5
$\begingroup$

We know that $W^{k,p}\hookrightarrow C^{k-\lfloor\frac{n}{p}\rfloor-1,\gamma}(\bar{\Omega})$ with $kp>n,\gamma=\lfloor\frac{n}{p}\rfloor+1-\frac{n}{p}$, where $n$ is the dimension of $\Omega$, $\Omega$ is a bounded domain in $\mathbb{R}^n$ with $C^1$ boundary. From Wikipedia. This can also be seen in C.L.Evan's "Partial Differential Equations".

However, when $n/p$ is an integer, the theorem does not state anything more about $\gamma=1$. Is there any counterexample to $W^{k,p}\hookrightarrow C^{k-\frac{n}{p}-1,1}(\bar{\Omega})$ when $n/p$ is an integer? I don't know how to construct it, Thanks for your attention!.

$\endgroup$
1
  • $\begingroup$ In the case $W^{1,n}$ how do you interpret $C^{-1,1}$? $\endgroup$ Apr 12, 2013 at 14:51

1 Answer 1

3
$\begingroup$

This response is closely related to my answer Here.

For the case $W^{2,n}$ we automatically get Holder regularity by applying Sobolev and then Morrey. However, we don't get Lipschitz. The following example "integrates" the counterexample to boundedness of $W^{1,n}$ functions.

Take a function $\psi$ supported on $B_2$ with $\psi$ linear and nonconstant on $B_1$ and $|\nabla \psi| < c$, and add dyadic rescalings together as follows. Consider $$u(x) = \sum_{i=1}^{\infty} h_i\psi(2^ix)$$ for some $h_i$ we will choose to get bounded $W^{2,n}$ norm but unbounded derivative. Note that $|D^2(h_i\psi(2^ix))|$ grows like $h_i2^{2i}$ and they are supported on disjoint dyadic rings of volume going like $2^{-in}$. Thus, to get bounded $W^{2,n}$ norm we want $$\sum_i h_i^n2^{in} < C.$$

To give unboundedness of the derivative we want $$\sum_i h_i2^{i} = \infty.$$

The natural choice for $h_i$ is $\frac{2^{-i}}{i}$, which gives the counterexample.

Remark: This function has size $~ 2^{-k}\sum_{i=1}^k \frac{1}{i}$ ~ $2^{-k}|\log\log(2^{-k})|$ at $|x| = 2^{-k},$ so a more explicit example might look something like $|x||\log\log(|x|)|$, which looks almost like a cone away from $0$ but the slope gets unboundedly high near $0$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.