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  "body": "\n## Definition\n\nLet $X$ be a [[Banach space|06GZ956657]] and let $L_n \\colon X \\to X$, $n \\ge 1$, be [[continuous linear maps|03TM135G0N]].\nLet $\\varphi(n) \u003e 0$ with $\\varphi(n) \\to 0$.\nThe sequence $(L_n)$ is *saturated* with *saturation order* $\\varphi(n)$ if both conditions hold.\n\n1. If $f \\in X$ and $\\norm{L_n f - f} = o(\\varphi(n))$, then $L_n f = f$ for all $n$.\n2. Some $f \\in X$ with $L_n f \\ne f$ for some $n$ satisfies $\\norm{L_n f - f} = O(\\varphi(n))$.\n\nThe *saturation class* is the set of $f \\in X$ with $\\norm{L_n f - f} = O(\\varphi(n))$, in [[big O notation|0TGV80E6A4]].\n\n## Remarks\n\nExample: the Bernstein operators $B_n$ on $C[0, 1]$ with the uniform norm are saturated with order $1/n$.\nFor $f(x) = x^2$ one has $B_n f(x) = x^2 + \\frac{x(1 - x)}{n}$, so $\\norm{B_n f - f} = \\frac{1}{4n}$, which is $O(1/n)$ but not $o(1/n)$.\nThe only $f$ with $\\norm{B_n f - f} = o(1/n)$ are the linear functions, which $B_n$ reproduces.\n\nNon-example: the Fourier partial sum operators $S_n$ on $L^2$ of the circle are not saturated for any order.\nA trigonometric polynomial $f$ of degree $k \\ge 1$ has $S_n f = f$ for $n \\ge k$, so $\\norm{S_n f - f} = o(\\varphi(n))$, but $S_0 f \\ne f$, so clause 1 fails.\n\nThe Fejér means of Fourier series are saturated with order $1/n$ in $C$ of the circle, and only constants are approximated with error $o(1/n)$.\nSaturation theorems typically identify the saturation class with a smoothness class, such as functions with Lipschitz first derivative for Bernstein operators.\n",
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