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  "body": "\n## Definition\n\nLet $X$ be a [[normed vector space|03J5RAKB30]] over $\\R$ or $\\C$.\nThe space $X$ is *strictly convex* if for all $x, y \\in X$ with $\\norm{x} = \\norm{y} = 1$ and $x \\ne y$,\n$$\\left\\| \\frac{x + y}{2} \\right\\| \u003c 1.$$\n\n## Remarks\n\nEquivalent forms: the unit sphere contains no nondegenerate line segment; and $\\norm{x + y} = \\norm{x} + \\norm{y}$ with $x, y \\ne 0$ implies $y = c x$ for some real $c \u003e 0$.\nExample: $\\R^2$ with the Euclidean norm.\nIf $\\norm{x} = \\norm{y} = 1$ and $x \\ne y$, the parallelogram law gives $\\norm{(x + y)/2}^2 = 1 - \\norm{(x - y)/2}^2 \u003c 1$.\nNon-example: $\\R^2$ with $\\norm{(a, b)}_1 = \\abs{a} + \\abs{b}$.\nFor $x = (1, 0)$ and $y = (0, 1)$ the midpoint $(1/2, 1/2)$ has norm $1$.\nEvery [[uniformly convex space|BEAMC5011M]] is strictly convex, but not conversely.\nFor instance, $\\ell^1$ with the norm $\\norm{x}_1 + \\norm{x}_2$ is strictly convex.\nIndeed, equality in the triangle inequality forces equality for $\\norm{\\cdot}_2$.\nBut this space is not reflexive, so it is not uniformly convex by the [[Milman–Pettis theorem|AJ5RG9XQYZ]].\n",
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