Let $D=(-1,1),$ $f(x)=1-1|x|,$
$g(x)=-\text{sgn}(x)$, $V=\{v \in C^{1}[-1,1]\mid v(1)=v(-1)=0\}.$
Prove there exists a sequence $v_{n} \in V$ such that $v_{n} \rightarrow f$ and $v'_{n} \rightarrow g$ in $L^2(D)$.
Prove that V isn't a Hilbert space w.r.t. the inner product $(u,v)_{1}= \int (uv + u'v') dx$.
I've found a sequence satisfying these conditions is
\begin{equation*} v_{k}(x)= \begin{cases} 1-|x| \;\; \text{for } x \leq \frac{-1}{k}\\ -\frac{k}{2}x^2 + 1-\frac{1}{2k} \;\; \text{for } \frac{-1}{k} \leq x \leq \frac{1}{k} \\ 1-|x| \;\; \text{for } x \geq \frac{1}{k} \end{cases} \end{equation*}
Now to prove the 2nd part of the question I need to show that $v_{k}$ is a Cauchy sequence in order to show that $V$ isn't complete and therefore not a Hilbert space. How can I do this? Thanks.