Comparison of two versions of fractional Sobolev spaces: do we have $W^{s,p}(\mathbb{R}^{n})=H^{s,p}(\mathbb{R}^{n})$?

There are two version of fractional Sobolev spaces.

Definition 1: (Via Gagliardo semi-norm) Let $1\leq p\leq +\infty$, $0<s<1$ and let $\Omega\subseteq \mathbb{R}^n$ be 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

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

Definition 2: (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 the 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 couple of questions.

1) What are the values 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 is true for $p=2$, that is that $$W^{s,2}(\mathbb{R}^{n})=H^{s,2}(\mathbb{R}^{n})$$

Do we still have equality or one sided 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 advantages and disadvantages of using one of 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.

• For $H^{s,p}$, did you mean to take the completion of this space? As currently defined, it is a subspace of Schwartz space that is not complete with the norm you gave it. – Itai Bar-Natan Dec 16 '17 at 17:17
• @ItaiBar-Natan in fact you are right it is the completion. The schwarz space is there only to insure that the Fourier makes sense:) – Guy Fsone Dec 16 '17 at 17:23
• This seems to be basically the same as the question you posted on math.SE: math.stackexchange.com/questions/2569423/… – Martin Sleziak Dec 17 '17 at 6:29

This is standard and well explained in several treatises: the space $W^{s,p}$ belongs to the scale of Besov spaces, while $H^{s,p}$ is in the scale of Triebel-Lizorkin spaces. The two scales satisfy mutual embeddings but the spaces concide only when they are $L^2$ based, while they are different when $p\neq 2$ (provided $s$ is not an integer). A good reference is the texbook by Runst and Sickel, Sobolev Spaces of Fractional order; or Triebel, Interpolation spaces (where the case of domains is analyzed in detail).