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  "body": "\n## Definition\n\nLet $X$ be a [[normed vector space|03J5RAKB30]] over $\\R$ or $\\C$.\nThe space $X$ is *uniformly convex* if the following holds.\nFor every $\\varepsilon$ with $0 \u003c \\varepsilon \\le 2$ there is $\\delta \u003e 0$ such that for all $x, y \\in X$ with $\\norm{x} \\le 1$, $\\norm{y} \\le 1$ and $\\norm{x - y} \\ge \\varepsilon$,\n$$\\left\\| \\frac{x + y}{2} \\right\\| \\le 1 - \\delta.$$\n\n## Remarks\n\nExample: an inner product space is uniformly convex.\nBy the parallelogram law, for $\\norm{x}, \\norm{y} \\le 1$ and $\\norm{x - y} \\ge \\varepsilon$,\n$$\\left\\| \\frac{x + y}{2} \\right\\|^2 = \\frac{\\norm{x}^2 + \\norm{y}^2}{2} - \\left\\| \\frac{x - y}{2} \\right\\|^2 \\le 1 - \\frac{\\varepsilon^2}{4},$$\nso $\\delta = 1 - \\sqrt{1 - \\varepsilon^2/4}$ works.\nNon-example: $\\R^2$ with the norm $\\norm{(a, b)}_\\infty = \\max(\\abs{a}, \\abs{b})$.\nFor $x = (1, 1)$ and $y = (1, -1)$ one has $\\norm{x - y}_\\infty = 2$ but $\\norm{(x + y)/2}_\\infty = \\norm{(1, 0)}_\\infty = 1$, so no $\\delta \u003e 0$ works for $\\varepsilon = 2$.\nThe spaces $\\ell^p$ and $L^p$ for $1 \u003c p \u003c \\infty$ are uniformly convex by [[Clarkson's inequalities|1V11QM5C07]]; $\\ell^1$ and $\\ell^\\infty$ are not.\nUniform convexity depends on the norm, not only on the topology: $\\R^2$ is uniformly convex with the Euclidean norm and not with the norm $\\norm{\\cdot}_\\infty$.\n",
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