Prove that this space is not Banach

Question

Let $\Omega\subset\mathbb{R}^n$ be an open, bounded set with boundary $\partial\Omega$ of class $C^1$. $$\mathcal{A}:=\{u\in C^2(\bar\Omega):u=0\text{ on }\partial\Omega \}$$ endowed with the scalar product $$(u,v)_{\mathcal{A}}:=\int_{\Omega}(\nabla u,\nabla v)_{\mathbb{R}^n}\,dx.$$ I have to prove that $\mathcal{A}$ equipped with the induced norm is a normed space, but it is not a Banach space. I can't find a proper counterexample. Any help?

Answer 1

As far as I understand you only need help in showing that $\mathcal A$ is not Banach. Таке а function $f\in C^1(\bar\Omega)$ such that $f=0$ on $\partial\Omega$ and $f$ is $C^2$ everywhere in $\bar\Omega$ except at a single point $x_0\in\Omega$, where the second derivatives do not exist. Then you can approximate $f$ in the norm defined by this inner product by functions $f_n\in\mathcal A$; thus, $\mathcal A$ is not complete in this norm and hence by definition not Banach.