366 CHAPTER 13. LEBESGUE MEASURE

13.10 The Brouwer Fixed Point TheoremThis seems to be a good place to present a short proof of one of the most important of allfixed point theorems. There are many approaches to this but one of the easiest and shortestI have ever seen is the one in Dunford and Schwartz [45]. This is what is presented here. InEvans [48] there is a different proof which depends on integration theory. A good referencefor an introduction to various kinds of fixed point theorems is the book by Smart [118]. Thisbook also gives an entirely different approach to the Brouwer fixed point theorem.

The proof given here is based on the following lemma. Recall that for A an n×n matrix,cof(A)i j is the determinant of the matrix which results from deleting the ith row and the jth

column and multiplying by (−1)i+ j. The following lemma is proved earlier. See Lemma16.3.1.

Lemma 13.10.1 Let g : U → Rn be C2 where U is an open subset of Rn. Then

n

∑j=1

cof(Dg)i j, j = 0,

where here (Dg)i j ≡ gi, j ≡ ∂gi∂x j

. Also, cof(Dg)i j =∂ det(Dg)

∂gi, j.

To prove the Brouwer fixed point theorem, first consider a version of it valid for C2

mappings. This is the following lemma.

Lemma 13.10.2 Let Br = B(0,r) and suppose g is a C2 function defined onRn which mapsBr to Br. Then g(x) = x for some x ∈ Br.

Proof: Suppose not. Then |g(x)−x|must be bounded away from zero on Br. Let a(x)be the larger of the two roots of the equation,

|x+ z(x−g(x))|2 = |x|2 +2(x,x−g(x))z+ z2 |x−g(x)|2 = r2. (13.10.23)

Thus, from the quadratic formula,

a(x) =−(x,(x−g(x)))+

√(x,(x−g(x)))2 +

(r2−|x|2

)|x−g(x)|2

|x−g(x)|2(13.10.24)

That under the square root is positive if |x| < r. What if |x| = r? In this case, sinceg(x) ∈ Br, you cannot have (x,(x−g(x))) = 0 because if so, you would have

r2 = (x,g(x)) = |x| |g(x)|cos(θ)

so r = |g(x)|cosθ where θ is the angle between the vectors x,g(x). The only way this canhappen is for g(x) = x and this is assumed not to occur. Thus what is under the square rootsign is always positive. It follows that a(·) is a C2 function because t →

√t is smooth on

t > 0. When |x|= r, one solution to 13.10.23 is z = 0. There is also a solution for negative zbased on geometric reasoning. Therefore, these are the two roots to 13.10.23, one negative