# Comparaison of two version of Fractional sobolev spaces: what do we have $W^{s,p}(\mathbb{R}^{n})=H^{s,p}(\mathbb{R}^{n})$?

There are two version of Fractional sobolev spaces .

Definition1: (Via Galiardo semi-norm) Let $1\leq p\leq +\infty$, $0<s<1$ and $\Omega\subseteq \mathbb{R}^n$ an open set. The fractional Sobolev space $W^{s,p}(\Omega)$ is defined to be

$$W^{s,p}(\Omega) = \left\{ u\in L^p(\Omega) : \frac{|u(x)-u(y)|}{|x-y|^{\frac{n}{p} + s}} \in L^p(\Omega\times\Omega) \right\}$$

equipped with the norm

$$\|u\|_{W^{s,p}(\Omega)} = \left( \int_\Omega |u|^p \; dx + \int_\Omega\int_\Omega \frac{|u(x)-u(y)|^p}{|x-y|^{n+ sp}} \; dx dy \right)^{1/p}.$$

Definition2:(Via Fourier Transform) For $s\in\mathbb{R}$, $1<p<\infty$, and $n\geq 1$, define the Sobolev space $H^{s,p}(\mathbb{R}^{n})$ by $$H^{s,p}(\mathbb{R}^{n}):=\left\{f\in\mathcal{S}(\mathbb{R}^{n}) : \|(\langle{\xi}\rangle^{s}\widehat{f})^{\vee}\|_{L^{p}}<\infty\right\}$$, equipped with norm $$\|f\|_{H^{s,p}}=\|(\langle{\xi}\rangle^{s}\widehat{f})^{\vee}\|_{L^{p}}$$

Where, $$\langle{\xi}\rangle^{s} =(1+|\xi|^2)^s$$

From these definitions I have a couples of questions

1) What are the value of $p\in[1,\infty)$ such that $W^{s,p}(\mathbb{R}^{n})$ and $H^{s,p}(\mathbb{R}^{n})$ coincide?

I have found that this true for $p=2$ that is $$W^{s,2}(\mathbb{R}^{n})=H^{s,2}(\mathbb{R}^{n})$$

Do we still have equality or one side inclusion for some $p\neq 2$ if yes which one ? if no please provide me with some counter example or reference.

2) Next I would like to know what are the advantage and disadvantage using one the spaces $W^{s,p}(\mathbb{R}^{n})$ and $H^{s,p}(\mathbb{R}^{n})$.

I know that the definition $W^{s,p}(\Omega)$ makes sense on any domain which not the case for $H^{s,p}(\Omega)$ due to the lack of Fourier transforms.

• But you can easily construct $H^{s,p}(\Omega)$ spaces. (By the way, these are usually called "Bessel potential spaces" when $p\ne 2$). This is done for example in the book on wave equations by Struwe. Dec 16, 2017 at 17:17
• @GiuseppeNegro I think the Bessel potential is already implicitly involve in the definition of $H^{s,p}$ Dec 16, 2017 at 17:21
• Yes, that's true. I mean, if you do an Internet search try using the keyword "Bessel potential", you should find the answers you are looking for. Dec 16, 2017 at 17:23
• This seems to be basically the same as the question you posted on MO: mathoverflow.net/questions/288619/… In my opinion, this answer gives a very reasonable advice about cross-posting. Of course, you might have a look also at other discussions on meta about cross-posting. Dec 17, 2017 at 6:30