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A Variational Inequality Theory Chapter | 4 81



                              b) If (M, ) ∈ P0 n , then condition (4.15) on M entails that for each x ∈
                           D( ) ∞ , x 
= 0, there exists α ∈{1,...,n} such that
                                                            2
                                           x α (λx + Mx) α = λx + x α (Mx) α > 0;
                                                            α
                           thus λI + M satisfies condition (4.13), and the result follows from Proposi-
                           tion 19.
                              c) Let (M, ) ∈ PS0 n .Let t ∈[0,1] and z ∈ B((1−t)I +tM, ). We claim
                           that z = 0. Suppose on the contrary that z 
= 0. From (4.16) we deduce that
                                   ∗
                           (D(  ∞ )) ={0}, and the second relation in (4.27) yields
                                                (1 − t)z + tMz + tλz = 0.

                           If t = 0, then z = 0, a contradiction. If 0 <t ≤ 1, then
                                                         (1 − t)
                                                  Mz =−       z − λz,
                                                           t

                                  ∗
                           so that ν =− (1−t)  − λ< 0 is a real eigenvalue of M, a contradiction to (4.19).
                                         t
                              The following theorem is the basic result of this section. It reduces the study
                           of the general class of variational inequalities VI(M,q, ) to semicomplemen-
                           tarity problems SCP ∞ (tM + (1 − t)I, )(t ∈[0,1]) involving the convex com-
                           binations of the matrix M and the identity matrix I. More precisely, we prove
                                                                       n
                           that if the couple (M, ) ∈ AC n , then for each q ∈ R , problem VI(M,q, )
                           has at least one solution. In other words, we prove that
                                                      AC n ⊂ Q n .

                           Let us first recall some basic results from Brouwer degree theory, which we will
                                                          n
                           use to prove the next result. Let D ⊂ R be an open bounded set. If f : D → R n
                                                                                    ¯
                           is continuous and 0 /∈ f(∂D), then the Brouwer topological degree of f with
                           respect to D and 0 is well defined (see e.g. [63]) and denoted by deg(f,D,0).
                           Let us recall some properties of the topological degree.

                           P1. Solution property: If 0 /∈ f(∂D) and deg(f,D,0) 
= 0, then there exists
                           x ∈ D such that f(x) = 0.
                                                                          n
                                                                     ¯
                           P2. Homotopy invariance property: Let ϕ :[0,1]× D → R ;(λ,x)  à  ϕ(λ,x),
                           be continuous and such that 0 /∈ ϕ(λ,∂D) for all λ ∈[0,1]. Then the map
                                                  λ  à  deg(ϕ(λ,·),D,0)

                           is constant on [0,1].
                           P3. Normalized property: If p ∈ D, then deg(id R − p,D,0) = 1.
                                                                   n
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