# Why the properties of $W^{m,p}(\Omega)$ cannot be extended to $W^{m,p}(\mathbb{R}^d)$

The text books on the topic of Sobolev Spaces and PDE etc., they treate the case $W^{m,p}(\Omega)$ with $\Omega\subset \mathbb{R}^d$ and $\Omega = \mathbb{R}^d \text{ or }\mathbb{R}^d_+$ separately.

And some results cannot be extended from the first case to the latter one.

Could anyone summarize the reasons why the extension is not generally valid?

I always find it difficult to wrap up the main result of big theorems after going through many proofs of lemmas, propositions.

• The main difference may be that $\Omega$ is bounded and $\mathbb R^n$ is not. As a consequence, $L^2(\Omega)$ embeds into $L^1(\Omega)$, and the same does not hold for $\mathbb R^n$. – gerw Apr 11 '13 at 18:25
• @grew It is not explicitly stated that $\Omega$ is bounded. The only requirement is that $\Omega$ is a open set in $\mathbb{R}^n$. – newbie Apr 11 '13 at 18:31

## 1 Answer

One of the reasons could be that you should be carefull when you deal with extentions of functions in Sobolev spaces: consider $f(x) = \chi_{(0,1)}(x)$. $f \in W^{1,1}((0,1))$ for example, but $f \notin W^{1,1}(\mathbb{R})$ because it has two jump discontinuities.