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  "body": "\n## Definition\n\nLet $X$ and $Y$ be real [[vector spaces|000AYPA9AJ]], each with a preorder $\\le$, such as a [[partial order|0QV2JS9RJ3]], compatible with the vector operations.\nCompatible means that $x \\le y$ implies $x + z \\le y + z$ and $tx \\le ty$ for all vectors $z$ and all real $t \\ge 0$.\nA [[linear map|0NRYHJPQG2]] $L \\colon X \\to Y$ is a *positive linear operator* if $x \\ge 0$ implies $L x \\ge 0$ for every $x \\in X$.\n\n## Remarks\n\nEquivalently, $L$ is monotone: $x \\le y$ implies $L x \\le L y$, since $L y - L x = L(y - x)$.\n\nExample: on $C[0, 1]$ with the pointwise order $f \\le g$ when $f(x) \\le g(x)$ for all $x$, the Bernstein operator\n$$B_n f(x) = \\sum_{k=0}^{n} \\binom{n}{k} x^k (1 - x)^{n-k} f(k/n)$$\nis positive, because all weights $\\binom{n}{k} x^k (1 - x)^{n-k}$ are nonnegative on $[0, 1]$.\n\nNon-example: differentiation $D \\colon C^1[0, 1] \\to C[0, 1]$ is linear but not positive.\nThe function $f(x) = 1 - x$ satisfies $f \\ge 0$, while $D f = -1 \u003c 0$.\n\nFor a compact Hausdorff space $K$ and a positive linear operator $L \\colon C(K) \\to C(K)$ with the uniform norm, $L$ is bounded with $\\norm{L} = \\norm{L 1}$.\nIndeed $-\\norm{f} \\cdot 1 \\le f \\le \\norm{f} \\cdot 1$ gives $\\abs{L f} \\le \\norm{f} \\, L 1$.\nA positive linear functional is the case $Y = \\R$; on $C(K)$ these are the integrals against finite positive measures.\n",
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