This
theorem states that if S is a
convex set in the
topological vector space X=\mathbb{R}^n, and x_0 is a point on the
boundary of S, then there exists a supporting hyperplane containing x_0. If x^* \in X^* \backslash \{0\} (X^* is the
dual space of X, x^* is a nonzero
linear functional) such that x^*\left(x_0\right) \geq x^*(x) for all x \in S, then :H = \{x \in X: x^*(x) = x^*\left(x_0\right)\} defines a supporting hyperplane. Conversely, if S is a
closed set with nonempty
interior such that every point on the boundary has a supporting hyperplane, then S is a convex set, and is the intersection of all its supporting closed half-spaces. 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, {{Math proof|title=Proof|proof= Define T to be the intersection of all its supporting closed half-spaces. Clearly S \subset T. Now let y\not \in S, show y \not\in T. Let x\in \mathrm{int}(S), and consider the line segment [x, y]. Let t be the largest number such that [x, t(y-x) + x] is contained in S. Then t\in (0, 1). Let b = t(y-x) + x, then b\in \partial S. Draw a supporting hyperplane across b. Let it be represented as a nonzero linear functional f: \R^n \to \R such that \forall a\in T, f(a) \geq f(b). Then since x\in \mathrm{int}(S), we have f(x) > f(b). Thus by \frac{f(y) - f(b)}{1-t} = \frac{f(b) - f(x)}{t-0} , we have f(y) , so y \not\in T. }} ==See also==