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  "body": "\n## Definition\n\nLet $\\psi \\in L^2(\\R)$ (see [[Lp space|SASP14MA1Z]]).\nFor $j, k \\in \\Z$ put\n$$\\psi_{j,k}(x) = 2^{j/2} \\, \\psi(2^j x - k).$$\nThe function $\\psi$ is an *orthonormal wavelet* if the family $(\\psi_{j,k})_{j, k \\in \\Z}$ is an [[orthonormal basis|0S9GGTR0EX]] of $L^2(\\R)$.\n\n## Remarks\n\nThe factor $2^{j/2}$ makes $\\norm{\\psi_{j,k}}_2 = \\norm{\\psi}_2$.\nThen every $f \\in L^2(\\R)$ has the expansion $f = \\sum_{j,k} \\ip{f}{\\psi_{j,k}} \\psi_{j,k}$, convergent in $L^2(\\R)$.\nIf $\\psi$ is also integrable, then $\\int_\\R \\psi(x) \\, dx = 0$.\nWeaker notions are also called wavelets: $\\psi$ such that $(\\psi_{j,k})$ is a Riesz basis or a frame, and the admissible functions of the continuous wavelet transform.\n\nExample: the [[Haar wavelet|JW5PA9XA7F]] $\\psi = 1_{[0, 1/2)} - 1_{[1/2, 1)}$ is an orthonormal wavelet.\nFor instance, $\\psi_{0,0}$ and $\\psi_{1,0}$ are orthogonal, because $\\psi_{1,0}$ lives on $[0, \\tfrac12)$, where $\\psi_{0,0} = 1$, and has integral $0$.\n\nNon-example: the Gaussian $\\psi(x) = e^{-\\pi x^2}$ is not an orthonormal wavelet.\nIts translates $\\psi_{0,0}$ and $\\psi_{0,1}$ are both positive, so $\\ip{\\psi_{0,0}}{\\psi_{0,1}} \u003e 0$ and they are not orthogonal.\n",
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