326 CHAPTER 13. DEGREE THEORY

8. Suppose C is a compact set in Rn which has empty interior and f : C→ Γ ⊆ Rn isone to one onto and continuous with continuous inverse. Could Γ have nonemptyinterior? Show also that if f is one to one and onto Γ then if it is continuous, so isf−1.

9. Let K be a nonempty closed and convex subset of Rp. Recall K is convex means thatif x,y ∈K, then for all t ∈ [0,1] , tx+(1− t)y ∈K. Show that if x∈Rp there existsa unique z ∈ K such that |x−z| = min{|x−y| : y ∈ K} . This z will be denotedas Px. Hint: First note you do not know K is compact. Establish the parallelogramidentity if you have not already done so,

|u−v|2 + |u+v|2 = 2 |u|2 +2 |v|2 .

Then let {zk} be a minimizing sequence,

limk→∞

|zk−x|2 = inf{|x−y| : y ∈ K} ≡ λ .

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.

10. In Problem 9 show that Px satisfies the following variational inequality. (x−Px) ·(y−Px)≤ 0 for all y ∈ K. Then show that |Px1−Px2| ≤ |x1−x2|. Hint: For thefirst part note that if y ∈ K, the function

t→ |x−(Px+ t (y−Px))|2

achieves its minimum on [0,1] at t = 0. For the second part,

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

Explain why(x2−Px2− (x1−Px1)) · (Px2−Px1)≥ 0

and then use a some manipulations and the Cauchy Schwarz inequality to get thedesired inequality.

11. Suppose D is a set which is homeomorphic to B(0,1). This means there exists acontinuous one to one map, h such that h

(B(0,1)

)= D such that h−1 is also one

to one. Show that if f is a continuous function which maps D to D then f has a fixedpoint. Now show that it suffices to say that h is one to one and continuous. In thiscase the continuity of h−1 is automatic. Sets which have the property that continuousfunctions taking the set to itself have at least one fixed point are said to have the fixedpoint property. Work Problem 7 using this notion of fixed point property. What abouta solid ball and a donut? Could these be homeomorphic?

12. Using the definition of the derivative and the Vitali covering theorem, show thatif f ∈ C1

(U,Rn

)and ∂U has n dimensional measure zero then f (∂U) also has

measure zero. (This problem has little to do with this chapter. It is a review.)

3268.10.11.12.CHAPTER 13. DEGREE THEORYSuppose C is a compact set in R” which has empty interior and f :C +I C R” isone to one onto and continuous with continuous inverse. Could I have nonemptyinterior? Show also that if f is one to one and onto I then if it is continuous, so isfi.Let K be a nonempty closed and convex subset of IR’. Recall K is convex means thatif x,y € K, then for allt € [0,1] ,ta+(1 —t) y € K. Show that if x € R? there existsa unique z € K such that |jw— z| = min{|x—y|: y € K}. This z will be denotedas Px. Hint: First note you do not know K is compact. Establish the parallelogramidentity if you have not already done so,ju—v|>+lu+ol? =2|ul*+2]o|?.Then let {z;} be a minimizing sequence,lim |z,—a@|° = inf {|a—y|:y CK} =A.k—y00Using convexity, explain why2 2L— Zk22Zk— 2m2L—Zm22ZK+Zm2=2 +2+eand then use this to argue {z,} is a Cauchy sequence. Then if z; works for i= 1,2,consider (z| + 22) /2 to get a contradiction.In Problem 9 show that Px satisfies the following variational inequality. (w—Pa) -(y—Px) <0 for all y € K. Then show that |Pa — Pa2| < |x) — @2|. Hint: For thefirst part note that if y € K, the functiont > |w— (Pa +t(y—Px))|”achieves its minimum on [0, 1] at t = 0. For the second part,(a — Pa) - (Paz — Pa) <0, (a2 —Par)- (Pa, — Par) <0.Explain why(a2 — Pay — (x1 — Px1))-(Px2— Px) >0and then use a some manipulations and the Cauchy Schwarz inequality to get thedesired inequality.Suppose D is a set which is homeomorphic to B(0,1). This means there exists acontinuous one to one map, A such that h (3 (0, 0) =D such that h~! is also oneto one. Show that if f is a continuous function which maps D to D then f has a fixedpoint. Now show that it suffices to say that h is one to one and continuous. In thiscase the continuity of h~ | is automatic. Sets which have the property that continuousfunctions taking the set to itself have at least one fixed point are said to have the fixedpoint property. Work Problem 7 using this notion of fixed point property. What abouta solid ball and a donut? Could these be homeomorphic?Using the definition of the derivative and the Vitali covering theorem, show thatif f ec! (U,R") and QU has n dimensional measure zero then f (OU) also hasmeasure zero. (This problem has little to do with this chapter. It is a review.)