876 CHAPTER 25. NONLINEAR OPERATORS

by monotonicity of A. Hence H(⃗

λ , λ⃗)≤ 0. Then from 25.7.47, for all µ⃗

H(

µ⃗, λ⃗ 0

)≤ H

(µ⃗0, λ⃗ 0

)≤ H (⃗µ0, µ⃗0)≤ 0

It follows that

m

∑i=1

µ i

⟨By+ vi,

n

∑j=1

λ0ju j−ui

⟩≤ 0

m

∑i=1

µ i

⟨By+ vi,ui−

n

∑j=1

λ0ju j

⟩≥ 0

where λ⃗ 0 ≡(

λ01, · · · ,λ

0n

). This is true for any choice of µ⃗. In particular, you could let µ⃗

equal 1 in the ith position and 0 elsewhere and conclude that for all i = 1, · · · ,n,⟨By+ vi,ui−

n

∑j=1

λ0ju j

⟩≥ 0

so you let x = ∑nj=1 λ

0ju j and this shows that Ty ̸= /0 because the sets Ku,v have the finite

intersection property.Thus T : K→P (K) and for each y ∈ K,Ty ̸= /0. In fact this is true for any y but we are

only considering y ∈ K. Now Ty is clearly a closed subset of K. It is also clearly convex. Isit upper semicontinuous? Let yk → y and consider Ty+B(0,r) . Is Tyk ∈ Ty+B(0,r) forall k large enough? If not, then there is a subsequence, denoted as zk ∈ Tyk which is outsidethis open set Ty+B(0,r). Then taking a further subsequence, still denoted as zk, it followsthat zk→ z /∈ Ty+B(0,r) . Now

⟨Byk + v,u− zk⟩ ≥ 0 all [u,v] ∈ G (A)

Therefore, from continuity of B,

⟨By+ v,u− z⟩ ≥ 0 all [u,v] ∈ G (A)

which means z∈ Ty contrary to the assumption that T is not upper semicontinuous. Since Tis upper semicontinuous and maps to compact convex sets, it follows from Theorem 25.4.4that T has a fixed point x ∈ T x. Hence there exists a solution x to

⟨Bx+ v,u− x⟩ ≥ 0 all [u,v] ∈ G (A)

Next suppose that K is only closed and convex but B is coercive and 0 ∈ D(A). Thenlet Kn ≡ B(0,n)∩K and let An be the restriction of A to B(0,n). It follows that there existsxn ∈ Kn such that for all [u,v] ∈ G (An) ,

⟨Bxn + v,u− xn⟩ ≥ 0

Then since 0 ∈ D(A) , one can pick v0 ∈ A0 and obtain

⟨Bxn + v0,−xn⟩ ≥ 0, ⟨v0,−xn⟩ ≥ ⟨Bxn,xn⟩