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Gruber P. Convex and Discrete Geometry

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Metric Projection<br />

C<br />

pC (x)<br />

x<br />

pC (y)<br />

y<br />

S<br />

4 Support <strong>and</strong> Separation 53<br />

Hx<br />

Fig. 4.1. Metric projection is non-expansive<br />

Let C be a closed convex set in Ed . For each x ∈ Ed , there is a unique point pC(x) ∈<br />

C closest to it. Since C is closed, the existence is obvious. To see the uniqueness,<br />

assume that there are points y, z ∈ C, y �= z, both having minimum distance from x.<br />

Then �y − x� =�z− x� <strong>and</strong> therefore � 1 2 (y + z) − x� < �y − x�, �z − x�, noting<br />

that y �= z. Since 1 2 (y + z) ∈ C by the convexity of C, this contradicts our choice<br />

of y, z. The mapping pC : x → pC(x) of Ed onto C, thus obtained, is the metric<br />

projection of Ed onto C with respect to the Euclidean norm.<br />

The following useful result is due to Busemann <strong>and</strong> Feller [183].<br />

Lemma 4.1. Let C ⊆ E d be a closed convex set. Then the metric projection pC :<br />

E d → Cisnon-expansive, i.e.<br />

Hy<br />

�pC(x) − pC(y)� ≤�x − y� for x, y ∈ E d .<br />

Given a hyperplane H in E d ,letH + <strong>and</strong> H − denote the closed halfspaces determined<br />

by H. Wesay,H separates two sets, if one set is contained in H + <strong>and</strong> the<br />

other one in H − .<br />

Proof. Let x, y ∈ E d . We consider only the case where x, y �∈ C (Fig. 4.1). The<br />

other cases are treated similarly. If pC(x) = pC(y), we are done. Assume then that<br />

pC(x) �= pC(y).LetS be the slab orthogonal to the line segment [pC(x), pC(y)] ⊆<br />

C <strong>and</strong> such that its boundary hyperplanes Hx <strong>and</strong> Hy contain pC(x) <strong>and</strong> pC(y),<br />

respectively. We claim that x <strong>and</strong> pC(y) are separated by Hx. Otherwise there is a<br />

point on [pC(x), pC(y)] <strong>and</strong> thus in C which is closer to x than pC(x), which is<br />

impossible. Similarly, y <strong>and</strong> pC(x) are separated by Hy. Taken together, this means<br />

that x <strong>and</strong> y are on different sides of the slab S. Hence �x − y� is at least equal to<br />

the width of S, that is �x − y� ≥�pC(x) − pC(y)�. ⊓⊔<br />

Support Hyperplanes, Normal Vectors <strong>and</strong> Support Sets<br />

Let C ⊆ E d be closed <strong>and</strong> convex. A hyperplane H = HC(x) is a support hyperplane<br />

of C at a point x of the boundary bd C of C, ifx ∈ HC(x) <strong>and</strong> C is contained

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