Can we prove that $x_n$ converges to a zero point if $F(x_n)$ converge to 0 when the kernel of $F$ is convex?

Question

Suppose $F$ is a continuous function and $C=\{x\in \mathbb{R}^n\mid F(x)=0\}$ is convex.

We have a sequence $\{x_n\}$ s.t. $F(x_n)\rightarrow 0$ as $n\rightarrow \infty$ and $F(x_n)\ne 0$, $\forall n$.

Can we prove that there is a point $x_0\in C$, such that $x_n\rightarrow x_0$ and $F(x_0)=0$?

Answer 1

No. Take $F:\mathbb R \to\mathbb R$ to be $$ F(x) = \begin{cases}0, & x\in[-1,1] \\ x^2-1, & x\notin[-1,1]\end{cases}.$$ Then $F$ is continuous and $C = [-1,1]$ is convex. Now take $x_n = (-1)^n\cdot(1+\frac 1n)$. We have $F(x_n)\to 0$ but $x_n$ does not converge.