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Separating Hyperplanes
Katta G. Murty, IOE 611 Lecture slides
The intersection of family of convex sets is always convex. The union of two convex sets may not be convex. The T. of A. Farkas Lemma are separating hyperplane theorems separating a polyhedral cone from a point outside it.
Given nonempty subsets K 1 , K 2 ⊂ Rn, a separating hy- perplane for them is H = {x : cx = α} satisfying
cx
≥ α ∀x ∈ K 1 ≤ α ∀x ∈ K 2 It is strict separating hyperplane if inequalities hold strictly as inequalities. May not exist for certain pairs. A necessary condition for exis- tence is (Interior of K 1 ) ∩ (interior of K 2 ) = ∅. However, even if K 1 ∩ K 2 = ∅, a separating hyperplane may not exist.
Ability to construct separating hyperplanes efficiently
has great significance in LP and in combinatorial op- timization.
Sum of two Convex Sets: Given two convex subsets, K 1 , K 2
of Rn, their sum, K 1 + K 2 = {x + y : x ∈ K 1 , y ∈ K 2 }.
The sum of two convex sets is always convex. THEOREM: K ⊂ Rn^ nonempty, closed, convex. b ∈ Rn, b 6 ∈ K. Then there exists a hyperplane separating b from K.
THEOREM: K ⊂ Rn^ convex, nonempty. b 6 ∈ K. Then K can be separated from b by a hyperplane.
COROLLARY: SUPPORTING HYPERPLANE THEOREM: b a boundary point of a convex set K ⊂ Rn. There exists a hyperplane through b containing K on one of its sides. Such a hyperplane is called a supporting hyperplane for K at its bound- ary point b.
−z ≤ 0 −Mz − q ≤ 0 zT^ (Mz + q) ≤ 0
- K = convex hull of integer solutions to: Ax ≤ d, each xj ∈ { 0 , 1 }. Given b, efficiently conclude that either b ∈ K or produce a hyperplane separating b from K.
