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  "body": "\n## Definition\n\nLet $V$ be a real [[topological vector space|T32W553W95]], such as $\\R^n$, and let $C \\subseteq V$.\nThe set $C$ is *strictly convex* if for all $x, y \\in C$ with $x \\ne y$, the [[open segment|CCZA91NE5X]] $(x, y)$ lies in the [[interior|0RZH6W98HK]] of $C$.\n\n## Remarks\n\nA strictly convex set is [[convex|02AS45T3H9]], since $(x, y) \\subseteq \\operatorname{int} C \\subseteq C$.\n\nExample: the closed unit disc $D = \\set{p \\in \\R^2 : \\abs{p} \\le 1}$.\nIf $x \\ne y$ in $D$ and $0 \u003c t \u003c 1$, then $\\abs{(1 - t)x + t y} \u003c 1$.\nIndeed, equality would need $\\abs{x} = \\abs{y} = 1$ and equality in the triangle inequality.\nFor the Euclidean norm this means that $x$ and $y$ point in the same direction, which forces $x = y$.\nSo the open segment lies in the open disc, the interior of $D$.\n\nNon-example: the closed square $[0, 1]^2$.\nThe points $(0, 0)$ and $(1, 0)$ lie in it, but the midpoint $(1/2, 0)$ is on the boundary, not in the interior.\n\nEvery open convex set is strictly convex, so the notion is of interest mainly for closed sets.\n",
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