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Assume that Then = epi(f ) for a convex operator f := X → Y · and h is increasing. (inf ◦ )(x) = inf {(hβ ) ◦ (f α)(x) : α, β ≥ 0, α + β = 1} = inf βh α x f β α : α, β ≥ 0, α + β = 1 . 10. There are two more important constructions with convex operators, namely, +-convolution and ∨-convolution. Consider two convex operators f1 : X1 × X → E and f2 : X × X2 → E . Denote epi(f1 , X2 ) := {(x1 , x, x2 , e) ∈ W : f1 (x1 , x) ≤ e}, epi(X1 , f2 ) := {(x1 , x, x2 , e) ∈ W : f2 (x, x2 ) ≤ e}, where W := X1 × X × X2 × E .

If p(x + y) ≤ p(x) + p(y) for all x, y ∈ X , then p is said to be subadditive. If 0 ∈ dom(p) and p(λx) = λp(x) for all x ∈ X and λ ≥ 0, then p is called positively homogeneous. Note that for a positively homogeneous operator we will always have p(0) = 0 since p(0) < +∞ and 0 = 0p(0) = p(0). For an operator p : X → E · the following statements are equivalent: E· (a) p is sublinear; (b) p is convex and positively homogeneous; (c) p is subadditive and positively homogeneous; (d) 0 ∈ dom(p) and p(αx + βy) ≤ αp(x)+ βp(y) for all x, y ∈ X and α, β ∈ R+ .

N ∈ A+ , n k=1 αk = IE , n ∈ N . Denote the right-hand side of the sought equality by U0 . It is clear that U0 is an operator-convex set containing U . Therefore, op(U ) ⊂ U0 . To prove the reverse inclusion we have to show that if U is an operator-convex set, then n + k=1 αk ◦ Tk ∈ U for every collections T1 , . . , Tn ∈ U and α1 , . . , αn ∈ A provided that nk=1 αk = IE . By way of induction on n, suppose that the previous +1 claim is established for some n ∈ N, n ≥ 2. Let S := nk=1 αk ◦ Tk , where n+1 + T1 , .