180 CHAPTER 6. FIXED POINT THEOREMS

for [ui,vi] ∈ G (A) , for i = 1,2, · · · ,n there exists a solution x ∈ K to the system ofinequalities

⟨z+ vi,ui− x⟩ ≥ 0, i = 1,2, · · · ,n

Hint: 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

⟨z+ vi,

n

∑j=1

λ ju j−ui

⟩(6.17)

Show that it is both convex and concave in both arguments. Then apply the min maxtheorem. Then argue that H

(⃗λ , λ⃗

)≤ 0 from monotonicity considerations. Letting(

µ⃗0, λ⃗ 0

)be the saddle point, you will have

H(

µ⃗, λ⃗ 0

)≤ H

(µ⃗0, λ⃗ 0

)≤ H

(µ⃗0, λ⃗

)H(

µ⃗, λ⃗ 0

)≤ H

(µ⃗0, λ⃗ 0

)≤ H (⃗µ0, µ⃗0)≤ 0

H(

µ⃗, λ⃗ 0

)≤ 0

Now choose µ⃗ judiciously while allowing λ⃗ 0 to be used to define x which satisfiesall the inequalities.

35. ↑It gets even better. Let Ku,v≡{x ∈ K : ⟨z+ v,u− x⟩ ≥ 0} . Show that Ku,v is compactand that the sets Ku,v have the finite intersection property. Therefore, there existsx ∈ ∩[u,v]∈G (A)Ku,v. Explain why ⟨z+ v,u− x⟩ ≥ 0 for all [u,v] ∈ G (A). What wouldthe inequalities be if −A were monotone?

36. Problem 33 gave a solution to the inequality ⟨ f − z,y− x⟩ ≤ 0,z ∈ Ax under the con-dition that A is upper semicontinuous. What are the differences between the result inthe above problem and the result of Problem 33. You could replace A with −A in theearlier problem. If you did, would you get the result of the above problem?

37. Are there convenient examples of monotone set valued maps? Yes, there are. Let Xbe a Banach space and let φ : X → (−∞,∞] be convex, lower semicontinuous, andproper. See Problem 28 for a discussion of lower semicontinuous. Proper means thatφ (x) < ∞ for some x. Convex means the usual thing. φ (tx+(1− t)y) ≤ tφ (x)+(1− t)φ (y) where t ∈ [0,1]. Then x∗ ∈ ∂φ (x) means that

⟨x∗,z− x⟩ ≤ φ (z)−φ (x) , for all z ∈ X

Show that if x∗ ∈ ∂φ (x) , then φ (x) < ∞. The set of points x where φ (x) < ∞ iscalled the domain of φ denoted as D(φ). Also show that if [x,x∗] , [x̂, x̂∗] are twopoints of the graph of ∂φ , then ⟨x̂∗− x∗, x̂− x⟩ ≥ 0 so that ∂φ is an example of amonotone graph. You might wonder whether this graph is nonempty. See the nextproblem for a partial answer to this question. Of course the above problem pertainsto finite dimensional spaces so you could just take any φ : Rn→ R which is convexand differentiable. You can see that in this case the subgradient coincides with thederivative discussed later.

18035.36.37.CHAPTER 6. FIXED POINT THEOREMSfor [ui,v;] © Y (A), for i= 1,2,--- ,n there exists a solution x € K to the system ofinequalities(c+vj,uj —x) > 0,i=1,2,---,nHint: Let P, be all A= (A1,--+,An) such that each A, > 0 and Y?_, Ax = 1. LetH: P, x P, > R be given byH(ii,4) Ea (oem aas a) (6.17)Show that it is both convex and concave in both arguments. Then apply the min maxtheorem. Then argue that 7 (4.4) <0 from monotonicity considerations. Letting(fio. io) be the saddle point, you will haveH (ii,40) < H (ilp,Ao) <H (Bip. )H(fi,40) < H (Zoo) <H (Go.flo) <0H (ii,4o) < 0Now choose ff judiciously while allowing ho to be used to define x which satisfiesall the inequalities.fit gets even better. Let K,, = {x € K : (z+v,u—x) > 0}. Show that K,,, is compactand that the sets K,,, have the finite intersection property. Therefore, there existsx € Nw vjeg(a)Kuy- Explain why (z+ v,u—x) > 0 for all [u,v] € Y (A). What wouldthe inequalities be if —A were monotone?Problem 33 gave a solution to the inequality (f — z,y—x) < 0,z € Ax under the con-dition that A is upper semicontinuous. What are the differences between the result inthe above problem and the result of Problem 33. You could replace A with —A in theearlier problem. If you did, would you get the result of the above problem?Are there convenient examples of monotone set valued maps? Yes, there are. Let Xbe a Banach space and let @ : X — (—ce,0] be convex, lower semicontinuous, andproper. See Problem 28 for a discussion of lower semicontinuous. Proper means that(x) < ce for some x. Convex means the usual thing. @ (tx+(1—t)y) < t@ (x) +(1 —t) @ (y) where t € [0,1]. Then x* € 0 (x) means that(x*,z—x) < 6(z)— (x), forall ze XShow that if x* € 0 (x), then @ (x) < co. The set of points x where @ (x) < co iscalled the domain of @ denoted as D(@). Also show that if [x,x*],[%,2*] are twopoints of the graph of 0@, then (%* —x*,£—x) > 0 so that 0@ is an example of amonotone graph. You might wonder whether this graph is nonempty. See the nextproblem for a partial answer to this question. Of course the above problem pertainsto finite dimensional spaces so you could just take any @ : R” — R which is convexand differentiable. You can see that in this case the subgradient coincides with thederivative discussed later.