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A Variational Inequality Theory Chapter | 4 79


                              The following result shows that the class AC n is nothing else but the class
                           of couples (M, ) whose generalized eigenvalues are positive.
                           Proposition 17. The couple (M, ) is of class AC n if and only if

                                                  σ ∞ (M, ) ⊂]0,+∞[.                  (4.36)

                           Proof. a) Let us first prove that if (4.36) holds, then the couple (M, ) is of
                           class AC n .Inviewof Remark 22, it suffices to check that for all t ∈]0,1],
                           we have B((1 − t)I + tM, ) ={0}. Suppose on the contrary that there exist
                                                                                      (1−t)
                           t ∈]0,1] and z 
= 0 such that z ∈ B((1−t)I +tM, ). Then setting μ =−  ,
                                                                                        t
                           we easily check that z ∈ B(M − μI, ), and this is a contradiction since μ ≤ 0.
                           b) Let us now check that if the couple (M, ) is of class AC n , then (4.36)
                           holds. Suppose on the contrary that there exist μ ≤ 0 and z 
= 0 such that z ∈
                                                  1     1
                           B(M −μI, ).Weset t =      =     . It is clear that t ∈]0,1] and z ∈ B(M +
                                                 1−μ   1+|μ|
                            (1−t) I, ), and thus z ∈ B((1 − t)I + tM, ), which is a contradiction.
                             t
                                                  n
                           Proposition 18. Let   : R → R be a proper convex lower semicontinuous
                           function with closed domain, and let M ∈ R n×n .If D( ) is bounded, then
                                                     (M, ) ∈ AC n .

                           Proof. Here D( ) is assumed to be bounded, and thus D( ) ∞ ={0}.If t ∈
                           [0,1] and z ∈ B((1−t)I +tM, ), then z ∈ D( ) ∞ , and the result follows.
                                                  n
                           Proposition 19. Let   : R → R be a proper convex lower semicontinuous
                           function with closed domain, and let M ∈ R n×n . Then we have

                                                 PD n ∪ P n ∪ PS n ⊂ AC n .

                           Proof. a) We prove that PD n ⊂ AC n .Let (M, ) ∈ PD n , t ∈[0,1], and z ∈
                           B((1 − t)I + tM, ). Then z ∈ D( ) ∞ and

                                                  (1 − t)z + tMz,z ≤ 0.

                           If t = 1, then  Mz,z ≤ 0 and (4.10) yield z = 0. If 0 ≤ t< 1, then
                                                           t
                                                    2
                                                 ||z|| ≤−      Mz,z ,
                                                         1 − t
                           and from (4.10) we deduce that necessarily z = 0. The result follows.
                              b) We prove that P n ⊂ AC n .Let (M, ) ∈ P n , t ∈[0,1], and z ∈ B((1 −
                           t)I + tM, ). We claim that z = 0. Suppose on the contrary that z 
= 0. We
                           claim that there exists some index k ∈{1,...,n} such that
                                                       2
                                                (1 − t)z + t(Mz) k z k > 0.
                                                       k
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