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  "body": "\n## Definition\n\nLet $I$ be an open [[interval|0TK8Y3J5RW]] of $\\R$, and let $f \\colon I \\to \\R$ be a [[differentiable function|0A61B60B5R]] whose derivative $f'$ is [[strictly increasing|011EJQWWJS]] on $I$.\nLet $J = f'(I)$; it is an interval, and $f' \\colon I \\to J$ is a bijection with inverse $g \\colon J \\to I$.\nThe *Legendre transform* of $f$ is the function $f^* \\colon J \\to \\R$ given by\n$$f^*(p) = p \\, g(p) - f(g(p)).$$\nThe map $f \\mapsto f^*$ is the *Legendre transformation*.\n\n## Remarks\n\nFor $p \\in J$ one has $f^*(p) = \\sup_{x \\in I} \\big(p x - f(x)\\big)$, and the [[supremum|0BTMQ04556]] is attained at $x = g(p)$.\nIndeed $x \\mapsto p x - f(x)$ has derivative $p - f'(x)$, which is positive for $x \u003c g(p)$ and negative for $x \u003e g(p)$.\nThe function $f^*$ is differentiable with $(f^*)' = g = (f')^{-1}$, and $(f^*)^* = f$.\nThe supremum formula, taken over all $x$, defines the convex conjugate of any function, which extends this definition.\n\nExample: $f(x) = e^x$ on $\\R$.\nHere $f'(x) = e^x$ is strictly increasing, $J = (0, \\infty)$ and $g(p) = \\ln p$.\nSo $f^*(p) = p \\ln p - p$ for $p \u003e 0$.\n\nNon-example: $f(x) = x^3$ on $\\R$.\nIts derivative $3x^2$ is not strictly increasing, since $3 \\cdot (-1)^2 = 3 \\cdot 1^2$, so the definition does not apply.\n",
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