In geometry, a supporting hyperplane of a set [math]\displaystyle{ S }[/math] in Euclidean space [math]\displaystyle{ \mathbb R^n }[/math] is a hyperplane that has both of the following two properties:[1]
Here, a closed half-space is the half-space that includes the points within the hyperplane.
This theorem states that if [math]\displaystyle{ S }[/math] is a convex set in the topological vector space [math]\displaystyle{ X=\mathbb{R}^n, }[/math] and [math]\displaystyle{ x_0 }[/math] is a point on the boundary of [math]\displaystyle{ S, }[/math] then there exists a supporting hyperplane containing [math]\displaystyle{ x_0. }[/math] If [math]\displaystyle{ x^* \in X^* \backslash \{0\} }[/math] ([math]\displaystyle{ X^* }[/math] is the dual space of [math]\displaystyle{ X }[/math], [math]\displaystyle{ x^* }[/math] is a nonzero linear functional) such that [math]\displaystyle{ x^*\left(x_0\right) \geq x^*(x) }[/math] for all [math]\displaystyle{ x \in S }[/math], then
defines a supporting hyperplane.[2]
Conversely, if [math]\displaystyle{ S }[/math] is a closed set with nonempty interior such that every point on the boundary has a supporting hyperplane, then [math]\displaystyle{ S }[/math] is a convex set, and is the intersection of all its supporting closed half-spaces.[2]
The hyperplane in the theorem may not be unique, as noticed in the second picture on the right. If the closed set [math]\displaystyle{ S }[/math] is not convex, the statement of the theorem is not true at all points on the boundary of [math]\displaystyle{ S, }[/math] as illustrated in the third picture on the right.
The supporting hyperplanes of convex sets are also called tac-planes or tac-hyperplanes.[3]
The forward direction can be proved as a special case of the separating hyperplane theorem (see the page for the proof). For the converse direction,
Define [math]\displaystyle{ T }[/math] to be the intersection of all its supporting closed half-spaces. Clearly [math]\displaystyle{ S \subset T }[/math]. Now let [math]\displaystyle{ y\not \in S }[/math], show [math]\displaystyle{ y \not\in T }[/math].
Let [math]\displaystyle{ x\in \mathrm{int}(S) }[/math], and consider the line segment [math]\displaystyle{ [x, y] }[/math]. Let [math]\displaystyle{ t }[/math] be the largest number such that [math]\displaystyle{ [x, t(y-x) + x] }[/math] is contained in [math]\displaystyle{ S }[/math]. Then [math]\displaystyle{ t\in (0, 1) }[/math].
Let [math]\displaystyle{ b = t(y-x) + x }[/math], then [math]\displaystyle{ b\in \partial S }[/math]. Draw a supporting hyperplane across [math]\displaystyle{ b }[/math]. Let it be represented as a nonzero linear functional [math]\displaystyle{ f: \R^n \to \R }[/math] such that [math]\displaystyle{ \forall a\in S, f(a) \geq f(b) }[/math]. Then since [math]\displaystyle{ x\in \mathrm{int}(S) }[/math], we have [math]\displaystyle{ f(x) \gt f(b) }[/math]. Thus by [math]\displaystyle{ \frac{f(y) - f(b)}{1-t} = \frac{f(b) - f(x)}{t} }[/math], we have [math]\displaystyle{ f(y) \lt f(b) }[/math], so [math]\displaystyle{ y \not\in T }[/math].
Original source: https://en.wikipedia.org/wiki/Supporting hyperplane.
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