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  "body": "\n## Statement\n\nLet $n \\ge 1$, let $S \\subseteq \\R^n$ be a [[convex set|02AS45T3H9]], and let $x_0$ be a point of the [[boundary|05D5EGMFYV]] of $S$.\nThen there is a [[supporting hyperplane|K7N565NKCV]] of $S$ through $x_0$.\nThat is, there is a vector $a \\in \\R^n$ with $a \\ne 0$ such that $a \\cdot x \\le a \\cdot x_0$ for all $x \\in S$, where $a \\cdot x$ is the [[dot product|0S52TNNTH9]].\n\n## Remarks\n\nProof idea: since $x_0$ is a boundary point of the convex set $S$, it is not an interior point of the closure $\\overline{S}$, so there are points $y_k \\notin \\overline{S}$ with $y_k \\to x_0$.\nFor each $k$, the nearest point of $\\overline{S}$ to $y_k$ gives a unit vector $a_k$ with $a_k \\cdot x \\le a_k \\cdot y_k$ for all $x \\in S$.\nA subsequence of the unit vectors $a_k$ converges to a unit vector $a$, and passing to the limit gives $a \\cdot x \\le a \\cdot x_0$ for all $x \\in S$.\n\nExample: for the square $[0, 1]^2$ and the corner $x_0 = (1, 1)$, every $a = (a_1, a_2) \\ne 0$ with $a_1, a_2 \\ge 0$ works, so the supporting line at a corner is not unique.\n\nConvexity is needed: for the annulus $\\set{x \\in \\R^2 : 1 \\le \\abs{x} \\le 2}$, the inner boundary point $(1, 0)$ has no supporting line, since every line through it has points of the annulus on both sides.\nA partial converse holds: a closed set with nonempty interior that has a supporting hyperplane at each of its boundary points is convex.\n\n## Explanations\n\n### L1 human\n\nTake a convex shape and any point on its edge.\nThe theorem says you can always lay a flat wall against the shape at that point, with the whole shape on one side.\nFor a smooth shape the wall is the tangent plane; at a corner there may be many possible walls.\nThis fails for shapes with dents, where the wall would have to cut through the shape.\n\n### L2 human\n\nThe theorem is the finite-dimensional, boundary-point form of the separation theorems; the general version is the [[Hahn–Banach separation theorem|6GWMGFGFJJ]].\nIt is used to prove that a compact convex set in $\\R^n$ is the convex hull of its extreme points.\nThat proof passes to the face cut out by a supporting hyperplane and inducts on dimension.\nIt gives the existence of subgradients of a convex function at interior points of its domain, by applying it to the epigraph.\nIn optimisation it is the geometric root of Lagrange multipliers and of the Karush–Kuhn–Tucker conditions.\n",
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