Analysis

16.5 Applications

With these theorems it is possible to give easy proofs of some very important and difficult theorems.

Definition 16.5.1 If f : U ⊆ ℝ^p → ℝ^p where U is an open set. Then f is locally one to one if for every x ∈ U, there exists δ > 0 such that f is one to one on B

As a first application, consider the invariance of domain theorem. This result says that a one to one continuous map takes open sets to open sets. It is an amazing result which is essential to understand if you wish to study manifolds. In fact, the following theorem only requires f to be locally one to one. First here is a lemma which has the main idea.

Lemma 16.5.2 Let g : B

→ ℝ^p be one to one and continuous where here B

is the ball centered at 0 of radius r in ℝ^p. Then there exists δ > 0 such that

g(0)+ B (0,δ) \subseteq g(B (0,r)).

The symbol on the left means:

Proof: For t ∈

, let

( ) ( ) h(x,t) \equiv g --x- - g --tx 1+ t 1+ t

Then for x ∈ ∂B

, h

≠0 because if this were so, the fact g is one to one implies

--x- = -- tx 1 + t 1 + t

and this requires x = 0 which is not the case since

= r. Then d

is constant. Hence it is an odd integer for all t thanks to Borsuk’s theorem, because h

is odd. Now let B

be such that B

∩ h

= ∅. Then d

= d

for z ∈ B

because the degree is constant on connected components of ℝ^p ∖ h

. Hence z = h

= g

− g

for some x ∈ B

. Thus

g(B (0,r)) \supseteq g (0 )+ B(0,δ) ■

Now with this lemma, it is easy to prove the very important invariance of domain theorem.

A function f is locally one to one on an open set Ω if for every x₀ ∈ Ω, there exists B

⊆ Ω such that f is one to one on B

Theorem 16.5.3 (invariance of domain)Let Ω be any open subset of ℝ^p and let f : Ω → ℝ^p be continuous and locally one to one. Then f maps open subsets of Ω to open sets in ℝ^p.

Proof: Let B

⊆ Ω where f is one to one on B

. Let g be defined on B

given by

g (x) \equiv f (x+ x0)

Then g satisfies the conditions of Lemma 16.5.2, being one to one and continuous. It follows from that lemma there exists δ > 0 such that

f (Ω ) \supseteq f (B(x0,r)) = f (x0 + B (0,r)) = g(B (0,r)) \supseteq g(0)+ B (0,δ) = f (x )+ B (0,δ) = B (f (x ),δ) 0 0

This shows that for any x₀ ∈ Ω,f

is an interior point of f

which shows f

is open. ■

With the above, one gets easily the following amazing result. It is something which is clear for linear maps but this is a statement about continuous maps.

Corollary 16.5.4 If n > m there does not exist a continuous one to one map from ℝⁿ to ℝ^m.

Proof: Suppose not and let f be such a continuous map,

T f (x) \equiv (f1(x),\cdot\cdot\cdot,fm (x )) .

Then let g

≡

^T where there are n − m zeros added in. Then g is a one to one continuous map from ℝⁿ to ℝⁿ and so g

would have to be open from the invariance of domain theorem and this is not the case. ■

Corollary 16.5.5 If f is locally one to one and continuous, f : ℝ^p → ℝ^p, and

lim |f (x)| = \infty, |x|\to \infty

then f maps ℝ^p onto ℝ^p.

Proof: By the invariance of domain theorem, f

is an open set. It is also true that f

is a closed set. Here is why. If f

→ y, the growth condition ensures that

is a bounded sequence. Taking a subsequence which converges to x ∈ ℝ^p and using the continuity of f, it follows f

= y. Thus f

is both open and closed which implies f must be an onto map since otherwise, ℝ^p would not be connected. ■

The next theorem is the famous Brouwer fixed point theorem.

Theorem 16.5.6 (Brouwer fixed point) Let B = B

⊆ ℝ^p and let f : B → B be continuous. Then there exists a point x ∈ B, such that f

= x.

Proof: Assume there is no fixed point. Consider h

≡ tf

− x for t ∈

. Then for

= r,

0 ∕\intf (x)- x,t \in [0,1]

By homotopy invariance,

t \to d(tf - I,B, 0)

is constant. But when t = 0, this is d

ⁿ≠0. Hence d

≠0 so there exists x such that f

− x = 0. ■

You can use standard stuff from Hilbert space to get this the fixed point theorem for a compact convex set. Let K be a closed bounded convex set and let f : K → K be continuous. Let P be the projection map onto K as in Problem 12 on Page 169. Then P is continuous because

≤

. Recall why this is. From the characterization of the projection map P,

≤ 0 for all y ∈ K. Therefore,

(x- Px,Py - P x) \leq 0, (y- Py,P x- P y) \leq 0 so (y - P y,Py- Px) \geq 0

Hence, subtracting the first from the last,

(y- Py - (x- Px),P y- P x) \geq 0

consequently,

|x- y||Py- Px| \geq (y - x,Py- Px ) \geq |P y- P x|2

and so

≤

as claimed.

Now let r be so large that K ⊆ B

. Then consider f ∘ P. This map takes B

→B

. In fact it maps B

to K. Therefore, being the composition of continuous functions, it is continuous and so has a fixed point in B

denoted as x. Hence f

= x. Now, since f maps into K, it follows that x ∈ K. Hence Px = x and so f

= x. This has proved the following general Brouwer fixed point theorem.

Theorem 16.5.7 Let f : K → K be continuous where K is compact and convex and nonempty, K ⊆ ℝ^p. Then f has a fixed point.

Definition 16.5.8 f is a retract of B

onto ∂B

if f is continuous,

( ------) f B(0,r) \subseteq \partialB (0,r)

and f

= x for all x ∈ ∂B

Theorem 16.5.9 There does not exist a retract of B

onto its boundary, ∂B

Proof: Suppose f were such a retract. Then for all x ∈ ∂B

, f

= x and so from the properties of the degree, the one which says if two functions agree on ∂Ω, then they have the same degree,

1 = d (I,B (0,r),0 ) = d(f,B(0,r),0)

which is clearly impossible because f⁻¹

= ∅ which implies d

= 0. ■

You should now use this theorem to give another proof of the Brouwer fixed point theorem.

The proofs of the next two theorems make use of the Tietze extension theorem, Theorem 3.12.6.

Theorem 16.5.10 Let Ω be a symmetric open set in ℝⁿ such that 0 ∈ Ω and let f : ∂Ω → V be continuous where V is an m dimensional subspace of ℝⁿ,m < n. Then f

= f

for some x ∈ ∂Ω.

Proof: Suppose not. Using the Tietze extension theorem on components of the function, extend f to all of ℝⁿ, f

⊆ V . (Here the extended function is also denoted by f.) Let g

= f

− f

. Then 0

and so for some r > 0, B

⊆ ℝⁿ ∖ g

. For z ∈ B

d(g,Ω,z) = d (g,Ω, 0) ⁄= 0

because B

is contained in a component of ℝⁿ ∖g

and Borsuk’s theorem implies that d

≠0 since g is odd. Hence

V \supseteq g (Ω) \supseteq B (0,r)

and this is a contradiction because V is m dimensional. ■

This theorem is called the Borsuk Ulam theorem. Note that it implies there exist two points on opposite sides of the surface of the earth which have the same atmospheric pressure and temperature, assuming the earth is symmetric and that pressure and temperature are continuous functions. The next theorem is an amusing result which is like combing hair. It gives the existence of a “cowlick”.

Theorem 16.5.11 Let n be odd and let Ω be an open bounded set in ℝⁿ with 0 ∈ Ω. Suppose f : ∂Ω → ℝⁿ ∖

is continuous. Then for some x ∈ ∂Ω and λ≠0, f

= λx.

Proof: Using the Tietze extension theorem, extend f to all of ℝⁿ. Also denote the extended function by f. Suppose for all x ∈ ∂Ω, f

≠λx for all λ ∈ ℝ. Then

0 ∕\in tf (x)+ (1- t)x, (x,t) \in \partialΩ \times [0,1]

0\in∕tf (x)- (1- t)x, (x,t) \in \partialΩ \times[0,1].

Thus there exists a homotopy of f and I and a homotopy of f and −I. Then by the homotopy invariance of degree,

d(f,Ω,0) = d (I,Ω,0),d (f,Ω,0) = d(- I,Ω,0).

But this is impossible because d

= 1 but d

ⁿ = −1. ■