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  "body": "\n## Definition\n\nLet $n \\ge 1$, let $s \\in \\R$ and let $1 \u003c p \u003c \\infty$.\nLet $\\mathcal{F}$ denote the [[Fourier transform|DPVM34ADZV]] on the space $\\mathcal{S}'(\\R^n)$ of [[tempered distributions|AES2T5JPFP]].\nThe *Bessel potential space* $H^{s,p}(\\R^n)$ is the set of all $f \\in \\mathcal{S}'(\\R^n)$ such that\n$$J^s f = \\mathcal{F}^{-1}\\big[ (1 + \\abs{\\xi}^2)^{s/2} \\, \\mathcal{F} f \\big]$$\nis a function in the [[Lebesgue space|SASP14MA1Z]] $L^p(\\R^n)$.\nIts norm is $\\norm{f}_{H^{s,p}} = \\norm{J^s f}_{L^p}$.\n\n## Remarks\n\nThe operator $J^{-s} = (I - \\Delta)^{-s/2}$ is the *Bessel potential* of order $s$.\nFor $s \u003e 0$ it is convolution with an integrable kernel $G_s$ whose Fourier transform is $(1 + \\abs{\\xi}^2)^{-s/2}$.\nSo for $s \u003e 0$, $H^{s,p}(\\R^n)$ is the set of functions $G_s * g$ with $g \\in L^p(\\R^n)$, and $\\norm{G_s * g}_{H^{s,p}} = \\norm{g}_{L^p}$.\n\nExample: for $s = 0$ the multiplier is $1$, so $J^0 f = f$ and $H^{0,p}(\\R^n) = L^p(\\R^n)$ with the same norm.\n\nNon-example: the constant function $f = 1$ is a tempered distribution, but it is not in any $H^{s,p}(\\R^n)$.\nIts Fourier transform is a multiple of the Dirac mass at $0$, and the multiplier equals $1$ at $\\xi = 0$.\nSo $J^s 1 = 1$, which is not in $L^p(\\R^n)$ for $p \u003c \\infty$.\n\nFor an integer $k \\ge 0$ and $1 \u003c p \u003c \\infty$, $H^{k,p}(\\R^n)$ equals the [[Sobolev space|M4CEZ5K9HP]] $W^{k,p}(\\R^n)$ with an equivalent norm.\nThe proof of this uses the boundedness of the Riesz transforms on $L^p$.\nFor $p = 2$, Plancherel's theorem gives $H^{s,2}(\\R^n) = H^s(\\R^n)$, the [[fractional Sobolev space|N6NX1VT837]].\nOther notations are $L^p_s$, $H^s_p$ and $W^{s,p}$.\n",
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