There are two versions 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 $$\langle{\xi}\rangle^{s} =(1+|\xi|^2)^{s/2}$$

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 $$W^{s,2}(\mathbb{R}^{n})=H^{s,2}(\mathbb{R}^{n})$$

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 counterexample or reference.

  1. 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 is not the case for $H^{s,p}(\Omega)$ due to the lack of Fourier transforms.

  • $\begingroup$ 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. $\endgroup$ Commented Dec 16, 2017 at 17:17
  • $\begingroup$ @ItaiBar-Natan in fact you are right it is the completion. The schwarz space is there only to insure that the Fourier makes sense:) $\endgroup$
    – Guy Fsone
    Commented Dec 16, 2017 at 17:23
  • 2
    $\begingroup$ This seems to be basically the same as the question you posted on math.SE: math.stackexchange.com/questions/2569423/… $\endgroup$ Commented Dec 17, 2017 at 6:29

1 Answer 1


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).


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