6.7. EXERCISES 179

identity if you have not already done so, |u−v|2 + |u+v|2 = 2 |u|2 +2 |v|2 . Thenlet {zk} be a minimizing sequence,

limk→∞

|zk−x|2 = inf{|x−y| : y ∈ K} ≡ λ .

Now using convexity, explain why∣∣∣∣zk−zm

2

∣∣∣∣2 + ∣∣∣∣x−zk +zm

2

∣∣∣∣2 = 2∣∣∣∣x−zk

2

∣∣∣∣2 +2∣∣∣∣x−zm

2

∣∣∣∣2and then use this to argue {zk} is a Cauchy sequence. Then if zi works for i = 1,2,consider (z1 +z2)/2 to get a contradiction.

31. In Problem 30 show that Px satisfies and is in fact characterized as the solution tothe following variational inequality. (x−Px,y−Px) ≤ 0 for all y ∈ K. Then showthat |Px1−Px2| ≤ |x1−x2|. Hint: For the first part note that if y ∈ K, the functiont → |x−(Px+ t (y−Px))|2 achieves its minimum on [0,1] at t = 0. For the secondpart,

(x1−Px1) · (Px2−Px1)≤ 0, (x2−Px2) · (Px1−Px2)≤ 0.

Explain why (x2−Px2− (x1−Px1)) · (Px2−Px1) ≥ 0 and then use a some ma-nipulations and the Cauchy Schwarz inequality to get the desired inequality. Thus Pis called a retraction onto K.

32. Browder’s lemma says: Let K be a convex closed and bounded set in Rn and letA : K → Rn be continuous and f ∈ Rn. Then there exists x ∈ K such that for ally ∈ K,

(f −Ax,y−x)≤ 0

show this is true. Hint: Consider x→P(f −Ax+x) where P is the projection ontoK. If there is a fixed point of this mapping, then P(f −Ax+x) = x. Now considerthe variational inequality satisfied. This little lemma is the basis for a whole lot ofnonlinear analysis involving nonlinear operators of various kinds.

33. Generalize the above problem as follows. Let K be a convex closed and boundedset in Rn and let A : K→P (Rn) be upper semi-continuous having closed boundedconvex values and f ∈ Rn. Then there exists x ∈ K and z ∈ Ax such that for ally ∈ K, (f −z,y−x)≤ 0 show this is true. Also show that if K is a closed convexand bounded set in E a finite dimensional normed linear space and A : K→P (E ′) isupper semicontinuous having closed bounded convex values and f ∈ E ′, then thereexists x ∈ K and z ∈ Ax such that for all y ∈ K,⟨ f − z,y− x⟩ ≤ 0. Hint: Use the con-struction for the proof of the Kakutani fixed point theorem and the above Browder’slemma.

34. This problem establishes a remarkable result about existence for a system of in-equalities based on the min max theorem, Theorem 5.12.5. Let E be a finite dimen-sional Banach space and let K be a convex and compact subset of E. A set valuedmap A : D(A) ⊆ K → E ′ is called monotone if whenever vi ∈ Aui, it follows that⟨v1− v2,u1−u2⟩ ≥ 0. The graph, denoted as G (A) consists of all pairs [u,v] suchthat v ∈ Au. This is a monotone subset of E ×E ′. Let z ∈ E ′ be fixed. Show that

6.7. EXERCISES 17931.32.33,34.identity if you have not already done so, |u—v|* +|u+ v|? = 2|u|? +2|v|?. Thenlet {z,} be a minimizing sequence,lim |z,—a|° = inf {|a—y|:y CK} =A.k-ooNow using convexity, explain why2L— Zk2zk 2m °22Zzk— 2m2L—Zm=22+2+eand then use this to argue {z;,} is a Cauchy sequence. Then if z; works for i = 1,2,consider (21 + 22) /2 to get a contradiction.In Problem 30 show that Pa satisfies and is in fact characterized as the solution tothe following variational inequality. (e—Px,y—Px) < 0 for all y € K. Then showthat |Pa, — Pa2| < |a, —a2|. Hint: For the first part note that if y € K, the functiont + |w—(Pa+t(y—Px))|” achieves its minimum on (0, 1] at t = 0. For the secondpart,(ay — Px) . (Px — Px) <0, (az — Px) : (Px — Px) <0.Explain why (a2 — Paz — (a1 — Px,))-(Px2— Px) > 0 and then use a some ma-nipulations and the Cauchy Schwarz inequality to get the desired inequality. Thus Pis called a retraction onto K.Browder’s lemma says: Let K be a convex closed and bounded set in R” and letA: K — R" be continuous and f € R”. Then there exists 2 € K such that for ally € K,(f —Av,y—x) <0show this is true. Hint: Consider « + P(f —Ax-+ <x) where P is the projection ontoK. If there is a fixed point of this mapping, then P(f — Ax +a) = a. Now considerthe variational inequality satisfied. This little lemma is the basis for a whole lot ofnonlinear analysis involving nonlinear operators of various kinds.Generalize the above problem as follows. Let K be a convex closed and boundedset in R” and let A: K + A(R”) be upper semi-continuous having closed boundedconvex values and f € R”. Then there exists x € K and z € Ax such that for ally €K, (f —z,y—2) <0 show this is true. Also show that if K is a closed convexand bounded set in E a finite dimensional normed linear space and A: K > Y(E’) isupper semicontinuous having closed bounded convex values and f € E’, then thereexists x € K and z € Ax such that for all y € K, (f —z,y—x) <0. Hint: Use the con-struction for the proof of the Kakutani fixed point theorem and the above Browder’slemma.This problem establishes a remarkable result about existence for a system of in-equalities based on the min max theorem, Theorem 5.12.5. Let EF be a finite dimen-sional Banach space and let K be a convex and compact subset of E. A set valuedmap A: D(A) C K —> E’ is called monotone if whenever v; € Au;, it follows that(v1 —v2,u; —u2) > 0. The graph, denoted as ¥ (A) consists of all pairs [u,v] suchthat v € Au. This is a monotone subset of E x E’. Let z € E’ be fixed. Show that