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Example text

C) Assume now x < y. From f (y) − f (x) f (y) − f (z) = lim ≥ lim f + (z) z x z x y−x y−z we obtain limz x f + (z) ≤ f + (x), hence limz x f + (z) = f + (x). Similarly, we get limz x f − (z) = f − (x) . 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.

Hence mα := αl0 + l1 is an affine function fulfilling mα ≤ f . Since l0 (x0 ) > 0, we have mα (x0 ) > r0 for sufficiently large α. We now come to a further important operation on convex functions, the construction of the conjugate function. Definition. Let f : Rn → (−∞, ∞] be proper and convex, then the function f ∗ defined by f ∗ (y) := sup ( x, y − f (x)), y ∈ Rn , x∈Rn is called the conjugate of f . 4. The conjugate f ∗ of a proper convex function f : Rn → (−∞, ∞] fulfills: (a) f ∗ is proper, closed and convex.

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)}. If hA (u) = ∞, both sets are empty. 4), namely if A(u) = ∅. If A(u) = ∅, then E(u) is a supporting hyperplane of A (at each point x ∈ A(u)) and A(u) is the corresponding support set. We discuss now the support function of A(u).