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  "body": "\n## Definition\n\nLet $H$ be a [[Hilbert space|HM2HHE3J51]] with [[inner product|0NB0AWGJBW]] $\\ip{\\cdot}{\\cdot}$.\nLet $(f_i)_{i \\in I}$ be a *frame* of $H$: there are constants $0 \u003c A \\le B \u003c \\infty$ with $A \\norm{f}^2 \\le \\sum_{i \\in I} \\abs{\\ip{f}{f_i}}^2 \\le B \\norm{f}^2$ for all $f \\in H$.\nThe *frame operator* of $(f_i)_{i \\in I}$ is the map $S \\colon H \\to H$,\n$$S f = \\sum_{i \\in I} \\ip{f}{f_i} f_i,$$\nwhere the series converges unconditionally in $H$.\n\n## Remarks\n\n$S$ is bounded, [[self-adjoint|0NEGN5CRSC]] and [[positive|77020CY5ZW]], since $\\ip{Sf}{f} = \\sum_i \\abs{\\ip{f}{f_i}}^2$.\nThe frame inequality reads $A I \\le S \\le B I$, so $S$ is [[invertible|CPAD5W1CGW]] with $\\norm{S^{-1}} \\le A^{-1}$.\nThe family $(S^{-1} f_i)_{i \\in I}$ is the *canonical dual frame*, and $f = \\sum_i \\ip{f}{S^{-1} f_i} f_i = \\sum_i \\ip{f}{f_i} S^{-1} f_i$ for all $f$.\nWith the analysis operator $T f = (\\ip{f}{f_i})_{i \\in I} \\in \\ell^2(I)$, one has $S = T^* T$.\nA frame is tight with bound $A$ exactly when $S = A I$.\n\nExample: for an orthonormal basis $(e_i)$, $S f = \\sum_i \\ip{f}{e_i} e_i = f$, so $S = I$.\n\nExample: in $\\C^2$ with the frame $(e_1, e_1, e_2)$, $S(x, y) = 2x e_1 + y e_2$, so $S = \\mathrm{diag}(2, 1)$, with spectrum $\\set{1, 2}$ equal to the optimal frame bounds.\n\nNon-example: in $\\ell^2(\\N)$ the family $(e_n / n)_{n \\ge 1}$ satisfies only the upper bound.\nThe map $f \\mapsto \\sum_n \\ip{f}{e_n/n} e_n/n$ is still defined, but it sends $e_n$ to $e_n/n^2$, so it has no bounded inverse; the family is not a frame and the reconstruction formula fails.\n",
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