Is $C^1[a,b]$ with the norm $\left \| f \right \|_1=(\int_{a}^{b}\left | f(t) \right |dt)+(\int_{a}^{b}\left | f´(t) \right |dt)$ a complete space? Is $C^1[a,b]$ with the norm   $\left \| f \right \|_1=(\int_{a}^{b}\left | f(t) \right |dt)+(\int_{a}^{b}\left | f´(t) \right |dt)$ a complete space?
I thought with parabolas based on this link , but the area is infinite $C([0, 1])$ is not complete with respect to the norm $\lVert f\rVert _1 = \int_0^1 \lvert f (x) \rvert \,dx$.
Thanks.
 A: To avoid abusing notation, let $\|\cdot \|_1$ represent the usual $L^1$ norm
and $\|f\|_* = \|f\|_1 + \|f'\|_1$.
Consider $C^1[-1,1]$ with the above norm, and let $g_n(x) = {2 \over \pi}\arctan nx$. Note that the $g_n$ are smooth, odd, strictly monotonically
increasing, $g_n(0) = 0$, $|g_n(x) \le 1$ for any $x$ and if $x \neq 0$, $\lim_n g_n(x) = \mathbb{sgn} \ x$. The discontinuity of the limit function
at $x=0$ is the crucial element here.
If $\epsilon>0$, $m \le n$ and $x \ge \epsilon $, then
$|g_n(x)-g_m(x)| \le |1-g_m(x)| \le |1-g_m(x)| \le |1-g_m(\epsilon)|$. Then
$\|g_n-g_m\|_1 = 2\int_0^1 |g_n(x)-g_m(x)| dx \le 2 \epsilon + 2 |1-g_m(\epsilon)|$, and so $g_n$ are Cauchy in the $\| \cdot \|_1$ norm.
Define $f_n(x) = \int_0^x g_n(t)dt$, and note that $f_n' = g_n$. Note that
the $f_n$ are smooth, even and hence $f_n'(0) = 0$. It is not hard to show, but irrelevant to this proof, that $\lim_n f_n(x) = |x|$.
Note that $|f_n(x)-f_m(x) | \le \int_{-1}^1 |g_n(x) - g_m(x)| dx = \|g_n-g_m\|_1$, hence $\|f_n-f_m\|_1 \le 2 \|g_n-g_m\|_1$ and so, for
any $ \epsilon>0$, we have
$\|f_n-f_m\|_* \le 3 (2 \epsilon + 2 |1-g_m(\epsilon)|) $ from
which it follows that $f_n$ are Cauchy.
Then $f_n$ is Cauchy, but has no limit in $C^1[-1,1]$. Very roughly,
it has no limit
with respect to $\|\cdot \|_*$ because the only continuous function to
which the $f_n$ can converge is $x \mapsto |x|$ and this fails to be
differentiable, let alone continuously differentiable, at $x=0$.
To show that there is no limit in $C^1[-1,1]$, we suppose $\|f_n -f\|_* \to 0$
with $f \in C^1[-1,1]$
look for a contradiction.
First, note that $f$ must be even, since the $ f_n$ are. To see this,
let $\phi(x) = f(-x)$ and note that $\|f-\phi\|_1 \le \|f-f_n\|_1 + \|f_n-\phi\|_1 = 2 \|f-f_n\|_1$, and since $n$ is arbitrary, we have $\|f-\phi\|_1 = 0$. Since $f, \phi$ are continuous, we have $f= \phi$
and so $f$ is even. 
Since $f$ is even, we have 
 $f'(0) = 0$. There is some $\delta>0$ such that if $|x| < \delta$, then
$|f'(x)| < {1 \over 2}$. However, $f_n'({\delta \over 2})=g_n({\delta \over 2}) \to 1$, which is
a contradiction.
A: For the sake of better notation, let's use $\Vert\cdot\Vert_1$ instead to denote the norm
$$\Vert f\Vert_1=\int_a^b|f(t)|dt$$
on $C^0[a,b]$, and $\Vert\cdot\Vert_{1,1}$ instead to denote the norm you are considering:
$$\Vert f\Vert_{1,1}=\Vert f\Vert_1+\Vert f'\Vert_1.$$
Take $f_n$ as in the link you provided, and set $F_n(t)=\int_a^t f_n(s)ds$. Note that
\begin{align*}
\Vert F_n-F_m\Vert_1\leq\int_a^b\int_a^t|f_n(s)-f_m(s)|dsdt=\int_a^b\int_s^b|f_n(s)-f_m(s)|dtds
\end{align*}
by Fubini, and then
$$\Vert F_n-F_m\Vert_1\leq\int_a^b(b-s)|f_n(s)-f_m(s)|ds\leq(b-a)\Vert f_n-f_m\Vert_1$$
so $\Vert F_n-F_m\Vert_{1,1}\leq (1+b-a)\Vert f_n-f_m\Vert_1$, so $(F_n)$ is Cauchy in $(C^1[a,b],\Vert\cdot\Vert_{1,1})$.
But the derivative map $D:(C^1[a,b],\Vert\cdot\Vert_{1,1})\to(C^0[a,b],\Vert\cdot\Vert_1)$ is continuous, and $D(F_n)=f_n$ does not converge, so $F_n$ does not converge as well.

EDIT: Let's explain the application of Fubini with some more details. The usual version of Fubini is that $\int_a^b\int_c^df(s,t)dsdt=\int_c^d\int_a^bf(s,t)dtds$.
Now suppose we have an integral of the form $\int_a^b\int_a^tf(s,t)dsdt$, where $f:[a,b]\times[a,b]\to\mathbb{R}$. 
Let $g(s,t)=\begin{cases}1&\text{ if }s\leq t\\0&\text{ otherwise}\end{cases}$.
When we fix $t$, the inner variable of the inner integral ranges from $a$ to $t$, and so it can be rewritten as
$$\int_a^tf(s,t)ds=\int_a^b g(s,t)f(s,t)ds$$
So we can rewritte the double integral as
$$\int_a^b\int_a^tf(s,t)dsdt=\int_a^b\int_a^bg(s,t)f(s,t)dsdt$$
Now apply Fubini:
$$\int_a^b\int_a^tf(s,t)dsdt=\int_a^b\int_a^bg(s,t)f(s,t)dtds$$
Loog at the inner integral in the RHS, i.e., for a fixed $s$, we have
$$\int_a^b g(s,t)f(s,t)dt=\int_s^bf(s,t)$$
by definition of $g$, and therefore
$$\int_a^b\int_a^tf(s,t)dsdt=\int_a^b\int_s^bf(s,t)dtds$$
We applied the above to $f(s,t)=|f_n(s)-f_m(s)|$.
