Extensions of Invexity to Nondifferentiable Functions
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4.1 Preinvex Functions Since invexity requires the differentiabilty assumptions, in [18] and subsequently in [244], the class of preinvex functions, not necessarily differentiable, has been introduced. For the reader’s convenience we recall Definition 3.2: A subset X of Rn is η-invex with respect to η : Rn → R if x, y ∈ X, λ ∈ [0, 1] ⇒ y + λη(x, y) ∈ X. It is obvious that the above definition is a generalization of the notion of convex set. It is to be noted that any set in Rn is invex with respect to η(x, y) ≡ 0, ∀x, y ∈ Rn . However, the only function f : Rn → R invex with respect to η(x, y) ≡ 0 is the constant function f (x) = c, c ∈ R. Definition 4.1. Let f be a real-valued function defined on an η-invex set X; f is said to be preinvex with respect to η if f [y + λη(x, y)] ≤ λf (x) + (1 − λ)f (y),
∀x, y ∈ X, ∀λ ∈ [0, 1]
(4.1)
It is obvious that the class of convex functions is strictly contained in the class of preinvex functions: take η(x, y) = x − y. The inclusion is strict as shown by the following example, taken from [245]. Consider the function f (x) = −|x|, x ∈ R. We show that f is preinvex with respect to x − y, if xy ≥ 0 η(x, y) = y − x, if xy < 0. Let x ≥ 0, y ≥ 0, in this case we have, for the preinvexity of f : −(y + λ(x − y)) ≤ λ(−x) + (1 − λ)(−y), relation which is obviously true for any λ ∈ [0, 1]. The same result holds, if x ≤ 0, y ≤ 0. Let now be x < 0 and y > 0 : we must have −|y + λ(y − x)| = −y − λ(y − x) ≤ λx − (1 − λ)y,
∀λ ∈ [0, 1].
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4 Extensions of Invexity to Nondifferentiable Functions
We obtain λ(2y) ≥ 0, so being y > 0, relation (4.1) holds for all λ ∈ [0, 1]. Consider now the last case, i.e., x > 0, y < 0. We must have −|y + λ(y − x)| = y + λ(y − x) ≤ −λx + (1 − λ)y,
∀λ ∈ [0, 1],
i.e., λ(2y) ≤ 0. Being y < 0, relation (4.1) holds for all λ ∈ [0, 1]. The function is therefore preinvex on R, but obviously it is not convex. Similarly to convex functions, it is possible to characterize preinvex functions in terms of invexity of their epigraph, however not with reference to the same function η (first of all, one is an n-vector, the other is an (n + 1)-vector). Theorem 4.2. Let f : X → R, where X ⊂ Rn is an η-invex set. Then f is preinvex with respect to η if and only if the set epi f = {(x , α) : x ∈ X , α ∈ R, f (x ) ≤ α} is an invex set with respect to η1 : epif × epif → R(n+1) , where η1 ((y, β), (x, α)) = (η(y, x), β − α), ∀(x, α), (y, β) ∈ epif. Proof. Necessity: Let (x, α) ∈ epif and (y, β) ∈ epif, that is, f (x) ≤ α and f (y) ≤ β. From the preinvexity of f, we have f [x + λη(x, y)] ≤ λf (y) + (1 − λ)f (x) ≤ (1 − λ)α + λβ,
∀λ ∈ [0, 1].
It follows that (x + λη(x, y), (1 − λ)α + λβ) ∈ epif,
∀λ ∈ [0, 1],
That is, (x, α) + λ(η(y, x), β − α) ∈ epif,
∀λ ∈ [0, 1].
Hence epif is an invex set with respect to η1 ((y, β), (x, α)) = (η(y, x), β − α). Sufficiency: Assume that epif is an invex set with respect to η1 ((y, β), (x, α)) = (η(y, x), β − α). Let x, y ∈ X and α, β ∈ R such that f (x) ≤ α and f (y) ≤ β. Then (x, α) ∈ epif and (y, β) ∈ epif. From the invexity of the set epif with r
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