If $f \in C^{2}(\Omega) \cap C(\overline{\Omega})$ and $f \equiv 0$ in $\partial\Omega$, then $ f \in C^2({\overline\Omega})$?

Question

If $f \in C^{2}(\Omega) \cap C(\overline{\Omega})$ and $f \equiv 0$ in $\partial\Omega$, then $ f \in C^2({\overline\Omega})$?

It is true that $\overline\Omega = \Omega \cup \partial\Omega$, then, since $0$ is $C^2$ in any domain, can I say that $f$ will be $C^2$ in $\overline{\Omega}$, since it will be $C^2$ in $\Omega$ and in $\partial\Omega$? I'm not sure that this is true, how can I properly prove this? Do I need to use that $f \in C(\overline{\Omega})$ for this?

($C^2(\Omega)$ is a function which second derivative exists and is continuous in $\Omega$ and $C(\Omega)$ is a function that is continuous in $\Omega$).

Answer 1

This already fails in dimension $1$. Let $\Omega=(0,1)$, so that $\overline{\Omega}=[0,1]$, and let $$f(x)=\sqrt{x(x-1)^2},\qquad x\in [0,1] $$ Then $f\in C^2(\Omega)\cap C(\overline{\Omega})$ and $f=0$ in $\partial \Omega=\left\{0,1\right\}$, but it cannot be extended to a twice continuously differentiable function on an open interval containing $[0,1]$ because $$\lim_{x\to 0^+}f'(x)=+\infty $$