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On the other hand, x= and hence f (x) ≤ 1 1 (x + αu) + 1 − 1+α 1+α (x − u), 1 1 f (x + αu) + 1 − 1+α 1+α which implies f (x) ≤ 1 α f (z) + c. 1+α 1+α We obtain α(f (x) − c) ≤ f (z) − f (x). Together, the two inequalities give |f (z) − f (x)| ≤ α(c − f (x)), for all z ∈ x + αU . f (x − u), 48 CHAPTER 2. CONVEX FUNCTIONS Now we discuss differentiability properties of convex functions. We first consider the case f : R1 → (−∞, ∞]. 2. Let f : R1 → (−∞, ∞] be convex. (a) In each point x ∈ int dom f , the right derivative f + (x) and the left derivative f − (x) exist and fulfill f − (x) ≤ f + (x).

This is equivalent to K ⊂ B(c ), for all K ∈ M, for some constant c > 0. Here, we can replace the ball B(c ) by any compact set, in particular by a cube W ⊂ Rn . The subset M is relative compact, if every sequence K1 , K2 , . . , with Kk ∈ M, has a converging subsequence. Therefore, the mentioned topological property is a consequence of the following theorem. 4 (Blaschke’s Selection Theorem). Let M ⊂ Kn be an infinite collection of convex bodies, all lying in a cube W . Then, there exists a sequence K1 , K2 , .

The support function hA : Rn → (−∞, ∞] of A is defined as hA (x) := sup x, y , x ∈ Rn . 1. For nonempty, convex sets A, B ⊂ Rn , we have: (a) hA is positively homogeneous, closed and convex (and hence subadditive). (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 .

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A course on convex geometry by Weil W.


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