How can I show that the embedding of $W^{1,1}(a,b)$ in $C[a,b]$ is not compact? How can I show that the embedding of $W^{1,1}(a,b)$ in $C[a,b]$ is not compact? 
I know that $W^{1,p}$ is compactly embedded in $C(\Omega \subset \mathbb R^n)$ for $p>n$. But since we now have $\Omega=[a,b]$ we have $n=1=p$. Why does it not work for $p=n$? 
 A: One way to show that the embedding is not compact it so find a sequence of function $(u_n)$ such that:
$$\|u_n\|_{W^{1,1}} \leq C$$
$$\|u_n-u_m\|_\infty \geq 1 \text{ for } n \neq m$$
the second condition preventing the sequence to have any convergent subsequence in $C[a,b]$..

Without loss of generality let us take $[a,b]=[0,1]$. Let $\phi$ a nice function with compact support in $(0,1)$ such that $\|\phi\|_\infty=1$. Then:
$$u_n(x)=\begin{cases}0 &\text{ for } 0 \leq x <\frac {1}{n+1}\\
\phi \left(n(n+1)\left(x -\frac{1}{n+1} \right)\right)  &\text{ for } \frac{1}{n+1} \leq x <\frac{1}{n}\\
0 & \text{ for } x > \frac{1}{n}\end{cases}$$
verifies:


*

*$$\|u_n-u_m\|_\infty = 1 \text{ for } n \neq m$$

*$$\|u_n\|_{L^1}= \frac{1}{n(n+1)}\|\phi\|_{L^1} \to 0$$
$$\|u_n'\|_{L^1}=\|\phi'\|_{L^1}$$
so $$\|u_n\|_{W^{1,1}} \leq C$$.



The idea behind this counter example is to use a gliding hump. But the reason why it work is because of the scaling, in $\Bbb R^n$ with $\psi_\lambda(t)=\psi(\lambda t)$:


*

*$$\|\psi_\lambda \|_{L^p}=\lambda^{-\frac{n}{p}}\|\psi\|_{L^p}$$

*$$\|\nabla \psi_\lambda \|_{L^p}=\lambda^{1-\frac{n}{p}}\|\nabla
\psi\|_{L^p}$$

*$$\| \psi_\lambda \|_{\infty}=\lambda^{0}\|\nabla
\psi\|_{\infty}$$


this a way to understand why the exponents in the whole space, and in a bounded domain for compactly supported function you can take $\lambda \to + \infty$. This impose $1-\frac{n}{p} \leq 1$ but when you have equality, i.e the critical case, even if you have embedding you can find function as the previous counterexample (which does not works for $p>n$).
A: Here is a simpler example: On $\Omega=(-1,1)$ define
$$
u_n(x) = \min(-1, \max(nx,1)).
$$
Then $\|u_n\|_{L^1}\le 2$, $\|u'\|_{L^1} = 1$, $u_n \to v$ in $L^1$ with
$v(x) = sign(x)$, which is not continous. 
If the embedding $W^{1,1}(\Omega)\hookrightarrow C(\bar \Omega)$ would be compact, then the limit of a converging subsequence in $C(\bar \Omega)$ has to coincide with $v$, which is a contradiction.
