Let $\mathcal{A}=\{ v \in C^2([a,b]): v(a)=v(b)=0 \}$ and let $(u,v)_{\mathcal{A}}=\int_{a}^{b} u'v' \, dx $. Finally, set $\| u \|_{\mathcal{A}}=\sqrt{(u,u)_{\mathcal{A}}}$ whenever $u \in \mathcal{A}$. Then prove that $(\mathcal{A},\| \cdot \|_{\mathcal{A}})$ is a normed vector space, but not complete.
Trivially, it is a n.v.s. For the last part:
Idea: find a function $v$ with is differentiable and vanishes at $a,b$, but not twice differentiable. Then construct a sequence $v_n \in \mathcal{A}$ converging to $v$. For example, let $v(x)=x|x|-x$, then $v(-1)=v(1)=0$, $v \in C^1([-1,1]$. but $v$ is not twice differentiable at zero. Now, let $v_n(x)=x\sqrt{x^2+n^{-1}}-x$, then everything works apart from the fact that $v_n$ does not vanish on the boundary. If I try to fix this, then something else goes wrong (for example, the sequence fails to converge wrt $\| \cdot \|_{\mathcal{A}}$ to $v$). Any other ideas to construct such counterexample?