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  "body": "\n## Definition\n\nLet $n \\ge 1$.\nLet $\\psi_0$ be a function in the [[Schwartz space|CG9JEATW3Q]] $\\mathcal{S}(\\R^n)$ with $\\psi_0(\\xi) = 1$ for $\\abs{\\xi} \\le 1$ and $\\psi_0(\\xi) = 0$ for $\\abs{\\xi} \\ge 2$.\nFor $j \\ge 1$ let $\\psi_j(\\xi) = \\psi_0(2^{-j} \\xi) - \\psi_0(2^{-j+1} \\xi)$, so that $\\sum_{j \\ge 0} \\psi_j = 1$ and $\\psi_j$ is supported in $2^{j-1} \\le \\abs{\\xi} \\le 2^{j+1}$.\nFor a [[tempered distribution|AES2T5JPFP]] $f$, let $\\Delta_j f = \\mathcal{F}^{-1}(\\psi_j \\, \\mathcal{F} f)$, where $\\mathcal{F}$ is the [[Fourier transform|DPVM34ADZV]]; each $\\Delta_j f$ is a smooth function.\nLet $s \\in \\R$ and $0 \u003c p, q \\le \\infty$.\nThe *Besov space* $B^s_{p,q}(\\R^n)$ is the set of tempered distributions $f$ with\n$$\\norm{f}_{B^s_{p,q}} = \\Big( \\sum_{j=0}^{\\infty} 2^{jsq} \\norm{\\Delta_j f}_{L^p(\\R^n)}^q \\Big)^{1/q} \u003c \\infty,$$\nwith the usual change to $\\sup_{j \\ge 0} 2^{js} \\norm{\\Delta_j f}_{L^p}$ when $q = \\infty$.\n\n## Remarks\n\nA different choice of $\\psi_0$ gives the same space with an equivalent quasi-norm.\nFor $p, q \\ge 1$ the space is a Banach space; otherwise it is a complete quasi-normed space.\nHere $L^p(\\R^n)$ is the [[Lp space|SASP14MA1Z]], with the same formula when $0 \u003c p \u003c 1$.\n\nExample: $B^s_{2,2}(\\R^n)$ is the space $H^s(\\R^n)$ of tempered distributions with $(1 + \\abs{\\xi}^2)^{s/2} \\hat f \\in L^2$.\nIndeed, by Plancherel, $\\norm{\\Delta_j f}_{L^2}^2 = \\int \\psi_j^2 \\abs{\\hat f}^2 \\, d\\xi$ up to a constant factor, and $2^{2js}$ is comparable with $(1 + \\abs{\\xi}^2)^s$ on the support of $\\psi_j$.\n\nNon-example: the Dirac delta $\\delta$ on $\\R^n$ is not in $B^0_{1,q}(\\R^n)$ for $q \u003c \\infty$.\nHere $\\Delta_j \\delta = \\mathcal{F}^{-1} \\psi_j$, and for $j \\ge 1$ this function is $x \\mapsto 2^{(j-1)n} (\\mathcal{F}^{-1} \\psi_1)(2^{j-1} x)$.\nSo $\\norm{\\Delta_j \\delta}_{L^1}$ is the same nonzero number for all $j \\ge 1$, and the sum defining the norm diverges.\nThe same computation shows that $\\delta \\in B^0_{1,\\infty}(\\R^n)$.\n\nFor $s \u003e 0$ not an integer, $B^s_{\\infty,\\infty}(\\R^n)$ is the Hölder space $C^{k,\\alpha}(\\R^n)$ with $k = \\floor{s}$ and $\\alpha = s - k$.\n",
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