of $\mathbb{R}$ and $f,g$ be real-differentiable functions on $U$ such that $fg=0$.

Is it possible that $f\neq 0$ and $g\neq 0$?

As the above argument about holomorphic case indicates, if there is an example, it would not be expressed by means of elementary functions. Hence I'm having trouble with finding one. What it would be? Thank you in advance.

Answer 1

Yes, take two disjoint open intervals $I,J$ and bump functions with support on each of them, which you can extend to all of $\Bbb R$, which is open and connected. This works in higher dimensions: you can obtain nonzero smooth functions $\Bbb R^k\to \Bbb R$ with $fg=0$.

Answer 2

The functions: \begin{align*}f(x)&=\begin{cases}\mathrm e^{-1/x^2}&\text{if } x>0\\ 0&\text{if }x\le0 \end{cases}& g(x)&=\begin{cases}0&\text{if } x\ge 0\\ \mathrm e^{-1/x^2}&\text{if }x < 0 \end{cases} \end{align*} are such an example, with $\mathcal C^\infty$ functions (but not real analytic).