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](http://en.wikipedia.org/wiki/Sobolev_inequality#Sobolev_embedding_theorem). 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!.



Sharpness of the Sobolev embedding theorem
This response is closely related to my answer [Here](https://mathoverflow.net/questions/124028/what-goes-wrong-for-the-sobolev-embeddings-at-kn-p/124331#124331).


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 $0$.


