Some Aspects of Extended General Variational Inequalities

and Applied Analysis 3 For given nonlinear operators T, g, h, we consider the problem of finding u ∈ H : h u ∈ K such that 〈 Tu, g v − h u 〉 ≥ 0, ∀v ∈ H : g v ∈ K, 2.1 which is called the extended general variational inequality. Noor 13–16 has shown that the minimum of a class of differentiable nonconvex functions on hg-convex set K in H can be characterized by extended general variational inequality 2.1 . For this purpose, we recall the following well-known concepts, see 7 . Definition 2.1 see 6, 13 . Let K be any set in H. The set K is said to be hg-convex if there exist two functions g, h : H −→ H such that h u t ( g v − h u ) ∈ K, ∀u, v ∈ H : h u , g v ∈ K, t ∈ 0, 1 . 2.2 Note that every convex set is an hg-convex set, but the converse is not true, see 6 . If g h, then the hg-convex set K is called the g-convex set, which was introduced by Youness 2 . From now onward, we assume that K is an hg-convex set, unless otherwise specified. Definition 2.2 see 24, 28 . The function F : K −→ H is said to be hg-convex, if and only if, there exist two functions h, g such that F ( h u t ( g v − h u )) ≤ 1 − t F h u tFg v ) 2.3 for all u, v ∈ H : h u , g v ∈ K, t ∈ 0, 1 . Clearly, every convex function is a gh-convex, but the converse is not true. For g h, Definition 2.2 is due to Youness 2 . We now show that the minimum of a differentiable hg-convex function on the hgconvex set K in H can be characterized by the extended general variational inequality 2.1 . This result is due to Noor 13–16 . We include all the details for the sake of completeness and to convey the main idea. Lemma 2.3 see 13–16 . Let F : K −→ H be a differentiable hg-convex function. Then u ∈ H : h u ∈ K is the minimum of hg-convex function F onK if and only if u ∈ H : h u ∈ K satisfies the inequality 〈 F ′ h u , g v − h u 〉 ≥ 0, ∀v ∈ H : g v ∈ K, 2.4 where F ′ u is the differential of F at h u ∈ K. Proof. Let u ∈ H : h u ∈ K be a minimum of hg-convex function F on K. Then F h u ≤ Fg v , ∀v ∈ H : g v ∈ K. 2.5 Since K is an hg-convex set, so, for all u, v ∈ H : h u , g v ∈ K, t ∈ 0, 1 , g vt h u t g v − h u ∈ K. Setting g v g vt in 2.5 , we have F h u ≤ Fh u tg v − h u . 2.6 4 Abstract and Applied Analysis Dividing the above inequality by t and taking t −→ 0, we have 〈 F ′ h u , g v − h u 〉 ≥ 0, ∀v ∈ H : g v ∈ K, 2.7 which is the required result 2.4 . Conversely, let u ∈ H : h u ∈ K satisfy the inequality 2.4 . Since F is an hg-convex function, for all u, v ∈ H : h u , g v ∈ K, t ∈ 0, 1 , h u t g v − h u ∈ K, and F ( h u t ( g v − h u )) ≤ 1 − t F h u tFg v , 2.8