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  "body": "\n## Statement\n\nLet $X$ be a real [[normed space|03J5RAKB30]] with [[continuous dual space|4A82S5FVRT]] $X^*$.\nLet $f \\colon X \\to (-\\infty, +\\infty]$ be a function that is not identically $+\\infty$, and let $f^*$ be its [[Fenchel conjugate|T7V2VPY61S]].\nThen for all $x \\in X$ and $y \\in X^*$,\n$$f(x) + f^*(y) \\ge y(x).$$\nMoreover, let $x$ be a point with $f(x)$ finite.\nThen equality holds if and only if $y$ is a subgradient of $f$ at $x$, that is, $f(z) \\ge f(x) + y(z - x)$ for all $z \\in X$.\n\n## Remarks\n\nSince $f$ is finite somewhere, $f^*$ never takes the value $-\\infty$, so the left side is a well-defined element of $(-\\infty, +\\infty]$.\nThe proof is one line: $f^*(y) \\ge y(x) - f(x)$ by the definition of $f^*$ as a supremum over $x$.\nEquality means that $x$ attains this supremum, which is the subgradient inequality rearranged.\nThe set of subgradients is the [[subdifferential|9C5WXYYQEK]] $\\partial f(x)$.\n\nExample: for $p \u003e 1$ and $q = p/(p - 1)$, the function $f(x) = \\abs{x}^p / p$ on $\\R$ has $f^*(y) = \\abs{y}^q / q$.\nThe inequality becomes Young's inequality $xy \\le \\abs{x}^p/p + \\abs{y}^q/q$, with equality exactly when $y = \\sgn(x) \\abs{x}^{p-1}$.\nFor $p = 2$ it is $xy \\le \\tfrac12 x^2 + \\tfrac12 y^2$, which is $(x - y)^2 \\ge 0$.\n\n## Explanations\n\n### L1 human\n\nThe conjugate $f^*$ is built so that this inequality holds: it is the smallest function with $f(x) + f^*(y) \\ge xy$ for all $x$.\nIt generalizes the inequality $xy \\le \\tfrac12 x^2 + \\tfrac12 y^2$ and Young's inequality for products.\nEquality picks out the pairs where $y$ is a slope of $f$ at $x$.\n\n### L2 human\n\nThe inequality gives weak duality in convex optimization: the value of a dual problem never exceeds the value of the primal problem.\nIts equality case links the conjugate with the subdifferential: $y \\in \\partial f(x)$ if and only if $f(x) + f^*(y) = y(x)$.\nFor proper convex lower semicontinuous $f$, the Fenchel–Moreau theorem then gives the symmetric statement $x \\in \\partial f^*(y)$, so $\\partial f^*$ is the inverse of $\\partial f$.\nIt is the source of many classical inequalities obtained by choosing $f$.\n",
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  "msc": [
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  "name": "Fenchel–Young inequality",
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