Product of two functions vanishes when one vanishes

Question

Let $f \in C^0([0, 1]) \cap C^1(0, 1)$ be given such that $f(0)=0$. Let $g \in H^1_{loc}(0, 1)$ be such that \begin{equation} fg \in C^0([0, 1]). \end{equation}

Can we then say that $(fg)(0) = 0$?

Answer 1

Counterexample: Let $f(x)=\sqrt{x}.$ Then $f\in C^0([0,1])\cap C^1((0,1))$ with $f(0)=0.$ Consider $g(x)=\frac{1}{f(x)}=\frac{1}{\sqrt{x}}$. Then $g\in H^1_{loc}(0,1)$ since the only problematic integrals occur when the lower bound is $0$. However, for any compact set $K\subset (0,1)$, $\min K \geq \epsilon$ for some $\epsilon>0$.

Obviously, $(fg)(x)\equiv 1 \in C^0([0,1]),$ but $(fg)(0)\neq 0.$