By J.N. Coldstream

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6, the latter is equivalent to the differentiability of hK at u. If all the support sets K(u), u ∈ S n−1 , of a nonempty, compact convex set K consist of points, the boundary bd K does not contain any segments. Such sets K are called strictly convex. Hence, K is strictly convex, if and only if hK is differentiable on Rn \ {0}. We finally consider the support functions of polytopes. We call a function h on Rn piecewise linear, if there are finitely many convex cones A1 , . . , Am ⊂ Rn , such that Rn = m i=1 Ai and h is linear on Ai , i = 1, .

4, there is a representation x = α1 x1 + · · · + αk xk with k ∈ N, xi ∈ A, αi > 0, and k ≥ 2, αi = 1. In case k = 1, we have x = x1 ∈ A. If x = α1 x1 + (1 − α1 ) and α2 x2 + · · · + αk xk α2 + · · · + αk α2 x2 + · · · + αk xk ∈ K. α2 + · · · + αk Since x is extreme, we obtain x = x1 ∈ A. Thus, in both cases we have x ∈ A, therefore ext K ⊂ A. 34 CHAPTER 1. CONVEX SETS In the other direction, we need only show that K = conv ext K. We prove this by induction on n. For n = 1, a compact convex subset of R1 is a segment [a, b] and ext [a, b] = {a, b}.

F) Let A be closed. e. e. A = −A). (g) Let A be closed. Then dA is a norm on Rn , if and only if A is symmetric, compact and contains 0 in its interior. (h) If A is closed, then dA is closed. 2. 2 47 Regularity of convex functions We start with a continuity property of convex functions. 1. A convex function f : Rn → (−∞, ∞] is continuous in int dom f . Proof. Let x ∈ int dom f . There exists a n-simplex P with P ⊂ int dom f and x ∈ int P . If x0 , . . , xn are the vertices of P and y ∈ P , we have y = α0 x0 + · · · + αn xn , with αi ∈ [0, 1], αi = 1, and hence f (y) ≤ α0 f (x0 ) + · · · + αn f (xn ) ≤ max f (xi ) =: c.