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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!.

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In the case $W^{1,n}$ how do you interpret $C^{-1,1}$? – Connor Mooney Apr 12 '13 at 14:51
up vote 3 down vote accepted

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 $0$.

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