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  "body": "\n## Definition\n\nLet $U \\subseteq \\R^n$ be open, let $1 \\le p \u003c \\infty$, and let $0 \u003c \\sigma \u003c 1$.\nThe *fractional Sobolev space* $W^{\\sigma,p}(U)$, or Sobolev–Slobodeckij space, is the set of $u \\in L^p(U)$ with finite Gagliardo seminorm\n$$[u]_{\\sigma,p,U} = \\Big( \\int_U \\int_U \\frac{\\abs{u(x) - u(y)}^p}{\\abs{x - y}^{n + \\sigma p}} \\, dx \\, dy \\Big)^{1/p}.$$\nHere $L^p(U)$ is the [[Lp space|SASP14MA1Z]].\nIt carries the norm $\\norm{u}_{W^{\\sigma,p}(U)} = \\big( \\norm{u}_{L^p(U)}^p + [u]_{\\sigma,p,U}^p \\big)^{1/p}$.\nLet $s = k + \\sigma$ with $k \\ge 1$ an integer and $0 \u003c \\sigma \u003c 1$.\nThe 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 $\\abs{\\alpha} = k$.\n\n## Remarks\n\nEach $W^{s,p}(U)$ is a Banach space.\nOn $\\R^n$ 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]].\nFor $0 \u003c s$ not an integer, $W^{s,2}(\\R^n) = H^s(\\R^n)$ with equivalent norms; for $0 \u003c \\sigma \u003c 1$ 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 $c = c(n, \\sigma) \u003e 0$.\nFor $s$ not an integer, $W^{s,p}(\\R^n)$ is the [[Besov space|95QBBPRJ9E]] $B^s_{p,p}(\\R^n)$.\nNegative orders are defined by duality, as spaces of [[tempered distributions|AES2T5JPFP]] or of distributions.\n\nExample: on $U = (-1, 1)$, let $u = 1$ on $(0, 1)$ and $u = 0$ on $(-1, 0)$, and let $0 \u003c \\sigma \u003c 1/2$.\nThe double integral is $2 \\int_0^1 \\int_{-1}^0 \\abs{x - y}^{-1 - 2\\sigma} \\, dy \\, dx$, which is finite because $\\int_0^1 \\int_0^1 (a + b)^{-1 - 2\\sigma} \\, da \\, db \u003c \\infty$ when $2\\sigma \u003c 1$.\nSo $u \\in W^{\\sigma,2}(-1, 1)$.\n\nNon-example: the same $u$ is not in $W^{1/2,2}(-1, 1)$.\nFor $\\sigma = 1/2$ the inner integral $\\int_{-1}^{0} (x - y)^{-2} \\, dy$ equals $\\frac{1}{x} - \\frac{1}{x + 1} \\ge \\frac{1}{x} - 1$, and $\\int_0^1 (\\frac{1}{x} - 1) \\, dx = \\infty$.\n",
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