Analysis

4.3 The Brouwer Fixed Point Theorem

S ≡

is a simplex in ℝⁿ. Assume _i=1ⁿ are linearly independent. Thus a typical point of S is of the form

n ∑ tixi i=0

where the t_i are uniquely determined and the map x → t is continuous from S to the compact set

.

To see this, suppose x^k → x in S. Let x^k ≡∑ _i=0ⁿt_i^kx_i with x defined similarly with t_i^k replaced with t_i, x ≡∑ _i=0ⁿt_ix_i. Then

∑n ∑n ∑n xk − x0 = tkixi − tkix0 = tki (xi − x0) i=0 i=0 i=1

Thus

k ∑n k ∑n x − x0 = ti (xi − x0), x − x0 = ti(xi − x0) i=1 i=1

Say t_i^k fails to converge to t_i for all i ≥ 1. Then there exists a subsequence, still denoted with superscript k such that for each i = 1,

,n, it follows that t_i^k → s_i where s_i ≥ 0 and some s_i≠t_i. But then, taking a limit, it follows that

∑n ∑n x − x0 = si(xi − x0) = ti(xi − x0) i=1 i=1

which contradicts independence of the x_i − x₀. It follows that for all i ≥ 1,t_i^k → t_i. Since they all sum to 1, this implies that also t₀^k → t₀. Thus the claim about continuity is verified.

Let f : S → S be continuous. When doing f to a point x, one obtains another point of S denoted as ∑ _i=0ⁿs_ix_i. Thus in this argument the scalars s_i will be the components after doing f to a point of S denoted as ∑ _i=0ⁿt_ix_i.

Consider a triangulation of S such that all simplices in the triangulation have diameter less than ε. The vertices of the simplices in this triangulation will be labeled from p₀,

,p_n the first n + 1 prime numbers. If is one of these simplices in the triangulation, each vertex is of the form ∑ _l=0ⁿt_lx_l where t_l ≥ 0 and ∑ _lt_l = 1. Let y_i be one of these vertices, y_i = ∑ _l=0ⁿt_lx_l. Define r_j ≡ s_j∕t_j if t_j > 0 and ∞ if t_j = 0. Then p will be the label placed on y_i. To determine this label, let r_k be the smallest of these ratios. Then the label placed on y_i will be p_k where r_k is the smallest of all these extended nonnegative real numbers just described. If there is duplication, pick p_k where k is smallest.

Note that for the vertices which are on

, these will be labeled from the list because t_k = 0 for each of these and so r_k = ∞ unless k ∈. In particular, this scheme labels x_i as p_i.

By the Sperner’s lemma procedure described above, there are an odd number of simplices having value ∏ _i≠kp_i on the k^th face and an odd number of simplices in the triangulation of S for which the product of the labels on their vertices, referred to here as its value, equals p₀p₁

p_n ≡ P_n. Thus if is one of these simplices, and p is the label for y_i,

∏n ∏n p(yi) = pi ≡ Pn i=0 i=0

What is r_k, the smallest of those ratios in determining a label? Could it be larger than 1? r_k is certainly finite because at least some t_j≠0 since they sum to 1. Thus, if r_k > 1, you would have s_k > t_k. The s_j sum to 1 and so some s_j < t_j since otherwise, the sum of the t_j equalling 1 would require the sum of the s_j to be larger than 1. Hence r_k was not really the smallest after all and so r_k ≤ 1. Hence s_k ≤ t_k.

Let S≡

denote those simplices whose value is P_n. In other words, if are the vertices of one of these simplices in S, and

∑n ys = tsixi i=0

r_ks ≤ r_j for all j≠k_s and

= . Let b denote the barycenter of S_k = .

∑n b ≡ --1--yi i=0n + 1

Do the same system of labeling for each n + 1 simplex in a sequence of triangulations where the diameters of the simplices in the k^th triangulation is no more than 2^−k. Thus each of these triangulations has a n + 1 simplex having diameter no more than 2^−k which has value P_n. Let b_k be the barycenter of one of these n + 1 simplices having value P_n. By compactness, there is a subsequence, still denoted with the index k such that b_k → x. This x is a fixed point.

Consider this last claim. x = ∑ _i=0ⁿt_ix_i and after applying f, the result is ∑ _i=0ⁿs_ix_i. Then b_k is the barycenter of some σ_k having diameter no more than 2^−k which has value P_n. Say σ_k is a simplex having vertices

and the value of is P_n. Thus also

k lk→im∞ yi = x.

Re ordering these if necessary, we can assume that the label for y_i^k is p_i which implies that, as noted above, for each i = 0,

,n,

si≤ 1, si ≤ ti ti

the i^th coordinate of f

with respect to the original vertices of S decreases and each i is represented for i = . As noted above,

yki → x

and so the i^th coordinate of y_i^k,t_i^k must converge to t_i. Hence if the i^th coordinate of f

is denoted by s_i^k,

k k si ≤ ti

By continuity of f, it follows that s_i^k → s_i. Thus the above inequality is preserved on taking k →∞ and so

0 ≤ si ≤ ti

this for each i. But these s_i add to 1 as do the t_i and so in fact, s_i = t_i for each i and so f

= x. This proves the following theorem which is the Brouwer fixed point theorem.

Proof: Let S be a large simplex containing K and let P be the projection map onto K. See Problem 12 on Page 169 for the necessary properties of this projection map. Consider g

≡ f. Then g satisfies the necessary conditions for Theorem 4.3.1 and so there exists x ∈ S such that g = x. But this says x ∈ K and so g = f. ■

The proof of this corollary is pretty significant. By a homework problem, a closed convex set is a retract of ℝⁿ. This is what it means when you say there is this continuous projection map which maps onto the closed convex set but does not change any point in the closed convex set. When you have a set A which is a subset of a set B which has the property that continuous functions f : B → B have fixed points, and there is a continuous map P from B to A which leaves points of A unchanged, then it follows that A will have the same “fixed point property”. You can probably imagine all sorts of sets which are retracts of closed convex bounded sets.