$||f||_1 =(\int_a^b [|f|^2+|f'|^2]dx)^{1/2}$. Is this normed space complete? Define $C_1^1[a,b]$ to be the space of continuously differentiable functions on $[a,b]$, with norm 
$$||f||_1 =\left(\int_a^b \left(|f|^2+|f'|^2\right) dx \right)^{1/2}$$ 
Is this normed space complete?
So far I have shown that this is a normed space by satisfying the four axioms of a normed space. Now I need to show that it is complete. That is, show that every Cauchy sequence in this space is convergent. I'm stuck on how to do this part. Any hints or solutions are greatly appreciated.
 A: Hint
Pointwise limit of differentiable functions needn't be differentiable.
A: I was working on a similar problem. Here is a way of showing that $C^1[a,b]$ is not a Banach space. We will assume $a=0$ and $b=1$; the method behind the proof below will motivate the general idea. Let $\{f_j\}_{j\geqslant 1}\subset C^1[0,1]$ be a sequence, defined as follows. 
$$f_j(t)=\left\{
                \begin{array}{ll}
                  \frac{1}{2(j+1)}(2t)^{(j+1)}, ~~0\leqslant t<\frac{1}{2}\\
                  (t-\frac{1}{2})+\frac{1}{2(j+1)}, ~~\frac{1}{2}\leqslant t\leqslant 1
                \end{array}
              \right.$$
It's easy to check that $f_j$ is differentiable on $(0,1)$ and continuous on $[0,1]$. Now we claim that $\{f_j\}_{j\geqslant 1}$ is Cauchy. 
Proof. Check that for any $m,n\in\mathbb{N}$ (assuming $m>n$),
\begin{eqnarray}
d(f_m,f_n) & = & \Vert f_m-f_n\Vert \\
           & = & \left(\int_0^1(f_m-f_n)^2+[(f_m-f_n)']^2dt\right)^{\frac{1}{2}} \\
           & = & \left(\int_0^{1/2}(f_m-f_n)^2+[(f_m-f_n)']^2dt+\int_{1/2}^1(f_m-f_n)^2+[(f_m-f_n)']^2dt\right)^{\frac{1}{2}}.
\end{eqnarray}  
The two integrals in the squareroot are easily bounded above by $\left[\frac{1}{2(n+1)}\right]^2+\left(\frac{1}{2}\right)^{2n+1}$, which approaches zero. Hence, for any $\varepsilon>0$ picked, there exists $N\in\mathbb{N}$ such that for all $m,n\geqslant N$ the expression above is strictly bounded above by $\varepsilon$. $\blacksquare$
We claim that $\{f_j\}_{j\geqslant 1}$ does not converge to a member of $C^1[0,1]$. What is $f=\lim_{j\rightarrow\infty}f_j$? It is given by
$$f(t)=\left\{
                \begin{array}{ll}
                  \lim_{j\to\infty}\frac{1}{2(j+1)}(2t)^{(j+1)}, ~~0\leqslant t<\frac{1}{2}\\
                  \lim_{j\to\infty}(t-\frac{1}{2})+\frac{1}{2(j+1)}, ~~\frac{1}{2}\leqslant t\leqslant 1
                \end{array}
              \right.$$  
In other words, it's the function $f(t)=0$ when $0\leqslant t<\frac{1}{2}$ and $f(t)=t-\frac{1}{2}$ when $\frac{1}{2}\leqslant t\leqslant 1$. It's straightforward to check that $f$ isn't differentiable on $(0,1)$. 
