---
tag: A7Q6581B1F
kind: definition
name: Fractional Sobolev space
msc: ["46E35", 35R11]
defines: N6NX1VT837
uses: [M4CEZ5K9HP, SASP14MA1Z, DPVM34ADZV, AES2T5JPFP, 95QBBPRJ9E]
sources:
  - url: https://en.wikipedia.org/wiki/Sobolev_space
---

## Definition

Let $U \subseteq \R^n$ be open, let $1 \le p < \infty$, and let $0 < \sigma < 1$.
The *fractional Sobolev space* $W^{\sigma,p}(U)$, or Sobolev–Slobodeckij space, is the set of $u \in L^p(U)$ with finite Gagliardo seminorm
$$[u]_{\sigma,p,U} = \Big( \int_U \int_U \frac{\abs{u(x) - u(y)}^p}{\abs{x - y}^{n + \sigma p}} \, dx \, dy \Big)^{1/p}.$$
Here $L^p(U)$ is the [[Lp space|SASP14MA1Z]].
It carries the norm $\norm{u}_{W^{\sigma,p}(U)} = \big( \norm{u}_{L^p(U)}^p + [u]_{\sigma,p,U}^p \big)^{1/p}$.
Let $s = k + \sigma$ with $k \ge 1$ an integer and $0 < \sigma < 1$.
The space $W^{s,p}(U)$ is the set of $u$ in the [[Sobolev space|M4CEZ5K9HP]] $W^{k,p}(U)$ with $D^\alpha u \in W^{\sigma,p}(U)$ whenever $\abs{\alpha} = k$.

## Remarks

Each $W^{s,p}(U)$ is a Banach space.
On $\R^n$ and for every real $s \ge 0$, one also defines $H^s(\R^n)$ as the set of $u \in L^2(\R^n)$ with $(1 + \abs{\xi}^2)^{s/2} \hat{u} \in L^2(\R^n)$, where $\hat u$ is the [[Fourier transform|DPVM34ADZV]].
For $0 < s$ not an integer, $W^{s,2}(\R^n) = H^s(\R^n)$ with equivalent norms; for $0 < \sigma < 1$ this follows from Plancherel's theorem, since $[u]_{\sigma,2,\R^n}^2 = c \int \abs{\xi}^{2\sigma} \abs{\hat u(\xi)}^2 \, d\xi$ for a constant $c = c(n, \sigma) > 0$.
For $s$ not an integer, $W^{s,p}(\R^n)$ is the [[Besov space|95QBBPRJ9E]] $B^s_{p,p}(\R^n)$.
Negative orders are defined by duality, as spaces of [[tempered distributions|AES2T5JPFP]] or of distributions.

Example: on $U = (-1, 1)$, let $u = 1$ on $(0, 1)$ and $u = 0$ on $(-1, 0)$, and let $0 < \sigma < 1/2$.
The double integral is $2 \int_0^1 \int_{-1}^0 \abs{x - y}^{-1 - 2\sigma} \, dy \, dx$, which is finite because $\int_0^1 \int_0^1 (a + b)^{-1 - 2\sigma} \, da \, db < \infty$ when $2\sigma < 1$.
So $u \in W^{\sigma,2}(-1, 1)$.

Non-example: the same $u$ is not in $W^{1/2,2}(-1, 1)$.
For $\sigma = 1/2$ the inner integral $\int_{-1}^{0} (x - y)^{-2} \, dy$ equals $\frac{1}{x} - \frac{1}{x + 1} \ge \frac{1}{x} - 1$, and $\int_0^1 (\frac{1}{x} - 1) \, dx = \infty$.