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  "body": "\n## Statement\n\nLet $a \u003c b$ and let $f \\colon [a, b] \\to \\R$ be [[continuous|0X3WGX95FE]].\nSuppose that\n$$\\int_a^b f(x) \\, \\eta'(x) \\, dx = 0$$\nfor every [[continuously differentiable|0VW90BBSJ4]] $\\eta \\colon [a, b] \\to \\R$ with $\\eta(a) = \\eta(b) = 0$.\nThen $f$ is constant on $[a, b]$.\n\n## Remarks\n\nLet $c = \\frac{1}{b - a} \\int_a^b f \\, dx$ be the mean of $f$, and set $\\eta(x) = \\int_a^x (f(s) - c) \\, ds$.\nThen $\\eta$ is $C^1$, $\\eta(a) = \\eta(b) = 0$ and $\\eta' = f - c$.\nThe hypothesis gives $\\int_a^b f (f - c) \\, dx = 0$, and $\\int_a^b c (f - c) \\, dx = 0$ by the choice of $c$.\nSubtracting, $\\int_a^b (f - c)^2 \\, dx = 0$, so $f = c$ by continuity.\n\nA common extension: if $f, g$ are continuous and $\\int_a^b (g \\eta + f \\eta') \\, dx = 0$ for all such $\\eta$, then $f$ is $C^1$ and $f' = g$.\nTo see it, apply the lemma to $f - G$, where $G(x) = \\int_a^x g(s) \\, ds$, after an integration by parts of $\\int_a^b g \\eta \\, dx$.\n\nExample: for $f(x) = x$ on $[0, 1]$ and $\\eta(x) = x(1 - x)$, one finds $\\int_0^1 x (1 - 2x) \\, dx = -1/6 \\ne 0$, as the lemma predicts for a non-constant $f$.\n\n## Explanations\n\n### L1 human\n\nIf $f$ gives zero against the slope of every function that starts and ends at zero, then $f$ cannot vary.\nThe reason is that one can build a test function whose slope is exactly $f$ minus its average.\n\n### L2 human\n\nThe extension above gives the Euler–Lagrange equation in integrated form, $L_p(x, y, y') = \\int_a^x L_z(s, y, y') \\, ds + C$, for a $C^1$ minimiser $y$, without assuming that $y$ has a second derivative.\nIt follows that $x \\mapsto L_p(x, y(x), y'(x))$ is $C^1$, which is the start of regularity theory in one variable.\nThe [[fundamental lemma|JNDZ86CV3Y]] is the version where $\\eta$ itself, instead of $\\eta'$, appears in the integral.\n",
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  "name": "du Bois-Reymond lemma",
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