Fractional Sobolev space

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. 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 $B^s_{p,p}(\R^n)$. Negative orders are defined by duality, as spaces of tempered distributions 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 $\int_0^1 (\frac{1}{x} - 1) \, dx = \infty$.