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Gruber P. Convex and Discrete Geometry

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4 Support <strong>and</strong> Separation 59<br />

(i)⇒(ii) For the proof of the convexity of C − D, let x, z ∈ C, y,w ∈ D <strong>and</strong><br />

0 ≤ λ ≤ 1. Then<br />

(1 − λ)(x − y) + λ(z − w) = � (1 − λ)x + λz � − � (1 − λ)y + λw � ∈ C − D<br />

by the convexity of C <strong>and</strong> D. Thus C − D is convex. Next, let S ={x : α ≤ u · x ≤<br />

β},α 0, separate {o} <strong>and</strong><br />

C − D. Then u · (x − y) ≤−γ , i.e. u · x + γ ≤ u · y for all x ∈ C <strong>and</strong> y ∈ D.<br />

Let α = sup{u · x : x ∈ C}. Then the slab {z : α ≤ u · z ≤ α + γ } separates C<br />

<strong>and</strong> D. ⊓⊔<br />

Theorem 4.4. Let C, D ⊆ E d be convex. Then the following hold:<br />

(i) Let C be compact, D closed <strong>and</strong> C ∩ D = ∅. Then C <strong>and</strong> D are strongly<br />

separated.<br />

(ii) Let relint C ∩ relint D =∅. Then C <strong>and</strong> D are separated.<br />

Proof. (i) By the assumptions in (i), we may choose p ∈ C, q ∈ D having minimum<br />

distance. Let u = q − p (�= o). Then the slab {x : u · p ≤ u · x ≤ u · q} separates C<br />

<strong>and</strong> D.<br />

(ii) By Proposition 3.1, the sets relint C <strong>and</strong> relint D are convex. Since, by<br />

assumption, these sets are disjoint, Proposition 4.1 shows that o �∈ E = relint C −<br />

relint D. We shall prove that o �∈ int cl E. Otherwise, there are points x1,...,xd+1 ∈<br />

E such that o is an interior point of the simplex with vertices x1,...,xd+1. Since E is<br />

convex, by Proposition 4.1 we have o ∈ E, which is the desired contradiction. Thus<br />

o ∈ bd cl E or o �∈ cl E. Since E is convex, cl E is convex by Proposition 3.1. It thus<br />

follows from Theorem 4.2, respectively, from (i), that {o} <strong>and</strong> cl E can be separated<br />

by a hyperplane. Hence, a fortiori, {o} <strong>and</strong> E = relint C − relint D can be separated<br />

by a hyperplane. Thus relint C <strong>and</strong> relint D can be separated by a hyperplane by<br />

Proposition 4.1. This, in turn, implies that cl relint C <strong>and</strong> cl relint D can be separated<br />

by a hyperplane. Now apply Proposition 3.1 to see that C <strong>and</strong> D can be separated by<br />

a hyperplane. ⊓⊔<br />

Simple examples show that, in general, disjoint closed convex sets C <strong>and</strong> D<br />

cannot be strongly separated, but if C <strong>and</strong> D are convex polyhedra, this is possible.<br />

This fact is of importance in optimization (Fig. 4.3).<br />

Strong <strong>and</strong> Weak Oracles to Specify <strong>Convex</strong> Bodies<br />

Before considering oracles, some definitions are in order: Let C be a proper convex<br />

body in E d <strong>and</strong> ε>0. Define the ε-neighbourhood of C or the parallel body of C at<br />

distance ε <strong>and</strong> the inner parallel body of C at distance ε by<br />

Cε = C + εB d = � x + εy : x ∈ C, y ∈ B d� , C−ε = � x : x + εB d ⊆ C � .

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