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  "body": "\n## Definition\n\nLet $a \u003c b$, let $n \\ge 0$, and let $x_0 \u003c x_1 \u003c \\cdots \u003c x_n$ be points of $[a, b]$.\nLet $\\ell_0, \\dots, \\ell_n$ be the [[Lagrange basis polynomials|DA3CQJ5MJ6]] of these nodes, so that $\\ell_j(x_i) = 1$ if $i = j$ and $0$ otherwise.\nThe *Lebesgue function* of the nodes is $\\lambda_n(x) = \\sum_{j=0}^{n} \\abs{\\ell_j(x)}$.\nThe *Lebesgue constant* of the nodes is\n$$\\Lambda_n = \\max_{a \\le x \\le b} \\lambda_n(x).$$\n\n## Remarks\n\nLet $I_n \\colon C[a, b] \\to C[a, b]$ send $f$ to its [[Lagrange interpolation|237KR6FRCK]] polynomial $\\sum_j f(x_j) \\ell_j$.\nWith the uniform norm, $\\Lambda_n$ is the [[operator norm|06EKW1AKXF]] of $I_n$.\n\nExample: nodes $-1, 0, 1$ in $[-1, 1]$.\nHere $\\ell_0 = x(x - 1)/2$, $\\ell_1 = 1 - x^2$ and $\\ell_2 = x(x + 1)/2$.\nFor $0 \\le x \\le 1$ this gives $\\lambda_2(x) = 1 + x - x^2$, and by symmetry $\\Lambda_2 = \\lambda_2(1/2) = 5/4$.\n\nNon-example: the number $\\max_x \\abs{\\sum_j \\ell_j(x)}$ is not the Lebesgue constant; it always equals $1$, because $\\sum_j \\ell_j = 1$.\nThe absolute values must be taken term by term.\n\nAlways $\\Lambda_n \\ge 1$.\nFor equally spaced nodes $\\Lambda_n$ grows like $2^{n+1} / (e \\, n \\log n)$.\nFor Chebyshev nodes $\\Lambda_n = \\frac{2}{\\pi} \\log n + O(1)$, and no choice of nodes does better than $\\frac{2}{\\pi} \\log n - C$ for an absolute constant $C$.\n",
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