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  "body": "\n## Statement\n\nLet $(\\Omega, \\Sigma, \\mu)$ be a [[measure space|P15VQS5DJ9]] and let $f, g \\in L^p(\\mu)$, the [[Lp space|SASP14MA1Z]].\nWrite $\\norm{\\cdot}$ for the norm of $L^p(\\mu)$.\n\n1. If $2 \\le p \u003c \\infty$, then\n$$\\Big\\| \\frac{f + g}{2} \\Big\\|^p + \\Big\\| \\frac{f - g}{2} \\Big\\|^p \\le \\frac{1}{2} \\big( \\norm{f}^p + \\norm{g}^p \\big).$$\n2. If $1 \u003c p \\le 2$ and $q = p/(p-1)$, then\n$$\\Big\\| \\frac{f + g}{2} \\Big\\|^q + \\Big\\| \\frac{f - g}{2} \\Big\\|^q \\le \\Big( \\frac{1}{2} \\norm{f}^p + \\frac{1}{2} \\norm{g}^p \\Big)^{q/p}.$$\n\n## Remarks\n\nThe first inequality follows from the same inequality for numbers $a, b$, integrated over $\\Omega$.\nFor numbers, with $\\alpha = (a+b)/2$ and $\\beta = (a-b)/2$, one has $\\abs{\\alpha}^p + \\abs{\\beta}^p \\le (\\abs{\\alpha}^2 + \\abs{\\beta}^2)^{p/2}$ since $p \\ge 2$.\nThe [[parallelogram law|026HPGYGS4]] gives $\\abs{\\alpha}^2 + \\abs{\\beta}^2 = (\\abs{a}^2 + \\abs{b}^2)/2$.\nConvexity of $t \\mapsto t^{p/2}$ then bounds $((\\abs{a}^2 + \\abs{b}^2)/2)^{p/2}$ by $(\\abs{a}^p + \\abs{b}^p)/2$.\nThe second inequality needs a finer argument.\n\nFor $p = 2$ both inequalities are the parallelogram law, with equality.\nClarkson proved them in 1936 to show that $L^p(\\mu)$ is [[uniformly convex|BEAMC5011M]] for $1 \u003c p \u003c \\infty$.\nIf $\\norm{f} = \\norm{g} = 1$ and $\\norm{f - g} \\ge \\varepsilon$, then for $p \\ge 2$ the first inequality gives $\\norm{(f+g)/2}^p \\le 1 - (\\varepsilon/2)^p$.\n\n## Explanations\n\n### L1 human\n\nIn a space of functions with the $L^p$ norm, two unit vectors that are far apart have a midpoint that is shorter than $1$ by a definite amount.\nClarkson's inequalities make this precise with explicit numbers.\nThey generalise the parallelogram law of Euclidean geometry.\n\n### L2 human\n\nThe inequalities are the standard route to uniform convexity of $L^p$ for $1 \u003c p \u003c \\infty$.\nBy the Milman–Pettis theorem, uniform convexity then gives reflexivity of $L^p$.\nThrough uniform convexity, they can also be used to prove the duality of $L^p$ and $L^q$ without the Radon–Nikodym theorem.\nThe spaces $L^1$ and $L^\\infty$ are not uniformly convex, and the inequalities have no analogue there.\n",
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