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Consider the half-line s starting in x, orthogonal to E and in direcˆ On s, we can find a point z with tion of that half-space of E , which contains K. g. we may choose a cube W large enough to contain ˆ and such that the point x¯ with {¯ K, x} := s ∩ E is the center of a facet of W . Now we choose a ball B with center z ∈ s in such a way that W ⊂ B, but x ∈ / B. Then z is the required point). ˆ there exists yz ∈ K ˆ with By definition of K, yz − z = max y − z ≥ x − z , y∈K ˆ Because of yx ∈ exp K, for all x ∈ Rn , we a contradiction.

B) hA = hcl A and cl A = {x ∈ Rn : x, u ≤ hA (u) ∀u ∈ Rn }. (c) A ⊂ B implies hA ≤ hB ; conversely, hA ≤ hB implies cl A ⊂ cl B. (d) hA is finite, if and only if A is bounded. (e) hαA+βB = αhA + βhB , for all α, β ≥ 0. (f ) h−A (x) = hA (−x), for all x ∈ Rn . (g) If Ai , i ∈ I, are nonempty and convex and A := conv i∈I Ai , then hA = sup hAi . i∈I (h) If Ai , i ∈ I, are nonempty, convex and closed and if A := then hA = cl conv (hAi )i∈I . i∈I Ai is nonempty, (i) δA∗ = hA . Proof. (a) For α ≥ 0 and x, z ∈ Rn , we have hA (αx) = sup αx, y = α sup x, y = αhA (x) y∈A y∈A and hA (x + z) = sup x + z, y ≤ sup x, y + sup z, y = hA (x) + hA (z).

This result now shows that the intersection of finitely many polytopes is again a polytope. 1 and the remark preceding it). There is however a modified version which holds true for exposed points. 4. Let K ⊂ Rn be compact and convex. Then K = cl conv exp K. Proof. e. a point with yx − x = max y − x . y∈K The hyperplane E through yx orthogonal to yx − x is then a supporting hyperplane of K. We have E ∩ K = {yx }, hence yx ∈ exp K. Let ˆ := cl conv {yx : x ∈ Rn }. K ˆ ⊂ K, thus K ˆ is compact. 2 there is a hyAssume that there exists x ∈ K \ K.

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