25.7. MAXIMAL MONOTONE OPERATORS 875
whenever y ∈ K. Then one argues that T is set valued, has convex compact values and isupper semicontinuous. Then one applies the Kakutani fixed point theorem to get x ∈ T x.
Since these sets Ku,v are compact, it suffices to show that they satisfy the finite intersec-tion property. Thus for {[ui,vi]}n
i=1 a finite set of elements of G (A) , it is necessary to showthat there exists a solution x to the inequalities
⟨ui− x,By+ vi⟩ ≥ 0, i = 1,2, · · · ,n
and then it follows from finite intersection property that there exists
x ∈ ∩[u,v]∈G (A)Ku,v
which is what was desired. Let Pn be all λ⃗ = (λ 1, · · · ,λ n) such that each λ k ≥ 0 and∑
nk=1 λ k = 1. Let H : Pn×Pn→ R be given by
H(
µ⃗, λ⃗)≡
n
∑i=1
µ i
⟨By+ vi,
n
∑j=1
λ ju j−ui
⟩(25.7.46)
Then this is both convex and concave in both λ⃗ , µ⃗ and so by Theorem 25.7.6, there existsµ⃗0, λ⃗ 0 both in Pn such that for all µ⃗, λ⃗ ,
H(
µ⃗, λ⃗ 0
)≤ H
(µ⃗0, λ⃗ 0
)≤ H
(µ⃗0, λ⃗
)(25.7.47)
However, plugging in µ⃗ = λ⃗ in 25.7.46,
H(⃗
λ , λ⃗)
=n
∑i=1
λ i
⟨By+ vi,
n
∑j=1
λ ju j−ui
⟩
=n
∑i=1
⟨By+ vi,
n
∑j=1
λ iλ ju j−λ iui
⟩
=n
∑i=1
⟨By+ vi,
n
∑j=1
(λ iλ ju j−λ iλ jui)
⟩
=
=0︷ ︸︸ ︷⟨By,
n
∑i=1
n
∑j=1
(λ iλ ju j−λ iλ jui)
⟩+
n
∑i=1
⟨vi,
n
∑j=1
(λ iλ ju j−λ iλ jui)
⟩The first term obviously equals 0. Consider the second. This term equals
∑i
∑j
λ iλ j⟨vi,(u j−ui)
⟩The terms equal 0 when j = i or they come in pairs
λ iλ j⟨vi,(u j−ui)
⟩+λ iλ j
⟨v j,(ui−u j)
⟩= λ iλ j
(⟨vi,(u j−ui)
⟩−⟨v j,(u j−ui)
⟩)= λ iλ j
(⟨vi,(u j−ui)
⟩−⟨v j,(u j−ui)
⟩)≤ 0