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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 $p=n$?

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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 $p>n$).

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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.

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