# Supporting hyperplane

In geometry, a supporting hyperplane of a set ${\displaystyle S}$ in Euclidean space ${\displaystyle \mathbb {R} ^{n}}$ is a hyperplane that has both of the following two properties:[1]

• ${\displaystyle S}$ is entirely contained in one of the two closed half-spaces bounded by the hyperplane,
• ${\displaystyle S}$ has at least one boundary-point on the hyperplane.

Here, a closed half-space is the half-space that includes the points within the hyperplane.

## Supporting hyperplane theorem

This theorem states that if ${\displaystyle S}$ is a convex set in the topological vector space ${\displaystyle X=\mathbb {R} ^{n},}$ and ${\displaystyle x_{0}}$ is a point on the boundary of ${\displaystyle S,}$ then there exists a supporting hyperplane containing ${\displaystyle x_{0}.}$ If ${\displaystyle x^{*}\in X^{*}\backslash \{0\}}$ (${\displaystyle X^{*}}$ is the dual space of ${\displaystyle X}$, ${\displaystyle x^{*}}$ is a nonzero linear functional) such that ${\displaystyle x^{*}\left(x_{0}\right)\geq x^{*}(x)}$ for all ${\displaystyle x\in S}$, then

${\displaystyle H=\{x\in X:x^{*}(x)=x^{*}\left(x_{0}\right)\}}$

defines a supporting hyperplane.[2]

Conversely, if ${\displaystyle S}$ is a closed set with nonempty interior such that every point on the boundary has a supporting hyperplane, then ${\displaystyle S}$ 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 ${\displaystyle S}$ is not convex, the statement of the theorem is not true at all points on the boundary of ${\displaystyle S,}$ 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,

Proof

Define ${\displaystyle T}$ to be the intersection of all its supporting closed half-spaces. Clearly ${\displaystyle S\subset T}$. Now let ${\displaystyle y\not \in S}$, show ${\displaystyle y\not \in T}$.

Let ${\displaystyle x\in \mathrm {int} (S)}$, and consider the line segment ${\displaystyle [x,y]}$. Let ${\displaystyle t}$ be the largest number such that ${\displaystyle [x,t(y-x)+x]}$ is contained in ${\displaystyle S}$. Then ${\displaystyle t\in (0,1)}$.

Let ${\displaystyle b=t(y-x)+x}$, then ${\displaystyle b\in \partial S}$. Draw a supporting hyperplane across ${\displaystyle b}$. Let it be represented as a nonzero linear functional ${\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} }$ such that ${\displaystyle \forall a\in S,f(a)\geq f(b)}$. Then since ${\displaystyle x\in \mathrm {int} (S)}$, we have ${\displaystyle f(x)>f(b)}$. Thus by ${\displaystyle {\frac {f(y)-f(b)}{1-t}}={\frac {f(b)-f(x)}{t}}}$, we have ${\displaystyle f(y), so ${\displaystyle y\not \in T}$.