Ch.1 Mathematical Preliminaries

Advanced Microeconomics, Spring 2016.

1. Fixed Point Theorem

Keishun Suzuki (Chiba University) 11 April, 2016 1

References • The following books must be helpful to understand the contents:  Border (1989). “Fixed point theorems with applications to economics and game theory,” Cambridge university press.  Fuente (2000). “Mathematical Methods and Models for Economists,” Cambridge university press.  Jehle and Reny (2010). Mathematical Appendix, in “Advanced microeconomic theory, 3rd edition,” Prentice Hall.  Mas-colell, Whinston and Green (1995). Mathematical Appendix, in “Microeconomic Theory,” Oxford University Press.  川越敏司 (2010). ブラウワーの不動点定理とスペルナーの補題, “行動ゲーム 理論入門,” NTT出版, pp.48-51.  二階堂副包 (1960). “現代経済学の数学的方法,” 岩波書店.

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Affine Combination Definition 1.1 Affine Combination An affine combination of points 𝒙𝒙𝟎𝟎 , … , 𝒙𝒙𝒏𝒏 in ℝ𝑚𝑚 is 𝒏𝒏

� 𝝀𝝀𝒊𝒊 𝒙𝒙𝒊𝒊 = 𝝀𝝀𝟎𝟎 𝒙𝒙𝟎𝟎 + ⋯ + 𝝀𝝀𝒏𝒏 𝒙𝒙𝒏𝒏 , 𝒊𝒊=𝟎𝟎

which 𝝀𝝀𝟎𝟎 ,…,𝝀𝝀𝒏𝒏 are coefficients satisfying ∑𝒏𝒏𝒊𝒊=𝟎𝟎 𝝀𝝀𝒊𝒊 = 𝟏𝟏.

𝒙𝒙𝟎𝟎

𝒙𝒙𝟏𝟏

𝒙𝒙𝟑𝟑

𝒙𝒙2

Example: 𝒙𝒙𝟎𝟎 = 𝟐𝟐, 𝟒𝟒 , 𝒙𝒙𝟏𝟏 = (𝟔𝟔, 𝟖𝟖) 𝒙𝒙𝟐𝟐 = (𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏)

⇒ 𝒙𝒙𝟐𝟐 = −𝟏𝟏 𝒙𝒙𝟎𝟎 + 𝟐𝟐𝒙𝒙𝟏𝟏

𝒙𝒙𝟑𝟑 = (𝟖𝟖, 𝟒𝟒) ⇒ 𝒙𝒙𝟑𝟑 = −𝟓𝟓 𝒙𝒙𝟎𝟎 + 𝟑𝟑𝒙𝒙𝟏𝟏

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Affine Independence Definition 1.2 Affine Independence A set {𝒙𝒙𝟎𝟎 , … , 𝒙𝒙𝒏𝒏 } is affinely independent if it does not contain 𝒙𝒙𝒊𝒊 that can be affine combination of other points, such that

𝒙𝒙𝒊𝒊 = 𝝀𝝀𝟎𝟎 𝒙𝒙𝟎𝟎 + ⋯ + 𝝀𝝀𝒊𝒊−𝟏𝟏 𝒙𝒙𝒊𝒊−𝟏𝟏 + 𝝀𝝀𝒊𝒊+𝟏𝟏 𝒙𝒙𝒊𝒊+𝟏𝟏 + ⋯ + 𝝀𝝀𝒏𝒏 𝒙𝒙𝒏𝒏 ,

  

which the sum of 𝝀𝝀𝟎𝟎 , … , 𝝀𝝀𝒊𝒊−𝟏𝟏 , 𝝀𝝀𝒊𝒊+𝟏𝟏 , … , 𝝀𝝀𝒏𝒏 is 𝟏𝟏.

𝒙𝒙𝟏𝟏 , 𝒙𝒙𝟐𝟐 are affinely independent ⇔ They are not the same point.

𝒙𝒙𝟏𝟏 , 𝒙𝒙𝟐𝟐 , 𝒙𝒙𝟑𝟑 are affinely independent ⇔ They are not on the same straight line. 𝒙𝒙𝟏𝟏 , 𝒙𝒙𝟐𝟐 , 𝒙𝒙𝟑𝟑 , 𝒙𝒙𝟒𝟒 are affinely independent ⇔ They are not on the same plane.

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Exercise Exercise 1.3 Assume that a set of points {𝒙𝒙𝟎𝟎 , … , 𝒙𝒙𝒏𝒏 } in ℝ𝑚𝑚 is affinely

independent. Then, ∑𝒏𝒏𝒊𝒊=𝟎𝟎 𝝀𝝀𝒊𝒊 𝒙𝒙𝒊𝒊 = 𝟎𝟎 and ∑𝒏𝒏𝒊𝒊=𝟎𝟎 𝝀𝝀𝒊𝒊 = 𝟎𝟎 imply that 𝝀𝝀𝟎𝟎 = ⋯ = 𝝀𝝀𝒏𝒏 = 𝟎𝟎. Prove this.

Exercise 1.4 (a) Show that {𝒙𝒙𝟎𝟎 , 𝒙𝒙𝟏𝟏 , 𝒙𝒙𝟐𝟐 , 𝒙𝒙𝟑𝟑 } in ℝ2 cannot be affinely independent.

(b) Show that if {𝒙𝒙𝟎𝟎 , … , 𝒙𝒙𝒏𝒏 } in ℝ𝑚𝑚 is affinely independent, then 𝒎𝒎 ≥ 𝒏𝒏.

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Convex Combination Definition 1.5 Convex Combination A convex combination of points 𝒙𝒙𝟎𝟎 , … , 𝒙𝒙𝒏𝒏 in ℝ𝑚𝑚 is 𝒏𝒏

� 𝝀𝝀𝒊𝒊 𝒙𝒙𝒊𝒊 = 𝝀𝝀𝟎𝟎 𝒙𝒙𝟎𝟎 + ⋯ + 𝝀𝝀𝒏𝒏 𝒙𝒙𝒏𝒏 , 𝒊𝒊=𝟎𝟎

which 𝝀𝝀𝟎𝟎 ,…,𝝀𝝀𝒏𝒏 are non-negative scalars satisfying ∑𝒏𝒏𝒊𝒊=𝟎𝟎 𝝀𝝀𝒊𝒊 = 𝟏𝟏.

𝒙𝒙𝟎𝟎

𝒙𝒙2

𝒙𝒙𝟏𝟏

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Convex Hull Definition 1.6 Convex Hull A convex hull of set 𝑨𝑨 ⊂ ℝ𝒎𝒎 , denoted 𝒄𝒄𝒄𝒄 𝑨𝑨, is the set of all convex combinations from 𝑨𝑨. Formally, 𝒄𝒄𝒄𝒄 𝑨𝑨 = {𝒙𝒙 ∈ ℝ𝒎𝒎 ; 𝒙𝒙 = ∑𝒏𝒏𝒊𝒊=𝟎𝟎 𝝀𝝀𝒊𝒊 𝒙𝒙𝒊𝒊 },

where 𝒙𝒙𝒊𝒊 ∈ 𝑨𝑨, 𝝀𝝀𝟎𝟎 ,…,𝝀𝝀𝒏𝒏 ∈ ℝ+ , and ∑𝒏𝒏𝒊𝒊=𝟎𝟎 𝝀𝝀𝒊𝒊 = 𝟏𝟏.

Example: 𝑨𝑨 = {𝒙𝒙𝟎𝟎 , 𝒙𝒙𝟏𝟏 , 𝒙𝒙𝟐𝟐 }

⇒ 𝒄𝒄𝒄𝒄 𝑨𝑨 = ∆𝒙𝒙𝟎𝟎 𝒙𝒙𝟏𝟏 𝒙𝒙𝟐𝟐

𝒙𝒙𝟎𝟎

𝒙𝒙𝟏𝟏

𝒙𝒙𝟐𝟐

Example: 𝑩𝑩 = {𝒙𝒙𝟎𝟎 , 𝒙𝒙𝟏𝟏 , 𝒙𝒙𝟐𝟐 , 𝒙𝒙𝟑𝟑 , 𝒙𝒙𝟒𝟒 } 𝒙𝒙𝟎𝟎

𝒙𝒙𝟑𝟑

𝒙𝒙𝟒𝟒

𝒙𝒙𝟏𝟏

𝒙𝒙𝟐𝟐

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Simplex Definition 1.7 (Closed) n-simplex An (closed) n-simplex, 𝑺𝑺𝒏𝒏 ,is the convex hull of 𝒏𝒏 + 𝟏𝟏 points

affinely independent. Formally, it is given by 𝒙𝒙𝟎𝟎 , … , 𝒙𝒙𝒏𝒏 ∈ ℝ𝒏𝒏+𝟏𝟏 +

𝒏𝒏 𝒏𝒏+𝟏𝟏 𝒊𝒊 , where 𝝀𝝀 , … , 𝝀𝝀 ∑ 𝑺𝑺𝒏𝒏 = 𝒙𝒙 ∈ ℝ𝒏𝒏+𝟏𝟏 ; 𝒙𝒙 = 𝝀𝝀 𝒙𝒙 ∈ ℝ and 𝟎𝟎 𝒏𝒏 + + 𝒊𝒊=𝟎𝟎 𝒊𝒊

∑𝒏𝒏𝒊𝒊=𝟎𝟎 𝝀𝝀𝒊𝒊 = 𝟏𝟏.

Definition 1.8 Unit n-simplex An unit n-simplex, denoted by ∆𝒏𝒏 , is given by ∆𝒏𝒏 =

𝒏𝒏

𝒚𝒚𝟎𝟎 , … , 𝒚𝒚𝒏𝒏 ∈ ℝ𝒏𝒏+𝟏𝟏 + ; � 𝒚𝒚𝒊𝒊 = 𝟏𝟏 , 𝒊𝒊=𝟎𝟎

that is the convex hull of 𝒏𝒏 + 𝟏𝟏 unit vectors 𝒆𝒆𝟎𝟎 , … , 𝒆𝒆𝒏𝒏 ∈ ℝ𝒏𝒏+𝟏𝟏 + .

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Example: unit 2-simplex ∆𝟐𝟐 =

𝒚𝒚𝟎𝟎 , 𝒚𝒚𝟏𝟏 , 𝒚𝒚𝟐𝟐 ∈ ℝ𝟑𝟑+ ; 𝒚𝒚𝟎𝟎 + 𝒚𝒚𝟏𝟏 + 𝒚𝒚𝟐𝟐 = 𝟏𝟏

= 𝒄𝒄𝒄𝒄 𝒄𝒄𝒄𝒄 {𝒆𝒆𝟎𝟎 , 𝒆𝒆𝟏𝟏 , 𝒆𝒆𝟐𝟐 }.

• ∆𝟐𝟐 is the convex hull of three affinely

𝒆𝒆𝟐𝟐

independent points 𝒆𝒆𝟎𝟎 , 𝒆𝒆𝟏𝟏 , 𝒆𝒆𝟐𝟐 such that

𝑒𝑒 0 = 1,0,0 , 𝑒𝑒1 = 0,1,0 , 𝑒𝑒 2 = 0,0,1 .

• ∆𝟐𝟐 is equilateral triangle in the plane.

𝒆𝒆𝟎𝟎

𝒆𝒆𝟏𝟏

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Simplicial subdivision Definition 1.9 Simplicial subdivision of an n-simplex A simplicial subdivision of an n-simplex 𝑺𝑺𝒏𝒏 is a finite collection

of simplexes {𝑺𝑺𝒊𝒊 : 𝒊𝒊 ∈ 𝑰𝑰} satisfying ⋃𝒊𝒊∈𝑰𝑰 𝑺𝑺𝒊𝒊 = 𝑺𝑺𝒏𝒏 and such that for

any 𝒊𝒊, 𝒋𝒋 ∈ 𝑰𝑰 , 𝑺𝑺𝒊𝒊 ∩ 𝑺𝑺𝒋𝒋 is either empty or equal to the closure of a common face.

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Example: is it simplicial subdivision? • Is the collection {𝒙𝒙𝟎𝟎 𝒙𝒙𝟐𝟐 𝒙𝒙𝟒𝟒 , 𝒙𝒙𝟏𝟏 𝒙𝒙𝟐𝟐 𝒙𝒙𝟑𝟑 , 𝒙𝒙𝟏𝟏 𝒙𝒙𝟑𝟑 𝒙𝒙𝟒𝟒 } simplicial subdivision of ∆𝟐𝟐 = 𝒙𝒙𝟎𝟎 𝒙𝒙𝟏𝟏 𝒙𝒙𝟐𝟐 ? → No.

 𝒙𝒙𝟎𝟎 𝒙𝒙𝟐𝟐 𝒙𝒙𝟒𝟒 ∪ 𝒙𝒙𝟏𝟏 𝒙𝒙𝟐𝟐 𝒙𝒙𝟑𝟑 ∪ 𝒙𝒙𝟏𝟏 𝒙𝒙𝟑𝟑 𝒙𝒙𝟒𝟒 = 𝒙𝒙𝟎𝟎 𝒙𝒙𝟏𝟏 𝒙𝒙𝟐𝟐 .  𝒄𝒄𝒄𝒄

𝒙𝒙𝟏𝟏 𝒙𝒙𝟐𝟐 𝒙𝒙𝟑𝟑

∩ 𝒄𝒄𝒄𝒄

𝒙𝒙𝟏𝟏 𝒙𝒙𝟑𝟑 𝒙𝒙𝟒𝟒

= 𝒄𝒄𝒄𝒄

𝒙𝒙𝟏𝟏 𝒙𝒙𝟑𝟑 .

 𝒙𝒙𝟏𝟏 𝒙𝒙𝟑𝟑 is the closure of a common face.

 𝒄𝒄𝒄𝒄 𝒙𝒙𝟎𝟎 𝒙𝒙𝟐𝟐 𝒙𝒙𝟒𝟒 ∩ 𝒄𝒄𝒄𝒄 𝒙𝒙𝟏𝟏 𝒙𝒙𝟐𝟐 𝒙𝒙𝟑𝟑 = 𝒄𝒄𝒄𝒄 𝒙𝒙𝟐𝟐 𝒙𝒙𝟑𝟑 . 𝒙𝒙𝟎𝟎

𝒙𝒙𝟐𝟐 𝒙𝒙𝟑𝟑

 𝒄𝒄𝒄𝒄 𝒙𝒙𝟐𝟐 𝒙𝒙𝟑𝟑 is not the closure of a face of 𝒙𝒙𝟎𝟎 𝒙𝒙𝟐𝟐 𝒙𝒙𝟒𝟒 .

𝒙𝒙𝟒𝟒

𝒙𝒙𝟏𝟏 11

Example: Equilateral subdivision of ∆𝟐𝟐 • The length of side in triangle ∆𝟐𝟐 is 2.

• Suppose that we divide each side into m equal intervals. Then, we can divide ∆𝟐𝟐 into small equilateral triangles.  m is any positive integer.

• The subdivision has 𝒎𝒎𝒏𝒏 2-simplexes that the length of side is 2/𝑚𝑚.

𝒎𝒎 = 𝟐𝟐

𝒎𝒎 = 𝟑𝟑

𝒎𝒎 = 𝟒𝟒

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Proper labeling • Let 𝑺𝑺𝒏𝒏 = 𝒄𝒄𝒄𝒄 𝒄𝒄𝒄𝒄 {𝒙𝒙𝟎𝟎 , 𝒙𝒙𝟏𝟏 , … , 𝒙𝒙𝒏𝒏 } be simplicially subdivided.

• Let 𝑽𝑽 denote the collection of all the vertices of all the subsimplexes of 𝑺𝑺𝒏𝒏 .

• For 𝒗𝒗 = ∑𝒏𝒏𝒊𝒊=𝟎𝟎 𝝀𝝀𝒊𝒊 𝒙𝒙𝒊𝒊 ∈ 𝑺𝑺𝒏𝒏 , let 𝝌𝝌 𝒗𝒗 = 𝒊𝒊: 𝝀𝝀𝒊𝒊 > 𝟎𝟎 .

 𝝌𝝌 𝒗𝒗 , a subset of index set 𝑰𝑰, is a set of the indices of vertices 𝒙𝒙𝒊𝒊 used in a convex combination 𝒗𝒗.

Definition 1.10 Proper Labeling

A function 𝝀𝝀: 𝑽𝑽 → {𝟎𝟎, … , 𝒏𝒏} satisfying 𝝀𝝀(𝒗𝒗) ∈ 𝝌𝝌(𝒗𝒗) is called a

proper labeling of the subdivision.

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Example: Which is properly labeled? 𝟐𝟐 𝟎𝟎 𝟎𝟎

𝒙𝒙𝟎𝟎

𝒙𝒙𝟐𝟐

𝒙𝒙𝟐𝟐 𝒙𝒙𝟒𝟒 𝟎𝟎

𝟑𝟑

𝒙𝒙

𝟏𝟏

𝒙𝒙𝟓𝟓

𝒙𝒙𝟒𝟒 𝟐𝟐

𝒙𝒙𝟑𝟑 𝟐𝟐 𝟏𝟏

𝒙𝒙𝟏𝟏

𝟎𝟎

𝒙𝒙𝟎𝟎

𝟐𝟐

𝟎𝟎

𝒙𝒙𝟓𝟓

𝟏𝟏

𝒙𝒙𝟏𝟏 14

Proper labeling in unit 2-simplex 𝚫𝚫𝟐𝟐 • If 𝝀𝝀 is proper labeling function, then we have the followings.

 A vertex of subsimplex that lies on the side 𝒙𝒙𝒋𝒋 𝒙𝒙𝒌𝒌 of 𝚫𝚫𝟐𝟐 is

labeled “𝒋𝒋" or “𝒌𝒌"since such vertex can be represented as a

convex combination of 𝒙𝒙𝒋𝒋 and 𝒙𝒙𝒌𝒌 .

 A vertex of subsimplex in the interior of 𝚫𝚫𝟐𝟐 is labeled an arbitrary integer from {𝟎𝟎, 𝟏𝟏, 𝟐𝟐}.

 A vertex of “original” unit 2-simplex, 𝒙𝒙𝒊𝒊 ∈ {𝒙𝒙𝟎𝟎 , 𝒙𝒙𝟏𝟏 , 𝒙𝒙𝟐𝟐 }, is labeled "𝒊𝒊𝒊 since 𝝌𝝌 𝒙𝒙𝒊𝒊 = 𝒊𝒊

.

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Complete Labeling Definition 1.11 Complete labeling We call a subsimplex completely labeled if labeling function 𝝀𝝀 assigns all the values 𝟎𝟎, 𝟏𝟏, … , 𝒏𝒏 on its set of vertices.

𝒙𝒙𝟎𝟎 𝟎𝟎

𝟐𝟐

𝟐𝟐 𝟐𝟐 𝒙𝒙

𝟎𝟎

𝟎𝟎 𝟎𝟎

𝟏𝟏

𝟐𝟐

completely labeled subsimplex

𝟏𝟏

𝒙𝒙𝟏𝟏 𝟏𝟏

16

Sperner’s Lemma Lemma 1.12 Sperner (1928) Consider a simplicially subdivided n-simplex 𝑺𝑺𝒏𝒏 which is

properly labeled by the function 𝝀𝝀. Then, we can find, at least, a completely labeled subsimplex in the subdivision. More precisely, there are an odd number of such subsimplexes.  Here, we will discuss the proof in the case of 𝑺𝑺𝒏𝒏 = 𝚫𝚫𝟐𝟐 only.

 For more general proof, see Border (1989) and 二階堂 (1960).

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The First Step of Proof Lemma 1.13 In the side 𝒙𝒙𝒋𝒋 𝒙𝒙𝒌𝒌 of 𝚫𝚫𝟐𝟐 , there are an odd number of subintervals labeled “𝒋𝒋𝒋𝒋𝒋.

Proof.

• The side 𝒙𝒙𝒋𝒋 𝒙𝒙𝒌𝒌 can be divided into subintervals, “jj”, “kk” or “jk”. • Let “A” and “B” be the number of jj and jk in the side 𝒙𝒙𝒋𝒋 𝒙𝒙𝒌𝒌 , respectively. j 𝒙𝒙𝒋𝒋

j

j

 In the example, A=2 and B=3.

k

k

j

k 𝒙𝒙𝒌𝒌

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The First Step of Proof (Cont’d) j

j

j

j

j

k k

k k

j

j

𝒙𝒙𝒋𝒋

k

• If we can count the number of vertex labeled “j” in the side of

𝒙𝒙𝒌𝒌

𝒙𝒙𝒋𝒋 𝒙𝒙𝒌𝒌 without caring about double-counting, then,

the number of vertex “j” = 𝟐𝟐𝟐𝟐 + 𝑩𝑩.

• Because 𝒙𝒙𝒋𝒋 , labeled j, was not double-counted, so if we denote “C” the number of j double-counted in 𝒙𝒙𝒋𝒋 𝒙𝒙𝒌𝒌 , then we have 𝟐𝟐𝑪𝑪 + 𝟏𝟏 = 𝟐𝟐𝟐𝟐 + 𝑩𝑩.

• The LHS of the equation is odd. Therefore, B must be also odd. ∎

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The Second Step • Consider the case of 𝒋𝒋 = 𝟎𝟎 and 𝒌𝒌 = 𝟐𝟐.

• Mark both sides of subinterval 02 with a dot respectively as illustrated in the right figure. 𝟐𝟐

𝟐𝟐

𝟎𝟎

𝟎𝟎

 The number of dots must be even.

 A completely labeled subsimplex ∆𝟎𝟎𝟎𝟎𝟎𝟎 has one dot in the interior.

𝟎𝟎

𝟐𝟐

𝟎𝟎

𝟏𝟏

 A subsimplex ∆𝟎𝟎𝟎𝟎𝟎𝟎 (also ∆𝟎𝟎𝟎𝟎𝟎𝟎) has two dots in the interior.

𝟏𝟏 𝟏𝟏

20

The Final Step • 𝚫𝚫𝟐𝟐 has odd number of subinterval 02 in the sides.

→ There are odd number of dots in the exterior of 𝚫𝚫𝟐𝟐 .

• The total number of dots is even.

→ There are odd number of dots in the interior of 𝚫𝚫𝟐𝟐 .

• Only ∆𝟎𝟎𝟎𝟎𝟎𝟎 has one dot among the subsimplexes. → There are odd number of ∆𝟎𝟎𝟎𝟎𝟎𝟎 in the 𝚫𝚫𝟐𝟐 .

→ Sperner’s lemma is proved. ∎

21

Brouwer Fixed Point Theorem Theorem 1.14 The Brouwer Fixed Point Theorem Let 𝑺𝑺 ⊂ ℝ𝒏𝒏+𝟏𝟏 be a non-empty, compact and convex set and let

𝒇𝒇: 𝑺𝑺 → 𝑺𝑺 be a continuous function. Then, there exists at least one fixed point of f in S. Formally, there exists at least one 𝒚𝒚∗ ∈ 𝑺𝑺 such that 𝒚𝒚∗ = 𝒇𝒇(𝒚𝒚∗ ).

• This lecture only provides the proof when 𝑺𝑺 is unit 2-simplex 𝚫𝚫𝟐𝟐 . • Let 𝒇𝒇𝒊𝒊 (𝒚𝒚) denote the ith coordinate of 𝒇𝒇(𝒚𝒚), for 𝒊𝒊 = 𝟎𝟎, … , 𝒏𝒏 and 𝒚𝒚 = (𝒚𝒚𝟎𝟎 , 𝒚𝒚𝟏𝟏 , … , 𝒚𝒚𝒏𝒏 ).  Notice: the first coordinate is subindexed “0”.

• Because 𝒇𝒇(𝒚𝒚) ∈ 𝑺𝑺, we have ∑𝒏𝒏𝒊𝒊=𝟎𝟎 𝒇𝒇𝒊𝒊 (𝒚𝒚) = 𝟏𝟏 for every 𝒚𝒚 ∈ 𝑺𝑺.

22

Exercises Exercise 1.15 Show that an unit 2-simplex, ∆𝟐𝟐 =

𝒚𝒚𝟎𝟎 , 𝒚𝒚𝟏𝟏 , 𝒚𝒚𝟐𝟐 ∈ ℝ𝟑𝟑+ ; 𝒚𝒚𝟎𝟎 + 𝒚𝒚𝟏𝟏 + 𝒚𝒚𝟐𝟐 = 𝟏𝟏 ,

is non-empty, compact and convex set. Exercise 1.16

Prove that the Brouwer fixed point theorem is true in the special case of 𝑺𝑺 = [𝟎𝟎, 𝟏𝟏]. (Hint: Intermediate value theorem) 23

The First Step of Proof Lemma 1.17 Consider an unit n-simplex in ℝ𝒏𝒏+𝟏𝟏 . For every 𝒌𝒌 = 𝟏𝟏, 𝟐𝟐, …, there

exists 𝒏𝒏 + 𝟏𝟏 points, 𝒚𝒚𝟎𝟎,𝒌𝒌 , 𝒚𝒚𝟏𝟏,𝒌𝒌 , … , 𝒚𝒚𝒏𝒏,𝒌𝒌 , all within a single 𝟏𝟏/𝒌𝒌 ball, such that,

𝒊𝒊,𝒌𝒌 𝒚𝒚𝒊𝒊,𝒌𝒌 ≥ 𝒇𝒇 𝒚𝒚 𝒊𝒊 𝒊𝒊

• The inequality says that the 𝒊𝒊th coordinate of the 𝒊𝒊th point is weakly greater than the 𝒊𝒊th coordinate of its image under f.

𝟐𝟐 𝟐𝟐 𝟐𝟐 𝟐𝟐  𝒚𝒚𝟐𝟐,𝟑𝟑 means the “0th” coordinate of point 𝒚𝒚 = 𝒚𝒚 , 𝒚𝒚 , … , 𝒚𝒚 𝒏𝒏 𝟎𝟎 𝟏𝟏 𝟎𝟎

which is in a ball of radius 𝟏𝟏/𝟑𝟑.

24

Example: unit 1-simplex (𝒏𝒏 = 𝟏𝟏) 𝒚𝒚𝟎𝟎 , 𝒚𝒚𝟏𝟏 ∈ ℝ𝟐𝟐+ ; 𝒚𝒚𝟎𝟎 + 𝒚𝒚𝟏𝟏 = 𝟏𝟏 .

∆𝟏𝟏 =

𝟏𝟏

𝒚𝒚𝟎𝟎𝟎𝟎

𝒆𝒆𝟎𝟎

𝒇𝒇(𝒚𝒚𝟎𝟎 )

𝒚𝒚𝟎𝟎𝟏𝟏

𝒆𝒆𝟏𝟏

𝒇𝒇(𝒚𝒚𝟏𝟏 )

𝒇𝒇

𝒚𝒚𝟎𝟎

𝒇𝒇(𝒚𝒚𝟏𝟏 )

𝟒𝟒𝟓𝟓∘

𝒇𝒇

𝒇𝒇(𝒚𝒚𝟎𝟎 )

𝒚𝒚𝟏𝟏

𝒇𝒇

𝒆𝒆𝟏𝟏

𝟏𝟏

𝒆𝒆𝟎𝟎

𝒚𝒚𝟎𝟎

𝟎𝟎 𝟎𝟎 𝒚𝒚𝟎𝟎 : 𝒚𝒚𝟎𝟎 > 𝒇𝒇𝟎𝟎 𝒚𝒚𝟎𝟎 , 𝒇𝒇𝟏𝟏 𝒚𝒚𝟎𝟎 > 𝒚𝒚𝟏𝟏

𝒚𝒚𝟏𝟏 : 𝒇𝒇𝟎𝟎 𝒚𝒚𝟏𝟏 > 𝒚𝒚𝟏𝟏𝟎𝟎 , 𝒚𝒚𝟏𝟏𝟏𝟏 > 𝒇𝒇𝟏𝟏 (𝒚𝒚𝟏𝟏 )

𝒚𝒚𝟏𝟏

𝒆𝒆𝟏𝟏

25

Example: unit 1-simplex (𝒏𝒏 = 𝟏𝟏)

𝟒𝟒𝟓𝟓∘

𝒆𝒆𝟏𝟏

𝒇𝒇(𝒚𝒚𝟏𝟏 ) 𝒇𝒇(𝒚𝒚𝟎𝟎 )

𝒆𝒆𝟎𝟎

𝒇𝒇

𝒚𝒚𝟎𝟎 𝒚𝒚𝟏𝟏

𝟏𝟏/𝒌𝒌 ball

𝒆𝒆𝟏𝟏 26

If Lemma 1.17 holds… 𝟏𝟏,𝒌𝒌 }∞ ,…,{𝒚𝒚𝒏𝒏,𝒌𝒌 }∞ , • We can make 𝒏𝒏 + 𝟏𝟏 sequences in S, {𝒚𝒚𝟎𝟎,𝒌𝒌 }∞ , {𝒚𝒚 𝒌𝒌=𝟏𝟏 𝒌𝒌=𝟏𝟏 𝒌𝒌=𝟏𝟏 𝒊𝒊,𝒌𝒌 ) for 𝒊𝒊 = 𝟎𝟎, 𝟏𝟏, … , 𝒏𝒏, such that the claim of Lemma 1.17, 𝒚𝒚𝒊𝒊,𝒌𝒌 ≥ 𝒇𝒇 (𝒚𝒚 𝒊𝒊 𝒊𝒊

holds for every 𝒌𝒌 = 𝟏𝟏, 𝟐𝟐, ….

• S is bounded because it is compact. Therefore, each sequence has a convergent subsequence, by the Bolzano-Weierstrass theorem. ′

′

• Because the 𝒌𝒌′ th points in the sequences, 𝒚𝒚𝟎𝟎,𝒌𝒌 , … , 𝒚𝒚𝒏𝒏,𝒌𝒌 , are within

distance 𝟏𝟏/𝒌𝒌′ from one another, as 𝒌𝒌′ → ∞, the subsequences must

converge to the same point, 𝒚𝒚∗ ∈ ℝ𝒏𝒏+𝟏𝟏 .

• S is closed because it is compact. Therefore, we have 𝒚𝒚∗ ∈ 𝑺𝑺.

27

If Lemma 1.17 holds… (Cont’d) 𝒊𝒊,𝒌𝒌 , • By taking the limit of the inequality in Lemma 1.17, 𝒚𝒚𝒊𝒊,𝒌𝒌 ≥ 𝒇𝒇 𝒚𝒚 𝒊𝒊 𝒊𝒊 𝒊𝒊,𝒌𝒌 . we obtain lim 𝒚𝒚𝒊𝒊,𝒌𝒌 ≥ lim 𝒇𝒇 𝒚𝒚 𝒊𝒊 𝒊𝒊 𝒌𝒌→∞

𝒌𝒌→∞

• From of the continuity, we have lim 𝒇𝒇𝒊𝒊 𝒚𝒚𝒊𝒊,𝒌𝒌 = 𝒇𝒇𝒊𝒊 (𝒚𝒚∗ ). Therefore, 𝒌𝒌→∞

𝒚𝒚∗ is a point that 𝒚𝒚∗𝒊𝒊 ≥ 𝒇𝒇𝒊𝒊 (𝒚𝒚∗ ) holds for 𝒊𝒊 = 𝟎𝟎, 𝟏𝟏, … , 𝒏𝒏.

• Because 𝒇𝒇(𝒚𝒚∗ ) is in S, we have ∑𝒏𝒏𝒊𝒊=𝟎𝟎 𝒚𝒚∗𝒊𝒊 = ∑𝒏𝒏𝒊𝒊=𝟎𝟎 𝒇𝒇𝒊𝒊 (𝒚𝒚∗ ) = 𝟏𝟏. Then, we obtain 𝒚𝒚∗ = 𝒇𝒇(𝒚𝒚∗ ) because 𝒚𝒚∗𝒊𝒊 must be equal to 𝒇𝒇𝒊𝒊 (𝒚𝒚∗ ) for 𝒊𝒊 = 𝟎𝟎, 𝟏𝟏, … , 𝒏𝒏.

• Consequently, if Lemma 1.17 is true, there is a fixed point 𝒚𝒚∗ in S and Theorem 1.14 is true.

28

The Second Step of Proof Lemma 1.17’ (𝒏𝒏 = 𝟐𝟐)

For every 𝒌𝒌 = 𝟏𝟏, 𝟐𝟐, …, there exists 𝟑𝟑 points, 𝒂𝒂, 𝒃𝒃, 𝒄𝒄, all within a single 𝟏𝟏/𝒌𝒌 ball, such that, 𝒂𝒂𝟎𝟎 ≥ 𝒇𝒇𝟎𝟎 𝒂𝒂 , 𝒆𝒆𝟐𝟐

𝒆𝒆𝟎𝟎

𝒆𝒆𝟏𝟏

𝒃𝒃𝟏𝟏 ≥ 𝒇𝒇𝟏𝟏 𝒃𝒃 ,

𝒄𝒄𝟐𝟐 ≥ 𝒇𝒇𝟐𝟐 (𝒄𝒄)

𝒃𝒃

𝒂𝒂

𝒄𝒄

29

The Third Step of Proof • We show the following statement: For any 𝒌𝒌, there is a ∆𝒂𝒂𝒂𝒂𝒂𝒂 that satisfies the conditions in Lemma 1.17’ and fits into 𝟏𝟏/𝐤𝐤 ball.  Assumption: No vertex in the subdivision is a fixed point. Consider a labeling rule:

𝑨𝑨 𝒆𝒆𝟎𝟎

𝒆𝒆𝟐𝟐

𝒚𝒚𝒊𝒊 > 𝒇𝒇𝒊𝒊 𝒚𝒚 ⟹ we can assign the label 𝒊𝒊 𝒇𝒇(𝑨𝑨)

𝒆𝒆𝟏𝟏

the vertex 𝒚𝒚.

Example: 𝑨𝑨 = 𝟏𝟏/𝟒𝟒, 𝟏𝟏/𝟒𝟒, 𝟏𝟏/𝟐𝟐 and 𝒇𝒇 𝑨𝑨 = 𝟎𝟎, 𝟐𝟐/𝟑𝟑, 𝟏𝟏/𝟑𝟑

⟹ 𝑨𝑨 can be assigned 0 or 2.

30

The Third Step of Proof (Cont’d) • If a labeling of each of the vertices in the subdivision satisfies the rule, then we call the labeling feasible.  There can be more than one feasible labeling of a subdivision.  For any point 𝒚𝒚, at least one 𝒊𝒊 ∈ {𝟎𝟎, 𝟏𝟏, 𝟐𝟐} must satisfy 𝒚𝒚𝒊𝒊 > 𝒇𝒇𝒊𝒊 (𝒚𝒚).

𝒆𝒆𝟐𝟐

𝒆𝒆𝟎𝟎

We can assign …  𝑨𝑨

𝑪𝑪

𝑨𝑨 𝑩𝑩

𝒆𝒆𝟏𝟏

→ 𝟎𝟎 or 𝟐𝟐.

 𝑩𝑩 →  𝑪𝑪 →

𝟎𝟎 or 𝟏𝟏.

𝟏𝟏 or 𝟐𝟐.

 𝒆𝒆𝟎𝟎  𝒆𝒆𝟏𝟏  𝒆𝒆𝟐𝟐

→ 𝟎𝟎 only. → 𝟏𝟏 only. → 𝟐𝟐 only.

31

The Final Step of Proof • Because ∆𝒂𝒂𝒂𝒂𝒂𝒂 can be completely labeled, 𝟎𝟎𝟎𝟎𝟎𝟎, Lemma 1.17’ is replaced with Lemma 1.17’’. Lemma 1.17’’ For any size of equilateral subdivision, there exists at least one completely labeled subsimplex ∆𝒂𝒂𝒂𝒂𝒂𝒂. • This lemma is identical to Sperner’s lemma, which is already proved. • So, the Brouwer fixed point theorem in the case of 𝒏𝒏 = 𝟐𝟐 is also proved.∎

32