MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Partial Differential Equation by Lawerence Evan p284 there is this theorem stated:

Let $U$ be a bounded open subset of $\mathbb{R}^n$ with $C^1$ boundary. Suppose $u\in W^{k,p}$ then if $k>n/p$ we have

$ u\in C^{\alpha, \gamma}(\overline{U}) $ where $\alpha = k-\left[n/p\right]-1$ and $\gamma = \left[n/p\right]+1-n/p$ if $n/p$ is not an integer and any $0<\gamma<1$ if $n/p\in\mathbb{N}$.

I have two questions:

  1. Does this result extend to $U$ being an open subset with only Lipschitz boundary?

2.Does the result also holds $k\not\in\mathbb{N}$? The author doesn't mention anyway that $k$ should be an integer but I just wanted to check.

Thank you in advance.

share|cite|improve this question
up vote 1 down vote accepted

1- The Sobolev embedding is proved first in the case $U=\mathbb R^n$, and then for a general $U$ by using a right inverse $j$ of the restriction operator $W^{s,p}(\mathbb R^n)\rightarrow W^{k,p}(U)$. When $U$ is a half-space, a convenient $j$ is Babitch's extension. When the boundary is smooth, use an atlas to reduce to the previous case.

The situation is however not easy if the boundary is only Lipschitz. If $U$ is a (hyper-)rectangle, then $j$ can be build by applying $n-1$ times the Babitch operator. Thus the Sobolev embedding holds true in this case, whatever $\alpha$ is. Still true if the boundary is smooth, up to some kinks with angle $\frac{\pi}{m}$ with $m$ an integer. It is less clear for more general domains with Lipschitz boundary.

2- Sobolev embeddings hold even if $k$ is not an integer. The definition of the Sobolev space is then a bit more involved. What really matters is that $\beta:=k-\frac{n}{p}$ is not an integer. Then $\alpha=[\beta]$ and $\gamma=\beta-\alpha$. And of course regularity of the boundary.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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