By Weil W.

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We mention without proof a couple of further properties of support functions, which are mostly simple consequences of the definition or the last two theorems. In the following remarks, A is always a nonempty closed convex subset of Rn . Remarks. (1) We have A = {x}, if and only if hA = x, · . (2) We have hA+x = hA + x, · . e. e. hA (x) = hA (−x), for all x ∈ Rn . (4) We have 0 ∈ A, if and only if hA ≥ 0. Let A ⊂ Rn be nonempty, closed and convex. For u ∈ S n−1 , we consider the sets E(u) := {x ∈ Rn : x, u = hA (u)} and A(u) := A ∩ E(u) = {x ∈ A : x, u = hA (u)}.

We assume the usual rules for addition and multiplication with ∞, namely: α + ∞ := ∞, α − ∞ := −∞, α ∞ := ∞, (−α)∞ := −∞, 0 ∞ := 0. for α ∈ (−∞, ∞], for α ∈ [−∞, ∞), for α ∈ (0, ∞], Definition. For a function f : Rn → (−∞, ∞], the set epi f := {(x, α) : x ∈ Rn , α ∈ R, f (x) ≤ α} ⊂ Rn × R is called the epigraph of f . f is convex, if epi f is a convex subset of Rn × R = Rn+1 . Remarks. (1) A function f : Rn → [−∞, ∞) is concave, if −f is convex. Thus, for a convex function f we exclude the value −∞, whereas for a concave function we exclude ∞.

Finally we define, for arbitrary a, a function g by x f − (u) dy. g(x) := f (a) + a We first show that g is convex, and then g = f . For z := αx + (1 − α)y, α ∈ [0, 1], x < y, we have z f − (s) ds ≤ (z − x)f − (z), g(z) − g(x) = x y f − (s) ds ≥ (y − z)f − (z). e. g is convex. As a consequence, g + und g − exist. Since g(y) − g(x) 1 = y−x y−x y f − (s) ds = x 1 y−x y f + (s) ds ≥ f + (x), x we obtain g + (x) ≥ f + (x). Analogously, we get g − (x) ≤ f − (x). By the continuity from the left of g − and f − , and the continuity from the right of g + and f + , it follows that g + = f + and g − = f − .

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A course on convex geometry by Weil W.


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