By Weil W.

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B) The extreme points of K are precisely the permutation matrices. Hint for (b): You may use the following simple combinatorial result (marriage theorem): Given a finite set H, a nonempty set D and a function f : H → P(D) with ˜ f (h) ≥ |H|, ˜ ⊂ H, for all H ˜ h∈H then there exists an injective function g : H → D with g(h) ∈ f (h), for all h ∈ H. 38 CHAPTER 1. 1 Properties and operations of convex functions In the following, we consider functions f : Rn → [−∞, ∞]. We assume the usual rules for addition and multiplication with ∞, namely: α + ∞ := ∞, α − ∞ := −∞, α ∞ := ∞, (−α)∞ := −∞, 0 ∞ := 0.

Am ⊂ Rn , such that Rn = m i=1 Ai and h is linear on Ai , i = 1, . . , m. 4. Let K ∈ Kn . Then K is a polytope, if and only if hK is piecewise linear. Proof. K is a polytope, if and only if K = conv {x1 , . . , xk }, for some x1 , . . , xk ∈ Rn . ,k which holds, if and only if hK is piecewise linear. ,k i = 1, . . , k. Conversely, if hK is linear on the cone Ai , then xi is determined by xi , · = hK on Ai . 58 CHAPTER 2. CONVEX FUNCTIONS Exercises and problems 1. Let f : Rn → R be positively homogeneous and twice continuously partially differentiable on Rn \{0}.

Then, z = (1 − α)x + α(x + u), hence f (z) ≤ (1 − α)f (x) + αf (x + u) ≤ (1 − α)f (x) + αc. This gives us f (z) − f (x) ≤ α(c − f (x)). 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.