Normed space but not complete As someone pointed out this could seems to be a duplicate of another question already posted. Is in fact the same question but I have provided also the effort of the solution and additionally I have shown something else, for instante the proof of the triangular inequality via de Cauchy-Schwarz inequality. Hence is definetly a duplicate but with a better description of the problem and the solution.
Let $\Omega \subset \mathbb{R}^{n}$ open and bounded and let $\mathcal{A}=\{ v \in C^{2}(\overline{\Omega}) \, | \,v=0\, on \, \partial\Omega\}$ with the following scalar product
\begin{equation}
(u,v)_{\mathcal{A}}=\int_{\Omega} (\nabla u, \nabla v)_{\mathbb{R}^{n}}dx
\end{equation}
where $(\cdot, \cdot)_{\mathbb{R}^{n}}$ is the canonical scalar product on $\mathbb{R}^{n}$. 
I have to prove that $(\mathcal{A}, \| \cdot \|_{\mathcal{A}})$ is a normed space but not a Banach space, where $\| u \|_{\mathcal{A}}=\sqrt{(u,u)_{\mathcal{A}}}$. Proving that $(\mathcal{A}, \| \cdot \|_{\mathcal{A}})$ is a normed space is quite straightforward, the only point that make me think was the triangular inequality that I solved using the Chauchy-Schwarz inequality, namely
\begin{equation}
|(u,v)_{\mathcal{A}}|\leq \|u\|_{\mathcal{A}}\|v\|_{\mathcal{A}},
\end{equation}
infact if I consider $\| u+v\|_{\mathcal{A}}^2$ I can proced in this way
\begin{split}
\| u+v\|_{\mathcal{A}}^2= & \| u\|_{\mathcal{A}}^2+2(u,v)_{\mathcal{A}} +\|v\|_{\mathcal{A}}^2 \\
\leq &\| u\|_{\mathcal{A}}^2+2\|u\|_{\mathcal{A}}\|v\|_{\mathcal{A}} +\|v\|_{\mathcal{A}}^2 \\
= & (\| u\|_{\mathcal{A}}+\|v\|_{\mathcal{A}})^2
\end{split}
Hence we have proved $\| u+v\|_{\mathcal{A}}\leq \| u\|_{\mathcal{A}}+\|v\|_{\mathcal{A}} $.
But I was stucked on the noncompleteness, I tried to argue as following but I am not sure if it is right. Let $n=1$ and $\Omega=(-1,1)$ and suppose to take the sequence $(u_n)_n\subset \mathcal{A}$, given by
\begin{equation}
u_n=\sqrt{1+\frac{1}{n^2}}-\sqrt{x^2+\frac{1}{n^2}},
\end{equation}
I can notice that $u_n \rightarrow u$ when $n\rightarrow \infty$ using for instance the uniform norm, where $u=1-|x|$ that does not belong to $\mathcal{A}$. Hence I found a sequence that converges to an element that is not in the space and consequentely $\mathcal{A}$ is not complete.
 A: The problem with the space $(\mathcal{A}, \|\cdot\|_{\mathcal{A}})$ is that the norm $\|\cdot\|_{\mathcal{A}}$ takes in account only the first derivative of the function, but the set $\mathcal{A}$ is the space of the functions with continuous second derivative that annihilate on the border of $\Omega$.
Hence we can put ourselves in the situation in which $n=1$ and $\Omega=(-1,1)$, hence the set $\mathcal{A}$ becomes $\{ v \in C^{2}([-1,1]) \, | \,v(x)=0\, when \, x=-1,1 \}$. Thus is enough to find a sequence that converges to a function that has the first derivative not continuous. Firstly I thought it was enough to take such a sequence
\begin{equation}
u_n(x)=1-\sqrt{x^2+\frac{1}{n^2}},
\end{equation}
but $u_n \notin \mathcal{A}$ because for instance $u_1(1)=1-\sqrt{2}\ne 0$. Hence I can adjust $u_n$ to find a sequence $(v_n)_n\subset \mathcal{A}$ that $v_n\notin C^{1}([-1,1])$. So I can use the following sequence
\begin{equation}
v_n(x)=\sqrt{1+\frac{1}{n^2}}-\sqrt{x^2+\frac{1}{n^2}}.
\end{equation} 
$(v_n)_n\subset \mathcal{A}$ in fact $(v_n)_n\subset C^{2}([-1,1])$ and annihilate itself in $x=-1,1$. But $v_{n}$ converges to $v=1-|x|$ as $n\rightarrow \infty$. As we know the function $v$ has the following derivative
\begin{equation}
v(x)=-\frac{x}{|x|},
\end{equation} 
that is not continuous since is not defined in $x=0$ and there is a leap from the left to the right limit.
Consequently we have found a counter example, and we can conclude that the space $(\mathcal{A}, \|\cdot\|_{\mathcal{A}})$ is a normed space but is not a Banach space because is not complete, namely there exists a sequence contained in $\mathcal{A}$ that converges to an element not in $\mathcal{A}$.
