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  "body": "\n## Definition\n\nLet $n \\ge 1$ and $\\alpha \u003e 0$.\nUse the [[Fourier transform|DPVM34ADZV]] $\\hat g(\\xi) = \\int_{\\R^n} g(x) e^{-2\\pi i x \\cdot \\xi} \\, dx$.\nLet $G_\\alpha$ be the integrable function on $\\R^n$ whose Fourier transform is\n$$\\widehat{G_\\alpha}(\\xi) = (1 + 4 \\pi^2 \\abs{\\xi}^2)^{-\\alpha/2}.$$\nFor $f \\in L^p(\\R^n)$ with $1 \\le p \\le \\infty$, the *Bessel potential* of order $\\alpha$ of $f$ is the [[convolution|2Z29JN3VES]]\n$$J_\\alpha f = G_\\alpha * f.$$\n\n## Remarks\n\nThe kernel $G_\\alpha$ exists, is positive, and has integral $\\widehat{G_\\alpha}(0) = 1$.\nSo $J_\\alpha$ is a bounded operator on each $L^p$ with norm at most $1$, by Young's inequality.\nOn Schwartz functions $J_\\alpha = (I - \\Delta)^{-\\alpha/2}$, where $\\Delta$ is the [[Laplacian|0E0B6RA8K0]], since $-\\Delta$ has symbol $4 \\pi^2 \\abs{\\xi}^2$.\nThe semigroup law $J_\\alpha J_\\beta = J_{\\alpha + \\beta}$ holds for all $\\alpha, \\beta \u003e 0$.\nNear $0$, $G_\\alpha(x)$ is comparable to $\\abs{x}^{\\alpha - n}$ when $0 \u003c \\alpha \u003c n$, and $G_\\alpha(x)$ decays like $e^{-c\\abs{x}}$ for some $c \u003e 0$ as $\\abs{x} \\to \\infty$.\n\nExample: for $n = 1$ and $\\alpha = 2$, the kernel is $G_2(x) = \\frac12 e^{-\\abs{x}}$.\nCheck: $\\int_{\\R} \\frac12 e^{-\\abs{x}} e^{-2\\pi i x \\xi} \\, dx = \\frac12 \\cdot \\frac{2}{1 + 4\\pi^2 \\xi^2} = (1 + 4\\pi^2 \\xi^2)^{-1}$.\nSo $u = J_2 f$ solves $u - u'' = f$ for nice $f$.\n\nNon-example: the kernel $\\abs{x}^{\\alpha - n}$ of the Riesz potential is not the Bessel kernel, since it is not integrable on $\\R^n$, while $G_\\alpha$ must be integrable with integral $1$.\n",
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