I have the space $X=\{u:[0,1]\to \mathbb{R}, u\in C^2, u(0)=u(1)=0\}$ with the norm: $$||u||=\left(\int_{0}^{1}(u')^2+\int_{0}^{1}u^2 \right)^{1/2}$$ and I'm asked to prove that it is not complete.
I have trying to find a sequence of Cauchy that is not convergent in X, and I have the following one:$$u_n(x)=\left\{ \begin{array}{lcc} x & {\rm if} & x \in [0,\frac{1}{2}-\frac{1}{(n+1)}] \\ \\ -x^2+x+\frac{(n-1)^2}{4(n+1)^2} & {\rm if} & x\in [\frac{1}{2}-\frac{1}{(n+1)},\frac{1}{2}+\frac{1}{(n+1)}] \\ \\ 1-x & {\rm if} & x \in [\frac{1}{2}+\frac{1}{(n+1)},1] \end{array} \right.$$ It converges to the function that is $x$ in $[0,1/2]$ and $1-x$ in $[1/2,1]$, which is not $C^2$ and therefore does not belong to $X$. How can I prove that it is a Cauchy sequence?
What you are implicitly doing is to find a bigger space $Y$ containing $X$ (and the induced norm on $X$ is the norm that you are fixed) which is complete. Then $X$ is complete only if it is closed in $Y$.
By the way, you have find a function $u$ not belonging to your space $X$, and such that $u_n\to u$ in that norm. This means that $u_n$ is not convergent in $X$, otherwise there could exists another $v\in X$ such that $u_n\to v$, and this imply trivially $u=v$, which is a contradiction, since $u\notin X$.
Now you have to prove $u_n$ is Cauchy. This is easy because $u_n\to u$, and so for each $\epsilon >0$ there exists $N\in \mathbb N$ such that $||u_n-u||\leq \epsilon $ for each $n\geq N$. But then, if you take $n,m\geq N$, then
$$ ||u_n-u_m||\leq ||u_n-u||+||u_m-u||\leq 2\epsilon$$
Thus, we have proved that for any $\epsilon > 0$ there exists $N\in \mathbb N$ such that $||u_n-u_m||\leq \epsilon$ for any $n,m \geq N$. This is exactly the property of a sequence to be Cauchy.
Therefore, you have found a Cauchy sequence $\lbrace u_n\rbrace $ of $X$ which is not convergent. In other words, $X$ is not complete.