On Global Univalence Theorems

Lecture Notes in Mathematics Edited by A. Dold and B. Eckmann

977

T. Parthasarathy

On

Global Univalence Theorems

Springer-Verlag Berlin Heidelberg New York 1983

Author

T. Parthasarathy Indian Statistical Institute, Delhi Centre ?, S.J.S. Sansanwal Marg., New Delhi 110016, India

AMS Subject Classifications (1980): 26-02, 26 B10, 90-02, 90A14, 90 C 30 ISBN 3-540-11988-4 Springer-Verlag Berlin Heidelberg New York ISBN 0-38?-11988-4 Springer-Verlag NewYork Heidelberg Berlin

This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to "Verwertungsgesellschaft Wort", Munich. © by Springer-Verlag Berlin Heidelberg 1983 Printed in Germany Printing and binding: Beltz Offsetdruck, Hemsbach/Bergstr. 2146/3140-543210

PREFACE This volume of lecture notes contains results on global univalent mappings. Some of the material of this volume had been given as seminar talks at the Department of Mathematics,

~hiversity of Kansas, Lawrence during 1978-79 and at

the Indian Statistical Institute, Delhi Centre during 1979-80. Even though the classical local inverse function theorem is well-known, C~le-Nikaido's global univalent results obtained in (1965) are not known to many mathematicians that I have sampled.

Recently some significant contributions have

been made in this area notably by ~arcia-Zangwill Soarf-Hirsch-Chilnisky

(1980).

(1979), Mas-Colell

(1979) and

Global univalent results are as important as local

univalent results and as such I thoughtit is worthwhile to make these results well-known to the mathematical community at large.

Also I believe that there are

very many interesting open problems which are worth solving in this branch of Mathematics.

I have also included a number of applications from different

disciplines like Differential Equations, Mathematical EconomiNcs, Mathematical Programming, Statistics etc.

Some of the results have appeared only in Journals

and we are bringing them to-gether in one ~lace. These notes contain some new results.

For example Proposition 2, Theorem 4

in Chapter II, Theorem 4, Theorem 5 in Chapter III, Theorem 2" in Chapter V, Theorem 8 in Chapter VI, Theorem 2 in Chapter VII, Theorem 9 in Chapter VIII are new results. It is next to impossible to cover all the known results on global univalent mappings for lack of space and time.

For example a notable omission could be the

role played by univalent mappings whose domain is complex numbers.

We have also

not done enough justice to the oroblem when a PL-function will be a homeomorphism in view of the growing importance of such functions.

We have certainly given

references where an interested reader can ~et more information. I am grateful to Professors : Andreu Mas-Colell, Ruben Schr~mm, Albrecht Dold and an anonymous referee for their several constructive suggestions on various parts

IV

of this material.

I am also grateful to Professor David Cale for the example given

at the end of Chapter II and Professor L. Salvadori for some useful discussion that I had with him regardin~ Chapter VII. Moreover I wish to thank the Indian Statistical Institute, Delhi Centre for ~roviding the facilities and the atmosphere necessary and conducive for such work. Finally I express my sincere thanks to Mr. V.P. Sharma for his excellent and painstakin~ work in t~ming several revisions of the manuscript, Mr. Mehar Lal who tyoed a preliminary version of this manuscriot and Mr. A.N. Sharma who heloed me in filling many s~rmbols.

1 DECKW~ER 1982

T. PART F~AK&T RY INDIAN STATISTICAL INSTITL~E DELIYI CKNTRE

CONTENTS PREFACE INTRODUCTION CFAPTER

CFAPTER

I

II

CFAPTER III

CFAPTER

C FAPTER

CFAPTER

IV

V

VI

CFAPTER VII

:

:

:

:

PRELIMINARIES AND STATE...WENTOF THE PROBLEM

1

Classical inverse function theorem

1

Invariance of domain theorem

3

Statement of the oroblem

3

P-N3[TRICES AND N-N~TRICES

I0

Is AB a P-matrix or N-matrix when A and B are P-matrices or N-matrices

12

F[NDAMENTAL GLOBAL LNIVALENCE RESLGTS OF CALE-NIKAIDO-INADA

:

17

Fundamental global univalence theorem

20

Univalent results in R ~

22

GLOBAL }DMEOMORFI{ISN~$ BETWEEN FINITE DIMENSIONAL SPACES

28

Plastock's theorem

29

More-Rheinboldt's theorem and its consequences

34

Light open mappings and homeomorphisms

36

: SCARF'S CONJECT~E AND ITS VALIDITY

:

6

Characterization of P-matrices

41

Carcia-Zangwill's result on univalent mappings

41

Mas-Colell's univalent result

44

GLOBAL INIVALENT RESULTS IN R 2 AND R 3

49

Univalent mappings in R 2

49

Univalent results in R ~

57

ON THE GLOBAL STABILITY OF AN AtTONOMOU3 SYSTEM ON THE PLANE

59

Proof of theorem 2

6O

Vidossich's contribution to Olech's oroblem on stability

65

VI

CFAPTER

VIII

CHAPTER

CFAPTER

IX

X

: ~TIVALENCE FOR MJIPPINCS WITH LEOh~IEF TYPE JACOBIANS

68

thivalence for dominant diagonal mappings

68

Interrelation between P-prooerty and M-property

71

Lhivalence for comoosition of two functions

74

: ASSORTED APPLICATIONS OF LNIVALENCE MAPPING RESb~TS

77

An aoolication in Mathematical Economics

77

On the distribution of a function of several random variables

79

On the existence and uniqueness of solutions in non-linear complimentarity theory

81

An aDolication of Hadamard's inverse function theorem to algebra

85

On the infinite divisibility of multivariate gamma distributions

86

: FLRT~R

~NERALIZATIONS AND REMARKS

9O

A generalization of the local inverse function theorem

9O

Monotone functions and univalent functions

92

On PL functions

93

On a ~lobal univalent result when the Jacobian vanishes

96

Injectivity of quasi-isometric mappings

99

REFERENCES

i01

INDEX

105

INTRODLDTION Let ~ be a subset of R n and let F be a differentiable function from ~ to R n. We are looking for nice conditions that will ensure the equation F(x) = y to have at most one solution for all y s R n.

In other words we want the equation F(x) = y

to have a unique solution for every y in the range of F. Classical inverse function theorem says that if the Jacobian of the mao does not vanish then if F(x) = y has a solution x*, then x* is an isolated solution, that is, there is a neighbourhood of x* which contains no other solution.

In the

global univalence problem, we demand x* to be the only solution throughout ~. It is a fascinating fact, why the ~lobal univalence problem had not been posed or at any rate solved before ~adamard in 1906, which of course is a very late stage in the development of Analysis.

It is funny and actually baffling, how much

misunderstanding associated with the global univalence problem survived right into the middle of the twentieth century.

A brief history of this may not be out

of place here. Paul Samuelson in (1949) gave as sufficient condition for uniqueness, that the Jaeebian should not vanish and it was pointed out by A. was faulty.

Turing that this statement

However Paul Samuelson's economic intuition was correct and in his

case all the elements of the Jacobian were essentially one-signed and this condition combined with the non-vanishing determinant, turns out to be sufficient to guarantee uniqueness in the large. Paul Samuelson (1953) then stated that non-vanishing of the leading minors will suffice for global univalence in ~eneral.

But Nikaido produced a counter

example to this assertion and he went on to show that global univalence prevails in any convex region provided the Jacobian matrix is a quasi-positive definite matrix. Later,

C~le proved that it is sufficient for uniqueness in any rectangnlar region

provided the Jacobian matrix is a P-matrix, that is, every principal minor is positive.

In fact this culminated in the well-known article of ~le-Nikaido

(1965)

which is the main source of inspiration for the present writer. I should mention two other articles.

The article of Banach-Mazur

(1934) gives

probably the first proof of a relevant result formulated with the demands of rigour still valid to-day.

The more recent article by Palais (1959) covers a much wider

area than the article of Banach and Mazur. There are several approaches one can consider to the global univalent problem. For examole the approach could be via linear inequalities, monotone functions or PL functions. inequalities.

Throughout we have followed more or less the approach through linear

VIMI

In most of the theorems the conditions for global univalence are very stringent and therefore often not satisfied in applications. conditions of the theorem in practice.

Another problem is to verify the

In general it is hard to obtain necessary

and sufficient conditions for global univalence results. further research in this area.

There is lot of room for

C~le-Nikaido's global univalent theorem is valid

even if the partial derivatives are not continuous whereas Mas-Colell's results as well as C~rcia-Zangwill's results demand the partial derivatives to be continuous. One of the major open problems in this area is the following:

Can continuity of

the derivatives in Mas-Colell's results be dispensed with (altogether or at least in part) or alternatively - are there counter example~ following:

Amother problem is the

Is the fundamental global univalent result due to C~le-Nikaido valid in

any compact convex region? As alrea~v pointed out in some of the applications complete univalence is not warranted but in which some weaker univalence enunciations can nevertheless be made. In this connection I would like to cite at least two important papers one by Chua and Lam and the other by Schramm. Because of the lack of a text on the global univalence and since the results are available only in articles scattered in various journals or in texts devoted to other subjects

(for example Economics), I felt the need for writing this notes

on global univalent mappings.

In the next ten chapters with the exception of the

first two chapters, various results on global univalent mappings as well as their applications are discussed.

Also many examples are given and several open problems

are mentioned which I believe will interest research workers. Prerequisites needed for reading this monograph are real analysis and matrix theory.

Fere are a few suggestions.

[i].

W.Rudin (1976), Principles of Nathematical Analysis, Third Edition (International Student Edition) McGraw-Hill, Koyakusha Ltd.

[2].

F.R. Gantmacher (1959), The Theory of Matrices Vols. I and II, Chelsea Publishing Company, New York.

[3].

C.R. Rap (1974), Linear Statistical Inference and its Applications, Second Edition, Wiley Eastern Private Limited, New Delhi (Especially Chapter I dealing with 'Algebra of vectors and matrices').

[4].

G.S. Rogers (1980), Matrix derivatives, Marcel Dekker, New York and Basel (Actually only chapters 13 and 14 have the Jacobian and its orooerties as their central topic while 11 and 12 refer to the general theory).

[5].

W.Fleming (1977), Functions of several variables, Second Edition, SpringerVerlag, Heidelberg-New York. Some knowledge of algebraic topology will be useful (especially degree theory)

and we have mentioned a few references in Chapter IV.

CHAPTER

I

PRELIMINARIES AND STATEMENT OF THE PROBLEM

Abstract :

In this chapter we will collect some well-known results like classical

inverse function theorem, domain invariance theorem etc for ready reference (without proof).

We will then give the statement of the problem considered in this monograph

cite a few results and make some remarks. Classical inverse function theorem : C Rn

to

Let F be a transformation from an open set

R n. We will say that F is locally univalent, if for every x s ~ there

exists a neighbourhood U x of x such that F I U

(=F restricted to Ux ) is one-one.

Inverse function theorem gives a set of sufficient conditions for F to be locally univalent.

We come across such problems in various situations.

for a given y, there exists an

x°

such that F(x o) = y.

For example, suppose

We may like to know whether

there are points x other than Xo, contained in a small neighbourhood around x ° satisfying F(x) = y. is unique locally.

Classical inverse function theorem asserts that the solution In order to state the inverse function theorem we need the

following. Definition : A transformation F is differentiable at t

if there exists a linear

O

transformation L (depending on t o) such t h a t lim h÷0

1 ~-~

[F(to+h)-F(to)-L(h)]

Here Ilhll stands for the usual vector norm.

=

0

The linear transformation L is called

the differential of F at to and is often denoted by DF(to).

Write F = (fl,f2,...,fn)

where each fi is a real-valued function from ~. We denote their partial derivatives ~f. 4 1 as fJ. = i ~x. " J Remark i : A transformation F is differentiable at t

if and only if each of its O

components f. is differentiable at t i

Remark 2 :

for i = 1,2,...,n. O

If F is differentiable at to, then the matrix of the linear transformation

L is simply the Jacobian matrix J of partial derivatives fJ(to).

Definition : C (q).

Call F a transformation of class q, q ~ 0

That is, for every

if each

fi

is of class

fi(i = 1,2,...,n) all the partial derivatives upto order

q exist and are continuous over its domain.

We are now ready to state the (local) inverse function theorem. Local inverse function theorem : set

~ ~'R n

into

an open set A o ~

R n.

Let F be a map of class C (q), q > i from an open

If the Jacobian at t o s ~ does not vanish, then there exists

~ containing t o such that :

(i)

FIA °

is one-one, that is, F restricted to A ° is univalent.

(ii)

F(A e) is an open set.

(iii) The inverse G of F[A ° is of class C (q). (iv)

JG(X) = (JF(t)) -I where F(t) = x, t s A o.

matrix evaluated at x.

Remark i :

Here JG(X) denotes the Jacobian

Proof of this may be found in Fleming

In one dimension the situation is simpler.

[17].

If F is a real-valued

function with domain an open interval ~, then F -I (=inverse map of F) exists if F is strictly monotone.

Also F will be strictly monotone if F'(t) ~ 0 for all t E ~,

and in fact G'(x) = F ' ~I JF(t) takes the place

where x = F(t). of F'(t).

In higher dimensions the Jacobian

The situation here is more complicated.

For

example, the non-vanishing of the Jacobian does not guarantee that F has a (global) inverse as in the univariate case.

However, if JF(to) does not vanish at to, we

can find a small neighbourhood A ° containing t o such that F restricted to A ° will have an inverse.

In other words we can only assert local inverse.

This is precisely

part of the statement of inverse function theorem. If one is interested in just the local univalence we have the following theorem (proof may be found in

[44]).

Local univalent theorem : connected subset of R n. (i)

Let F:~ C R n ÷ R n be a mapping where ~ is an open

We have the following:

If F is differentiable at a point t o s ~ and JF(to) # 0, then there is a

neighbourhood U of t o such that F(y) = F(to) , y s U ~ (ii)

y = to •

If F is continuously differentiable in a neighbourhood of an interior point

t o of ~

and JF(to) # 0, then there is a neighbourhood U of t o where F is univalent,

that is, F(y) = F(z), y,z s U ~

y = z.

We are now ready to state the following: Theorem on invarianoe of interior points :

Let F:O ÷ R n be a differentiable map

with non-vanishing Jacobian, where ~ is an open region in R n .

Then the image set

F(~) is also an open region. For a proof see Nikaido

[44].

This result is true not only for differentiable

mappings with nonvanishing Jacobians but also for homeomorphic mappings from a

region of R n into R n.

That is the content of the following classical theorem due

to Brouwer. Invariance

of domain theorem:If ~ is open in R n and F:~ ÷ R n is one-one and

continuous,

then F(~) is open and F is a homeomorohism.

Definition

:

For a proof see

A mapping F:~ ÷ R n is called a local homeomorphism

a neighbourhood

of t is mapped homeomorphically

if for each t s ~,

by F onto a neighbourhood

It is clear that if F:~ ÷ R n is a continuously non-vanishing

[30].

differentiable

Jacobian it follows from local inverse-function

univalent theorem that F is a local homeomorphism.

of F(t).

function with

theorem or local

We will introduce one more

definition. Definition

:

Let F:~ ÷ R n be a continuous mapping where ~ is an open region in R n

with the property that each y s F(~) has a neighbourhood of F-I(v)

is mapped homeomorphically

onto V by F.

map and (~,F) is called a covering space for F(~).

In this case, the cardinal number

n of the set F-l(y) is the same for all y s F(~). F is called a finite covering,

Remarks

:

It is well-known

V such that each component

Then F is called a covering

If n is a finite integer,

or more specifically,

then

an n-covering.

that every covering map F:~ ÷ R n is a homeomorphism

is connected and that every homeomorphic

if

onto function F:~ ÷ R n is a covering map

and every covering map is a local homeomorphism.

However the converse is not true.

A local homeomorphism need not be a covering map and a covering map need not be a homeomorphic function.

onto function.

A 1-covering map is necessarily

The following result is well-known

A theorem on covering space :

Let X and Y be connected,

spaces (for example X = Y = Rn). F is a homeomorohism [ Here F:X ÷ Y

Furthermore

a homeomorphic

onto

[48]. locally Dathwise connected

suppose Y is simply connected.

Then

of X onto Y if and only if (X,F) is a covering space of Y.

is a map from X to Y].

We need this result especially in chapter IV where sufficient conditions are given in order that a map F from R n to R n will be a hemeomorphism results on degree theory, we freely use from chapter VI in references

for degree theory are

Statement of the problem

:

[ 48].

onto R n.

For

Other good

[13,59,63].

Let F : ~ r - R n ÷ R n be a differentiable

F to be globally one-one throughout ~.

What conditions

map.

We want

should we impose on the

map F and the region ~ so that F is globally one-one ? Remark i : univariate

Non-vanishing case.

of the Jacobians

alone will not suffice except in the

See the example of Gale and Nikaido given in chapter III.

4

Remark 2 :

Even in R I non-vanishing

for global univalence.

of the derivative

is not a necessary condition

For example f(x) = x 3 is globally univalent

whereas its derivative vanishes at x = 0.

throughout R I

In general it appears difficult or

hopeless to derive necessary conditions whenever global univalence prevails. We will cite now a few typical results to give the reader some idea about this monograph.

Fundamental

global univalence theorem

be a differentiable

:

(Gale-Nikaido-lnada)

: Let F : O C R

mapping where ~ is a rectangular region in R n.

n ÷ Rn

Then F is

globally univalent in ~ if either one of the following conditions holds good. (a)

J(x) (= Jacobian of F at x) is a P-matrix for every x E O.

(b)

J(x) is an N-matrix and the partial derivatives

A global univalent theorem in R 3

[ Parthasarathy ]

are continuous

:

for all x s ~.

Let F be a differentiable

map

from a rectangular region O C R 3 to R 3 with its Jacobian J having the following two properties

for every x E ~:

(a)

diagonal entries are negative and off-diagonal

(b)

Every principal minor of order 2 x 2 is negative.

Then F is univalent

Plastock's

theorem

entries are positive.

in ~.

:

Let F:R n ÷ R n

Suppose J does not vanish at any x s R n. f

o then F is a homeomorphism

inf

be a continuously

differentiable

map.

If (I/lIJ(x)-llI)dt

=

11x11=t

of R n onto R n.

In fact F is a diffeomorphism.

(Here llx[l stands for the usual Euclidean distant norm and IIAII = suP11Aull A an

for

n x n matrix and u an n vector with norm one). In order to state

Definition

:

disconnected

MoAuley's theorem we need the following definition.

Call a continuous mapping F:O ÷ R n light if F-I(F(x)) for each x E ~.

is totally

[ Here we will assume ~ to be a unit ball].

open if for each U open in ~, F(U) is open relative to F(O).

Call F

Denote by S F the

set of points x E ~ such that F is net locally one-one at x.

McAuley's

Theorem

:

Suppose that F is a light open mapping of a unit ball ~ in

R n onto another unit ball B in R n such that (I) F -I F(8~) = 8~ (3) FIS F is one-one

8B

(4) for each component C of B-S F there is a nonempty V in

C open relative to B such that FIF-I(v) Scarf's conjecture

(2) F(8O) =

:

Let F:O C R n ÷ R n

is one-one.

Then F is a homeomorphism.

be continuously

differentiable

on a compact

rectangle ~ with det J(x) ~ C for every x s ~

for every x s ~.

(= boundary of ~).

Further suppose J(x) is a P-matrix

Then F is one-one

throughout ~.

This conjecture was proved by three different set of researchers Garcia-Zangwill, Mas-Colell and Scarf et al.

This result is an significant generalization of

Gale-Nikaido's theorem.

Schramm's theorem :

Let ~ be an x-simple domain in the (x,y)-plane, ~ its boundary.

Let F = (f,g) : ~ ÷ R 2 be a differentiable map, ~ the minimum and B the maximum of f on Z.

Suppose the Jacobian of F is an NVL matrix for each z s ~ and for each

u s (~,~),~uppose at most two points z s ~ satisfy f(z) = u. to

~ \ (A(~)UA(~))

Remark i :

is univalent where A(u) = (z:z s ~ a n d

Then F restricted f(z) = u~.

Results obtained so far on global umivalence are not complete and we

have mentioned several interesting open problems throughout the monograph.

For

example it is not known whether Gale-Nikaido's result holds good in any compact convex regions.

In chapter VIII and IX we have given various applications of

univalent results in other areas like differential equations, Economics, Mathematical programming, Algebra etc.

Remark 2 : All the theorems cited above with the exception of McAuley's theorem depend on the choice of a fixed coordinate system. conditions on the Jacobian matrix.

This is so because we place

Though one may argue that this may not be the

most natural approach to the problem under consideration, the present writer feels that this method yields useful results in many problems that arise in practice. See Chapter VII and Chapter IX in this connection.

Also in some special cases the

matrix conditions turn out to be necessary as well - see for example theorem I and theorem 6 in chapter VIII.

Also, the present writer feels that it is not difficult

to check these matrix conditions in a given problem.

CHAPTER

II

P-MATRICES AND N-MATRICES

Abstract

:

In this chapter we will give a geometric

We will give some properties

of N-matrices.

characterization

global univalence results due to Gale, Nikaido and Inada. interrelation between P-matrices

of P-matrices.

These facts we need later to prove We will also see the

and positive quasi-definite

matrices.

examine the question whether P-property holds good under multiplication P-matrices

Finally we (sum) of two

- this kind of result is useful in determining when the composition

F o G (sum, F+G) of two univalent Let A be an

functions

is univalent.

n × n matrix with entries real numbers.

matrices with complex entries. if the associated quadratic

We will not consider

If A is a symmetric matrix then A is positive definite

form x'Ax ~ 0, for any x different from O.

denotes the transpose of the vector x.

Here prime

It is well known that a symmetric matrix A

is positive definite if and only if every orincipal minor of A is positive. we drop the symmetric assumption results?

In other words,

for every x ~ 0.

from A.

In such situations

Suppose

can we prove similar

suppose A has the following property,

namely x'Ax ~ 0

They can we assert that every principal minor of A is positive?

Another interesting question is to characterize matrices whose principal minors are positive.

Next we will answer these questions.

Characterization not necessarily

of P-matrices

:

We will start with a few definitions.

Let A be a

symmetric real n × n matrix.

Definition

:

Call A a P-matrix if every principal minor of A is positive.

Definition

:

Call A a positive quasi-definite

Definition

:

Call A an N-matrix if every principal minor of A is negative.

N-matrices

are divided into two categories:

(i)

matrix if x'Ax ~ 0 for every x # 0. Further

An N-matrix is said to be of the first category if A has at least one positive

element. (ii)

An N-matrix is said to be of the second category if every element of A is

non-positive. Definition

:

Call A a Leontief-type matrix if the off-diagonal

non-positive. We will make a few quick remarks.

entries are

Remark i :

Every positive quasi-definite matrix is necessarily a P-matrix (we will

give a proof of this fact after characterizing the class of P-matrices). converse is not necessarily true as the following example shows.

But the

12 Let A = [ 0 1 ]"

Then (Au,u) = Ul+2UlU2+U2 2 2 (where u = (Ul,U2)) and (Au,u) = 0 whenever u I = - u 2. Thus A is a P-matrix but not positive quasi-definite. positive quasi-definite matrix if and only if ( - ~' In this example ( - - ) Remark 2 :

=

I [i

Also observe that A is a ) is a positive definite matrix.

i i ] is a singular matrix.

The following example shows that every positive quasi-definite matrix 2 2 Let A = [ 3 2 ] Clearly (Au,u) = Ul+5UlU2+8u2 > 0

need not be positive definite. for any u # O.

Hence A is positive quasi-definite but not a positive definite matrix

as A is not symmetric. Remark 3 :

First category N-matrices share some properties in common with P-matrices

as we shall see below. category N-matrices.

However there is a nice characterization for symmetric second In order to do that we need the following definition.

matrix A, merely positive definite if (i) x'Ax< 0

Call a

there exists some vector x such that

and (ii) whenever x'Ax < 0, this will imply Ax < 0 or Ax > 0-in other words

Ax is onesigned.

The result then is the following.

the second kind then A is merely positive definite. (simple) negative eigenvalue.

If A is a symmetric N-matrix of Furthermore A has exactly one

Proofs of these results may be found in Rao

[62 ].

We are now ready to prove some results on P-matrices. Theorem I :

Let A be a P-matrix or anN-matrix of the first category.

Then the

system of linear inequalities Ax<

0

and

x>O has only the trivial solution x = O. Game theoretic interoretation of theorem i :

Theorem I says that the minimax value

of the matrix game A (as well as the minimax value of every principal submatrix C of A) is positive, brovided A is a P-matrix or an N-matrix of the first kind. can he seen as follows. or equal to zero. columns). A'y ~ 0

This

Suppose yon Neumannvalue of the matrix game is less than

(We will assume minimizer chooses rows and maximizer chooses

We have a probability vector y for the minimizer such that y'A ~ 0 (prime denotes transpose).

If A is a P-matrix or an N-matrix so is A'.

Thus we have got a nontrivial non-negative vector y satisfying A'y ~ 0 dicts theorem i and consequently value of A must be positive. value of A' is positive.

See

or

which contra-

It is also clear that

[49, 54 ] for details regarding game theory and

for results relating game theory and M-matrices.

[61 ]

We follow the proof as given in

Nikaido

[44].

Proof of Theorem I :

First we will prove when A is a P-matrix.

induction principle.

For n = I, clearly Theorem i, is true.

We will use

So assume theorem 1

for n = k, prove that it holds good for n = k+l (Here n refers to the order of the square matrix A).

Let x > 0.

allX I . . . .

°

That is,

+ a12x 2 + .-. + al,k+ I Xk+ I .

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

%+l,lXl+~k+l,2X2+

.

.

.

.

.

.

.

.

.

.

.

.

.

j 0

.

+ %+l,k+l~k+l < 0

Since aii > 0, we can increase if necessary x i such that one of the inequality becomes an equality.

We will assume without loss of generality

allX I + a12x2 + ... + al,k+ 1 Xk+ 1 a21x I + a22x2 + ... + a2,k+ 1 Xk+ 1 . . . .

o

.

.

.

.

.

.

.

.

.

.

.

.

.

.

ak+l,lXl+ak+l,2X2+ Using the first equality, resulting

inequalities

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

=

0

<

0

.

...+ ak+l,k+iXk+l j 0 .

{+l,lX2 +

+ %+l,k+l Xk+l i

0

all where a~.12 =--xail ao.lj - alj where i,j = 2,3,...,k+i. is a P-matrix of order k.

allX I = 0.

x i = 0 ¥ i ~ 2. Substitu-

But all > 0 and hence x I = 0.

Now assume A is an N-matrix of first category.

a < 0, d < 0, and ad-bc < 0. A is of first category, A-IAx = x < 0.

This means b

b > 0 and c > 0.

But x > 0 by hypothesis,

This

Clearly order of an N-matrix Suppose A =

a [c

bd ] where

and c should be of the same sign. Consequently A -I > 0.

therefore x = 0.

Since

If Ax < 0 then

This proves the theorem

As before assume the result for n = k where k > 2 and prove it holds

good for n = k+l.

As A has at least one positive element, we can imitate the proof

verbatim given for P-matrices

till we get the matrix C = (a~j)i,j = 2,3,...,k+i.

Observe that det A = all det C. that det C > 0.

Since all < 0, det A = all det C < O, it follows

In fact one can check that C is a P-matrix.

the first part of the proof x i = 0 ¥ i ~ 2. equality.

C = (a~j)

the proof of theorem i when A is a P-matrix.

of first category should be at least 2 x 2 .

when n = 2.

Plainly the matrix

Hence by induction hypothesis,

ting this in the first equality, terminates

The

one can eliminate x I from the other inequalities.

can be written as

This terminates

Hence it follows from

Since all # 0,x I = 0 from the first

the proof of theorem i for N-matrices

of first kind.

Remark I :

Geometrically,

theorem I says the following:

Any non-trivial non-negative

vector cannot be mapped to a vector in the negative orthant when A is a P-matrix or an N-matrix of the first kind. Remark 2 :

Theorem I is valid for any matrix A which has non-negative

is A -I ~ 0.

Characterization

class of matrices.

inverse - that

results are available in the literature for such

In particular

if A is a Leontief type matrix then A -I ~ 0 if and

only if there exists some x > 0 such that Ax ~ 0. Remark 3 :

A result on linear inequalities

For any given matrix D not necessarily alternatives

holds.

asserts the following

Either the inequalities x'D < 0

the inequality Dy ~ 0

has a non-negative

trivial solution. poverse,

This statement

is the following,

has only a

is equivalent to the fact that A has a left

is due to Charnes et. al.

inequalities

solution or

conclusion of theorem i can be

For any matrix A, suppose the system Ax ~ O, x ~ 0

that is there exist non-negative

observation

has a semipositive

solution.

In view of this result on linear inequalities, viewed as follows:

E See 18, pp. 49].

a square matrix exactly one of the following

matrices N,M such that NA = I+M.

[ 8].

This

Another feature of the result on linear

which says that von Neumann value of a P-matrix game

is positive. Theorem 2 :

Suppose A is a P-matrix or an N-matrix of first category.

Then there

exists a positive vector Yo ~ 0 such that Ay ° ~ 0. Proof

:

From theorem i, the matrix D = (A,-I) has no semi-positive

x'D < 0 and consequently for some y ~ 0.

solution x with

from the above remark it follows that Dy = (A,-l)y > 0

Here y = (yl,Y2,...,yn,

Yn+l' Y n + 2 ' ' ' " Y 2 n )"

Then clearly Ay ° ~ z ~ 0 where z = (Yn+l' Y n + 2 ' ' ' " Y 2 n )"

Define Yo = (YI'Y2"'Yn)

This terminates

the proof

of theorem 2. Corollary I :

Let S n = {x : X ~

n 2 ~ x i = i}. i=l

O,

or an N-matrix of first category.

Let A be a P-matrix or

Then there exists an a ~ 0 such that

for every x s S .

Max (Ax)i>a_ l~i~n

n

Proof :

From theorem I, fer every x s S n it follows that Ax has at least one

coordinate strictly positive. Min x ~ S

Since

Max (Ax) i is continuous l~i~n

Max (Ax) = Max (Ax°) i for some x ° E S l~i~n i l
Set ~ =

in x and S n is compact, Max (Ax°).. l
This

n

will satisfy the requirements

of the corollary and the proof is complete.

In order to give a characterization following definition:

theorem for P-matrix,

we introduce the

10

Definition :

Let A be an n × n matrix.

The matrix A is said to reverse the sign of

a vector x c R n if (Ax) i x i ~ 0 for all i = 1,2,...,n.

(Here as usual (Ax) i denotes

the i-th coordinate of the vector Ax). Theorem 3 :

Let A be an n x n real matrix.

Then A is a P-matrix if and only if A

does not reverse the sign of any vector except the zero vector. Proof (Necessity) zero vector x.

:

Let A be a P-matrix.

If x ~ O

Suppose A reverses the sign of a non-

then (Ax) i x i ~ O for all i, will imply A x ~

O.

In

other words we have a semipositive solution to the system of inequalities A x ~ O and x ~ O, but this contradicts theorem I. diagonal matrix with diagonal entries j ~ L or j c L.

Let L = (j:xj < O) and let D be the

+I or -i in the j-th column according as

Trivially Dx > O and Dx is not a zero vector for x W O.

Also

D-IA D is a P-matrix and it reverses the sign of a non-negative vector Dx which is not possible as shown before. (Sufficiency).

This completes the necessity part.

Let A be any n x n matrix which does not reverse the sign of

any non-zero vector.

Note that every principal sub-matrix C of A also cannot

reverse the sign of any non-zero vector. is positive.

We will now prove,

As such it is enough to show that det A

Suppose det A ~ O , then (since det A is equal to the product of the

eigenvalues) there will exist a real eigenvalue ~ < 0. Let Ax = Xx where x is the _~2 corresponding eigenvector. Clearly (Ax)ix i- x i ~ O. In other words A reverses the sign of a non-zero vector which is a contradiction.

This terminates the proof

of theorem 3. One obvious consequence of theorem 3 is the following corollary. Corollary 2 :

Suppose A is a non-singular matrix.

Then A is a P-matrix if and only

if A -I (= inverse of A) is a P-matrix. Proof :

Suppose A is a P-matrix and A -I is not a P-matrix.

such that A -I reverses the sign of x.

Then there exists x ~ O,

This will imoly A will reverse the sign of

y = A-Ix # 0 and hence a contradiction (via theorem 3) to the supposition that A is a P-matrix.

This co~oletes the proof of corollary 2.

We will now identify several well known classes of P-matrices. Proposition I :

(i)

Any positive quasi-definite matrix is a P-matrix (ii) Any

matrix having a positive dominant diagonal is a P-matrix and (iii) Any matrix that satisfies Stolper-Samuelson condition (A matrix A is said to satisfy Stolper-Samuelson condition if A ~ O, A is non-singular and A -I is of Leontief type) is a P-matrix. Proof :

(i)

Let A be positive quasi-definite.

a vector x ~ 0 such that (Ax) i x i ~ O. x'Ax < O

If A is not a P-matrix there exists

[ This is a consequence of theorem 3].

which contradicts our hypothesis regarding A.

Hence

Therefore every positive

11

quasi-definite matrix is a P-matrix. be taken up last.

We will now prove (iii).

Proof of (ii) will

Let A be a matrix with A > 0, A non-singular and A -I Leontief type.

We will first prove that the yon Neumann value associated with A -I is positive. Suppose value of A -I is less than or equal to zero. of a probability vector y such that yA -I < O.

This will imply the existence

[ Here we assume that the rows are

chosen by the minimizer and the columns by the maximizer]. or y _< 0

which is a contradiction.

a probability vector x

Hence value of A -I > 0.

such that A -I x

O

Since A > O, yA-IA _< 0

> 0.

O

That is, there exists

Since A -I has off-diagonal entries

non-positive it follows that A -I has positive dominant diagonal property.

[We say

that a matrix B of order n × n has positive dominant diagonal property if there exists a strictly positive vector d = (dl,d2,...,d n) where each d. > 0 such that n 1 bii d i > ~ Ibijld j j=l j#i good.

for every i = 1,2,...,n].

For the moment assume (ii) holds

Then A -I is a P-matrix and the corollary after theorem 3 tells us that A is

a P-matrix.

Thus we need to prove only (ii) to comDlete the proof of proposition i.

Let A be a matrix having a positive dominant diagonal. That is, dl>0,d2>0,..., n dn ~ 0 such that a..~1 di >j=l ~ laijld"J Clearly a..m~ > 0 for i = 1,2,...,n. First j#i we will prove that det A ~ 0.

Then we will show det A is positive.

Suppose

n

j=l

a.. x. = 0 for i = 1,2, . lJ J

Define

,n .where x . .

rxil 0

=

max

l~l -

l 0.

di

dk

n = - ~ akj x.. j=l J j ~k

akk x k

det

a~

~ 0,

A # 0.

diagonal Note

that

clearly

l%T

Now

then det

0 + + ~.

say .

In case ~ = 0, then x will be zero vector which will contradict the

fact that x is a non-zero vector.

Since

(Xl,X2, .. .,xn) is a non-zero vector.

~ 0 %.

it will

(A+pI)

If det impossible.

shown has

that

this

Hence

det

a positive

is a continuous

A ~ 0

We have

n Then lakkXk I < I ~ akj xjl < ~ ~klakjldj < 0 akk dk. --j=l -- j j ~k

That is 0 ~ 0 which

be

A + pl also

Suppose 0 > 0.

will det

dominant

function imply

A > 0.

is impossible.

A > 0.

det

in

a positive

diagonal

for

0 and

that

a positive

dominant

every

det

(A + pe I) = 0 for

If A has

This shows that

If A has

p > 0.

(A +

some

dominant

pl) ÷ + ~ as

0 O > 0 which diagonal,

principal submatrix C of A also has the same property and consequently P-matrix.

This terminates the proof of the proposition.

enough to identify several classes of P-matrices. symmetric N-matrices of the second kind.

is every

A is a

[Thus we are fortunate

We are able to characterize only

The situation is far from complete in the

case of non-symmetric N-matrices ].However one can prove the following elementary

12 result for N-matrices of order 3 × 3 of the first kind Preposition 2 :

[53].

Let A be an N-matrix of the first kind with order 3 × 3.

Then

A will contain exactly four positive elements and five negative elements. Proof :

Since A is an N-matrix of the first kind and since A is of order 3 × 3, it

is clear that the number of positive elements is either 2 or 4 or 6. 2 positive elements.

Suppose A has

Then A will be of the form : +

A

Clearly det A > 0

=

which is impossible.

the off-diagonal entries are positive.

Similar contradiction will be reached if Thus it follows that A contains exactly four

positive elements and five negative elements.

This terminates the proof of

proposition 2. Remark i :

If A is a positive definite or a quasi-positive definite matrix and if

Q is a non-singular matrix then it follows that Q'AQ, is a positive definite or a quasi-positive definite matrix.

Such a result is not valid in general for P-matrices

(though it is true if Q is a diagonal matrix with all entries +I or -I. This fact is used during the course of our proof of Theorem 3). Let A = [ 1 0 -42 ] and Q =

[04

Ii ] "

Then Q'AQ =

[ 146 -12_i] "

Clearly A is a P-matrix while Q'AQ is not a

P-matrix. Remark 2 :

If A is an N-matrix of the first kind and if Q is a diagonal matrix

with all entries +I or -I then Q'AQ is an N-matrix but it may be of second kind. Open problem :

Let A be a positive definite matrix.

Can one find a diagonal matrix

D with diagonal entries strictly positive such that DA-I is a (weak) N-matrix? [We call a matrix B, a weak N-matrix if every principal minor of B is non-positive]. If the answer to this question is in the affirmative, this will settle an old conjecture of Paul Levy regarding the infinite divisibility of multivariate gamma distribution - See Paranjape

[51].

A final remark on N-matrices of the first kind:

If A is an n x n, N-matrix of

the first kind, then every principal minor of order less than or equal to n-i of A -I is positive.

(Note that det A -I < O since det A < 0).

Is AB a P-matrix or N-matrix when A and B are P-matrices or N-matrices ?

In this

section we are interested in examining whether the product of two P-matrices (or N-matrices) will be a P-matrix (or N-matrix).

In this connection we have the

13

following

[ 53].

Theorem 4 :

(i)

There exist two P-matrices (N-matrices) A and B such that AB is

not a P-matrix (N-matrix). (ii)

If A and B are 2 x 2 Leontief type P-matrices (N-matrices of the same kind)

then AB is a P-matrix. (iii)

If A and B are 3 x 3 Leontief type P-matrices then AB is a P-matrix but AB

need not be of the Leontief type. (iv)

If A and B are n x n Leontief type B-matrices and AB is also of Leontief type

then AB is a P-matrix.

However this result is not valid if we do not assume that AB

is of Leontief type when n > 4 [ see (iii) above]. Proof :

(i) A =

-~ ], B =

[0i

[2i

0] 1

Then AB = [-~

-~]

is not a P-matrix.

If we assume all the entries to be positive then clearly AB is a P-matrix if A and B are 2 × 2 P-matrices.

However such a result is not true for 3 x 3 P-matrices.

Let

111

I 2

,

4

=

12

i

2

i

i

1

2

Then

AB

=

13 i 7

~7

9

ii

21

29

Clearly A and B are positive-definite matrices but AB is not a P-matrix as the principal minor

4] is negative. 9

[

(ii) Proof of this is elementary and we omit it. necessary.

However a word of caution is

If A,B are 2 x 2 matrices with A, an N-matrix of the first kind, B and

N-matrix of the second kind then AB need not be

neither a P-matrix nor an N-matrix

as the following example demonstrates.

[-i i

AB = (iii)

[-2 - ]

Let

A =

2] B = -i '

[-3 -I

-7] -2

then

is neither a P-matrix nor an N-matrix.

We will prove that AB is a P-matrix provided A and B are 3 x 3 Leontief

type P-matrices.

A

=

Let -al 2

-al 3 ~

[~ i i

a22

-a23|

L-a31

-a32

a33 j

Here a.. and b.. are all non-negative. ij

Ij

'

B

=

bll

-bl 2

-bl 3

-b21

b22

-b23

L-b31

-b32

b33

It is clear that det(AB) = det A. det B > 0

and further the diagonal entries are positive in AB. every principal minor of order 2 is positive.

We need only to check that

Let us take the leading principal

14

minor C of order 2 from AB.

C

: la c

where

Then

+

a13

b31

b

+

a13

b32]

+

a23

b31

d

+

a23

b32j

la

all a12 L -a21

c Clearly Then

ad - bc > 0.

a22

bll

-b21

-b21

b22

Also note b : -all b12 -a12 b22 < 0

and c = -a21bll-a22b21 < O.

det C = (ad-bc) - a13 b32 c - a23 b32 b + a23 b32 a + a13 b31 d

>

O.

Similarly one can show the other two principal minors of order 2 in AB are positive. Thus we have shown that AB is a P-matrix. (iv)

We will show that AB is a P-matrix if A,B are Leontief t~/pe P-matrices and if

AB is of Leontief type.

Since A and B are Leontief type P-matrices it follows that

A -I and B -I are non-negative.

Consequently

(AB) -I satisfies Stolper-Samuelson

(AB) -I is non-negative.

In other words

condition and therefore AB is a P-matrix.

We

will now give a counter example to show that this result is false if we drop the assumption AB is of Leontief type when n > 4.

A

=

Let,

110ii iI li l°il i3i 0

~

0

0

0 0

_0

Then,

AB

:

-

and B

=

2

0

I

-4

i

0

0

0

7/8

2/8

0

0

-4

0

-I

0

0

-

One can check that A and B are Leontief type P-matrices but AB is not of Leontief type. 2,

Also note that AB is not a P-matrix as one of the principal minors of order

i [ 7/8

3 2/8 ]

is not positive.

This terminates the proof of theorem 4.

As already pointed out elsewhere, the motivation for proving theorem 4 is simply

15

that it helps us to determine when F o G will be univalent given that F and G are differentiable and univalent in R n.

Gale-Nikaido's fundamental result asserts

global univalence of a differentiable mapping F when the Jacobian of the map F is a P-matrix.

Note that the Jacobian associated with F o G is equal to the product

of the Jacobian matrices associated with F and G. in another chapter.

We will prove the following result

If F and G are differentiable maps from R 3 (or R 2) to R 3 (or R 2)

and if their Jacobians are P-matrices everywhere then F o G is a P-function and consequently one-one throughout R 3 (or R2).

From theorem 4 we may be able to

conclude such a result is valid in R n for any n provided we assume that the Jacobian associated with F o G is of Leontief type. If A and B are 2 × 2 Leontief type P-matrices then A+B is of Leontief type but need not be a P-matrix as the following simple example demonstrates• •

i ] is not a P-matrix.

Let A =

[~

-~]

Such results will be

useful to determine when F+G is one-one given that F and G are one-one, differentiable functions.

Observe the following result which is easy to prove.

If A and B

are

n x n Leontief type P-matrices and if there exists a non-negative vector x ~ O such that (A+B)x > 0

then A+B

is a Leontief type P-matrix.

[ Here (A+B)x > 0 means,

every component of the vector (A+B)x is positive]• It may not be out of place to mention other interesting equivalent properties of P-matrices. Theorem 5 :

We will state it in the form of a theorem without proof

Let A be an n × n

real matrix.

[ 56].

Then each of the following conditions

is equivalent to the statement : A is a P-matrix (i)

Every real eigenvalue of each principal submatrix of A is positive.

(ii)

A+D is nonsingular for each nonnegative diagonal matrix D

(iii) For each x ~ 0 there exists a nonnegative diagonal matrix D such that x,ADx > 0 (iv)

A does not reverse the sign of any nontrivial vector.

(v)

For each signature matrix S there exists an x > 0

Remark

such that SASx > 0.

I : We have already shown that (iv) is equivalent to the statement that A

is a P-matrix.

We can use this fact to prove the equivalence of other statements.

When A is of Leontief type, proof can be given using ideas from game theory-interested readers should refer to Raghavan available for N-matrices. problems: (I)

[ 61].

Such a beautiful characterization is not

We will close this chapter by mentioning two open

Suppose A is a P-matrix.

Does it mean A is positive stable?

[ Call

a matrix A positive stable if the real part of each eigenvalue of A is positive]. Clearly this is so if either A is a Leontief type P-matrix or if A is a 2 × 2 P-matrix.

(2) Suppose A is a positive definite matrix.

Does there exist a diagonal

matrix D with entries strictly positive such that DA-I is a (weak) N-matrix (that

16

is principal minors of (DA-I) is non-positive).

Again this result is true if A is

a 2 × 2 positive definite matrix. David Gale has communicated to the author the following example which answers the first question raised above in the negative.

A

Then A has eigenvalues

Let

lit!f0

=

l+s ,

E-½ + i ~ / 2

with negative real parts if s < ½

and

of course A is a P-matrix. We also wish to add that if A is a Dositive definite or quasi-positive definite matrix then A is positive stable. definite case.

We will indicate the proof in the quasi-positive

In fact as already remarked elsewhere

if and only if (A+A')/2 is a P-matrix. eigenvalues of (A+A')/2.

Let ~m

and

Let ~ be any eigenvalue of A. Xm-< Real part of ~ --< ~M

Since

A

~ > 0, it follows that A is positive stable. m

is quasi-positive definite

XM be the minimal and maximal Then it is well-known that

CHAPTER

III

FUNDAMENTAL GLOBAL UNIVALENCE RESULTS OF GALE-NIKAIDO-INADA

Abstract:

In this chapter we will derive fundamental global univalence results of

Gale-Nikaido-lnada for differentiable maps in rectangular regions whose Jacobian matrices are either P-matrices or N-matrices.

Using mean-value theorem of differen-

tial calculus we will obtain global univalence results for differentiable maps whose Jacobian matrices are quasi-positive definite matrices, in convex regions. result is due to Gale and Nikaido.

This

Finally we present two new results on univalent

mappings in R 3 . Let F be a differentiable mad from ~ R

n to R n.

We are interested in finding

suitable conditions such that F is univalent throughout ~.

Non-vanishing of the

Jacobian matrices will not suffice as we shall see below. approaches to the problem under consideration.

There are at least two

One approach places topological

assumptions on the map and the other places further conditions on the Jacobian matrices.

We will study the former in the next chapter and the latter in the

present chapter.

Global univalence theorems :

In order to state and prove results on univalent

mappings we will adopt the following notation and terminology. Gale and Nikaido's original paper. = {x:x ~ R n, a i ~ x i ~ bi).

We closely follow

Let ~ be a rectangular region in R n, where

Here ai,b i are real numbers where we may allow some

or all of them to assume -~ or +~.

Let F:~ ÷ R n be a mapping defined

by

F(x) = (fl(x), f2(x),...,fn(X)) where each fi(x) is a real-valued function defined on ~.

Then F is said to be differentiable in ~ just in case every fi has a total

differential ~fij(x)dxj in D - in other words for x, as~, fi(x)=fi(a)+ 2f + o(llx-all)(i=l,2

.....

n),

where o ( t l x - a l l ) / l l x - a l l

÷

0

as

x ÷ a.

The Jacobian matrix J of the mapping F is given by Jex) =

(a)(x.-a.)

j ij

a

a

llfij(x)ll.Differen-

tiability of F clearly implies continuity of F and it also implies partial differentiability of f!si and in fact fij(x) = ~fi(x)/~xj. the boundary points.

However, one has to be careful at

Yet the Jacobian can be defined by means of the coefficients of

the total differential at those points.

Now we are ready to prove a non-linear

counterpart of theorem I of chapter II.

Theorem i :

Let F be a differentiable mapping from ~ to R n.

matrix J(x) of the mapping F is a P-matrix for every x s ~.

Suppose the Jacobian Then for any a and x in

~, the inequalities F(x) < F(a) and x > a have only one solution namely x = a.

18

Proof :

We prove this theorem by induction on n, the common dimension of both the

argument and image vectors x,F(x). X = {x:x s ~ Clearly a ~ X.

and

Plainly theorem is true for n = I. F(x) i F ( a ) ,

x h a} .

We are through if we show that X contains only a.

we first prove that a is an isolated ooint of X. F(x)-F(a)

lim X

+

II

_ J(a)

llx_al I

we have

x-a

~

II = 0.

there is a positive number 6 > 0 by corollary i, chapter II

such that for any x > a, some coordinate of J(a) in any nei~hbourhood

7 x-a 1~

of a, some component of F(x)-F(a)

This shows that a is an isolated point of X.

Clearly b > a.

With that in mind,

As F is differentiable,

a

Since J(a) is a P-matrix,

in ~.

Let

Define y r - x

as follows

~ > 0.

Consequently,

must be oositive for x > a

Suppose b c X with b ~ a.

.

Y = (x : a < x < b, F(x) < F(a)) Plainly Y is compact and since a is an isolated ooint, Y - (a) is also compact. be a minimal element of Y - {a) in the sense that no other element y E Y -

Let

{a)

fulfills y ~ x (Note that such an x can be obtained by minimizing the sum of the components

of y when y ranges over Y - {a)).

As ~ s Y - {a), only two possibilities

can occur (i) x > a, (ii) x > a and x has some comoonent equal to the corresponding component of a.

Case (i) :

x > a.

Theorem 2 of cha~ter II ensures the existence of a vector u

satisfying u < 0 and J(x) u < 0 (for J(x) is a P-matrix). As u < 0, ~

a

and ~ is a rectangular

Moreover by differentiability

F(x(t))-F(x) tliull

we olease by letting t aoproach 0. t,F(x(t))

_ j(~) ~ _ ~

Case (ii)

:

Since J(x) u < 0 it follows for small positive

the minimality of x.

respectively.

Note that x(t) < x

This rules out the oossibility

Now we will apply induction principle.

generality that

small positive t,x(t)s~.

can be made as small as

< F(~) < F(a) and consequently x(t) s Y - (a}.

this contradicts

Define x(t) = x + tu.

region for sufficiently

We will assume without loss of

Xl = al where Xl and a I are the first coordinates We now define a new differentiable = ((x2,x3,...,Xn)

and

of case (i).

of x and a

mapping G:~ ÷ R n-I

where

: (al,x 2 ..... Xn) s ~}

and gi(x2,x3,...,Xn) Jacobian matrix of G is plainly a principal P-matrix

.

Further

= fi(al,x2,x3,...,Xn) , i = 2,3 .... ,n submatrix of J(x) and hence it is a

19

gi(x2,x 3 .-.,Xn ) ~ x. i

gi(a2,a3,..-,a n)

> a. --

for

i = 2,3,...,n.

1

Therefore we must have xi : ai for i = 2,3,...,n by induction hypothesis. contradicts x # a.

Hence case (ii) is also impossible.

This

This terminates the proof

of theorem i. For N-matrix we have the following [44].

Theorem 2 :

Let F be a continuously differentiable mapping from 9 ÷ R n (n > 2).

Suppose J(x) is an N-matrix of the first category for each x s 9.

Then F(x) < F(a)

and x > a imply x : a.

Proof :

Proof of this theorem is almost the same as that of theorem i except that

one has to exercise caution for two reasons:

(i) induction will start when n = 2

(ii) Althouth every principal submatrix of an N-matrix is also an N-matrix they need not be of the same category.

In other words we not only have to show that the

Jacobian of the new mapping G to be an N-matrix but also to be of the first category. (i)

We will now Drove theorem 2 when n = 2.

all the partial derivatives are continuous. fll J(x)

As F is continuously differentiable

Consider the Jacobian matrix J(x) : fl

=

21

f22~

As J(x) is of first category, fll < 0, f22 < 0 and f21 > 0, f12 > 0.

As all the

partial derivatives are continuous, they will be of definite sign throughout x s ~. Define G = (f2,fl). G(x) < G(a). (ii)

Jacobian matrix of G is a P-matrix.

Clearly F(x) ~ F(a) iff

Using theorem I, we can complete the argument.

We will now show that the Jacobian of the map G which appears under

case (ii) of theorem I, is an N-matrix of the first kind in order to complete the induction argument of the present theorem. theorem i.

The rest of the proof is same as that of

Define J(x2,x3,...,x n) = principal minor of J(x) got by omitting the

first column and the first row where x = (al,x2,...,xn). Suppose every element of J is non-positive.

Clearly J is an N-matrix.

If flj(Xl,X2,...,xn) > 0 for j=2,3,...,n

then flj(X) > 0 for every x s ~ - this is due to the fact that all partial derivatives should be of definite signs (as already pointed out in (i)). But since Xl = al'xi ~ ai, i = 2,...,n with strict inequality for atleast one i ~ 2, we have fl(x) > fl(a) contradicting F(x) ~ F(a).

Therefore flj (Xl'X2'''"Xn) ~ 0 for some Jo"

Define

a vector v which has zero everywhere except at the Joth coordlnate where it is i. This vector would be a nontrivial solution to J(x)v < 0 which contradicts theorem i

20

of chanter II.

Hence Jmust be s~ N-matrix of the first kind.

This completes the

discussion for theorem 2.

Remark :

Theorem I (as well as 2) can be proved under somewhat weaker restrictions

on the Jacobian matrix.

For example theorem i holds good if for every x, J(x) as

well as its principal submatrices and their transpose matrices considered as matrix games have positive minimax values.

We have already pointed out that every P-matrix

has this proDerty but the converse is not true.

For examole the matrix [~

~] is

not a P-matrix but every game that can be constructed out of this in the manner described as above has oositive value.

In order to Drove the main result due to

Gale Nikaido and Inada we need the following definition.

Definition :

A mapping F:O + R n is a P-function if for any x,y s ~, x W y, there

is an index k = k(x,y) E {l,2,...,n} such that (xk-Yk)(fk(x)-fk(y)) > 0. a P-function is a I-i function

Trivially

[42].

Fundamental Global Univalence Theorem (Gale-Nikaido-lnada)

:

differentiable mapping where ~ is a rectangular region in R n.

Let F:~ ÷ R n be a Then F is globally

univalent on ~ if either one of the following corLditions holds good. (a)

J(x) is a P-matrix

V x c

(b)

J(x) is an N-matrix and the partial derivatives are continuous for all x s ~. In fact, F is a P-function under condition (a).

Proof :

(under condition a) : We assume that J(x) is a P-matrix ¥ x s ~.

show that F is a P-function, using induction principle.

We will

For n = i, result is obvious.

We will assume the result to be true for n-I and show that it is true for n. x,y s ~ be such that x ~ y.

Suppose (xi-Yi)(fi(x)-fi(y)) ~ 0.

that x i ~ Yi for all i = 1,2,...,n.

Let

We will also assume

For if x i = Yi for some i, we can construct

a mapping G similar to that as in theorem I, and then use induction hypothesis to exhibit an index k such that (xk-Yk)(fk(x)-fk(y)) > 0.

Let D be a diagonal matrix

whose i-th diagonal entry is +i or -i according as x i > Yi or x i < Yi"

Let H be

a mapping from D(~) ~ R n defined as follows: H(z) = DoFo D(z) for every z ~ D(~). Plainly D(~) is a rectangular region and the Jaeobian matrix of the mapping H is again a P-matrix.

Let x* = D(x) and y* = D(y).

Since (xi-Yi)(fi(x)-fi(y)) ~ 0

and x i ~ Yi for all i, it follows that H(x*) ~ H(y*) where x* ~ y*, but this contradicts theorem I.

This terminates the proof of the first part.

Proof (under condition b) :

We will now prove univalence when J(x) is an N-matrix.

Suppose F(x) = F(y) and x ~ y.

Let D be a diagonal matrix with i-th diagonal entry

+ i or -i according as x i ~ Yi or x i < Yi"

As in part (a), let H be a mapping from

21

D(~) ÷ R n defined by H = D o F o D.

It is clear that the Jacobian JH of H will be

an N-matrix, the partial derivatives will be continuous and they will have definite signs.

Also throughout, JH will be of the same category.

then H(x*) = H(y*) and x* = D(x) ~ y *

If JH is of first category,

= D(y) imply (from theorem 2) that x* = y* or

x = y which contradicts ~ u r assumption x ~ y. member of JH will be strictly negative.

If JH is of second category then every

If x* ~ y*, then H(x*) ~ H(y*), with the

inequality strict in some coordinate, contradicting the fact that H(x*) = H(y*). Hence as before x* = y* or x = y contradicting x ~ y.

This finishes the proof of

part (b) and the theorem.

Remark I : an N-matrix.

Continuity of the partial derivatives is crucially used when J(x) is It is not clear whether the continuity of the partial derivatives can

be omitted from the theorem.

Remark 2 :

We have assumed ~ to be a rectangular region in theorem 1,2 and the

fundamental theorem.

How far can one relax this assumption?

assume O to be a convex region?

Will it suffice if we

The following theorem gives a partial answer, when

the Jacobian is everywhere a positive quasidefinite matrix - in this case proof is also extremely simple which uses mean value theorem of differential calculus of a single variable.

Theorem 3 :

Let ~ C R

n be a convex set and F:~ ÷ R n be a differentiable map, with

its Jacobian everywhere positive (or negative) quasidefinite in ~.

Then F is

univalent.

Proof :

We will prove only for positive quasi-definite case as the proof for negative

quasi-definite case is similar. that F(a) ~ F(b). assumption.

Suppose a,b s ~ with a ~ b.

Let x(t) = t a +

By convexity, x(t) s D for t E [O,i].

f.(b))(l > t > 0). i

We will directly show

(l-t)b = b+ t(a-b) = b+ th where h = a-b ~ 0 by Let @(t) = ~i hi(fi (x(t)) -

Differentiating with respect to t, we have @'(t)= Z ~h.h.f..(x(t)) j i 1 J lJ

which is positive everywhere by positive quasidefiniteness of the Jacobian. ~(0) = O, ~(i) ¢ O.

Since

That is 2hi(fi(a)-fi(b)) # 0 or f.(a)1 ~ fi (b) for some i.

Consequently F(a) ~ F(b).

This proves F is univalent.

As an application we have the following corollary.

Corollary (Noshira, 1934) :

Let g(z) be an analytic (complex) function of a complex

variable z in a convex region ~ of the complex plane. has positive real part in ~.

Proof :

Then g(z) is univalent in

Suppose its derivative g'(z) ~.

Let g(z) = u(x,y) + i v(x,y) where i2 = - I and z = x+ iy.

Then g'(z) =

22 ~u

~v

~x+ i ~ .

Note~1~ > 0.

Now, consider the mapping F : ~ ÷ R 2 defined by

F(x,y) = (u(x,y), v(x,y)).

Its Jacobian is given by

J=

?v

?v

However from the Cauchy-Riemann equations we have ~u ~x

~v ~u ~y ' ~y

~v ~x

Hence

JJ2

xu°I

is a P-matrix for

~U

~--> 0. 4x

This implies that J is positive quasidefinite and from theorem 3 one can conclude that F is univalent.

This terminates the proof of the corollary.

The following example shows that theorem 3 may not hold good in a non-convex region.

Let g(z) = z + ~ (z # 0). Z

Let ~ be the common exterior portion of two

circles of radius @ having their centres at ½ and - ½

respectively.

One can

verify that the Real part of g'(z) is positive in ~, as

Re(g'(z))

= ([Z-½12-¼)(Iz~212-¼)

+ (Ira(z)) 2

lzr 4 However g ( i )

= g(-i)

simply connected. subregion

= 0 and i, In fact

of ~ containing

-i

~ ~.

T h e o r e m 3 may f a i l

i n t h e a b o v e e x a m p l e we c a n t a k e i and -i where theorem 3 will

even if

the region

is

any simply connected

fail.

In the next chapter we will prove univalence results which place analytical conditions on the map F.

Also we will prove an important result of More and

P~heinboldt which uses results from degree theory.

Univalent results in R 3 : We will present two univalent results for differentiable maps in R 3.

One result is valid for rectangular regions and the other for any

convex region in R 3 [53]. We will start with a simple example. g(x,y,z) = x 2 y 2 and h(x,y,z) = y2+ z 2.

2 2 Let F = (f,g,h) where f(x,y,z) = x + z , Let ~ be any convex region in the strictly

positive orthant which includes points of the form (4, ½, z). J =

Ix°:I 2x

2y

m 0

2y

2z~

and

J+ 2J'

-

In this example,

[i x 2y y

2

23 Observe that J is a P-matrix but J+ J'

is not a P-matrix since the 2 x 2 leading

minor is (equal to 4xy-x 2) is negative at x = 4,y=½.Thus the Jacobian matrix is not positive-quasidefiniteandit is not difficult to check that the map F is one-one in any convex region in the strictly positive orthant.

In the example note that f

does not depend on y, g does not depend on z and h does not depend on x. Theorem 4 : Let F = (f,g,h) be a differentiable map from a convex region D c R 3 to R 3. z.

Suppose f does not depend on x,g does not depend on y and h does not depend on Further suppose the partial derivatives fy, fz' gx' gz' hx and hy are positive

throughout D.

Then F is univalent over ~.

Proof : Jacobian matrix J of F can be written as :

I O J =

Suppose F(a) = F(b) with a W b.

fy

gx

0

hx

hy

Let ~ =

(al,a2,a3) and b = (bl,b2,b3).

discuss essentially two cases (i) a i ~ b i and

a3 ~ b 3.

f~i1

for

We need to

i = 1,2,3 and (ii) a I ~ bl, a 2 ~ b 2

(Other cases can be reduced to one of these two cases).

Case (i) : a i ~ b i

for i = 1,2,3.

independent of the first variable.

Note that f(al,a2,a 3) = f(bl,a2,a 3) since f is We will assume without loss of generality a 2 ~ b 2.

Then, since fy ~ 0 and fz ~ 0, f(al,a2,a 3) = f(bl,a2,a 3) ~ f(bl,b 2,a 3)_>f(bl,b2,b3). In other words f(al,a2,a3) ~ f(bl,b2,b3) which contradicts our assumption that

F(a)

=

F(b).

Case (ii) : a I ~ bl, a 2 ~ b 2 a2 ~ b2, a 3 ~ b 3. matrix D =

i _

and

a 3 ~ b 3.

First we will prove when a I ~ bl,

As in the proof of the fundamental theorem define a 3 x 3 diagonal and the map H:D(~) ÷ R 3 when H(u) = D o F o D(u).

It is

0

not difficult to check that the first row of the Jacobian matrix associated with H will have the following property.

First entry will be identically zero while the

second and third entries will be negative.

Let D(a) = a* and D(b) = b*.

Then

H(a*) = H(b*) with a* ~ b* and as in case (i) we can argue to show that hl(a*) ~hi(b*) (Here h I is the first component function of H) which leads to a contradiction. In case a I ~ bl, a 2 > b2, a 3 = b 3 then clearly f(al,a2,a 3) = f(bl,a2,a 3) > f(bl,b2,a 3) = f(bl,b2,b 3) which again contradicts our assumption that F(a) = F(b). This terminates the proof of theorem 4.

24

Remark I :

One interesting feature of this result as well as the next is that we

have not explicitly assumed that the Jacobian is non-vanishing.

Automatically this

is satisfied because of our other conditions.

Remark 2 :

We do not know whether a similar result is true in R n for n > 3.

At the

present we do not have a counter example in R 4. We are ready to state our next theorem [53].

Theorem 5 :

Let F be a differentiable map from a rectangular region 2 C R

3 to R 3

with its Jacobian J having the following two properties for every x E 2 :

(i)

diagonal entries are negative and off-diagonal entries are positive.

(ii)

Every principal minor of order 2 x 2 is negative.

Then F is univalent in the

rectangular region. Before proving this result we would like to make the following observations. First note that the Jacobian is not an N-matrix.

It can be seen as follows.

Signs

of the entries of J can be written out explicitly because of condition (i).

Y ~ ÷ In other words -J is of Leontief type matrix.

Also one can easily check by expanding

through first row that determinant J > O - this is a consequence of condition ( i ) and (ii).

Second, we need to prove theorem i in this situation.

crucially depends on theorem i of chapter II.

Lemma i : (i) (ii)

Proof of theorem I

We first prove the following.

A be a 3 x 3 matrix with the following two properties.

diagonal entries are negative and off-diagonal entries are positive. Principal minors of order 2 x 2 are negative. Ax < 0

Proof :

and

x > 0

Suppose A x ~

A

=

0,

x ~ 0

and

all

a12

a13

a21

a22

a23

~31

a32

a33

I

which is clearly impossible

So we will assume x. > 0 l

Then the system of inequalities

has only the trivial solution

since

x # 0.

Let

x = 0.

x = (Xl,X2,X3). a11

i

I

If x 3 = 0 then L a21

I

ll

a12|

21

a22~

for i = 1,2,3.

!

We have,

is an N-matrix of the first kind.

25

allXl

+

a 12x2

+ a 13x3

a21x I +

a22x2

+

< 0 --

a23x 3 <

0 .

Since, al3x 3 ~ 0 and a23x 3 ~ 0, we have, allX I + a12x2 <_ 0 and a21x I + which is again impossible. This terminates the proof of lemma i.

Lemma 2 :

a22x2 <_ 0,

Let F be a differentiable map from a rectangular region ~ C R 3 with its

Jacobian J having the following two properties for every x s D : (i) diagonal entries are negative and off-diagonal entries are positive order 2 x 2 is negative.

(ii) every principal minor of

Then F(x) <_ F(y) and x > y imply x = y.

Proof of lemma

2 is similar to the proof of theorem i (and theorem 2) and hence omitted.

Proof of theorem 5 :

Let F = (f,g,h).

Suppose F(a) = F(b) with a ~ b.

If a ~ b

then from lemma 2, we infer a = b which leads to a contradiction.

Let a=(al,a2,a 3) and b = (bl,b2,b3). If ai=b i for some i, we can define a new map G in R 2 as we have done in the proof of theorem I where the Jaeobian of G will be an N-matrix of order 2 x 2 of the first kind and using this fact we will be able to show that a = b which leads to a contradiction.

any i=1,2,3.

From now onwards we will assume that

Let a I < bl, a 2 ~ b 2 and a 3 > b 3.

Let D =

i

a.1 ~ b.1 for and

0 H = D o F o D.

It is easy to check that every element in the first row of the

Jacobian matrix of the map H is strictly negative. Then H(a*) = H(b*) and a* > b*.

Let H = (hl,h2,h3).

this contradicts the fact that H(a*) = H(b*). reached in other cases.

Remark :

Let D(a) = a*, D(b) = b*. Hence hl(a*) < hl(b*) and

Similar contradictions will be

This terminates the proof of theorem 5.

It is not known whether theorem 5 remains true in convex regions.

Also

it is not clear how to formulate theorem 5 in higher dimensions. We will now give two examples.

The first example demonstrates that the

positivity of the leading minors will not suffice for global univalence as originally suggested by Paul Samuelson, while the second example shows that P-property Jacobian matrices)

Example i :

(of the

is not a necessary condition for global univalence.

Let F = (f,g) where f(x,y) = e2X-y2+ 3 and g(x,y) = 4y e2X-y 3.

= {(x,y): -2 ~ x,y ~ 2}.

J

=

Then the Jacobian J of F is given by

12e2x ~y

e 2x

-2y l 4e2X_3y2

Let

26

Clearly 2e 2x and det J = 8e 4x + lOy 2 e 2x one-one as F(0,2) = F(O,-2) = (0,0).

are positive throughout R 2 and F is not

Trouble arises in this example because the

function 4 2 ~ 3 y 2 that appears in the Jacobian matrix changes sign. due to Gale and Nikaido

x3

Example 2 :

This example is

[19]. X

Let F = (f,g) where f(x,y) = 3

2

Y, g(x,y) = x+ y.

Here the

Jacobian J is given by

Plainly J is not a P-matrix when x 2 < ½.

However F is one-one throughout R 2 as

I Let A = [0

the following argument shows.

1i ].

Then AJ =

[x~½ 0i ]

is a P-matrix.

In other words the map G = (f+ g,g) is one-one in R 2 by Gale-Nikaido's theorem and consequently F is one-one in R 2.

[Note that in the first example AJ can never be

a P-matrix in ~ for any non-singular matrix A]. If theorem i can be proved for convex regions then Gale-Nikaido's theorem will remain valid for convex regions.

The following two problems appear to be challenging

open problems (i) Does theorem I remain true for compact convex regions?

(2) Does

Inada's theorem on global univalenee remain valid for rectangular regions when the Jacobian matrix is an N-matrix and if we drop the assumption that the partial derivatives are continuous.

In other words does theorem 2 remain valid for diffe-

rentiable map F which is not necessarily of order C(1)? A result due to Kestelman asserts the following: Let F be a continuously differentiable map from ~ + R n where ~ is an open set in R n with J(x) non-singular for all x s D. nonempty interior.

Let K be a compact subset of ~ with

Then FK(= F restricted to K) is one-one if ~K(= boundary of K)

is connected and if F~K (= F restricted to ~K) is one-one. assertion see [31]. example shows.

The assertion is false if 3K is not connected as the following

Let K be the set in the complex plane defined as follows:

K Let F(z) = e z.

For a proof of this

=

{~ : l z l

± 1

Clearly F(O) = F(2~i) : I and

or

Iz-2~il ! ½}

•

K is not connected.

In Gale-Nikaido's theorem ~ is a rectangular region - we will assume without loss of generality ~ to be a compact rectangular region. where n > 2.

Then clearly 3D is a connected set.

differentiable with positive Jacobian throughout ~.

We will also assume ~ C

Rn

Also suppose F is continuously In view of Kestelmants result,

in order F to be one-one it is enough if we check that F is one-one on the boundary 80.

This naturally raises the following important question:

Is F univalent if we

simply assume that the Jacobian is a P-matrix for only those x which belong to the

27

boundary of D ? Scarf has conjectured that this question will have an affirmative answer.

Indeed this conjecture has been verified recently by three sets of

researchers: (i) C.Garcia and W.Zangwill (ii) G.Chichilnisky, M,Hirsch and H.Scarf and (iii) A, Mas-Colell.

This will form the subject matter of a subsequent chapter.

Garcia-Zangwill's proof of this conjecture depends on norm-coerciveness theorem while the proof of Mas-Colell depends on the Poincare index theorem of differential topology.

It is not clear how one can use Kestelman's result directly to give an

alternative proof of Scarf's conjecture. We pose another related question: "Is F univalent when the Jacobian is an N-matrix for every x s ~ ? answer in R n for n ~ 3.

".

This is certainly true in R 2 but we do not know the

CHAPTER

IV

GLOBAL HOMEONORPHISMS BETWEEN FINITE DIMENSIONAL SPACES

Abstract :

It is well-known from covering space theory that global homeomorphism

problem can be reduced to finding conditions for a local homeomorphism to satisfy the line lifting property.

We will show that this property is equivalent to a

limiting condition (which in many cases easy to verify) which we call by L.

We will

use this condition L to derive several results on global homeomorphisms due to Roy Plastock.

We will prove an approximation theorem due to Nere and Rheinboldt and

this result will then be used to prove Gale-Nikaido's theorem under weaker assumptions. In the last section we will prove a result

due to

NcAuley for light open mappings.

We will end this chapter with an old conjecture of Whyburn.

Line lifting property equivalent to condition (L) : We will start with a few definitions.

Let F be a map from R n to R n,

Definition : A continuous mad F is said to be proper if F-I(K) is compact whenever K is compact.

Definition :

Let ~ R

n be open and connected.

if for each line L(t) = (l-t)y I + x

Then F:q ÷ R n lifts lines in F(~)

ty 2 (0 < t < i) in F(~) and for every point

s F-l(yl ) there is a path P (t) such that P (0) = x

Remark : PJt)

and F(P (t)) = L(t).

If F is a local homeomorphism and F lifts lines in F(~), then the path

in the above definition is unique for each ~. Let ~ be open and connected in R n.

Let F:~ ÷ R n be continuous.

We now

introduce the condition (L).

Condition (L) :

Whenever P(t), 0 < t < b, is a path satisfying F(P(t)) = L(t) for

0 ~ t < b (where L(t) = (l-t)Yl+ ty 2 is any line in Rn), then there is a sequence t i ÷ b as i ~ ~ such that

limit P(t i) exists and is in ~. i÷~

We need the following well-known results for the proof of Plastook's theorem

[55]. Lemma :

Let X and Y be connected, locally pathwise connected spaces where X and

YC

Furthermore let Y be simply connected.

R n.

Then F is a homeomorphism of X

onto Y if and only if F is a covering map of X onto Y.

2g

For definitions regarding simply connected regions see [33]~

Theorem (Hermann) :

Let O C R n be open and connected, F:~ + R n.

In order that

F is a covering map of ~ onto F(O) it is necessary and sufficient that (i) F is a local homeomorphism and (ii) F lifts lines in F(D). For a proof of this theorem see Hermann [28] or Plastoek (p. 170-171,

[55]).

We now have the following result due to Plastock.

Plastook's Theorem :

Let F : ~ C R

n ÷ Rn

be a local homeomorphism.

Then condition

(L) is both necessary and sufficient for F to be a homeomorphism of D onto R n.

Proof :

In view of Hermann's theorem, to prove the sufficiency, it is enough if

we show that F lifts lines in F(D).

Let L(t) be any line in F(D) with L(O) = y and

let x E F-l(y). We can find an s > 0 and a oath P(t)(=F-I(L(t))), F(P(t)) = L(t) for 0 < t < E.

Let c ( < l )

0 < t < s with P(O) = x and

be the largest number for which P(t) can

be extented to a continuous oath for 0 < t < c and satisfyin~ F(P(t)) = L(t), 0 ~ t < c.

Let z =

limit P(t i) and observe that this limit exist for the map F t. ÷ e 1

satisfies condition (L).

By continuity, F(z) = L(c).

on which F is a homeomorphism.

Let U be a nei~hbourhood of z

There exists N o such that P(t i) c U for i ~ N o,

Also there exists a g > 0 and a oath Q(t) defined for c-g
By the maximality of c,

By virtue of Hermann's theorem F

We need only show that F(~) = R n in order to apply

the above lemma and thus conclude that F is a homeomorphism of ~ onto R n. y s R n.

Choose Yl s F(D) and let L(t) = (l-t)y I + ty.

So let

If we retrace the steps of

the first part of our proof, we find a path P(t), 0 < t < I, so that F(P(t)) = L(t) on 0 < t < I.

In particular F(P(1)) = L(1) = y, and so F(~) = R n.

This completes

the proof of Plastock's theorem.

Remark :

The same proof goes through even in infinite dimensional Banaeh spaces.

We will now deduce Browder's theorem from this theorem.

Browder's Theorem :

F:R n ÷ R n

is a homeomorphism if and only if F is a local

homeomorphism and a closed map.

Proof :

We will only check the sufficiency part.

condition (L).

We will show that F satisfies

Suppose Pit) is defined on 0 ~ t < b and satisfies F(P(t)) = L(t)

30

for 0 < t < b. is closed.

Let S = closure of {P(t):0 < t ~ b}.

Since F is a closed map F(S)

Thus since L(t) s F(S), for all t < b, by continuity L(b) e F(S).

means for some x e S, F(x) = L(b).

This

That is we can find {t i) such that P(t i) ÷ x.

Since t i is a bounded sequence, without loss of generality we will assume t i ÷ t o. We claim t o = b.

By continuity

(L) is satisfied.

Remark I :

L(t o) = L(b) and hence t o = b.

It is easy to see that Hadamard's theorem as well as Banach-Mazur's

theorem is a consequence of Browder's theorem. F:R n ÷ R n phism.

Thus condition

Now Browder's Theorem is a consequence of Plastock's theorem.

Their result simply states that

is a homeomorphism if and only if F is a proper map and a local homeomor-

It is easy to see that every proper map is a closed map.

Remark 2 :

If F is a local homeomorphism and if F is also closed then F is proper.

We will now give an analytic condition which will in turn imply condition (L). In order to do that we need the following. valued continuous function on R n.

Definition :

Let W(x) be a strictly positive real-

Let P(t) be a path in R n of class C (I) on

0~t~b.

The arc length of P with weight W is L~(P)

=

b f W(P(t))liP'(t)lldt O

Here P'(t) stands for the derivative of P(t) and il'll stands for the usual matrix norm.

Definition : implies

R n is complete with respect to arc length with weight W if Lb(p) < oo

limit t÷b

Remark :

P(t) exists and is finite whenever Pit) is a C (I) path on 0 <_ t < b.

The above definition is equivalent to the usual notion of R n being complete

with respect to the conformal metric induced by the tensor ds 2 : [W(x)] 2 dx 2. [28].

See

Definition given above is meaningful even in infinite dimensional Banach

spaces.

We now have the following theorem due to Plastock.

Theorem 2 :

Let F:R n ÷ R n be a continuously differentiable map.

of F does not vanish at any x s R n.

f o

inf

Suppose Jacobian J

If

~i/11JIxl-lII~dt

: ~

,

1[x11:t

then F is a homeomorphism of R n onto R n.

In fact F is a diffeomorphism (that is F

is one-one onto and its inverse is also differentiable).

Proof :

Define the weight function W(x) = (i/llJ(x)-111).

Let u(s) :

inf W(x). lixi1=s

31

We are given that

~

u(s)ds = ~.

We will first show that R n is complete with respe-

O

ct to arc length with weight W. Let P(t) s C(1)[O,b) and suppose L~(P) < ~ 0

= t o_< t I _ < " ' ~ t N = g o f

Let 0 < 6 < b.

For any partition

[0,6], let t i _< t i _< ti+ I be that number for which

sup ]JP'It)][ = liP'(ti)] I. Here P'(t) stands for the derivative of P(t). t i~t--
=

w(P(t))iIP'(t)II dt O

limZW(P(ti))llP'(ti)[[(ti+ >

l-ti)

limZ w(p(ti))(ilP(ti+l)[I-[iP(ti)l[) 6 f w(P(t))dIIP(t)l[ O

6 W(P(t))d IIP(t) rlis defined since

this last equality following from the fact that O

g(t) = ]lP(t) i] is of bounded variation on [0,6]. >

Now we have that

6 b ~ W(P(t))IBP'(t)II dt ~ ~ W(P(t))d[lP(t)ll O

O

6

f inf w(x) dlIP(t)ll o TlxIl=IIPItlIT iTP(6)Ti u(s)ds . = Since

S

So u(IrP(t)II)d[IP(t)II

=ll~(0)liP

u(s)ds = ~, it follows that {P(t)}0
Note that

O

sup {s:u(s) > 0} = ~.

Since u(s) is nonincreasing, we have that W(x) is bounded

from below on any bounded set. number a > 0, for all 0 ~ t < b.

In particular, W(P(t)) is bounded from below by some Let t i ÷ b, (ti ~ ti+l).

n

Then,

n

IIP(ti+l)-P(ti)ii

~

i=l

<

~ i=l

sup IIP'(t)il(ti+l-ti) ti
tn+ i ~ IIP'(t)li dt tI b W(P(t))llP'(t)Ildt < 0

Therefore we can find an x such that P(t i) ~ x as t i ÷ b.

Thus lira P(t i) exists for

32

any increasing sequence t i ÷ b, and is in fact unique for such sequences

(since from

any two such sequences we can form a new increasing sequence containing the original ones as subsequences).

Hence limit t÷b

P(t) exists as every sequence P(t i) has a

unique limit point which is independent of the sequence t i. Clearly F is a local homeomorphism. show that F satisfies condition

In view of Plastock's theorem, we need only

(L).

Suppose P(t) is defined on 0 < t < b and satisfies F(P(t)) = L(t) for 0 j t < b. It suffices to show that F satisfies condition

(L) only for those paths P(t) that

are constructed in the proof of Plastock's theorem.

Furthemore,

if P(t) is such a

path, from the local inverse function theorem, it follows that P(t) is continuously differentiable on 0 < t < b. and get J(P(t))P'(t)

Since F(P(t)) = L(t) on 0 < t < b, we use chain rule

= L'(t)(= z say).

That is, P'(t) = J(P(t))-Iz for 0 < t < b.

Observe that,

Tb (p) O

b

=

~

W(P(t))IIP'(t)II dt

0

=

Thus F satisfies condition

b S o

i

]ij(p(t))-izlid t

1lJ(P(t))-llI

(L) and consequently from Plastock's theorem, we can

conclude that F is a homeomorphism.

In fact it is easy to see that F -I is differen-

tiable and therefore F is a diffeomorphism of R n onto R n.

This terminates the proof

of theorem 2.

Corollary i :

Let F:R n + R n

for every x c R n.

be a 0 (I) differentiable map.

Suppose

Idet J(x) 1 > a > 0

Then F is a diffeomorohism of R n onto R n if any of the following

conditions is met. (a)

IIJ(x) I] ~

M

for every x s R n.

(b)

F is quasi-conformal

- that is there exists M such that IIJ(x)ll

IIJ(x)-iI

< M

for every x s R n.

Proof : condition

We will prove under condition (b).

(a).

A similar proof can be given under

We need the following fact for the proof : If L:R n ÷ R n is an

invertihle linear operator then ]det L 1 l(L-ix,y) I j Ilxll l]yll for all x,y s R n.

IILIIn-I (n-l)-(n-l)/2

[For a proof of this see Dunford and Schwartz Part II, p. 1020].

From this, we have that Idet J(x) I l(J(x)-Iz,w)~

c(n) IIz]I llwll IIJ(x) lln-I

33

where c ( n )

=

(n-1)-(n-1)/2

Let

z

be such that

llzll

-- 1 ~

w --

J(x)-1 z.

Then

the above inequality becomes,

ldet J(x) l llJ(x)-lzl 2 ~ c(n)llJ(x)-lzll Since

ldet

llJ<x)ll

$ ( x ) l > ~ > 0 and

<

IiJ(x)lln-1

M~

I1J(x) -lzll

±

llJ~x)-llt

i

c (n)Mn-i/a

II7.11:1

for all

That is

c

(n)~-ll~ .

It is easy to verify that the condition imposed in theorem 2 is satisfied. is a diffeomorphism of R n onto R n.

Corollary 2 [32]:

Definition :

Thus F

This terminates the proof of corollary I.

A continuous local k-extension F:R n ÷ R n is a homeomorphism.

Call F:R n ÷ R n

a local k extension if for every z s R n there exists a

neighbourhood U of z such that IIF(x)-F(y)II ~

Proof of Corollary 2 :

kIlx-yll

on U.

We will show that F satisfies condition (L). Let

(Xo,Y) s R n x Rn,L(t) = (l-t)F(Xo)+ ty, 0 w t ~ i and P:[O,b) ÷ R n a path such that F(P(t)) = L(t) for 0 ~ t ~ b, and P(O) = x . --

every 0 ~ t O ~ b

Since F is a local k-extension, for

O

there exists s ~ 0 such that It-tel < s, lS-tol ~ E

I IF(P(s))-F(P(t))II ~

kIiP(s)-P(t)II.

imply

Therefore,

IrP(s)-P(t)IT ~ ~i 11~(P(s))-F(P(t))II I

= ~

ll~IXol-yII.Tt-sl

It follows that P is a local M-Lipschitz map and consequently

-- M. Tt-sl.

lim t+b

P(t) exists.

In

other words F satisfies condition (L). Also observe that F is a local homeomorphism from Brouwer's theorem on the invariance of domain.

Thus F is a homeomorphism onto

R n from Plastock's theorem. Corollary 2 is due to R.Radulescu and S.Radulescu

Theorem 3 :

[60].

We now have the following

Let (Fm) be a sequence of homeomorphisms of R n onto R n converging

uniformly to F where F is a map from R n ÷ R n.

If F is a continuously differentiable

map with non-vanishing Jacobian then F is a homeomorphism of R n onto R n.

Proof :

Observe that each F

m

is a proper map and therefore F is a proper map.

Thus from Hadamard's theorem it follows that F is a homeomorohism of R n onto R n. Fowever one can also prove that F is onto as follows.

Let a s R n.

Then for every

34

m there exists x m such that Fm(x m) = a for each Fm is onto R n. sup IIFm(X)-F(x)II x s Rn Thus F(x m) ÷ a

as

m + ~.

~

Note that

l[Fm(Xm)-F(Xm)[].

Since IIall < ~,{x m} is a bounded sequence.

of generality let x m + Xo as m + ~.

Then F(x o) = a.

Without loss

This yields the desired result

We will now give examples to show the sharpness of the results obtained so far in this chaoter.

The first two examples indicate that theorem 2 may fail if we omit

the analytic condition.

Third example indicates that the full force of theorem 3 may

not be valid if the convergence is not uniform.

Let F(Xl,X 2) = (tan -I Xl,X2(l+ x~) 2) be a map from R 2 to R 2.

Example i :

the Jacobian J is given by

Here

[i 0 I i+ x~

J (x)

=

XlX2(l+x ll+x2 Since J is a P-matrix, F is one-one but not onto.

By looking at the characteristic

polynomial of J(x), it is not hard to check that 2 = i/(I+ x~) is an eigenvalue, and so

2

2

= (l+x l) is an ei~envalue of J(x) -I. Fence I IJ(x)-IIl ~ i/(l+Xl).

I o

inf

I

IIxlI=t

df

IIJ(x)-lll

< I -

Now

d__!t<

o

l+t 2

Thus in this example analytic condition is violated and the map F is univalent but not onto R 2. xI x1 Let F(Xl,X 2) = (e cos x2, e sin x2). This map is neither one-one x1 nor onto (it omits 0). Observe that lIJ(x) ll = e .(Here x = (Xl,X2)) 1/llJ(x)-lII~ xI e . So co Example 2 :

inf l lxll:t Example 3 :

i l IJ(x)-iIl

< e -t --

and

S e -t dt < co o

Let fm(X) = ex - (e-X/m) be a map from R I to R I.

Clearly each fm(X)

is a homeomorphism of R 1 but the limiting function f(x) = ex is not onto R 1 though it is one-one.

However convergence is not uniform over R 1.

A theorem of More and Rheinboldt and its consequences :

The following theorem is

due to More and Rheinboldt and the proof depends on degree theory-in particular the

35

homotopy invarianee theorem of degree theory is used.

In this section as well as

at several other places we use results from degree theory. reference from the point of view of mathematical Cronin

[13], Ortega-Rheinboldt

Theorem 4 :

A good source of

analysis would be Berger

[48], Rado-Reichelderfer

[59] and Rothe

[2],

[63].

Let F:D ÷ R n be a continuously differentiable map with J(x) nonsingular

for every x s ~ where ~ is an open subset of R n.

Suppose, for every s > 0, the map

Gs:~ ÷ R n defined by Gs(x) = F(x) + sx is one-one in D.

Then F is one-one.

Proof :

We can choose open sets

U

Suppose F(a) = F(b) = c for some a ~ b •

and ~

of a and b respectively with Ua g~ ~

choose Ua and ~ one-one). x = b.

= ~ and Ua' ~ r

such that F restricted to Ua as well as ~

In other words if F(x) = c for some x s Ua U ~

~.

We can also

is one-one

(locally

it will imply x = a or

Hence deg (F,D,c) is well defined for D = = Ua' D = ~

as well as D = Ua ~ .

(Roughly speaking deg (F,D,c) = ~#= of the solutions in D satisfying F(x) = e, x s D). For any one of these three sets consider the homotopy Ha:~ x [0,I] ÷ R n

defined by

Ha(x,t) = (l-t) F(x) + t(x-a+ c) (as well as the homotopy Yb defined by ~(x,t)

=

(l-t) F(x) + t(x-b+ c)). We will now prove that Ha(X,t) ~ c for x E ~D (= boundary of D)

and t s [O,I].

any x E SD.

Plainly Ha(X,O) = F(x) ~ c and H(x,t)

Suppose for some t s (0,I) and x s SD,H(x,t)

(l-t)F(x) + t(x-a+ c) = c. F(a) +

(~

)a

This implies that F(x) +

= x - a+

= c.

(t

c ~ o for

This means

)x = c +

(t

)a =

but this contradicts the assumption that F(x) + sx is one-one.

Ha(x,t) ~ c, ~(x,t)

¢ c for x s ~D, t s [O,i].

theorem of degree theory,

Hence

Invoking the homotopy invariance

(see chapter VI in [48]) it follows that deg(F,D,c)

=

deg(x-a+ c,D,c)

=

deg(x-b+ c,D,c)

This immediately tells us that deg(F,Ua,C) = deg(F,~,c) But by assumotion deg(F,U a V ~ , c ) this leads to a contradiction.

. = deg(F,UaU~,c)

= deg(F,Ua,C) + deg(F,~,c)

since Ua g~ ~

= i. = ~ and

Hence F is univalent on ~ and this terminates the

proof of the theorem.

Another proof of theorem 4 (due to Mas-Colell) is, F(a) = F(b) = c, a ~ b.

Take U , ~

with U ~

:

Suppose F is not as desired, that ~

= 4, such that for small s,

there are functions Xa(S), Xb(S) satisfying F(Xa(~))) + ~Xa(S) = c and F(Xb(~)+~b(c)=e. These functions exist locally by the implicit function theorem and the regularity of J(a), J(b) and we have a contradiction for Xa(S) ~ Xb(S).

See also pp 218-222 in [2].

The following corollary follows from the fundamental theorem of Gale-Nikaido and

36

Theorem 4.

Corollary 3 :

Let {Fm} be a sequence of differentiable functions defined over

where ~ is an open rectangle in R n, and let the Jacobian matrix for each n and x, be a P-matrix.

Suppose Fm(X) converges to F(x) for each x s ~.

F is differentiable with nonvanishing Jacobian for each x s ~.

Further suppose Then F is one-one

on ~.

Remark :

Let ~ = R n and suppose each F

m simple example shows that it need not be.

is onto R n. Is F also onto? The following RI Let ~ = and Fm(X) = tan -I x+ x/m.

Clearly each F m is a homeomorphism onto R I but the limiting function F(x) = tan -I x is one-one but not onto R I.

Proof :

From Gale-Nikaido's fundamental theorem it follows that each F

P-function.

is a n Since F n converges to F pointwise, F+ s I is a P-function for each s > 0.

In other words F + s i is injective for each s > 0 and furthermore for each x s there is an open set Ux such that F(y) = F(x) for each y s U

implies that u = x.

Hence we can conclude from the proof of theorem 4 that F is one-one on ~. We need the following definition for the next theorem.

Definition :

Call a matrix A weakly positive quasidefinite if det A > 0 and the

quadratic from is positive semidefinite.

A weakly negative quasidefinite

matrix is defined analogously.

Theorem 5 :

Let F:~ ÷ R n be a continuously differentiable map with ~ an open convex

region in R n. everywhere.

Proof :

Further suppose Jacobian matrix J(x) is weakly positive quasidefinite Then F is one-one.

Let Gs(x) = F(x) +sx

quasidefinite.

where

E > 0, it is easy to see that Js is positive

Now theorem 5 is a consequence of Gale-Nikaido's result and theorem

4 of the present chapter.

Light open mappings and homeomorphisms due to McAuley

:

In this section we will prove two results

[38] where we place purely topological conditions under which the

map F becomes a homeomorphism.

We do not even assume F to be differentiable.

While

Gale-Nikaido's result depends on the choice of a fixed coordinate system, the problem under consideration namely "FI~ is univalent" fixed coordinate system.

NcAuley's

result eliminates this particular difficulty.

We will start with a few definitions. F to be a continuous map from ~ ÷ R n.

does not depend on the choice of a

Throughout this section we will assume

37

Definition :

Call F a l i ~ t mapping if F-I(F(x)) is totally disco~ected for each

x~.

Definition :

Call F an open map if for each U open in ~, F(U) is open relative to

F(9). Remark 1 :

Local homeomorphisms are special light mappings while covering mappings

are indeed open local homeomorphisms.

The latter are simple cases of light open

mappings.

Remark 2 :

The study of light open mappings has deep motivations in the topological

nature of analytic functions defined in domains in the complex plane.

For example

a non-constant function w = g(z), analytic in a region D of the (complex) z-plane which takes D into the w-plane is strongly open. g(U) is open in the w-plane.

That is if U is open in D, then

Furthermore, g is light.

discrete (has no limit point).

In fact the set g-lg(x) is

For more details see [38].

Let F be a light mapping from ~ to R n.

We shall say that the singular set S F

of F is the set of points x s ~ such that F is not locally one-one at x, that is, there is no set U open in ~ and containing x such that [I0] refer to S F as the branch set of F.

FIU

is one-one.

The singular set S F is closed.

Some authors We shall

now give two sufficient conditions under which a light mapping F of ~ Dnto F(~) is open.

(i)

Floyd's condition :

For each continuous mapping g:[O,l] + F(~) and for each

x s F-l(g(o)), there is a continuous map h:[O,l] + ~ such that g = F o h and h(0) = x.

(ii) McAuley's

condition :

The singular set S F of F has the property that F(S F)

fails to separate any set V open relative to F (~) and F (SF) contains no nonempty open set. For a proof of (i) see Floyd [Ann. Math. Vol. 51,(1950)] and for a proof of (ii) see [38]. Our aim in this section is to provide conditions under which a light open mapping F of a compact subset ~ of R n into R n is a homeomorphism.

It is not difficult

to check that F(int ~) = int ~ and F(boundary ~) = boundary F(~) if such a mapping were to be a homeomorphism.

Theorem 6 : (McAuley).

We are ready to state our theorems.

Suppose ~ is a compact set in R n with 3D # @ and int D #

and F is a continuous light open mapping of ~ into R n such that (I)

F(int ~) = int F(~),

(2) F ( ~ ) = ~F(~),

(3) the singular set S F has the property

that F(S F) does not contain a non-empty set open relative to F(~),

(4) F(S F) does not

separate F(~), and (5) there exists a non-empty U in ~ open relative to ~ such that

38

FIU

is one-one and F-IF(u) = U.

Theorem 7 (McAuley)

:

Then F is a homeomorphism.

Suppose that F is a continuous light open mapping from a

unit ball ~ in R n onto another unit ball B in R n such that (i) F-IF(~) (2) F ( ~ )

=

=

~,

~B, (3) FIS F is one-one, and (4) for each component C of B \ F ( S F) there

is a nonempty V in C open relative to B such that FIF-I(v) is one-one.

Then F is

a homeomorphism.

Remark I :

If F is a differentiable map with non-vanishing Jacobian, then S F = 4-

Furthermore if ~ is compact then clearly F(int ~) = int F(~) and F($~) = ~F(~) - for a proof of this assertion see (p 447-448 in [38]).

Remark 2 :

It is not clear whether Gale-Nikaido's univalent theorem proved in the

last chapter is a consequence of theorem 6.

In fact if F and ~ satisfy the conditions

imposed on Gale-Nikaido's theorem, it is easy to verify that conditions and (4) given in theorem 6 are met.

(1),(2),(3)

Nowever it is not at all obvious how to check

condition (5) of theorem 6.

Proof of theorem 6 : nondegenerate. not empty.

Let P denote the set of all y in F(~) such that F-l(y) is

It is clear that F is a homeomorphism iff P is empty.

Suppose P is

It is not hard to check that the set P is open relative to F (~) and

contains a non-empty open set since F is an open map. Let A = F ( S F ) ~ P .

We will prove that A is closed and consequently compact.

Suppose that y is a limit point of A but y s F(~)\A.

Since F(S F) is closed, it

follows that y is a limit point of P\F(SF).

Clearly F-l(y) is degenerate and let

x = F-l(y).

In other words there exists a neighb-

Also F is locally one-one at x.

ourhood N x of x such that F-IF(Nx ) = N x.

Thus F(N x) g~ (PkF(SF)) = 4.

limit point of P\F(SF). Since F(S F) U P (F(~)\P)

FIN x is one-one, F(N x) is open relative to F(~), and

Thus the set A is closed and consequently compact.

is closed, we have B = F(~)\(F(S F) ~.J P) = (F(~)\F(SF))g~

is open in F(~).

From condition (5) of theorem 6, it follows that each

of B and F(~)\F(S F) is nonempty. is connected.

This contradicts the fact that y is a

Observe that from condition (4), F(~)\F(S F)

If we show that F(~)\F(S F) ~ P is open relative to F(~) then we arrive

at a contradiction to the fact that F(~)\F(S F) is connected. that (F(~)\F(SF)) g~ P =

Hence it will follow

~ and condition (3) will now imply that P = 4, contrary

to our assumption that P is not empty.

Thus the proof of the theorem 6 will be

complete if we show that (F(~)XF(SF)) g~P is open relative to F(~). Since F(int ~) = int F(~) and F ( ~ ) F(S F g~int ~).

= ~F(2), we have that F(S F) ~ int F(~) =

Now, F(S F) = F(SFg~ ~ ) U F ( S F ~ int ~).

Since P is open in F(~) and

F(S F) is closed, it follows that (F(~)\F(SF))g~ P is open relative to F(~). terminates the proof of theorem 6.

This

39

The following corollary which is needed for the proof of theorem 7 is an immediate consequence of theorem 6.

Corollary 4 : Suppose ~ is a compact subset of Rn with int D # @, ~ = closure of (int D), int F(~) is connected, and that F is a light open mapping of D into R n such that (i) F I(int D) is locally one-one, (2) F(int ~) = int F(~), (3) 8F(~)=F(8O), and (4) there is U open relative to ~ such that F -I F(U) = U and FIU

is one-one.

Then F is a homeomorphism.

Proof : It is not hard to check that

8D O SF, F(~) \F(S F) is connected, and F(S F)

contains no nonempty set open relative to F(~).

The hypothesis of theorem 6 is

satisfied and the corollary follows.

Proof of theorem 7 : Conditions (i) and (2) imply that, F(int ~) = int B.

Since B

is locally connected, we can find at most a countable number of components CI, C2, C3,...,

of B-F(S F) and moreover, if there is an infinite number of components,

then diameter C.l ~ O, that is, the sequence {Ci) is a null sequence. F-I(c i) is connected.

Let K i = F-I(ci).

Note that F(K i) = Ci.

For each i,

These facts follow

from condition (4) of theorem 7 and a theorem of Whyburn ([7~], (7.5); P. 148). Moreover {Ki) is a null sequence. Observe FIK i is locally one-one since K i ~ S F = @. (where D denotes the closure of D), and F-I(c i) = Ki"

Furthermore, F(K i) = Ci Also it is not hard to check

that, F(SK i) = 8F(Ki),SF(int Ki ) = int F(Ki) , and FIK i is a light open mapping of K. onto C.. 1 1 Apply corollary 4 to each of Ki"

Thus, FIK i is a homeomorphism.

Set S=UK i.

Since FIS F is one-one, S F contains no set open relative to ~ (the unit ball in Rn in the hypothesis), indeed, F(S F) contains no set open relative to B.

Note that

each point of F(S F) is a limit point of B-F(SF).

Thus B = U Ci"

F(S) = B and FIS is a homeomorphism of S onto B.

We conclude that S = ~ and that

F is a homeomorphism.

Remark i :

It follows that

This terminates the proof of theorem 7.

Condition (5) in theorem 6 (as well as condition (4) in theorem 7) is

crucial for the proof of theorem 6 (theorem 7).

Also it appears difficult to

check this condition in a given problem.

Remark 2 : In order to apply these results it is not even necessary unlike Gale-Nikaido's theorem for F to be differentiable (everywhere).

These results hold

good in more general spaces - see [38]. The following example shows that condition (5) is crucial for the proof of theorem (6). Without this condition theorem 6 may fail.

40

Example : Let F = (f,g) with f(x,y) = X2-y 2 and g(x,y) -- 2xV. 2 2 x + y < i}. Here the Jacobian of F is given by

Let ~ = {(x,y) :

-2y J

= 2y

and d e t J = 4 ( x 2 y 2) # 0 i f f F is not locally one-one. check conditions

2x

(x,y) # ( 0 , 0 ) .

Clearly origin is the only point where

In other words S F = {(0,O)).

Also it is not hard to

(i) - (4) of theorem 6 are satisfied, but condition (5) is not

satisfied.

Remark 3 :

One can prove the following result using theorem 6.

compact subset of R n with int ~ ~ 4.

Suppose ~ is a

Furthermore, suppose F is a local homeomorphism

of ~ into R n such that (i) F(~) is connected and (ii) there is some set U in ~ open relative to ~ such that F-IF(u) = U and F IU

is one-one.

Then F is a homeomorphism.

It is clear from this remark as well as corollary 4 that one can derive extremely useful results from theorem 6 (as well as theorem 7).

We will again make use of

this result in the last chapter. We will close this chapter with an old conjecture of Whyburn.

Suppose F is a

continuous light open mapping of In = [0,I] x [O,i] x ... x [0,I] (cartesian product of unit interval taken n times) onto In such that F -I F(~I n) = ~In and FIll n is one-one.

Does it follow that F is a homeomorohis~

But the answer is not known for n > 3. see McAuley

[38] or Whyburn

[74].

The answer is 'yes' when n = 2.

For more details regarding this conjecture

However this conjecture is true (via Kestelman's

result quoted in the previous chapter) if we further suppose that (i) F is continuously differentiable and (ii) the Jacobian matrix of F is non-singular for every x s In.

CHAPTER

V

SCARF'S CONJECTURE AND ITS VALIDITY

Abstract

:

Nikaido's

In this chapter we will nrove a substantial

generalization

of Gale-

theorem on univalent mappings in which we assume P-property only on the

boundary of the rectangular region and this was conjectured by Scarf.

We will

give two different proofs one due to Garcia and Zangwill and the other due to Mas-Colell.

Proof of Carcia and Zangwill uses the norm-coerciveness

theorem whereas

Mas-Colell uses results from degree theory.

There is a subtle difference between

these results and Cale-Nikaido's

theorem.

fundamental

The difference

lies in the J

fact that the proofs of Garcia-Zangwill whereas

Gale-Nikaido's

a C (I) fanction.

It is not clear whether

Garcia-Zangwill

result holds good if we assume F to be a differentiable

This seems to be an interesting open problem in this area. remains still unanswered convex set ~ C

~

demand F to be a C ~I) function

result holds good if we assume F to be a differentiable

function not necessarily or Mas-Colell's

and Mas-Colell

R n to R n.

is the following:

function.

Another problem which

Suppose F is a C ~I)" ~ map from a compact

Sunnose Jacobian of F is a P-matrix for every x c ~.

Does it imply F is one-one in ~ ?

Garcia-Zangwill's

result on univalent mapnings

:

In order to state the result we

need the following definition.

Definition to R n.

:

Let ~ be a compact rectangle

Call F norm-coercive

in R n with boundary

if llF(x) ll ÷ ~

~

as x approaches

theorem is known and a proof may be found in [48, p. 136].

and F a map from ~.

The following

Let ~o stand for the

interior of ~.

Norm-Coerciveness

Theorem

:

Let F:~ ° + R n be a continuously

coercive mapping with det J(x) positive for every x ~ D.

differentiable

Then F is one-one

norm(and

in fact F is a homeomorphism).

Remark

:

In finite dimensional

spaces,

to the fact that the map is proper. We will now state and nrove

Theorem I :

Let F:~ ÷ R n be continuously

(= boundary of ~).

in equivalent

[2].

(Scarf's Conjecture).

with det J(x) > 0 for every x s D~. x ~ ~

concept of norm-coerciveness

For a proof see Berger

differentiable

on a bounded rectangle

D

Further suppose J(x) is a P-matrix for every

Then F is univalent on ~.

[We will assume ~ = {x:a.<x.
42 for i = 1,2,...,n}]. Idea of the proof can be explained simply as follows.

We will construct a

function G which will coincide with F exceDt near the boundary ~ . function G will be norm coercive.

Also this

We will use norm-coerciveness theorem to conclude

that G is univalent which in turn will imply that F is also univalent.

One such G

can be defined as follows: gi(x)

=

fi(x)+di(x i)

for

i = 1,2 .... ,n

where

di(x i) =- (max(O,

I

1

i

1 ))2 + (max(O, b.-x.

61

x.1 -a.1

1

61

1

))2 .

b.-a. Here 61 is any positive number less than min ( the function

It is not hard to see that

di(x i) has the following properties:

(±)

di(x i) = 0

(ii)

d!(xi)1 > 0 if

(iii)

Idi(xi) I ÷ ~

Also

12-----!i).

if

ai+

6l ~ x

i ~ b i - 61

a i < x i < a i + 61 if either

or b i - 61 < x i < b i

x i + ai

or

and

x i + b i-

it is easy to see that G = (gl,g2,...,gn) coincides with F for all x which are

at least a distance 61 from the boundary.

Also G is norm coercive.

Further since

fi's are continuously differentiable functions and ~ is compact we can find a 62 such that J(x) will be a P-matrix for all x which are at most a distance 62 from the boundary.

Define

6 = min{61,62}.

Proof of theorem i : contradiction.

Now we prove theorem i.

Suppose x # y with F(x) = F(y) then we will arrive at a

We will analyse two cases (i) x and y are interior points,

least one of them is a boundary point.

(i) by modifying F slightly as will be shown below. (i) is impossible.

(ii) at

In fact case (ii) can be reduced to case

Construct a G as described above.

We will now prove that case Since 61 as well as 62 can be

chosen as small as we please, we will choose them so small so that both x and y are more than a distance of 6 from the boundary. G(x) = F(x), G(y) = F(y) and F(x) = F(y). and it is norm coercive.

This will imply G(x) = G(y) since

Clearly G is continuously differentiable

If x s ~ is at least a distance of 6 from ~ ,

Jacobian

of G(x) = Jacobian of F and hence det of the Jacobian of G(x) is positive. is within 6 of

If x

3~, Jacobian G(x) is a P-matrix and hence its determinant is positive.

In other words conditions of the norm coerciveness theorem are satisfied and hence G should be one-one which contradicts G(x) = G(y) for x # y. one.

Hence F must be one-

We will now demonstrate that case (ii) can be reduced to case (i).

F(x) = F(y) for two distinct points x and y in ~ not both in ~o. in ~o close to x and y respectively. component is given by

Suppose

Choose x' and y'

Define H:~ ÷ R n as follows and its ith

43

bi(Xl-Y~) hi(x) = fi (x) - (x~ - y~) where F(x') = b and F(y') : b'.

b~(Xl-X ~) (y~-x{)

Clearly H is continuously differentiable and

~fi (x) ~hi (x) ~x. J |

3fi (x)

[

~-~I

b.-b! l 1 + ~'--TYI-/Yifyl Xl

j = i .

Since we can make the term (bi-b~)/(v'-X'~l i ) arbitrarilv~ close to zero by selecting x' and y' close to x and y respectively it follows that H will satisfy the conditions of theorem I.

Also observe that H(x') = N(y') which is imDossible from case (i).

This terminates the proof of theorem i.

Remark :

During the course of the proof of theorem I we have made use of the

following result:

If D I is an n x n diagonal matrix with non-zero diagonal entries

only in positions i s I and if P is an arbitrary n x n matrix then determinant (P+ D I) =

Z K~I

( ~ dk)M K where MK'S are the minors of the matrix P. k~K

Here M K is

the minor resulting from P omitting the rows i and columns i for i s K.

Me

= det P and M(l,...,n % . = i by definition.

Also

As this formula may be a bit confusing.

Here is an example for n = 3, I = {1,2,3}. det(P+ D I) = det P+ dlMl+ d2M2+ d3M3+ dld2Ml2+ d2d3M23 + dld3Ml3+ dld2d 3 We make use of this result to conclude that the determinant of the Jacobian matrix of G is positive near the boundary.

This is where the proof will breakdown if we

imitate this to show that a map F will be one-one if we assume that the Jacobian is an N-matrix for all x s ~ open problem.

in theorem I.

This problem appears to be an interesting

We believe that theorem i should be true under the assumption that

the Jacobian is an N-matrix for all x s ~

and that the determinant of the Jacobian

is everywhere negative. Note that theorem i is a generalization in a certain direction of Gale Nikaido's fundamental theorem when F is continuously differentiable function with its Jacobian everywhere a P-matrix.

However one should also note that in Gale Nikaido's theorem

we do not assume that the partial derivatives to be continuous as in Theorem I. is not clear whether theorem I remains true when F is a differentiable but not a C (I) function. From theorem I one can deduce the following:

It

44

Theorem 2 :

Let F : R n ÷ R n

bounded rectangle.

be continuously differentiable.

~r- R n

be a

Suppose that the determinant J(x) > 0 ¥ x E R n and further

suppose J(x) is a P-matrix ¥ x s R n \ ~ .

Proof :

Let

Then F is one one.

Suppose F(x) = F(y) with x ~ y.

x,y and 9 in its interior.

Let ~' be a compact rectangle containing

Then F satisfies all the conditions of theorem I on ~'

and consequently F is one one on ~' contradicting our original assumption.

This

terminates the proof of theorem 2. At this juncture it is worth mentioning that Garcia and Zangwill prove

theorem

i (as well as theorem 2) under slightly weaker assumptions on the Jacobian matrix on the boundary.

Let l(x) = {i:x i ~ a i or x i = bi}.

We say that F has the S

property if J(x) K ( ~ J(x) i) > 0 for all x s ~ and K ~ l ( x ) . Garcia and Zangwill isK prove univalence when F satisfies the S property. This S property permits certain principal minors to be negatiw~ on the boundary.

We will now give an example of a

one-one function F whose Jacob~Lan is not everywhere a P-matrix. function defined as follows: f2(x,y) = x+y.

Let F:R 2 ÷ R 2 be a

F = (fl,f2), where fl(x,y) = x 3 / 3 + x ( y 2 - ½ ) - y and

Then

I

x2+ y2 ½

J (x,y) =

L

2xy_l I Plainly J(x,y)is not a P-matrix.

1

Also det ~x,y)=(x-y)2+ ½ > 0 V (x,y) E R 2.

1 Nowever J(x,y)is a P-matrix if x 2 y 2 > ½.

Consequently F satisfies the conditions of theorem 2 and hence F is one-one throughout R 2 .

Mas-Colell's univalent result :

Theorem 3 :

We will now prove the following theorem of Mas-Colell.

Let F:~ ÷ R n be a continuously differentiable function where ~ is a

compact convex polyhedron of full dimension.

For every nonempty subspace

let ~L: Rn + L denote the perpendicular projection map.

If for every x s

L~R

n

~ and

subspace L C R n spanned by a face of K (that is, the translation to the origin

of

the minimal affine space containing K)which includes x, the map H L DF(x) : L ÷ L has a positive determinant

(that is the linear map H L- DF(x) preserves orientation

where DF(x) stands for the derivative map of F at x) then F is one-one on ~ and consequently a homeomorphism. Several remarks will be in order now. convexDo~hedral set. theorem on univalence

Here ~ is assumed to he simply a compact

As such this theorem includes Gale-Nikaido's fundamental mappings.

Conditions imposed on this theorem are coordinate

free, in the sense that their formulation does not rely on a previous choosing of coordinates.

Proof of this theorem will depend on the following known results from

45

degree theory.

Theorem (a) :

Let F,G:C C

of an open bounded set C.

R n ÷ R n be two continuous maps whereC is the closure Suppose there exists a homotopy H:SC x [0,I] such that

N(x,O) = F(x), H(x,l) = G(x) for all x s 8C.

If y s R n such that H(x,t) ~ y for all

x E 8C, t ~ [0,i], then deg(F,C,y) = deg (O,C,y). For a proof see Schwartz 1964, p. 93 Non-linear functional analysis, Lecture notes Courant Inst. of Math. Sei., New York.

Theorem (b) :

Let F:D ~ R n + R n

be continuously differentiable on the open set D

and C an open bounded set such that C ~

D.

Given a coordinates system define

A = {x s C, Jacobian at x is singular}. If y ~ F(SC U A ) is empty and deg(F,C,y)

=

then either F={xsclF(x)=y}

0 or F consists of finitely many points x I , . . . , x m and

m

deg(F,C,y) =

~ sign det J(xJ). (For a proof see p. 159, [48]). j=l

As in theorem 1, it suffices to prove that F is one-one restricted to n ° = interior of ~; otherwise we can always enlarge the domain of F which will contain ~ in its interior and similar to ~ and satisfying all the other conditions of theorem 3.

For every x s R n let s(x) c 9 be the foot of x, that is, s(x) is the

unique element of minimizing

l lx-sll

for s s ~.

Plainly s(x) = x for x ~ ~.

We now

extend F:~ ÷ R n to the whole of R n by letting a function F:R n ÷ R n be defined as F(x) = F(s(x)) + x-s(x).

For any y s F(~) define Fy(X) = F(x)-y.

We will now state

and prove two lemmas that are needed for a proof of theorem 3.

{x:llxll =

Lemma i :

Let S r =

r) and B r = {x:Ilxll < r} be the sphere and ball of

radius r.

Then for any y s F(Q), and r sufficiently large, Fy restricted to S r has

degree one, that is, it is homotopic to the identity in S

r

with respect to R N

R n ' ~ {0}.

Proof :

Clearly the homotopy bridge H(x,t) is given by H(x,t) = t Fy(X)+ (l-t)l(x)

where t E [O,i], l(x) = x.

We have to only check that H(x,t) ~ 0 for any t s [O,i]

and x s Sr, if r is sufficiently large.

We will verify this by simply showing that

X.Fy(X) > 0 for any y E F(~) and x s S r when r is sufficiently large. choose any r >

max l lF(z)-z-Yll = s. zs~,ysF (~)

In fact

Then

x.~ (x) = l lxl 12 - x. (s(x)+ y-F(s(x))) Y

> llxll 2 -

ITxll

This finishes the proof of lemma i.

IIs(x)+ y-F(s(x)ll ~

r 2-rs

> O.

46

Lemma 2 :

Let K~be a polyhedron and F satisfy the hypothesis of theorem 3.

A = (x e Rn:F

is not continuously differentiable at x).

Let

Then if x ~ A, ID#(x) I

(= determinant of the linear map DF(x)) is positive.

Proof :

Let x ~ A.

Then x-s(x) is perpendicular to a single face of ~, which, of

course, includes s(x).Let L be the subspace spanned by this face and L ± orthogonal to L.

the subspace

For small v s L,s(x+ v) = s(x)+ v and so F(x+ v) = F(s(x)+v)+x+v-s(x);

hence DF(x)v = DF(s(x))v. Consequently DF (x)v ~ v.

For v s L ~

, s(x+v) = s(x) and so, F(x+v)=F(s(x))+x+v-s(x).

Choose an orthogonal coordinate system whose k first

coordinates generate L, J(x), the matrix of DF with respect to this coordinate system, takes the form

where JkF(S(X)) are the first k columns of J(x) (= Jacobian matrix for F).

Therefore

IDF(x) I = IJ(x) I = IJkk F(s(x)) I where Jkk F(s(x)) are the first k rows of JkF(S(X)). However Jkk~(S(X)) is the matrix of HL.DF(s(x)):L ÷ L and by hypothesis

IJkkF(S(X))I>0

and this terminates the proof of the lemma. We are now set to prove theorem 3.

Proof of theorem 3 "

Let A be the set (as defined in lemma 2) of those points x s R n

at which F is not continuously differentiable.

This set contains no open set.

This

is a consequence of the fact that s(x) is continuously differentiable except at those x which are contained in a finite number of hyperplanes.

Since F is lipschitzian,

F(A) contains no open set. Choose r > 0, sufficiently large so that lemma i holds good. Fvl(O) ~

A = @.

Also suppose

In other words we are assuming Fvl(0) lies entirely in the region

of differentiability.

From theorem (a) and theorem (b) it follows that

deg(Fy,Br,0)

~

deg(l, Br, 0)

=

Z sign det J F (x) = I. Y

Here the summation is extended over those x which belong to B r with Fy(X) = 0. Note that F-I(0) • B r ~ @ for any y s F(~). Also note that we are writing Y [As such we will assume for this proof a coordinate system is given to

det JFy(X). us].

Invoking lemma 2, we infer F-I(0) g~ B is a singleton set. y r As remarked earlier we will only show that F restricted to interior ~ = ~o is

one-one.

Suppose there are Xl,X 2 s ~o with x I ~ x 2 and F(x I) = F(x2).

disjoint open sets U I ~ Xl, U 2 ~ x 2

with F(U I) g~ F(U 2) ~ @ open.

tains no open set there is a y s F(U I) ~ F(U 2) such that y ~ #(A).

One can find

Since F(A) conConsequently

47

FyI(O)~A

= ¢.

Note that F-I(Y) C Fyl(0) t'~ B r and the latter set is not a

singleton set as F-l(y) has at least two elements.

This contradiction establishes

that F restricted to ~o must be one-one and this terminates the proof of theorem 3. Mas-Colell's proof of theorem 3 makes use of Poincare-Y~pf index theorem [see pages 35-41 John Milnor's Topology from the differentiable view point (1965), the U~iversity Press Virginia].

Instead we use theorem (a) and theorem (b).

We

now have the following theorem due to Nas-Colell.

Theorem 4 :

Let ~ C R n be a compact, convex set of full dimension with smooth

boundary $~ (= C I boundary) and F : ~ ÷ R n For each x s ~ ,

let T

x

a continuously differentiable function.

denote the tangent plane at x.

If the Jacobian J(x) has a

positive determinant at each x ~ ~ and if for each x E $~, J(x) is positive quasi-definite on Tx (that is is positive for every v s Tx with v # 0), then F is one one (and consequently a homeomorphism).

Proof :

We will first show that if x E ~

has a positive determinant. 3.

and L C Tx is a subspace then H L . D F ( x ) : L ÷ L

In other words this theorem is a consequence of theorem

Assume that we are given an orthogonal coordinate system such that the first

k coordinates generate L and the n-th is perpendicular to Tx and let J(x) denote the Jacobian matrix.

Then Jn_l,n_l(X), the matrix obtained by omitting the n-th

row and n-th column (by hypothesis) is positive quasidefinite. matrix is a P-matrix [see examples of P-matrines in chapter II]. Jkk(X) the matrix of HL.DF(x) determinant.

: L ÷ L

Fowever, any such This applies to

and yields the fact that Jkk(X) has a positive

Now one can complete the proof by approximatin~ ~ by a polyhedran ~'

and the above fact implies that the hypotheses of theorem 3 are satisfied for ~'. Hence F is i-I on ~.

Remark I :

This completes the proof of theorem 4.

Will theorem 4 be true if one assumes that J(x) is weakly positive

quasi-definite on T ? It is easy to see that the map F will be one-one on ~o = x interior of ~, by noting that the maps G = F(x) + s l(x) are one-one over ~ where s > 0 and l(x) = x. boundary.

Also the map G will be one-one throughout ~ including the s It is not clear whether F is one-one throughout ~. In other words this

raises the following question:

Let ~ be a closed convex region with nonempty

interior and let F:~ + R n be a one-one map throughout ~o = (interior of ~). other conditions should be imposed on ~ ~.

What

and F so that F will be one-one throughout

[see Kestelman's result given in chapter III].

In fact what we want is an

approximation theorem similar to theorem 4 in chapter IV which will hold good for closed region ~ (with nonempty interior).

In particular we would like to know

whether theorem 4 holds good when J(x) is weakly positive quasidefinite on Tx. Remark 2 : Nas-Colell's conditions on J(x) at the boundary of ~ neither imply nor

48

are implied by the fact J(x) is a P-matrix. One can give a slight generalization of theorem 2.

For that we need the

following

Definition :

Let A be an

n x n

matrix.

Call A a weak P-matrix if det A > 0 and

every other principal minor is non-negative. Theorem 2' [53] : rectangle in R n.

Let F:R n ÷ R n be continuously differentiable.

suppose J(x) is a weak P-matrix for all x s R n \ ~ . Proof :

Let ~ be a bounded

Suppose that the determinant J(x) > 0 for every x E R n and further Then F is one-one.

We will assume without loss of generality that D is a compact rectangle.

Let G (x) = F(x) + sl(x) when l(x) = x. every s > 0.

From theorem 2, each GE(x) is one-one for

Note that GE(x) ÷ F(x) as E + 0.

Hence we can conclude from More-

Rheinboldt's result that F(x) is one-one in R n. This terminates the proof of theorem 2' In fact theorem 2 can be further generalized as follows.

We allow determinant

J(x) to vanish on a set S of isolated points in R n. Theorem 2"[53]

:

Let n # 2 and let F:R n ÷ R n be continuously differentiable.

be a bounded rectangle in Rn.

Suppose that the determinant J(x) is positive for

all x except on a set S of isolated points where it vanishes.

Further suppose that

every principal minor of J(x) is non-negative for all x s R n \ ~ . Remark : the proof.

Let

Then F is one-one.

We will not attempt to prow~ this result but indicate the main steps in Using (Lemma 2, pp. 244-245 in [20]) we can construct a function G having

the following property (i) det G > 0 for every x except on a set of isolated points and (ii) G will be norm-coercive.

Now invoking a result of Chua and Lam (theorem

2.1, pp. 602-208 in [9]), we can conclude that G is one-one and consequently F is one-one.

Since Chua-Lam's theorem is valid only for n ~ 2, we have to impose this

condition in our theorem 2".

For n = 2, Chua-Lam's theorem is not true but we do

not know whether theorem 2" is true or false when n = 2.

It is worthwhile to look

at the following example which serves as a counter example in R 2 to theorem 2.1 in [9]. Let F=(f,g) where f(x,y) = x2-y 2 and g(x,y) = 2xy. Then the Jacobian is given by J = [ ~

-2Y2x] and det J = 4 ( x 2 y 2) which vanishes only when x = y = 0.

However F is not univalent since F(I,I) = F(-I,-I). But the diagonal entries of J do change sign and as such this will not be a counter example to theorem 2". We close this chapter by mentioning once again the following open problem: Can continuity of the derivatives in Mas-Colell's result be dispensed with or alternatively - are there counter examples ?

CFAPTER

VI

GLOBAL ~IVALENT RESULTS IN R 2 AND R 3

Abstract :

In this chapter we prove global univalent results obtained by Gale

Nikaido and Schramm when F is a map in R 2 .

Some extensions are indicated in R 3 .

Howeve~ the following two interesting open problems posed by Gale-Nikaido remain unanswered : (i) Suppose F is a differentiable map from a rectangular region ~ C R 3 to R 3.

Suppose the Jacobian is non-vanishing and every entry in the Jacobian is

non-negative.

Is F one-one?

gular region ~ C R

3 to R 3.

(ii) Suppose F is a differentiable map from a rectanSuppose every principal minor of the Jacobian is

non-vanishing for every x s ~.

Is F one-one ?

Gale and Nikaido have shown that

both (i) and (ii) together imply that F is one-one in any rectangular region in R 3. We have shown that (i) together with the assumption that the diagonal entries are identically zero will imply that F is one-one in any open convex region in R3-this result supplements the result obtained by Gale and Nikaido.

We can weaken our

assumptions in rectangular regions in R 3 using Garcia-Zangwill's result given in the previous chapter.

Univalent mappings in R 2 :

There are four results which we oresent in this section

which are due to Gale, Nikaido and Schramm.

The results are rather fragmentary but

are presented, for they suggest possible generalizations - in certain directions. In the first two results we relax conditions on the Jacobian matrix while in the third ~ is Assumed to be a region bounded by an x-curve. Let F:~ ÷ R 2 be a mapping given by F(x,y) = (f(x,y), g(x,y)), is a region in R 2.

(x,y) s ~ where

The first result concerns itself with one-signed principal

minors [19].

Theorem i : (a)

Let ~ be an arbitrary rectangular region either closed or not closed.

Suppose F has continuous partial derivatives and suppose none of the principal minors of the Jacobian vanishes. (b)

Then F is univalent.

Let ~ be an open rectangular region.

continuous.

Suppose the Jacobian does not vanish and no diagonal entries of the

Jacobian matrix change signs.

Proof (a) :

Then F is univalent.

Since F has continuous partial derivatives, every principal minor will

keep the same sign throughout ~. negative. f,

Let the partial derivatives of F be

That is, they will be everywhere positive or

We will assume without loss of generality that the diagonal entries

gy of the Jacobian matrix to be positive.

In case the Jacobian is also positive

50

then from Gale-Nikaido's fundamental theorem, F is univalent. is negative.

This means fx gy < fy gx"

be of the same sign.

Suppose the Jacobian

Since fx,gy > 0 it follows fy and gx should

In case fy and gx > 0 consider the map F(x,y) = (g(x,y),f(x,y)).

Then the Jacobian of F is a P-matrix and F is univalent in other words F is univalent. In case fy < 0, gx < 0 consider the map C(x,y) = (-f(x,y), -g(x,y)). Jaeobian of G is an N-matrix of the first kind. hence) F is univalent.

Then the

In this case also (G is univalent and

Similar proofs can be given for (b) also.

This terminates

the proof of theorem i. In the next theorem we will assume one-signedness of the entries in some row of the Jacobian matrix to assert univalence.

Theorem 2 (a) :

Let ~ be an arbitrary rectangular region.

partial derivatives and the Jacobian never vanishes.

Suppose F has continuous

If there are real numbers a

and b such that afx + bg x and afy + bgy do not change sign and one of them does not vanish, then F is univalent. (b)

~

Let ~ be an open rectangular region and let F be differentiable.

Jaeobian is everywhere either positive or negative.

Suppose the

If there are real numbers a

and b not both of them zero such that af x + bgx and afy + bgy do not change sign in ~, then F is univalent.

Proof :

(a)

We will assume a ~ 0 for by hypothesis one of the functions afx + bgx,

afy+ bgy does not vanish.

Observe that the mapping F(x,y) = (a f(x,y)+ bg(x,y),

g(x,y)) is univalent if and only if the original mapping F is univalent.

Jacobian

of F is given by

=

I afx + bgx

afy + bgy1

Jr

!

L

gx

gy

Then det J~ = a det JF" Since afx + bg x and afy + bgy do not change sign and one of them does not vanish and since the partial derivatives are continuous, we will assume afx+bg x ~ 0 and afy + bgy > 0.

In view of this analysis we may finally assume that the original

mapping F has the following properties f ~ 0 and f ~ 0. x y-Now suppose that F is not univalent, so that F(p,q) = F(r,s) = (a,B) for some distinct points (p,q), (r,s) s ~. will imply

Clearly q ~ s; for if q = s then p ~ r and this

fx = 0 for some point contradicting our assumption fx ~ 0 "

fixed y satisfying q ~ y ~ s [We assume q ~ s]. = f(r,s) ~ f(r,y). f(x,y) = ~.

Choose any

Since fy -~- 0, a = f(p,q) ~ f(p,y),

By the continuity of f we have an x between p and r such that

Since fx ~ 0, for each y in[q,s] there will exist a unique x = ¢(y) such

51

that f(¢(y),y) .= ~.

Since f(x,y) has continuous derivatives, ¢ is also continuously

differentiable, its derivative is given by ¢'(y) = -fy(¢(y),y)/fx(¢(y),y). G(y) = g(¢(y),y).

Define

Then G is continuously differentiable and its derivative is

given by, G'(y) = IJl/fx evaluated for x = ¢(y) where IJl denotes the determinant of the Jacobian.

Since IJI/fx # 0

, G is strictly monotonic in [q,s].

contradicts the fact that G(q) = G(s) = B.

This

Therefore F must be univalent.

This

terminates the proof of part (a) of theorem 2.

Proof (b) :

This can be reduced to case (a) as follows.

In any case we can assume

f > 0 and f > 0 throughout ~ (which is assumed to be an open rectangle). x -y -Suppose F(p,q) = F(r,s) for two distinct points (p,q), (r,s) belonging to ~.

Since

is an open set we can find real numbers ~i,Bi (i = 1,2) and an open set U containing (p,q) and (r,s) such that U = {(x,y) : ~ i ( x < ~i

and

~2 < y < B2}

Since a I < p, r < BI and ~2 < q,s < B2 for a suitably chosen small positive number s we can find an open subset ~

of U containing (p,q),

~i < x < BI, ~2 - S~l
(r,s) where ~

= {(x,y)} :

We will now define a new mapping

H:A ÷ R 2 where A = {(x,y):~ I < x < Bl,a 2 - E~ I < y < B2 - E6 I} and H(x,y)= (h(x,y), k(x,y)) with h(x,y) = f(x, sx+y) and k(x,y) = g(x, sx+y). (x,sx+y) ~ ~

iff (x,y) s A.

The Jacobian matrix JH of H is given by

fx(X,~X+y) + Efy(X,cx+y)

fy(X,~x+y) I

gx (x, sx+y) + Sgy (x, sx+ y)

gy (x, sx+y)

C l e a r l y det JH i s nonvanishing.

Note that

Also note t h a t fx and fy both cannot be zero f o r Thus

det J # 0 and consequently we have fx(X,~x+y) + Sfy(X,sx+y) > 0 and fy > O. h

> 0 and h > 0 for V(x,y) s A. Clearly the mapping H satisfies all the conditions x y -stipulated in part (a) of theorem 2 and hence H is univalent in A. But A contains

two distinct points

(p,q-sp) and (r,s-sr) which are mapped to the same point under H

and hence we arrive at a contradiction.

Therefore F is one one in ~ and this

terminates the proof of theorem 2.

Remark :

It is not clear how to formulate theorem 2 in higher dimensions

(when n > 2).

To state the next theorem we need the following [68].

Definition :

An x-simple domain in the (x,y)-plane is the interior domain determ-

ined by an x-simple curve, which is a Jordan curve cut in two points at most by every line parallel to the x-axis, except possibly by lines passing through points on the curve which have extremal values of y. y-simple domain are defined.

Analogously y-simple curves and

52

4 Observe that an x-simple curve can be put in the form U ~. such that for i i suitable mappings si:[c,d] ÷ R with c < d and s I < s 2 in (c,d) we have ~l={(sl(y),y): c < y < d},~ 2 = ~x,c):sl(c) < x < s 2 ( c ) ) ~ 3 = ~4 = {(x'd):Sl(d) < x < s2(d)}.

(s2(y),y) : c < y < d},

The following definition helps us to identify one

of the two subarcs seoarated by two points on a Jordan curve.

Definition : ~\{P'Q}

Let P ~ Q be two points on a Jordan curve ~ and ~i,~2 the subarcs with

= ~I ~

~2"

When ~ is a subset of X i for i = i or i = 2 define (P,Q,X)=~i;

when U is a proper subarc of ~ which contains

Definition :

A square matrix over the reals is called an NVL matrix when its leading

minors do not vanish.

Theorem 3 :

We are now ready to state and orove the following [68].

Let ~ be an x-simple domain in the (x,y)-olane,~ its boundary.

F = (f,g) : ~ ÷ R 2 ~.

Xi for i = I or i = 2 define (D'P'Q)=Xi"

Let

be a differentiable map,~ the minimum and B the maximum of f on

Suppose the Jacobian of F is an NVL matrix for each z s ~ and for each u s (~,~),

suppose at most two points z s Z satisfy f(z) = u. (a)

the sign of

(b)

defining A(u) = (z:z E ~ and f(z) = u}, both A(~) and A(B) are subarcs of ~ and

(c)

F restricted to

Remarks :

f x

Then

in ~ is constant

~\(A(a) UA(B))

is univalent.

When conditions on univalence are arranged in the partial order of

stringency of either the local conditions or the boundary conditions the above result will occupy an intermediate position - when we compare it with the GaleNikaido's fundamental theorem and Kestelman's result.

Clearly every P-function is

an NVL function and NVL functions have invertible derivatives.

Contrary to a

conjecture of Samuelson (see Gale-Nikaido's example) NVL functions are not unconditionally injective. Also note that theorem 3 asserts the univalence of F in the whole of ~, which is not necessarily a rectangular region and which need not be convex.

We will now

prove theorem 3.

Proof (a) :

Let c,d,Sl,S2,~l,~2,~ 3 and ~4 have the same meaning as defined above.

By the Darboux property sign fx(.,y) is a function X ef y only, for c < y < d.

We

will now show that the set {y:y c (c,d) and X(y) = i} is either empty or open in (c,d).

Let Yo be an element with X(y o) = i.

This means fx(.,yo ) > 0 where yoS(C,d).

Since s 2 > Sl, we have f(s2(Yo),y o) > f(sl(Yo),Yo).

As f is continuous for ly-yol

small enough f(s2(y),y) > f(sl(y),y) and hence fx(.,y) > 0.

This proves

(a) of

53

theorem 3-

Without loss of generality,

assume f

to be positive in ~.

Consequently

X

f(.,c) and f(.,d) are non-decreasing. We will now prove

(b). From (a) we have A(~) ~

meets Z3 at the end points at most.

Z, A(~) meets ZI and A(~)

Anologous assertions hold good for A(B).

To

show that A(~) is an arc, note first that f restricted to Z has exactly two solutions to f(z) = u when ~ < u < ~ since f-u changes sign on Z.

[In other words if f(zl)= ~,

f(z 2) = ~ where Zl,Z 2 are on ~ and since one can travel in two different directions from z I to z2, f(z) = u has at least two solutions on ~. f(z) = u can have at most two solutions]. z2sA(~) the arc ~ = (£i ~

However by hypothesis,

Now we shall orove that whenever ZlSA(~) ,

~2 ~ ~4,Zl,Z2) belongs to A(~).

Let 0 < s < ~ - ~ and let

~s,~ a be the subarcs of Z on which f < ~ + s, f ~ ~ + c, respectively. zI s ~ ' z2 s X similarly.

and ~ C

This proves

We will now prove

~ •

Since s is arbitrary f(~) = ~ .

A(~) is treated

(b) of theorem 3. (c). Since the given region is bounded by a Jordan curve and

since fX is assumed to be positive to prove the following:

Therefore

(without loss of generality)

in 2, it is sufficient

(i) For ~ < u < ~ we have A(u)={(h(u,y),y):y.(u)
with unique functions y.:(~,B) ÷ [c,d],y*: (~,~) + [c,d] and h(u,.): [y. (u) ,y* (u) ] ~ R and (ii) the functions ~(u,.): [y. (u) ,y* (u) ] ÷ R defined by ~(u,y) = g(h(u,y),y) strictly monotone when a ~ u < ~.

We will first prove

are

(i). Note that u-points

Zl,Z 2 of f restricted to Z separate subarcs Vl,V 2 of ~ with f < u on vl,f > u on v 2. It is clear that the y coordinates of z I and z 2 have to be different for, first f(.,y) is strictly increasing in (c,d) and second f(.,c), f(.,d) are nondecreasing and each has two u-points at most (by hypothesis). and larger of the y-coordinates ~2 = (sI(Y*)'Y*)'

Let y. and y* be the smaller

of Zl,Z 2 respectively.

~3 = (s2(Y*)'Y*)'

Let ~i = (sI(Y*)'Y*)'

~4= (s2(Y*)'Y*)'AL = {(sI(Y)'Y):Y* < y < y*}

and A R = (s2(y),y): y. < y ~ y*}.

Clearly A L ~

each y s [y.,y*] exactly one point

(h(u,y),y) s ~ satisfies f(g(u,y),y) = u.

changes sign on none of (~2,~4,~4), = u.

To prove

Hence for As f-u

(x,y) s ~ satisfies

at (u,y) for ~ < u < ~,

(As in the proof of theorem 2) we can conclude that ~y(U,y) =

~ 0 since Jacobian is an ~U~L function.

strictly monotone.

Remark :

(~i,~3,~2), no other point

(ii) note that h is differentiable

y.(u) < y < y*(u). (J/fx)(h(u,y),y)

~i' AR ~ ~2 as fx > 0.

Th~s shows that a(u,.) is

This terminates the proof of theorem 3.

Theorem 3 does not ensure the univalence of F on ~

simple example shows.

Let f(x,y) = x(l-y) and g(x,y) = y and

The mapping F = (f,g) transforms

~

the hypothesis of theorem 3. f(x,y) = x2y and g(x,y) = xy 2.

Let

~ = [1,2] x [0,i].

into the triangle with vertices

(0,i) the upper side of ~ shrinking into (0,I). the rectangle fx = (l-y) = IJI > 0.

as the following

(I,0), (2,0),

Also note that in the interior of

We will now give an example which will satisfy ~ = [0,i] x [0,i] with ~ interior of

~.

Let

Clearly the map F = (f,g) transforms two adjoining

sides of [0,i] x [0,I] into (0,0) and satisfies the conditions of theorem 3.

The

54

following corollary is helpful in drawing conclusions about the behaviour of F on ~.

Corollary

:

Suppose F satisfies the set of conditions of theorem 3 from which the

following condition is deleted: satisfy f(z) = u".

"For each u s (~,B) let two points z s £

Suppose F restricted to ~ is not one-one.

at most

Then for all u in

some subinterval of (~,B) the equation f(restricted to £) = u has more than two solutions. Proof of the corollary follows from theorem 3.

Remark :

To illuminate the corollary,

and Nikaido.

Let F(x,y) = (e2X-y23,

let us again look at the example due to C~le 4e2Xy-y 3) over

~ = [O,i] x [-2,2].

Here

£i = {(O,y):-2 < y < 2}, £3 ={(l,y):-2 < y < 2~£ 2 = {(x,c):O < x < l,o = - 2} and £4 = {(x,d):O < x < i, d=2} .

It turns out both

F(£ I)

Jordan curves symmetrical with respect to the x-axis.

and F ( £ 2 ~ £

3 ~£4

) are

Also F(£ I) lies in the

interior determined by F ( £ 2 ~ £

3 ~£4

the curves.

in a certain interval the equation fl£ = u has more

Thus for each

u

) except for the origin which belongs to both

than two solutions.

Theorem 4 :

In addition to conditions of theorem 3, suppose that for both Yo = c

and Yo = d either sl(Y o) = s2(Y o) or f(.,yo ) has a nonvanishing derivative in (Sl(Yo) , s2(Yo)) and F is NVL on £%£2\£4 .

Then F is one-one on ~ (= interior plus

the boundary).

Proof :

In view of theorem 3(c) it is enough to prove that F is one one on both

A(~) and A(B).

We will prove theorem 4 when Sl(C) = s2(c), Sl(d) = s2(d). A similar

proof can be given in the other case. A(~)C

£I or A(~) = {(sl(y),y)

From theorem 3(b) we can conclude that

: y.(~) < y < y*(~)} for some y.(~) and y*(~) s R.

Assume y.(~) < y*(~) and define 7(~,y) = g(sl(y),y).

In order to complete the proof

of the theorem it is enough if we show that 7y(~,.) # 0 in (y.(~),y*(~)). Yo s (y.(~), y*(~)) and let f and g have partial derivatives

Let

(P,Q) and (M,N) at

z° = (sl(Yo),Yo). From the local implicit function theorem it follows that s!(Yo)l exists and it is given by s~(y o) = - ~ • Also 7y(~,y o) = (PN~Q}____~)# O as F is NVL at z . o

This terminates the proof of theorem 4.

(Note that F'(z o) is unique since P = fx,M = gx' Q = - Ps~ and N = ((Pyy+QM) /P). We will now present a result due to Nikaido

[45].

Let F = (f(x,y),g(x,y))

be a

mapping from R 2 to R 2 where f and g have continuous partial derivatives throughout R 2"

55 Theorem 5 :

Suppose there are 4 positive numbers ml,m 2 and MI,M 2 such that

m I ! Ifx[ !

M1

and m2 ~ throughout R 2 .

fxgy-fygx I ~ M 2

Then the system of equations f(x,y)

:

aI

g(x,y)

=

a2

has exactly one solution in R 2 for any given constants al,a 2.

In other words F

is one-one and onto R 2.

Proof :

Observe that the existence and uniqueness of a solution of a single

equation f(x) = a in the single unk1~own x are clearly true if m _< If'(x) l < M(- ~ < x < ~)

for some positive m and M.

If this condition is satisfied, one can

see from the mean value theorem in calculus that either lim x÷+oo

lim x÷+ao

f(x) = + ~ or

f(x) = ~ ~ holds true, depending on the sign of the derivative f' (x).

H~nce,

f(x) can be equated to a at some value of x because of the continuity of f(x). Moreover the solution is unique from the montonicity of f(x). Suppose the conditions of theorem are met.

This means given a I and y,f(x,y)

= a I has a unique solution from the first paragraph of the proof. exists a function ¢(y) = x, such that f(¢(y),y) = a I.

That is there

Also one can check by the

local implicit function theorem that @ has continuous derivative @' with respect to y which will satisfy, fx(¢(y),y)¢'(y) + fv(¢(y),y) = 0.

Let G(y) = g(¢(y),y).

Then

u

G'(y) = gx(@(y),y)@'(y)+

gy(@(y),y).

f Hence it follows that G'(y) = - gx f-~x+ gy

m2 M2 ~ii <- IG' (Y) I <~ m-~

From the hypothesis,

Again from the first paragraph it follows

that G(y) = a 2 has unique solution for any given constant a 2. one-one and onto.

This shows that F is

This terminates the proof of this theorem.

This result can be generalized to higher dimensions and the proof goes through with very few changes.

Theorem 6 :

Let F = (fl,f2,...,fn) be a mapping from R n to R n, with continuous

partial derivatives

~f. fij _ ~x.i in the whole of R n.

Suppose that there are 2n

positive numbers m k and Mk, l < k < n such that the absolute values of the upper left-hand corner principal minor determinants

satisfy

5B

fll mk!

• ""

flk

absolute value of

J fkl

in the whole of R n.

"'"

M k (k = 1,2,...,n)

fkk

Then the system of equations fi(xl,x2,...,Xn) = a i (i = 1,2,...,n)

has exactly one solution in R n for any given constants a.. In other words F is i one-one and onto and consequently F is a homeomorohism of R n onto R n.

Counter example :

We will now present a counter example to illustrate that the truth

of theorem 5 depends very heavily on the assumption that the domain of F is the whole of R 2.

Consider the system of equations f(x,y)

=

x2-y2+3

=

0

g(x,y)

=

4x2y-y 3

=

0 .

Let ~ = {(x,y) : ½ < x < 2, -3 < Y < 3} • Also note that

2x

-2y 1

J=

,

8xy 244 > 18x 3 + 10xy21 > I

~ > L2xl > i

and

-3y2+ 4x 2

in the region ~.

But

f(l,2) = g(l,2) : 0 = f(l,-2) = g(l,-2). This example is due to Nikaido,

(and Cale-Nikaido's examole given elsewhere is a

slight modification of this example).

Theorem 5 is similar in nature to that of

theorem 3 in the sense that-we impose conditions on the leading minors and not on every principal minor of the JaOobian matrix.

It is not clear whether theorem 6

can be deduced from the results presented in chapter IV. We will now pose an interesting open problem in R 2 on univalent mappings. F be a continuously differentiable mao from R 2 to R 2. and (ii) det J(x) > 0

for every x s R 2.

Suppose

Is F univalent ?

Let

(i) trace J(x) < 0

In other words if for

every x,J(x) has only characteristic roots with negative real parts, is F one-one? Note that none of the results proved here covers this interesting case.

However

the result is not true if (i) trace J(x) < 0 and (ii) det J(x) < 0 as the following example shows.

Let F(x,y) : (f,g) when f(x,y) = -2e x + 3y2-I and g(x,y) = yeX-y 3. x 2 : - 2e 2x < 0 and trace J = - e -3Y < 0 and F maps

Then it is easy to see that IJl

(0,I) as well as (O,-i) to the same moint

(0,0).

57

t~ivalent results in R 3 :

The following result is due to Gale and Nikaido though

a proof of which is never published.

The proof which we give here is incomplete

and we invite the readers to furnish a complete proof.

Theorem 7 :

Let F : ~ C R

3 + R3

be a C (I) map where ~ is a compact rectangular region.

Suppose every entry in the Jacebian matrix is positive throughout ~.

Further suppose

that det J(x) > 0

Then F is

and all other principal minors are non-vanishing.

one-one. It is clear that principal minors of order one are positive.

If every principal

minor of order 2 is positive, then the Jacobian matrix is a P-matrix throughout and consequently F is one-one.

If every principal minor of order 2 is negative then

the Jacobian of the map -F is an N-matrix and once again from Gale-Nikaido's fundamental theorem -F is one-one or F is one-one. following.

The difficult case appears to be the

Suppose the leading principal minor of order 2 is negative and the other

principal minors are positive.

In this case we are unable to show that F is one-one.

Powever we can prove the following result.

Theorem 8 :

Let F : D ~ R

3 + R3

Let det J > 0 for every x E ~.

be a C (1) map where ~ is a compact rectangular region, Further suppose diagonal entries of the Jacobian are

0 and the off-diagonal entries are positive throughout ~ .

Proof :

Let F = (f,g,h).

Then F is one-one in ~.

Then the Jacobian of F is ~iven by

J =

0

fy

gx

0

hx

hy

z

where all the partial derivatives are positive by hypothesis on ~ .

Let G = (g,h,f).

Then clearly Jacobian G is non-vanishing throughout ~ and it is a P-matrix on ~ . Nence C~rcia-Zangwill theorem will imply that G is one-one or F is one-one.

This

terminates the proof of theorem 8. The following theorem complements theorem 7 of Gale-Nikaido in R 3 [53].

Theorem 9 :

Let F : ~ C R

3 ÷R 3

be a C (I) map where ~ is an open rectangular region

with determinant of the Jacobian positive throughout ~.

Further suppose diagonal

entries of the Jacobian are zero and the off-diagonal entries are non-negative throught ~.

Proof :

Then F is one-one in ~.

Let F = (f,g,h) where

(f,g,h)

are real-valued maps from ~.

map from ~ to R 3 defined as follows: Gs(x,y,z) = (g(x,y,z) + sx,h(x,y,z) + ey, f(x,y,z) + sz)

Let G6 be a

58

where E is any positive number.

Then it is easy to check that the Jacobian of the

map Gs is a P-matrix for every ~ > 0 and further GC ÷ (g,h,f) for every (x,y,z) ~ ~. F~nce F is univalent in ~.

This terminates the proof of theorem 9.

In fact theorem 9 holds good in any open convex region.

Theorem I0 :

Let F : O ~ R 3 -~ R 3 be a C (I) map where ~ is an ooen convex region in

R 3 with det J > 0

throughout ~.

Further suppose the diagonal entries are zero and

the off-diagonal entries are non-negative throughout D. Proof :

Then F is univalent in ~.

For every s > 0, let Gs be the map defined as follows:~(x,y,z) = (f+sz,

g+sx, h+Ey).

Then the Jacobian of G sis given by @

fy

fz

gx+ s

0

gz

L hx

hy+S

0

Now one can use argument similar to proof of theorem 4 in chapter III to conclude that G is one-one in ~.

Now an application of More-Rheinboldt's result yields the

fact that F is univalent in ~.

Remark :

It is not clear whether any of the univalent results proved above remain

ture in R 4. i0.

This terminates the proof of theorem i0.

We are unable to get any counter example in R 4 for theorems 9 and

In R 3 the most challenging problems appear to be the following:

is a differentiable map from a rectangular region ~ C

R 3 to R 3.

is non-vanishing and every entry is non-negative throughout ~.

Is F one-one in ~ ?

(2) Suppose F is a differentiable map from a rectangular region ~ C R every principal minor of the Jacobian is non-vanishing throughout ~. in ~ ?

(3)

Suppose F is a differentiable map from

(i) Suppose F

Suppose the Jacobian

3 to R 3.

Suppose

Is F one-one

a convex region ~ t o

R 3.

Suppose every entry in the Jacobian is non-negative and every principal minor is non-vanishing throughout ~.

Is F one-one in ~?

Note that in R 2 all these results

hold good. In R 2 the following problem appears to be unresolved.

Let ~ R

and closed rectangle whose sides are not oarallel to the axes.

2 be a bounded

Suppose the Jacobian

J of a continuously differentiable map F:~ ÷ R 2 is positive throughout ~ and on the boundary of ~ it is a P-matrix.

Is F one-one in ~ ?

This result is true if the

Jacobian is a positive definite or quasi-positive definite matrix on the boundary of ~.

It is also true if F is one-one on the boundary of ~.

CHAPTER ON T ~

Abstract :

VII

GLOBAL STABILITY OF AN AUTONOMOBS SYSTEM ON THE PLANE

In this chapter we consider an old problem of Olech (which is equivalent

to a global univalent problem in R 2) on the global stability of an autonomous system on the plane and extend a result of Olech using some of the recent global univalent theorems

(see [52]).

of the results.

We present also a variety of examples to show the sharpness

In the second half of this chapter we present the results obtained

by Vidossich [71] in this direction.

Introduction :

Consider an autonomous system (S) ... x' = F(x),

(Here orime denotes

derivative with respect to time point t) where x = (Xl,X2) , F(x) = (fl(xl,x2), f2(xl,x2)).

Suppose x = 0 = (0,0) is a critical point of (S) and assume that the

Jacobian matrix J at each point x of R 2 has characteristic roots with negative real parts.

That is assume (i) determinant of J > 0 and (ii) trace of J < 0 for every

x s R 2.

[Of course we do assume F to be a continuously differentiable function

throughout R2].

Is then the solution x = 0 of (S) global asymptotically stable, or

in other words does each solution curve of (S) approach the critical point O as t ÷ ~ ?

It is not very hard to see that this problem is equivalent to the following

global univalent problem [47].

Global Univalent Problem : Let F:R 2 ~ R 2 be a continuously differentiable function. Suppose the Jaeobian matrix J of F has the following two properties for every x in R2:(i) det J > 0 and (ii) trace of J < O.

Is then F globally univalent or one-one

throughout R 2 . It is net known whether global univalent problem has an affirmative answer.

If

it has, then one can conclude that the solution x = (O,O) of (S) is globally asymptotically stable.

However in [47] it is shown that x = 0 is globally asymptoti-

cally stable if (S) satisfies

(i), (ii) and if F is a one-one map.

We have the

following theorem due to Hartman, Markus and Yamabe and Olech [25, 34, 47].

Theorem I :

Let the autonomous system (S) satisfy (i) and (ii).

Let F(0) = 0.

Then the solution x = 0 is globally asymptotically stable if any one of the following conditions is met. [MY] one of the four partial derivative identically on R 2. [HI

~f. ~. i J

(i = 1,2, and j = 1,2) vanishes

if the symmetric part of the Jacobian matrix is negative definite on R 2.

6O

[0]

Remark :

~fl ~f2 R2 ~fl ~f2 Either ( - ~ i ) ( ~ 2 ) # 0 on or ( T ) (-~i)

R 2. #

0

on

Condition (0) is automatically satisfied if either [MY] or [HI is

satisfied•

We now have the following theorem which includes theorem i [52].

Theorem 2 : Let the autonomous system satisfy (i) and (ii) and F(O) = (0,0) and let F be a continuously differentiable function throughout R 2

Write F = (f,g)

Suppose any one of the following sets is bounded in R 2. (a)

D

=

{(x,y)

:

~f • ~~y g< ~--~

o)

(b)

E

=

{(x,y)

:

~f ~-y " ~~xg < O}

(c)

Q

=

((x,y)

:

~~x! .

~_~g ~y > 0) .

Then the solution (x,y) = (0,0) is globally asymptotically stable.

(Note that we

have slightly altered the notation; instead of writing fl,f2 we have written (f,g) and instead of writing (Xl,X2) we have written (x,y)). Proof of theorem 2 uses results from Olech [47] More and Rheinboldt [42] and Garcia-Zangwill [20]. Then we will give several examples to illustrate the sharpness of the results.

In particular we will give an example where theorem 2 is applicable

but not theorem i.

Proof of Theorem 2 : We will give a proof of theorem 2.

In view of Theorem 3,

(PP. 395 in [47]) it is enough if we show that the mapping F is globally one-one under each of the assumptions (a), (b) and (c).

We will write fx,fy instead of

~f ~x

Sf etc. Suppose (a) holds. That is we are given that the set D=[(x,y):fxgy<0] ~Y is bounded. We will assume without loss of generality that D(as well as E and G) are compact rectangles. Note that

We can always take the closure of the set D if necessary.

fx gy ~ 0 throughout R 2 \ D .

throughout R 2 \ D

and (8) fxgy = 0

We will now consider two cases (~) fxgy > 0

for some point in R 2 ~ D .

If (a) holds good,

since fx + gy < 0 by hypothesis it follows that fx < 0 and gy < 0 throughout R 2 ~ D . Invoking theorem 3 (PP. 246 in [20]) we may conclude that -F = (-f, -g) is globally univalent or that F is globally univalent in R 2 . If (~ holds good, then we define F (x,y) for every s > 0 F ( x , y ) = F ( x , y ) - (Ex,Ey).

as follows.

61

Observe that the JE = Jacobian of F

=

I fx-s

fY I

gx s

2

- S(fx+gy) - gxfy ~ 0

for every s ~ 0

trace of Js = fx + gy - 2s ~ 0.

Now det JE =

fxgY +

gy-~ since fx + gy ~ 0

and det J ~ 0.

Also

Furthermore product of the diagonal entries of Js

is given by

(fx-~)(gy-E) = fxgy - ~(f+gy) + 2 which is strictly positive throughout R2\ D.

In other words F s (x,y) satisfies all

the conditions of theorem 3 of Garcia-Zangwill [20] and consequently Fs(x,y) is globally univalent in R 2 for every s ~ 0.

That is -Fs(x,y) = -F(x,y) + (sx,sy) is

globally univalent in R 2 for every s ~ 0.

Also Jacobian of -F is non-vanishing and

R 2 is an open set.

Thus we may infer from theorem 5.9 of More and Rheinboldt [42]

that -F(x,y) or F(x,y) is globally univalent in R 2.

This terminates the proof under

condition (a). Suppose (b) holds good.

If the set E is bounded, since the Jacobian of F is

positive it follows that D ~ E or the set D is bounded and we are in case (a). Suppose (c) holds good.

Since fxgy ~ 0

we have fy gx ~ 0 throughout R 2 ~

G.

throughout R 2 ~

G and since det J ~ 0,

As F is of C (I) class and R 2 ~ G is a connected

set, it follows that f and gx will keep the same sign throughout R 2 ~ G. Suppose Y 2 fy ~ 0 and gx ~ 0 throughout R ~G. Let H = (g,-f). Then the Jacobian JH of H is a P-matrix - (that is every principal minor of JH is p~s~tive) throughout R2~G. Hence from Garcia-Zangwill's result it follows that N is globally univalent or F is one-one throughout R 2. throughout R 2 ~ G .

A similar proof can be given when fy ~ 0 and gx ~ 0

Thus we see that each one of the conditions (a), (b) and (c)

imply that the map F is univalent in R 2 and consequently (x,y) = (0,0) is ~lobally asymptotically stable.

This terminates the proof of theorem 2.

We will now deduce theorem I from theorem 2.

In order to do that it is enough

to verify that condition [0] implies one of the conditions (a), (b) and (c) of theorem 2.

If fx gy ~ 0

throughout R2~ condition (c) is trivially satisfied as G 2 is an empty set. If f g 0 throughout R , condition (a) is met as D is an empty x y 2 ~ R2 set. If f g ~ 0 throughout R it will imply f g 0 is and D will be an empty yx xy set. Suppose f g ~ 0 throughout R 2. We w~ll consider two cases (i) fy ~ 0 and

y x(2) fy ~ 0 and gx ~ 0. gx ~ 0 in R 2 and Then condition (a) is met for the map H. Then again condition (a) is met for L.

Under case (i) define a new map H = (-g,f). Under case (2) define a new map L=(g,-f). We may conclude from theorem 2, (x,y)=(0,0)

is globally asymptotically stable. One has the following corollary which can be deduced from theorem 2.

62

Corollary i :

Let the autonomous system satisfy (i) and (ii) and F(0) = (0,0).

Let F = (f,g) be a C (I) function.

Supoose fxgy ~ 0 throughout R 2.

Then the

solution (x,y) = (0,0) is globally asymptotically stable.

Remark :

Clearly corollary i includes Markus and Yamabe's result as well as

Hartman's result.

Also Olech's condition given in theorem i need not be satisfied.

We will utilize this corollary to construct an example where theorem i may not be applicable.

Examples:

In this section we will give several examples to illustrate the limita-

tions or the sharpness of the results known for the global asymptotic stability of the solutions for the autonomous system in the plane.

In the first two examples

Olech's condition given in theorem i will not be satisfied but conditions given in theorem 2 will be satisfied.

Fourth example will illustrate that condition (ii)

trace of J < 0 cannot be weakened to trace of J ~ 0, to get the asymptotic stability results.

If condition (i) is replaced by (i') det J < 0 throughout R 2 and retain

condition (ii) then one can have more than one critical points - we will demonstrate this in the third example.

The next two examples will show that it is possible to

have det J = 0 at some points and yet (0,0) to be a globally asymptotic solution. In other words condition

(i) or the map F to be one-one need not be necessary.

In

these two examples we will use the following result due to Nartman and 01ech (Theorem 2.1, pp. 155 in [27]).

Theorem 3 :

Let F be a continuously differentiable function from R 2 to R 2.

F(0) = 0 and F(x) # 0 whenever x # 0. solution of (S). F evaluated at x.

Suppose

Suppose x = 0 is a locally asymptotic stable

Let ~l(X) and ~2(x) be the characteristic roots of the Jacobian of Suppose~l(X) +

~2(x) ~ 0 for all x s R 2 and further suppose

Ixl If(x)] > constant > O whenever Ixl > constant > 0.

Then x = 0 is globally

asymptotically stable. Our last example will illustrate that theorem 3 may fail if ~l(X) + X 2 (x) admits positive values.

x2 Example i :

Let fl(Xl,X2) = - x I + log(l+x~) + x 2 and f 2 ( x l , x 2 ) = - x

- e

+ i.

Here 2x I

-I+

(

2 )

1

i+ x I J

= x2

2 - Bx I

Clearly det J > 0 and trace J < 0.

Note that

-e

~fl ~x I

~f2 2~x = 0

if

xI = i

and

63

~fl 8x2

~f2 8Xl

= 0 if x I = 0.

8fl 8xI

Also

8f2 8x2 ~ 0

throughout

R 2.

Here all the

conditions of corollary i are satisfied (but 01ech's condition given in theorem i is not satisfied).

Example 2 :

Consequently x = 0 is globally asymptotically stable.

9

Let

fl(Xl'X2 ) = - [ ~3-+ Xl(X~L - ½)-x2]

f2(Xl,X2)

=_

[xfx2].

Here the Jacobian J is given by 2

2

-(Xl+X 2 - ½1

-2XlX2+l]-l

J= -I Clearly det J > 0, trace J < 0. bounded set. applicable.

In this example the set D given in theorem 2, is a

2 In fact D = ~f~,Xl,X 2, : x I + x 2 2 _< ½}.

Here also theorem i is not

We can conclude from theorem 2 that x = 0 is globally asymptotically

stable.

Example 3 : This example is a slight modification of Gale-Nikaido given in (pp 82, [19]). Let fl(Xl,X2)

=

Xl 2 - 2 e + 3 x2 - i

f2(xl,x2)

=

x2

eXl

- x~ .

xI

Xl

Xl 2 e -3x 2

x2e Clearly det J < 0 and trace J < O.

However F = (fl,f2) is not globally univalent

since F(O,I) = F(0,-I) = (0,0). This example shows that global univalent problem is false if condition (i) is replaced by the condition det J < 0 throughout R 2. already remarked global univalent problem still remains open. of

the same example will serve

A modification

as a counter example if one attempts to prove

global univalent problem in R 3 under (i) and (ii).

Example 4 :

If we weaken condition (ii) to trace J ~ 0

then global asymptotic

As

84

stability may not hold good as the following example shows. -Xl+2X 2 and f2(xl,x2) = - Xl+X 2.

Set fl(Xl,X2) =

Here

Clearly det J > 0 and trace J < 0 throughout R 2.

However here x = 0 is not even

locally asymptotically stable though it is stable (see theorem 9.1, pp. 411 in [4]).

Example 5 :

The following example will show that condition

(i) namely det J > 0

is not a necessary condition for x = 0 to be globally asymptotically stable. fl(Xl,X2)

=

- Xl(l+ x~)

f2(xl,x2)

=

- x2(l+ x~).

Jfix

Let

and

12xlx21

[-2XlX2 2 22= Here det J = (I+ x 2)(I+ x 2) - 4XlX 2

0 if x I ± i and x 2 = ± I.

Consequently condition

(i) is violated.

It is not hard to check that x = 0 is locally asymptotically stable. = 2 (see pp. 440 in [4]). Also one can easily check that ~I + ~2 - (2 + x ~ + x 2) < 0.

Hence all the conditions of theorem 3 are satisfied and thus x = 0 is globally asymptotically

stable.

Alternatively one can use theorem 9.5 in pP. 439, [4] to

demonstrate that x = 0 is globally asymptotically stable.

Example 6 :

The following example shows that x = 0 may be globally asymptotically

stable but that F need not be globally one-one in R 2. x3 fl(Xl,X2) = - ~ +

Let

2XlX ~

f2(xl,x2) = - x~

J

x I + 2x 2

4XlX 2

0

2 -3x 2

=

1

65

det J = 3x2(2x 2 -

x I) ~ 0 if x 2 # 0 and 2x~ ~

x~

and trace J ~ 0.

F = (fl,f2) is not one-one since F(1,½) = F(-I,½) = (0,-½). IIF(x) ll ~ constant ~ 0 whenever

llxll ~ constant ~ 0.

Also

One can check that

All the conditions of

theorem 3 are satisfied and thus x = 0 is globally asymptotically stable.

[One

can check that x = 0 is locally asymptotically stable by Liapounou's second method for details see pp. 440 in (4)].

Example 7 :

The following example demonstrates that theorem 3 is probably the best

available result in R 2 for global asymptotic solution. fl(Xl,X2)

=

f2(xl,x2)

= _ x2 •

Let s ~ 0

and ~ ~ 0.

Let

sx I - ox~

Then the Jacobian

Ji

iiJ

In this example x = 0 is locally asymptotically stable but not globally asymptotically stable.

Also note that ~l(X) + ~2(x) = - i + s -2 ~Xl, may take both positive and

negative values.

In other words one of the conditions of theorem 3 is not satisfied.

In fact other conditions are met.

This shows that in general we may not be able

to relax the condition "~I + X2 ~ 0"

if we want globally asymptotic solution.

Vidossich's contribution to 01ech's problem on stability :

Vidossich's gives

another set of sufficient conditions under which F is one-one.

In fact we have the

following [71].

Theorem 4 :

Let F:R n ÷ R n

be of class C (I) and all eigenvalues of the Jacobian J(x)

of F have negative real parts, then F is one-one if any of the following conditions is met: (a)

there exist two positive constants ~ and B such that trace J(x) ~ - ~ IIJ(x) II ~ B

(b)

for every x E R n

and

or alternatively,

for each y s R n, there exists s ~ 0 such that for the topological degree Y for s ~ Sy, B(O,s) being the open ball with

we have deg(F,B(O,s),y) = I centre 0 and radius

Remarks :

s.

Theorem 4 is true for any n.

When (a) is satisfied one can show that F

66 is proper which will in turn imply that F is one-one - this can be seen from Hadamard's theorem or from Caccioppoli [6]. Also condition (b) is satisfied if there exists ~

o

> 0

and

0 ~ k < i such that

Ilx-r(x)II ~ kll~ll, (llxllm%). This fact follows from the homotopy invariance of the topological degree by considering l-k(l-F),

(0 < ~ < i).

Proof of theorem 4 : We will prove under condition (b). Suppose F is not one-one. This means for some

Yos

Rn' F-I(yo) will contain at least two points say

u,v.

Since F is locally one-one and R n is separable, F-l(y o) contains at most countably many points.

There exists s > 0 such that

> maX{ay o, 1 1 u l 1 , 1 1 v l l ~ and no points of F-l(y o) has norm equal to s, since otherwise F-l(yo) would contain a subset with the same cardinality of a non-trivial interval of R contrary to the countability of F-l(yo).

Since the closure of B(O,c) is compact and F-l(yo) fhB(O,E)

is discrete it follows that F-l(yo) lh B(O,c) must be finite : let x I .... ,xm points.

Clearly Ilxil I < a

for i = 1,2 .... ,m.

be its

Hence we can find for each i an

open set U i of x i such that the closed sets Ui are pairwise disjoint and U i ~ B(O,s). Let C-(x)

Then, F(x) = Yo

-- x +

F (x)-y °

if and only if G(x) = x.

.

Observe that G(x) has no fixed point in

m

B(O,s)~M

U i.

By the additivity of the topological degree, we have m

deg(l-G, B(O,~),0)=

~ dega-a, Ui,0) i=l

where the right hand side is well defined since no point of F-l(yo ) belongs to ~U i. We will now show that deg(l-G, Ui,0 ) = i, i = 1,2, ...,m. JG(Xi) = Jacobian of G. (~-l)x= JF(Xi)X. -i < 0

Let ~ be an eigenvalue of

Then there exists x ~ 0 such that (l+JF(Xi))x = kx

or

It follows that (~-i) is an eigenvalue of JF(Xi) and by hypothesis

or X < i.

It implies that JG has no positive eigenvalue greater than one.

Therefore a well-known theorem of Leray-Schauder deg(l-G, Ui,0) = i for i = 1,2 .... ,m.

(See pp 162-163 [48]) implies that

It is known that

deg(g,B(O,s),y) = deg(g-y, B(0,s),0) for any continuous function g and any point y provided that the equation g(x) = y has no solution x with

Iixll

--

~.

Therefore we have :

67

deg(F,B(O,s),y o)

=

deg(F-Yo, B(O,s),0)

=

deg(l-G, B(O,E),O) m

--

X degII-G, ~i,0) i=l m .

But m _> 2, since u,v s F-l(y o) g~ B(O,s), hence we have a contradiction to our assumption

(b).

This terminates the proof of theorem 4.

We now have the following theorem on stability.

Theorem 5 :

The solution x = 0 of the autonomous equation x' = F(x) in the plane

is globally asymptotically

stable if F is of class C (I), F(0) = 0, the eigenva!ues

of J(x) all have negative real parts and F satisfies

Proof :

(a) or (b) of theorem 4.

Follows from theorem 4 and Olech's theorem.

Remark i :

Example 6 shows that the solution x = 0 may be globally asymptotic stable,

without any of the conditions imposed by theorem 4 being satisfied.

Remark 2 :

Global asymptotic stability in R n has been studied by Hartman [25] and

Hartman and Olech [27].

Hartman considers the case when (J+ J')/2 is negative

definite while Nartman and Olech places conditions on the eigenvalues of the Jacobian matrix.

Interested readers should refer to their works for further details.

We close this chapter by inviting the readers to prove the global univalent problem or give a counter example.

CHAPTER

VIII

UNIVALENCE FOR MAPPINGS WITH LEONTIEF TYPE JACOBIANS

Abstract

:

In this chapter we prove several results on univalence

Leontief type Jacobians. Gale-Nikaido,s

for mappings with

The first result is in some sense a sort of converse to

theorem on univalence.

Here we prove a result due to Gale-Nikaido

and this says that if F and F -I are differentiable

and if F -I is monotonic

increasing

then the Jacobian of F is a P-matrix provided the Jacobian matrix of F is of Leontief type.

The second result due to Nikaido

says that there exists a unique solution to

F(x) = 0 provided its domain is non-negative

orthant and the Jacobian matrix is of

Leontief type satisfying certain uniform diagonal dominance property. present related results on M-functions Rheinboldt.

Then we

and inverse isotone maps due to More and

Finally we give some results on the univalence of the composition of

maps F and G when their Jacobians are of Leontief type.

In particular we show that

F o G is a P-function when F and G are maps from R 3 to R 3 with their Jacobians Leontief type P-matrices

throughout.

We give an example to show that F o G need

not be a P-function in R 4.

Univalence for dominant diagonal mappings section.

:

Two results will be presented

The first result shows that if both F and F -I are differentiable

F -I is monotonic

and if

increasing then the Jacobian matrix of F has to be a P-matrix,

provided the Jacobian matrix of F is of Leontief type. and Nikaido.

in this

This result is due to Gale

The second result due to Nikaido says that there exists a unique

solution to F = 0 provided the domain of F is ~ = non-negative

orthant and the

Jacobian matrix is of Leontief type satisfying certain uniform diagonal dominance property.

We will say that a matrix A is of Leontief type if the off-diagonal

entries of A are nonpositive. initiated by well-known

Such matrices play a prominent role in studies

economist Leontief.

that the following two conditions (i)

A is a P-matrix

(ii)

There exists a vector x > 0

For Leontief type matrices we have seen

are equivalent.

such that

Ax > 0.

In chapter II we have defined the notion of dominant diagonal.

That is, a

matrix A is said to possess a dominant diagonal if there is a set of positive numbers n d.(i = 1,2 ..... n) such that a..d. > ~ la..Id, 1 ii i ~--7 i$ condition is equivalent

for all

i = 1,2 ..... n.

This

to the condition that there exists a set of positive numbers

69 c. (j = 1,2,...,n) such that J n

aj j cj

>

~ laijlc i i=l

for all

j = 1,2 ..... n.

i#j Equivalence follows from the fact that any one of the above conditions implies that A is a P matrix. If A, whose elements are functions defined on a set ~ has dominant diagonal throughout the set ~, the weights d. in general will be a function of w s ~. di(w) ~ d i for all w s ~

then we say A has a uniformly dominant diagonal.

If

We say A

has a uniformly dominant diagonal in the strong sense if there are positive numbers d i and ci, i = 1,2,...,n such that d A ~

c where d = (dl,d2,...,dn) , A is an n x n

matrix and c = (Cl,C 2 ..... Cn). [Entries of A are functions of w].

We are now ready

to state the following theorems.

Theorem i :

Let F : ~ ÷ R n be a differentiable mapping where D is an open region

in R n and Jacobian matrix is of Leontief type.

Suppose F -I is differentiable and

monotonic increasing (that is F(a) < F(b) implies a < b).

Then the Jacobian matrix

of F is a P-matrix.

Theorem 2 :

Let F:R+n ÷ R n be a differentiable map whose Jacobian matrix is of (Here R+n = non-negative orthant). Suppose Jacobian matrix of F has a uniformly dominant diagonal in the strong sense. Further suppose fi(x) ~ 0

Leontief type.

whenever the i-th component of x is zero. F(x)).

(Here fi(x) is the i-th component of

Then F(x) = 0 has a unique solution.

In other words F(x) = a has a unique

solution for every a s R+n . Proof of Theorem i :

Since ~ is open in R n, by the invariance of domain, F(~) is

also an open set in R n.

Let b be any point of ~ such that b* = F(b).

any strictly positive vector.

Let u* be

Since F(~) is open, there exists s > 0 such that

x*(~) = b * + X u * s F(~) for all ~ with l~I < s.

Let F-l(x*(~)) = x(~).

Then since

F-I is differentiable and monotonic increasing, x(~) is differentiable a n d ~ x ( ~ ) for

IXI

< s.

0

Differentiating x*(X) = F(x(~)) at ~ = 0, we have J(b)x'(0) = u* > 0.

Since J(b) is of Leontief type, and since x'(0) > 0 and u* > 0 it follows that J(b) has to be a P-matrix. at b).

(Here as usual J (b) stands for the Jacobian matrix evaluated

This terminates the proof of theorem i.

Proof of theorem 2 : chapter VI.

Proof of this result is similar to the proof of theorem 5 in

Proof is based on induction on n, which will establish both the existence

and uniqueness of the solution of the system of equations F(x) = 0.

If n = i,

we are looking for a solution of a single equation of a scalar variable fl(x) = 0.

70

We are given two positive constants d I and c I such that d I fll(X) > c I for all x ~ 0.

This implies the first derivative fll(X) of fl(x) is positive and in

fact fll(X) > (Cl/dl). and fl(x) ÷ + ~ fl(x) = 0

Hence it follows that fl(x) is strictly monotonic increasing

as x ÷ ~.

Since from hypothesis fl(0) _< 0, it follows that

has a solution as fl is continuous and the solution is unique as fl is

strictly monotonic. Assume the result to be true for n-l.

Let us write F(x) = 0 explicitly as

follows: fi(xl,x2,...,Xn_l,Xn) = 0

for

i = 1,2,...,n-I

fn(Xl,X2,...,Xn_l,Xn) = 0 Fix xn and consider the system of equations in the (n-l) variables Xl,X2,...,Xn_l. The system fi(xl,...,Xn_l,Xn) = 0 for i = 1,2,...,n-i satisfies all the hypothesis of the theorem including the strong uniform dominant diagonal property. end, let Jn-i denote principal that of the original system.

To this

submatrix of order n-i in the upper left corner of Since (dl,...,dn)J ~ (Cl,...,c n) for some d ~ 0,e ~ 0

and since the off-diagonal entries of J are non-positive it follows that (dl,...,dn_l)Jn_l ~ (Cl,..-,Cn_l). Hence by induction hypothesis the system of equations fi(xl,...,Xn_l,Xn) = 0 for i = 1,2,...,n-l, for every fixed non-negative Xn, has a tmique solution, Xl,...,Xn_ I which will be functions of xn.

Let xj = gj( Xn), j = 1,2,...,n-I which of course will

satisfy fi(gl(Xn) , g2(Xn),...,gn_l(Xn),Xn) = 0,i = 1,2,...,n-I for all x n_> 0.

Let g(Xn) = fn(gl(Xn),g2(Xn) ,...,gn_l(xn),xn).

Now it is easy to

conclude that F(Xl,...,xn) = 0 has a unique solution if and only if g(xn) = 0 is uniquely solvable in the non-negative unknown x n. to one-variable case.

Thus we have reduced the problem ~ cn - these

We will now show that g(0) ~ 0 and d n ~

two facts will complete the proof,

n

Obviously g(0) = fn(gl(0),...,gn_l(0),0) ~ 0. dJ > c

and

Observe

J is of Leontief type.

Hence it follows that j-i is non-negative.

Thus

(dl,d2,...,dn)J ~ (Cl,...,Cn) or (dl,...,d n) ~(Cl,...,Cn)J-i n

Now dn > 111"=cifmn

where (fin) is the n-th column of j-l.

fin~ 0 or

and

Since

e i ~ 0, dn ~ en fnn = On

det Jn-i det J

71 d

det J > c n det J n - i n

Note that dg dxn

_

det J det Jn-i

Hence g(Xn) = 0 will have a unique solution and consequently F (x) = 0 will have a unique solution.

This terminates the proof of theorem 2.

Interrelation between P-property and M-property :

We will now introduce the

notion of M-function similar to the notion of P-function introduced in chapter III and prove that if F is a differentiable map over the rectangle and if the Jacobian is a P-matrix of the Leontief type then F must be an M-function.

Definition :

Consider a mapping F : ~ C ~

x ~ y

Rn ÷ Rn

whenever

with the following properties

(a)

F(x) ~ F(y)

x,y s

(b)

the functions T~:(t s Rllx+ te j E ~) + R I defined by Tij(t) = fi(x+teJ) are monotonic decreasing as functions of t where i # j and e

i

= (I,0,0,...,0), e 2

=

(0,I,O,...~0) etc. [Here f. as usual denotes 1

the ith component of F]. Then F is said to be an M-function.

Property (a) is usually called inverse isotone

and (b) is called off-diagonally antitone.

This notion is a nonlinear generalization

of Leontief type P-matrices introduced and studied by More and Rheinboldt.

We are

ready to prove the following theorem.

Theorem 3 :

Let F:~ C R n ÷ R n

be a differentiable map on the rectangle ~.

the Jacobian matrix of F is a P-matrix of the Leontief type for every x s ~.

Suppose Then

F is an N-function, in the sense described above.

Remark :

It is known from Gale-Nikaido's result that F is a P-function.

that F is off-diagonally antitone.

Also note

Theorem 3 is an immediate consequence of the

following result.

Theorem 4 :

Any off-diagonally antitone P-function F:~ ~ R n ÷ R n on a rectangle

is an M-function.

Proof :

We need only to verify property (a) of the above definition of an M-function.

That is, we have to establish that F is inverse isotene. some x,y s ~.

Let A = (i:x i > yi }.

empty and let A = {l,2,...,m}.

Suppose F(x) < F(y) for

If A is empty we are through.

Define

Suppose A is not

72

gi(tl,t2 .... ,tm) = fi(tl,t2 ..... tm,Ym+l,...,Y n) where i = 1,2,...,m.

Note that

gi(Yl,y2, ---,Fro) = fi(y) ~_ fi(x) ~_ gi(xl,x2 ..... Xm). Here the last inequality follows from the fact that F is off-diagonally antitone and the first inequality by hypothesis.

Hence we can conclude that

(Yi-Xi)(gi(Yl ..... ym)-gi(x I ..... Xm)) ~_ 0, Set x' = (Xl,X2,...,Xm, Ym~l,...,yn ).

Clearly x' ~ y

i = 1,2,...,m.

and

(Yi-x~)(fi(y) - f±(x,)) ~ 0 ~i = 1,2 ..... m and (Yi-X.~)(fi(y) - fi(x')) = 0)i = m + l ..... n

but this contradicts the fact that F is a P-ftunction. of theorem 4.

Remark :

This terminates the proof

Now theorem 3 follows from theorem 4.

We need the fact that ~ is a rectangle to conclude that x' s ~.

It is not

clear whether this result is valid if ~ is a convex region and not necessarily a rectangular region.

We have lot of freedom when

~ is a rectangular region.

One

can restate theorem I as follows, for convex maps in terms of property (a).

Theorem 5 : set ~.

Let F : ~ C R

n ÷ Rn

be convex and differentiable on the open convex

Then F is inverse isotone if and only if the Jacobian matrix J(x) is inver-

tible for each x s ~

and J(x) -I ~ 0

for each x E ~ (that is every entry in the

inverse matrix is non-negative).

Proof :

Suppose F is inverse isotone.

x s ~ and suppose J(x)h ~ 0.

That is F(x) ~ F(y)

implies

x ~ y.

Let

Since ~ is open, there is a 0 ~ 0, such that x + Oh s ~.

m

It fellows from the convexity of F F(x+eh)-F(x) ~ 0J(x)h ~ 0 The last inequality follows because 0 ~ 0

and

is inverse isotone, x+ eh ~ x

In other words J(x)h ~ 0 implies h > 0.

or

h ~ 0.

J(x)h _> 0

Consequently it follows that J(x) is non-singular.

by assumption.

Now we will prove J(x) -I _> 0.

In order to do that, let y_> 0 then we have J(x)(J(x) -I y) ~_ 0 implies J(x) -I y > 0.

Since F

which in turn

Hence J(x) -I _> 0.

Conversely suppose J(x) -I ~ 0

for every x s ~.

Let F(x) _> F(y) for some x,ys~

From the convexity of F it follows that J(x)(x-y) >_ F(x)-F(y) >_ 0.

That is

73

J(x)(x-y)

> 0

or J

(x) -1 J (x) (x-y) = x-y ~ 0.

This terminates the proof of

theorem 5.

Remark

:

If in addition we assume J(x) is of Leontief type in theorem 5 then F

is inverse isotone if and only if J(x) is a P-matrix for each x s ~. isotone concept is equivalent provided F -I exists.

Inverse

to fact that the mapping F -I is monotonic

increasing

If we drop the assumption of convexity of F as well as

convexity of ~ we have the following theorem.

Theorem 6 :

Let F : ~ C

R n ÷ R n be inverse isotone and differentiable

set ~.Suppose the Jacobian J(x) of F is of Leontief type.

on the open

Then J(x) -I is a

P-matrix for any x E ~ at which J(x) is non-singular.

Proof

:

Suppose x s D with J(x) non-singular.

In order to show J(x) is a P-matrix

it is enough if we show J(x) -I ~ 0 since J(x) is of Leontief type. we will prove the following J(x)h> 0 be seen as follows.

This can

First we will consider the case J(x)h~ O. limit t÷0

Hence F(x+ th)-F(x)

As in theorem 5

for some h s R n will imply h ~ 0.

[F (x+ th)-F (x) ]

~ 0 for sufficiently

isotonicity that h > 0.

=

J(x)h > 0 .

+

small t > 0 and consequently by the inverse

Now consider J(x)h > 0 for some h E R n.

i h k = h + ~ J(x) -I e~ k = 1,2 ....

Define

Here e denotes the vector with all entires one.

Clearly J(x)h k > 0 which in turn implies from the first part h k > O, for all k and hence h ~ 0 since h k ÷ h as k ÷ ~ proof of theorem 6.

This shows J(x) -I > 0.

Observe that theorem 6 is essentially

This terminates

the

the same as theorem i.

For open rectangular regions one can prove the following theorem.

Theorem 7 : set ~.

Let F:~ r- R n ÷ R n

be a differentiable

map on the open rectangular

Suppose J(x) is of Leontief type and suppose J(x) is non-singular

for each

x s ~ .

Then F is inverse isotone if and only if J(x) is a P-matrix for every x s ~.

Proof

If F is inverse isotone then from theorem 6, it follows that J(x) is a

:

P-matrix for every x c ~. fundamental

Now suppose J(x) is a P-matrix.

theorem F is univalent

by induction argument on n.

in ~.

From Gale-Nikaido's

We will prove that F is inverse isotone

When n = i, clearly F is inverse isotone.

In general,

if F(a) ~ F(b) where a = (al,a2,...,a n) and b = (bl,b2,...,bn),

then for some k we

have a k ~

We may assume without

bk

this is a consequence of theorem i, chapter III.

less of generality k = I that is a I ~ b I. Jaeobian matrix are non-positive,

Since the off-diagonal

we have for i ~ i,

entries of the

74

fi(bl,a2 .... ,an ) ~ fi(al,a2,--.,an ) ~ fi(bl,b2,...,bn )Here the second inequality follows from F(a) ~ F(b). G:~n_l c

We now define a new map

Rn-I ÷ R n-I by the rule G(x2,x3~...,Xn) = (f2(bl,X2 .... ,Xn),...,fn(bl,X2,

...,Xn)) where ~n-I is the image of ~ under the projection (Xl,X2,...,xn) ÷ (x2,...~Xn).

Now G(a2,...~an) ~ G(b2,...,bn) and the Jacobian matrix of G is again

a P-matrix of the Leontief type.

Fence by induction hypothesis a. ~ b. for i ~ 2. i

We already have a I ~ b I and hence a i ~ b i for every i.

--

i

This proves F is inverse

isotone and this terminates the proof of Theorem 7.

Univalence for composition of two functions :

In this section we will prove in R 3,

that if F and G are differentiable functions whose Jacobian matrices are Leontief type P-matrices then F o G is a P-function and hence univalent in R 3. can prove the following result in Rn.

However one

If F and G are differentiable functions and if

Jacobian matrices of F and G are Leontief type P-matrices and further if the Jacobian of F o G is also of Leontief type then F o G is a P-function and hence univalent in Rn.

The results we prove in this section are similar to theorem 3.4

in More and Rheinboldt [42]. We are ready to prove the following [53].

Theorem 8 :

Let F and G be C (I) differentiable maps from R n ÷ Rn.

Jacobians of F and G are Leontief type P-matrices. the composition map F o G is of Leontief type.

Suppose

Further suppose Jacobian of

Then F o G is a P-function and hence

univalent in R n.

Proof :

Let H = F o G.

Observe that the Jacobian JH of H can be computed by means

of the relation JH = JF JG where the elements of JF as well as JG will be evaluated at the appropriate points. P-matrix throughout R n.

Theorem 9 :

From theorem 4, chapter II we can conclude that JH is a

Invoking Gale°Nikaido's theorem, we have the desired result.

Let F and G be C (I) differentiable maps from R 3 to R 3.

Jacobians of F and G are Leontief type P-matrices.

Suppose

Then F o G and G o F are

P-functions and hence univalent in R 3.

Proof : We will prove F o G is a P-function.

Let H = F o G.

From theorem 4,

chapter II, we can conclude that the Jacobian of H is a P-matrix and consequently F o G is a P-function.

Similar proof may be given for the univalence of G o F.

From theorem 2, chapter V we can deduce the following:

75

Theorem I0 :

Let F and G be C (I) differentiable maps from R n to R n.

compact rectangle in R n. Rn\

~ respectively.

Let D be a

Further suppose Jacobians of F and G are P-matrices in

Then F o G is univalent in H n.

We gave an example in chapter II (see theorem 4) to show that the product of two Leontief type P-matrices in R 4 need not be either a P-matrix or a Leontief type matrix.

That example will correspond to the following functions in R4: F(x,y,z,w)

:

i (x-z, [ y-w, z, w)

O(x,y,z,w)

(x-y, -x+ 2y, -4y+ z, -x+w).

(Fo G) (x,y,z,w)

(x+ 3y-z,7 x +

Then o

~-

w, -4y+ z, -x+ W ) .

In this example Jacobian of F and G are Leontief type P-matrices. of F o G is a non-singular

However Jacobian

(constant) matrix which is not a P-matrix.

easy to see in this example that F o G is one-one throughout R 4. following question:

Also it is

This raises the

Suppose F and G are differentiable C (I) function in R 4 whose

Jacobian matrices are Leontief type P-matrices.

Is F o G a P-function in R47

The

answer is "no" and the example given above is a counter example and it can be seen as follows.

Jacobian matrix of F o G is a constant matrix A given by

A

=

1

3

? B-

-8-

2

-i

o

o

-1

0

-4

1

o

-i

0

0

i

Since A is not a P-matrix, then exists a non-trivial vector u ° = (x°, yO, zo, wO), such that (Au°)~ u~ ~ 0

for i = 1,2,3,4 [Here prime denotes the transpose vector

and u~ stands for the ith component of the vector u°]. I

Let 0 be the zero vector.

Note that (F o G)(u °) = Au °'

Clearly u ° ~ 0 and (F o C) (0) = O.

Hence it follows

that F o G is not a P-function. We will now deviate a little bit and start with two examples.

The first is an

example of a differentiable P-function whose Jacobian is not a P-matrix. F:R 2 ÷ R 2 where F = (f,g) with f(x,y) = x3-y and g(x,y) = x+y 3. check that F is a P-function throughout R 2.

However Jacobian

One can easily J

of

3x 2 J

-i

= I

is not a P-matrix at x = y = O.

3y 2

I

=

Consider

i + 9x2y 2 > 0

F

given by

76

The second example is the following. This says that if the Jacobian is a weak P-matrix in a rectangular region, F need not be a P-function. 0 Here J = [-I

Consider F = (f,g) where f(x,y) = y, g(x,y) = y-x.

1 i ] is a weak P-matrix but F is not a P-function.

Gale-Nikaido's

fundamental

theorem that this result is true if J is a P-matrix.

We will now introduce the concept of P

Definition

:

A mapping F : ~ c

there is an index k = k(x,y) For Po-funetions

Theorem ii :

Let F : ~ c

o

I :

function which is weaker than P-function

R n ÷ R n is a P -function if for any x,y ~ ~, x # y, o such that (xk-Yk)(fk(x)-fk(y)) ~ 0 and x k # Yk"

one can prove the following theorem

Rn ÷ Rn

be a differentiable

for every x s ~ where ~ is an open rectangle.

Remark

We know from

[42].

P -function with det J(x) > 0

Then F-l°is again a P -function. o

We will not attempt to give a proof but we will indicate the proof.

Let D be any diagonal matrix with entires non-negative.

Then one can check that

F(x) + Dx is a Po-function which will in turn imply J(x) + D is a weak P-matrix. other words F(x) + Dx is univalent

for every such D.

In

This in turn implies that

F -I is a P -function. o Remark 2 :

A differentiable

Jacobian.

Po-function need not possess a weak P-matrix as its

It will only be a P -matrix, o non-negative.

that is, every principal minor is

We will close this chapter with a conjecture of Nore-Rheinboldt

[42]:

For

any continuous,

injective P -function F:~ c R n ÷ R n on an open rectangle ~,F -I o is also a Po-function. This result certainly holds good for the linear case and

for F-differentiable

P -functions. In order to settle this conjecture it is enough o if one can show that F(x) + Dx is one-one for every diagonal matrix D with non-

negative entries.

CHAPTER

IX

ASSORTED APPLICATIONS OF I~IVALENCE MAPPING RESULTS

Abstract :

In this chapter we will give various applications of the univalence

results proved in the earlier chapters. results are quite handy and useful.

There ar~ several areas where univalence

The first application will deal with a problem

in Mathematical Economics where we will give a set of sufficient conditions due to Nikaido and Mas-Colell which will ensure factor price equalization.

The second

application deals with the distribution of a function of several independent random variables.

As a third application we will consider a ~roblem in nonlinear complim-

entarity theory due to Kojima and Megiddo. theorem to Algebra.

Next we give an application of ~damard's

In the fifth we consider the problem of deciding whether a

certain multivariate gamma distribution is infinitely divisible. weak N-matrices play an important role.

In this situation

There arel various other applications

(for

example to nonlinear net-work theory) but we will not attempt to exhaust all of them for lack of time and space.

[We have already seen a nice application of univalent

results in stability theory in chapter VII].

kn application in Mathematical Economics : As a first application we will consider a well known nroblem in mathematical economics first suggested by Samuelson which in a sense prompted ~le-Nikaido to ~rove their fundamental result on univalence mappings.

In this problem one seeks conditions implying that countries facing

the same prices for goods in foreign trade will have the same factor prices.

To be

precise we will give a set of sufficient conditions which will not only ensure factor price equalization but also a set of equilibrium factor prices under any given set of good prices.

Put differently, the condition ensures the complete invertibility

of the determination of good prices by factor prices, giving rise to the inverse unique determination of factor prices by arbitrary good prices. We will now explain the terminology that will be used.

Let wj > 0, Pi > 0

denote the price of the jth factor and the ith good respectively where j,i=l,2,...,n. Let ci(wl,w2,...,w n) be n cost functions which hay@ continuous non-negative partial derivatives cij = 3oi/3w j

and c i 's are positively homogeneous of degree one, that is

ci(Xw I .... ,Xwn) = ~ci(Wl,...,w n) for all X > 0 and wj > 0.

Let ~ij = cij wj/ci -

this quantity is the relative share of the jth factor in the ith good sector, because

n ~ j=l

c.. wj = e i. iJ

[The last equality follows from the fact that c.'s are 1

positively homogeneous of degree one].

By 'equalization of factor prices' we mean

78

the uniqueness of solution of the system of equations ci(wl,w2,...,w n) = Pi (i = 1,2,...,n). Nikaido

We are ready to state the complete invertibility theorem due to

[45].

Complete invertibility theorem :

Suppose ci(wl,w2,...,w n) are defined for all

w. > 0, take positive values everywhere, have continuous partial derivatives c.. 3 ~J which are non-negative everywhere and c.'s are positively homogeneous of degree one. i Further suppose the relative shares matrix A = (aij) where ~ij = cij wj/ci has the upper left-hand corner principal minors whose absolute values are bounded from below by some number

absolute value of det

all .

.--

C~kl

"'"

alk 1 h @,(k = 1,2,...,n)

.

&kk

Then for any given set of positive good prices Pi > 0, there exists a unique set of factor prices wj > 0 satisfying ci(wl,w2,...,w n) = Pi for i = 1,2,...,n. X.

Proof :

Set f. = log c. and w. = e J i 1 j functions defined on R n

In other words we are defining n new xI

fi(xl,x2,...,Xn)

= log ci(e

x2 , e

Xn) ,...,e

.

It is clear that the system of equations ci(wl,...,w n) = Pi has a unique solution for Pi > 0 if and only if the system of equations fi(xl,x2,...,Xn) a i = log Pi has a unique solution in R n.

the conditions imposed on theorem 6 in chapter VI. 3f i ~x. 3

= a i where

We will now verify that fi's satisfy Note that

x. xI x2 x xI xn - e J cij(e ,e ,...,e n)/ci(e ,...,e ) X.

so that

f.. = ~.. if evaluated for w. = e 3 in R n. Hence ~ given in the theorem m3 13 3 will serve as a lower bound. Also observe by homOgeneity, 2 a.. = i for j ~J e. = Z c.. w. and further by hypothesis ~.. > 0. i 13 3 iJ -the principal minors of (fij) are bounded above.

Fence 0 < a.. < I and consequently -- 13 -Therefore the system fi = ai has

a unique solution for any given Pi > 0 from theorem 6 chapter VI.

This terminates

the proof of the complete invertibility theorem.

Remark :

Mas-Colell has substantially generalized the complete invertibility

theorem.

He shows that the restrictions on the principal minors are irrelevant;

that matters is that that the determinant of laijln~m be uniformly bounded away from zero.

For details see Mas-Colell

[36].

all

79

We will now look at another example from input-output model recently considered by Chander.

Chander's model is similar to the model considered earlier

by I.W. Sandberg [1973, Econometrica pp. 1167-1182].

We assume that the economy is

divided into n industrial sectors each of which produces a single kind of good that is traded, consumed and invested in the economy.

The interrelations in such an

economy may be described by the system of equations

xi

-

n ~ a..(x ) = c. j=I ij j l

for

i = 1,2,...,n

where x i denotes the quantity of good i produced in the ith sector, aij(x j) represents the total amount of good i used as input for producing x. units of good j. Hence, J for each i the total amount of good i available for final consumption, export and investment is x i -

n ~ j i

demand vector.

a..(xj). ij

The vector (Cl,C2,...,c n) is called the final

Here the problem is to find the vector x given the demand vector c.

Also one would be interested in computing x. (i)

We will make two assumptions:

for each i and j, aij(.) is continuously differentiable on [O,~), aij(O) = 0

and a!.(a) > 0 ij --

for all ~ > 0

where prime denotes the derivative.

(ii) There

n exists Pi > 0, i = 1,2,...,n and v > 0 such that Pi~ ~ Pi a!.(~) + v for all i=l iJ ~ [0,~) and j=l,2 ..... n. We now have the following:

Theorem :

(for input-output model)

:

Under assumptions (i) and (ii) the mapping

A : R n ÷ R n where A(x) = ( nZ aij(xj), i = 1,2,...,n) j=l

is a contraction mapping.

Furthermore for every c s R~, there exists a unique x s Rn+ such that x-A(x) = c and for any x(0) s R+, n the sequence (x(t)) 0 defined by the Jacobi iterates x(t+l)

=

A(x(t)) + c, t > 0.

converges to x. For a proof of this result one can refer to Chander [7].

Alternatively one can

use Nikaido's result on uniform diagonal property in the strong sense.

Here we are

making stronger assumption and this enables us to compute the vector x to any degree of accuracy that we want.

On the distribution of a function of several random variables : statistics we come across problems of the following type.

Usually in

If XI,X2,...,Xn are

mutually independent random variables then we would like to know the distribution of a function u(XI,X2,...,X n) of the random variables XI,X2,...,Xn or we may even be interested in the joint distribution of F(XI,X2,...,Xn) = (uI(XI,X2,--.,Xn), u2(XI,X 2 .... ,Xn)...Un(Xl,...,Xn)).

To fix ideas and show how univalence results are

useful in such a situation we will work out a simple example.

Let XI,X2,X 3 be

80 mutually independent random variables each having a gamma distribution with B = i. Then the joint distribution of XI,X2,X 3 is given by 3 : i=iH ~

~(Xl'X2'X3)

=

I

~i-i -x. xi e l, 0 _< x.1 -< ~

0

and ~.I > 0

otherwise.

Let Ul(Xl,X2,X 3)

=

Xl/(Xl+X2+X 3)

u2(xl,x2,x 3)

=

xJ(xl+x2+x 3)

u3(xl,x2,x 3)

=

xI+ x2 + x3

and

Problem is to find the joint distribution of (UI,U2,U 3) (we are using capital letters to denote random variables).

In order to do that, first we have to check

whether the map F = (Ul,U2,U3) is one-one in the effective domain of (Xl,X2,X3). In this example we have to verify whether F is one-one in the positive orthant of R 3.

We now write down the Jacobian matrix

J

=

u3-x I 2 u3

xI 2 u3

xI 2 u3

x2 2 u3

u 3-x 2 2 u3

x2 2 u3

i

i

i

One can easily check that the Jaeobian matrix is a P matrix for every x in the positive orthant of R 3.

In fact in this example one can easily write down the

inverse map. In fact F-l(ul,u2,u 3) = (UlU3, u2u3, u3(l-Ul-U2)) where (Ul,U2,U 3) is a point in the range of F. Now one can write down the joint distribution of UI, U2, U3 by means of the following formula. Y(Ul,U2,U 3) whenever u I > O, u 2 > 0, Ul+U 2 < i

and

I %(UlU 3, u2u 3, u3 (l_Ul_U2)) = ~-~ 0 < u3 < ~

and Y = 0 otherwise.

Explicit

expression for ~(Ul,U2,U 3) is given by a2-1 u2

al+~2+~3-I ~ - I u3 uI

~3-i (l-Ul-U 2)

-u e

3/(r(al)r(~2)r(a3))

81

In particular joint distribution h of (UI,U2) is given by

h(Ul,U 2) =

r (~i+~2+~3) ~i-I ~2-i ~3-i u2 r(~l)F(o2)r(~3) Ul (l-Ul-U 2)

when 0 < Ul,U 2 and u I + u 2 < i

an@

h = 0 otherwise.

Random variables UI,U2 that

have a joint distribution of this form are said to have a Dirichlet distribution with parameters ~i' ~2' ~3"

One can easily check that the marginal of U I or U2

is a beta distribution.

On the existence and uniqueness of solutions in Nonlinear Complimentarity theory: Let F : R n ÷ R n be a continuous mapping where R+ is the non-negative orthant of R + n n" We are interested in finding a non-negative vector z s R+n such that F(z) E R+n and z.F(z) = 0. F(z).

~re

z.F(z) stands for the inner product between the two vectors z and

We call this problem the complimentarity problem associated with F.

The

CP (= Complimentarity Problem) associated with F is said to be globally uniquely solvable if for any vector q s R n the solution.

CP associated with F(.) + q has a unique

We will follow the approach of Megiddo and Kojima to give a solution

to the problem under consideration. G is a map from R n to R n.

Define G(x) = F(x+) + x- where ii i

x +i

Let G be an extension of F defined as follows.

if

xi > 0

if

xi < 0

=

x and

Let F:R~ ÷ R n be a continuous mapping.

x

i

if

xi < 0

if

xi > 0

=

0

Then the CP associated with F is globally

uniquely solvable if and only if for every q c R n, there is a unique x = x(q) s R+n such that F(x) + q s Rn+ and fi(x) + qi = 0 for those i where x.1 > 0. This then is equivalent to the existence of a unique z = z(q) s R n such that F(z+) + q = - z- or or G(z) = - q. Theorem i :

Now we have the following.

Let F:R~ ÷ R n be a coni~inuous madding.

Then the CP associated with F

is globally uniauely solvable if and only if extension G of F is a homeomorphism of R n onto itself.

Remark :

Suppose F:R n ÷ R n is a co~atinuous mapping such that the CP associated with

F is globally uniquely solvable.

T~en the solution of the CP associated with F(.)+ q

is a continuous function of the vector q for every q E R n.

This is a consequence

of the fact that, if z = G-l(q ) then x = z+ is a solution for the CP associated with F(.) + q.

Now we shall prove the following theorem giving sufficient conditions

on the map F such that the CP associated with F+ q will have at most one solution for every q s R n.

82

Theorem 2 : (a)

If F:R~ + R n is a continuously differentiable function such that

the Jaeobian associated with F is a P-matrix for every x e R+n then the CP associated with F+ q has at most one solution for every q ~ R n. (b)

If F:R~ ÷ R n

is a differentiable mapping such that all the principal miners

of the Jacobian matrix of F are bounded between 6 and 6 -1 for some 0 < 6 < I, then the CP associated with F is globally uniquely solvable.

Proof (a) :

From the fundamental theorem of Gale and Nikaido it follows that F

is a P-function in R+n.

That is for every x # y we have max(xi-Y i) (fi(x)-fi(y)) > O.

Consequently there can be at most one solution to the complimentarity problem, x~ This can be seen as follows.

O, F(x) _> 0

and

x.F(x)

= 0

Suppose x # y with

x _> O, F(x) _~ 0

and

x.F(x)

=

0

y_~ o, F(y) 3 o

and

y.F(y)

:

0 .

Consider for any i, (xi-Yi) (fi (x)-fi (y)) =

xifi(x) + Yifi(y) - xifi(y) - Yifi(x)

= - xif i(y) - Yifi(x) < 0 This contradicts the fact that F is a P-function. also a P-function for any q s R n. most one solution. Proof (b) :

Since F is a P-function F+ q is

Consequently CP associated with F+ q also has at

This terminates the proof of (a) of theorem 2.

We will now prove that the extension map G of F is a homeomorphism.

Then from theorem i it will follow that F is globally uniquely solvable. G(z) = F(z+) + z- for any z ~ R n.

Recall

Observe that the mappings x ÷ x+ and x + x- are

differentiable in points x such that x. ~ 0 (i = 1,2,...,n), it follows that G is i differentiable in the interior of every orthant of R n. However we can approximate G by differentiable functions G ( x ) functions.

Specially, for any ~ > 0 I t(~,~) =

0 (~+ ~)2/4~

and then apply Gale-Nikaido's theorem to these and any real number ~ let if

~<_-~

if

- ~ < ~ <

if

6<

83

Denote by ui(x,~) = t(xi,~) , i = 1,2,...,n, and u = (Ul,U2,...,u n) and define G ( x ) = F(u(x,~))-u(-x,~).

One can check that the principal minors of the Jacobian

matrices of the functions G~(x) lie between 6 and 6 -1 (6 > 0).

First we will show

that G is onto.

is onto.

Note that from theorem 6 of chapter VI each G

given any q s R n we can find for each ~, an {x~} is bounded. G is one-one.

x ~ such that G ( x ~) = q.

Hence it follows that for some x,G(x) = q.

From Gale-Nikaido~s theorem each G

one-one in the interior of every orthant in R n. globally one-one in R n.

Now we will show that

is one-one.

Consequently G is

We will now establish that G is

To achieve this end, for any x s R n, denote !(x)={i:xi=0}.

Let x,y ~ R n be such that G(x) = G(y). that x = y.

That is

If ~ s (O,i] ,

We shall prove by induction on ll(x) I+ ll(y) I

Suppose first that ll(x) I +

cardinality of the set l(x)).

II(y)] = 0 (Here ll(x) I stands for the

This means x and y are interior points of some orthants;

that is, x s int Q S , y s int QTwhere S and T are subsets of N = (l,2,...,n} with QS = {x:x s E n, x. > 0 F i s S and x. < 0 for all i s N N S ] e t c . Thus we can find i -i -u s Int QS and v s int QT such that G (u) = G(x) and G ( v ) = G(y) for some ~ > 0. Since G

is univalent in R n, u = v and consequently S = T.

Hence it follows that

x = y since G is univalent in the interior of every orthant. Assume, by induction, that G(x) = G(y) and II(x) l + Suppose x ~ y with G(x) = G(y) and ll(x) I + suppose II(x) I ~ II(y) l.

ll(y) I ~ k

ll(y) I = k+ i.

imply

We shall show first G is locally univalent.

a neighbourhood of x such that

y ~ N x and l(u) C

x = y.

Without loss of generality, Let N x denote

l(x) for every u s N x.

We claim

that G is univalent in N . Let u and v be two distinct points of N . We distinguish x x First l(u) = l(x). In this case there exists an orthant QS such that

two cases.

x,u,v s QS and since G is univalent in every orthant, G(u) # G(v). l(u)

~

l(x).

In this case ll(u) l +

hypothesis implies G(u) # G(v).

II(v) l ~ 211(x) I ~ k + l ,

so that the induction

Fence G is univalent on N .

It follows from the invariance theorem of d o m a i n t h a t contains G(y).

Second,

G(Nx) is an open set which

Since G is continuous at y, there exists w ~ N

such that l(w) =

and G(w) s G(Nx).

Thus, we can find a u s N x such that G(w) = G(u).

[l(w) l +

k+ i, it follows from the induction hypothesis that w = u which is

ll(u) l <

a contradiction.

This shows that G is one-one throughout R n.

morphism of R n onto R n.

Since

F~nce G is a homeo-

Consequently from theorem l, F is globally uniquely solvable.

This completes the proof of part (b) of theorem 2.

Remark :

Proof of part (b) of theorem is due to Megiddo and Kojima.

As a corollary

one can prove the following result due to Samuelson, Thrall and Wesler.

Corollary :

Let F(z) = Az + b be an affine mapping from R n into itself.

Then the

CP associated with F is globally uniquely solvable if and only if all the principal minors of A are positive.

84

Proof

:

"Tf" part follows

from theorem 2.

We will only prove the "only if" part.

That is we will assume F is globally uniquely We will show that A does not reverse Suppose

there exists

solvable

and show that A is a P-matrix.

the sign of any nontrivial

vector

in R n.

an x s R n, x # 0 such that xi(Ax) i ~ 0 for i = 1,2,...,n.

us write Yi = (Ax)i - ith component

of the vector Ax.

Let

Write y~ = Yi if Yi > 0

and y~ = 0 if Yi ~ 0; yi = 0 if Yi ~ 0 and yi = - Yi if Yi ~ 0.

Similarly

we write

4 dxZ Then, +

-

+

Yi = Yi-Yi

'

4 i-x

xi =

-

+

yi,y i ~ 0

,

,x

> 0

-

and

yiy i = 0

for

i = 1,2,...,n

and

x.x.

for

i = 1,2,...,n

+i-

--

= 0 -I-

-

-

for each i = 1,2,...,n,

i = 1,2,...,n.

Hence y+.x + = y-.x- = 0 [Here dot refers to the inner product

that xiY i = 0 = xiY i for

Since y = Ax, we have y+Ax - + = y--Ax- = qo

the vectors].

x # 0, x+ # x-.

it follows

4-

Since xiY i < 0

between

Hence one can conclude =

y+

and

x+.y+

=

o

Ax- + qo

=

Y-

and

x-.y-

=

0

qo

Ax+

qo has a unique

has two solutions

which contradicts

solution.

(say).

Since

from,

~++qo

Ax+

.

1

our assumption

Fence A is a P-matrix.

that CP associated

This terminates

with

the proof

of the corollary. In theorem 2(b) we have shown that the CP-associated provided

all the principal

minors of the Jacobian

and 6 -1 for some 0 < ~ < I. may fail without

The following

this condition.

example

with F is globally

solvable

matrix of F are bounded between demonstrates

that this result

Let F = (f,g,h) be a map from R +3 to R 3 where

f(xl,x2,x 3)

=

xI x2 -x 2 x (e - e - e + l)e 3

g(xl,x2,x 3)

=

x2 xI -x I x + l)e 3 (e - e - e

h (Xl,X2,X 3 )

=

x3 .

In this case Jacobian matrix turns out to be xI e

x e 3 -x I

xI

J

=

(-e

x2 (-e

+e

0 Clearly all the principal

x~ )e 5

-x + e x3

e

x e

0

x 2)e 3

x x I -x 2 x2 e 3(e -e -e + i) x x 2 x I -x I e 3(e -e -e + I)

I

minors of J are all bounded below by one but some principal

85 Xl minors

(for example e

x3 e ) are not bounded above.

theorem 2(b) are not met.

Thus the conditions

imposed on

We will now show that the CP associated with F is not

globally uniquely solvable.

Let q = (e,e,l) and G = F(Xl,X2,X3)- q.

associated with G is not solvable:

Then the CP

Suppose the CP associated with G has a solution.

This means that that there exists a z e R+3 such that G(z) s R+3 with z.G(z) = 0. Let z = (Zl,Z2,Z3).

Since G(z) s __R~' it follows that z3-1 _> 0

or

z 3 _> i.

z.G(z) = 0, and z 3 >_ I third coordinate of G(z) = O or z3-1 = 0 or z 3 = i.

Since Now

we have, zI (e

z2

-z 2

- e

- e

+ l)e > e

and z2 (i÷ e since O(z) s R+3.

zI

-z I

- e

- e

)e >_e

From these inequalities we have, zI e

z2 -e

-z 2 -e

>

0

>

0

and

z2 e

zI -e

-z I -e

-z I Adding these two inequalities

we get, -(e

thus we arrive at a contradiction has a solution.

-z 2 + e

) > 0 which is impossible and

to the supposition that the CP associated with G

Thus in this example F is not globally uniquely solvable.

Note that the CP associated with F+ q for any q ~ R 3 is feasible, exists z _> 0 such that F(z) + q > 0. (oo,~,oo).

An application of Hadamard's

This is met here since

inverse function theorem to Algebra

result of Hadamard we will demonstrate with the structure of a commutative

[22] :

x(~y)

=

~xy

(ii)

x(y+ z) = xy + xz

(iii)

xy = 0

(iv)

xy : yx.

~

By using a

More precisely we

consider the possibility of defining an operation of "multiplication"

(i)

(x,y) ÷ xy

:

where ~ is a scalar.

x = 0

or

y = 0

We now have the following theorem.

Theorem 3 : satisfies

For

n > 3

(i) - (iv).

=

that Euclidean n-space cannot be endowed

division algebra when n > 3.

of R n which obeys the following axioms

that is, there

limit F(l,l,a) a + oo

there is no operation of multiplication

on R n which

86

Remark

:

It is clear that when n = i or 2, we can define a multiplication

satisfying

(i) through

(iv) and also associative

possible to do so in R n for n > 3.

law.

Theorem 3 says that it is not

We will use a special case of a theorem due to

Fadamard to prove theorem 3 - this ~roof is due to Gordon

Hadamard's theorem n £ 3.

:

operation

Let F:R n - {0} ÷ R n - {0}

Then F is a diffeomorphism

be a

of R n - {0} onto

proper and the Jacobian of F never vanishes.

[21, 22].

C (I) differentiable

map with

R n - {0} if and only if F is

[For a proof of Hadamard's

theorem

see [21]]. The following two examples when n = i or n = 2.

show that Fadamard's theorem may fail to hold good

When n = I, f(x) = x 2 is a proper map with derivative non-zero

for x ~ 0 and clearly it is not one-one. 2 2 f(xl,x 2) = x I - x 2 and g(xl,x2) = 2XlX 2. and its Jacobian is non-vanishing

When n = 2, define F(Xl,X2) = (f,g) where One can easily check that F is proper

in R~-{0}.

Also F is not one-one since F (i,i) =

F(-1,-l). Remark

:

Proof of Hadamard's

connected when

theorem depends on the fact that Rn-{0}

is simply

n > 3.

Proof of theorem 3 :

(Proof due to Gordon)

We will show that axioms

(iv) imply that the map x ÷ x 2 is a homeomorohism when n > 3 absurd since axiom

(i) requires that

(-x) 2 = x2).

Let F be the map from R n - {0} to R n - (0} where F(x) = x 2. well-defined map because of axiom check that F is continuous

(i) through

(which is obviously

(iii).

and proper.

This is a

Using the relevant axioms one can easily Using axiom

(iv) we can compute dFx(V)

(-- the differential F and x operating on v) and it is seen that dFx(V)

=

lim h÷0

{ ~i (F(x+hv)-F(x))}

= xv+

vx = 2xv.

Hence axiom (iii) implies that dF is non-singular for all x s Rn-(0} so that x F(x) = x 2 is a homeomorphism via Fadamard's theorem which of course leads to a contradiction.

Hence there cannot exist a multiplication

operation in R n for n >_ 3

satisfying the four axioms.

This terminates

the proof of theorem 3.

On the infinite divisibility

of multivariate

gamma distributions

:

in this section

we will consider the problem of deciding whether a certain multivariate distribution Paranjape.

is infinitely divisible.

We will give a sufficient

gamma

condition due to

In this situation weak N-matrices play an important role.

Let X = (XI,X2,...,x ~) be a multivariate

normal random vector with zero mean

vector and positive definite variance covariance matrix E.

The characteristic

87

2 2 2 xI x2 x function of ( 2 ' 2 .... ' P2 ) is given by hp (t) = h(tl,t2,...,tp) = II-T21-½ where T is a diagonal matrix with diagonal elements itl,...,itp, i = -~.

Call hp(t)

infinitely divisible if (hp(t)) a is a characteristic function for every a > 0.

Paul

Levy conjectured that h (t) is not infinitely divisible. However Vere-Jones proved P h2(t) is infinitely divisible, and Moran and Vere-Jones established hp(t) is infinitely divisible if ~ = (i-0)I + 0Epp with 0 > 0, I = identity matrix and E matrix with every entry equal to one. Also Griffiths obtained a necessary and PP sufficient conditions for h3(t) to be infinitely divisible. We will present a result due to Paranjape which in some sense unifies the known results. posed by Levy is still open.

The problem

We will now prove the following:

Infinite divisibility theorem : If for a set of positive constants Cl,C2,...,Cp, the matrix (diag (Cl,C2,...,Cp)Z -I - I) is a weak N-matrix (that is all the principal minors are non-positive) then hp(t) is infinitely divisible.

Remark : If f(t) is a characteristic function then for 0 < ~ < i, (l-~)/(l-~(f(t)) is infinitely divisible.

If Cl,C2,...,Cp are positive constants, then h2(t)p can be

written as hp2(t)

=

P P H gj(tj)Iz-ll H c./det (I+A) j=l i=l !

where gj(tj) = (l-icjtj)-I and A = (diag(cl,c2,...,Cp)Z-l-l)diag(gl(t l),g2(t2),...,

gp(tp)). Proof :

(of infinite divisibility theorem).

Observe that det (I+ A) = I + ~ tr.A i=l m

where tr.A denotes the sum of all principal minors of order i, I < i < p of the matrix A.

Write P = diag(cl,c2,...,Cp)~ -I - I.

A typical rth order principal minor

of A is equal to r j=l

~. (tk) P(kl,k 2 ..... k r) J J

where P(kl,k2,...,k r) is the (kl,k2,...,kr)-th principal minor of P. I = ~ r=l

~ (-P(kl,k2,...,kr)). (kl,...,k r)

of the theorem, it follows that

I > 0.

Set

Since each P(kl,k2,...,kr) ~ 0 by hypothesis

Therefore

88 p

r

det(I+ A) = i-~ [

~

(-P(k I . . . . . kr)/k ) fI gk.(tk.)]

r=l (kl,...,kr) Since -P(kl,...,kr)/k ~ 0

and

j=l

Z(-P(kl,k2,...,kr))/k

~

j

= I, it follows that the

expression in the right hand side which appears within the brackets in the expansion of det (I+ A), is a multivariate characteristic function. hE(t )=det (E-1)

Define

P

H c./det(l+ A). i=l z P

Since hp(O) = 1, h~(O) = 1

and consequently det(E -I)

ci/(1-~) = i.

Since the

i=l numerator is positive, I-X > 0

or X < i. Thus h*(t) has the representation of the P form (l-~)/(l-~f(t)) where f(t) is a characteristic function and therefore h*(t) P is infinitely divisible. Consequently h~(t) is infinitely divisible or hp(t) is infinitely divisible. divisibility of

This terminates the proof of the theorem on infinite

hp(t). i

Remark : The following result due to Rao gives an equivalent condition for a matrix A to be a weakly N-matrix.

A nonpositive symmetric matrix A(# 0) is merely

positive subdefinite if and only if it is a weakly N-matrix. Rao [62] x,x'Ax < 0

For a proof see

[Call a real symmetric matrix A positive subdefinite if for any vector implies Ax < 0

or

Ax > 0.

A positive submatrix which is not positive

semidefinite is called a merely positive subdefinite matrix].

Examples :

We will prove

h2(t) is always infinitely divisible by actually checking

the condition given in the theorem.

If ~-I = I~

one can easily check, diag (ci,c2)~-i-I = Lb0

1

bla I

i I cI = ~ , c 2 =

choose

Clearly

principal

/c minors are non-positive.

Hence h2(t) is infinitely divisible.

we will look at Griffith's result in three dimensions where

Z = (l-p)l + PE33 =

p

I

p

p

p

i

As another example

89

1-p 2 (1-p)2(1+ 2p)

T.- 1 =

p2 -p (l-p) 2 (1+ 2p)

p2 -p (1-p)2(l+ 2p)

1-p 2

p2 -p

p2 -p (l-p) 2 (1+ 2p)

(l-p) 2 (Z+2p)

(l-p) 2 (1+ 2p)

p2 -p

1-p 2

(l-p) 2 (l+ 2p)

(z-p) 2 (i+ 2p)

p2 -p (1-p)2 (1+ 2p)

Choose cI = c2 = c 3 = (l-p)2 (I+ 2p)/(l-p2). weak N-matrix provided p > 0. is of the form (l-p)l + PE33 h3(t) is infinitely divisible.

Therefore

Then (diag(cl,c2,e3)~-l-l) will be a

h3(t)

where p ~ O.

will be infinitely divisible provided

In general it is not known whether

CPAPTER

X

FIRTI~ER 6ENERALIZATIONS AND REMARKS

Abstract :

In this chapter we will first discuss a generalization of local inverse

function theorem due to Clarke and Pedamard's theorem due to Pourciau when F is a h

~

Lipschitzian function but not necessarily a C ~I) function. the notion of monotone functions.

We will then discuss

Next we will say something about PL functions.

Finally we will discuss global univalent results when the Jacobian is allowed to vanish.

Main aim of this chaoter is to indicate oossible generalizations in global

univalent results.

A generalization of the local inverse function theorem :

Here we will describe a

nice generalization of the local inverse function theorem due to Clarke. F : R n ÷ R n satisfy a Lipschitz condition in a neighbourhood of a point

Let Yo

in Rn"

That is for some constant d, for all x and y near Yo' we have

where II'II denotes as usual the Euclidean norm. the partial derivatives exist.

Let J denote the Jacobian whenever

We will metrize the vector space M of n x n matrices

with the norm,

11AII

= max laij I

A

=

where

Definition :

(aij), I < i < n

and

I < j < n.

The generalized Jaeobian of F at Yo denoted by ~F (yo), is the convex

hull of all matrices A of the form,

A

=

lim n÷~

J (yn)

where Yn converges to Yo and F is differentiable at

Remark I :

Yn

for each n.

It follows from Rademacher's theorem that F is almost everywhere

differentiable near Yo"

Furthermore J(y) is bounded near Yo since F satisfies a

Lipschitz condition in a neighbourhood of Yo"

Remark 2 :

The generalized Jacobian 8F(y o) is a nonempty compact convex subset,

in the vector space M of matrices.

91

Remark 3 :

If F is C (I), 8F(y o) reduces to J(yo ) .

Definition :

The generalized Jacobian 8F (yo) is said to be of maximal rank if every

A in ~F(Yo ) is of maximal rank. We are ready to state the theorem due to Clarke.

Ceneralized local inverse function theorem :

If ~F(y o) is of maximal rank, then

there exist onen sets U and V of Yo and F(Yo) respectively, and a Lipschitzian function C:V ÷ R n (i)

G(F(u))

=

~

for every

u s U

(ii)

F(G(v))

=

v

for every

v c V

For a proof of this result see Clarke [12].

In view of remark 3, it is clear that

when F is C (I), Q is necessarily C (I). Also observe that it is not sufficient to assume that J is of maximal rank whenever it exists, as the function Ixl, (n=l) demonstrates.

A simple example to which the above theorem applies, n = 2, is the

followinz: F(x,y) = (IxI+y, 2x+ly]) near 3F(O,O)

:

s

[ [2

(0,0).

Here one can check t h a t ,

1

t ] : - 1 < s < i, -i < t < I]

Immediately one can raise the following question in view of this result due to Clarke : Is it possible to formulate a ~lobal univalent result when F satisfies Lipschitz condition but F not necessarily a C (I) function ?. In fact Pourciau recently has shown that Hadamard's theorem holds good for locally Lioschitzian maps.

In order

to state the theorem due to Pourciau [58] we need to slightly extend the concept of generalized Jacobian that was given earlier. Let F:R n ÷ R n.

Call F locally LiDschitzian nrovided each point x has a

neighbourhood U where some positive number M satisfiesIIF(Zl)-F(z2)ll < for all Zl,Z 2 E U. measure.

MIIZl-Z211

As already remarked F' (x) exists a.e. with respect to Lebesgue

Moreover almost every x is a Lebes~ule point of the derived mapping F'.

By definition such x satisfy

l~mit s + 0

1 ~(B~(x)) /

IIF'(z)-F'(x)fld~(z) = 0

BE(x)

~mre £(Be(x)) stands for the Lebesgne measure of Be(x) centre x).

(= Ball of radius e with

Let L(F') stand for the set of all these Lebesgue points and let p E R n.

Then the generalized derivative 8F(p) of F at o is the non-empty, comoact convex subset

~ A6(F,p) of L(Rn,Rn), where A6(F, p) denotes the collection 6>O Co----n{F'(x) : x e B6(p) ~

L(F')} .

This extra condition allows us to ignore null sets in forming the generalized

92

derivative.

With this notion of generalized derivative, Pourciau has proved the

local inverse function theorem stated in chaoter I for Lipschitz functions.

For

details see [57].We are now ready to state the following:

Lipschitzian F~damard Theorem : into R n and let M > O.

Suppose F is a locally Lipschitzian map from R n

If ~F(D) is invertible and IIA-III < M for each p ~ R n

and each A in 8F(p), then F is a homeomorDhism from R n onto R n. For a proof of this result see [58].

Remark I :

It appears that some of the results proved in chapter IV, in particular

Plastock's results might be suitably formulated for locally Lipschitzian maps.

Remark 2 :

F~damard's theorem is generalized to Banach spaces by Caccioppoli [6]

and Plastock has proved some of his results for Banach soaces.

As such can we

assert that Lipschitzian Fadamard's theorem holds good for Banach soaces?

Remark 3 :

Is it possible to formulate Mas-Colell's or Oarcia-Zangwill's global

univalent result for locally Lioschitzian maos ?

Monotone functions and univalent mappings :

In the introduction we have mentioned

as one of the approaches for tacklingglobal univalent problem, is via monotone functions.

In this section we will present an example of that aoproach.

Definition : x,y E R n.

A mapping F:Rn-~ R n

is monotone if (F (x)-F (y))'(x-y) > O

Call F strictly monotone if (F(x)-F(y))'(x-y) > 0

for every

whenever x # y.

Call

F uniformly monotone if there is a ~ > 0 so that (F(x)-F(y))'(x-y) > 6 (x-y)(x-y) for all

x,y E R n.

[Here as usual prime denotes transpose]. This concept is a natural non-linear generalization of positive definiteness. The following proposition is easy to prove.

Proposition i :

Let ~ be an open convex set in R n.

Let F:O ÷ R n

be a C (I) map.

Then (i)

F is monotone if and only if J is

positive semi-definite for all

(ii)

If J is quasi-positive definite for all x E ~, then F is strictly monotone on ~.

(iii) F is uniformly monotone on ~Q if and only if there is a ~ > 0 so that h'Jh > ~h'h for all x ~ ~ and h ~ R n.

x s ~.

93

Proof :

We will prove (i) and (iii) simultaneously.

monotone.

Then for any x E ~

and

h'J(x) h =

_>

Suppose F is uniformly

h ~ R n, we have h'lim t+O lim t+O

[F(x+ th)-F(x)]

t -2 ~IIthll 2 = ~h'h

If F is monotone then $ = 0 and J(x) is positive semi-definite. h'J(x)h> ~h'h.

Conversely suppose

Then from the mean-value theorem we have, (x-y)'F(x)-F(y))

=

i S (x-y)'J(y+ t(x-y))(x-y)dt 0 ~(x-y)'(x-y)

so that F is monotone or uniformly monotone depending on whether ~ = 0 or

6 ~ O.

Finally~ if J(x) is quasi-positive definite for all x s ~, and x ~ y, then the above integrand is positive for all t E [C,l] and hence F is strictly monotone.

This

terminates the proof of Proposition i. Proof of this proposition is taken from Ortega and Rheinboldt [48].

Remark :

Observe that (ii) of Proposition I is given in Chapter III.

However the

conditions do not suffice for the existence of solutions as the examole F (x) = ex in one-dimension shows.

We may guarantee existence, however, by strengthening the

monotonicity assumption.

Theorem i :

If F:R n ÷ R n

is a C (I) map and uniformly monotone on R n then F is

one-one and onto R n and consequently F is a homeomorohism. From the given hypothesis one can easily verify thatlIJ(x)-iII ~ x s R n.

[Here the norm of the matrix is taken as the Z2 norm].

theorem will imply that F is one-one and onto.

i ~

for all

Now F~damard's

For details of the proof see

142 or 48].

Remark i :

If F is continuous and montone on the open, convex set ~, then for any

b ~ R n, the solution set S = ~x ~ ~ : F(x) = b) is convex if it is not empty.

Remark 2 :

Suppose that F,G : R n ÷ R n

F is uniformly monotone. R n onto R n.

are both C (I) maps and montone and that

Then the map H defined by : H = F+ G is a homeomorphism of

In particular if A is a quasi-positive definite matrix, H = A + G is a

homeomorphism.

(See problem E

On PL-funetions

:

5.4-5 in [48]).

Because of the growing importance (as well as simplicity) of

94

piecewiselinear Kojima-Saigal

functions, in this section we will prove a result due to

[32].

In order to do that we need some preliminaries.

Let S be a closed convex polyhedral subset of R n, and let ~ be a class of closed

convex polyhedral subsets of S which partition S.

We will assume E contains

only finitely many members.

Definition : (a)

Call (S,Z) a subdivided polyhedron of dimension n if :

elements of Z are n-dimensional convex and polyhedral and are called pieces.

For

(b)

any two

members of Z are either disjoint or meet on a common face.

(c)

the union of the pieces in Z is S.

our result S = R n.

Let (Rn,Z) be a subdivided polyhedron, and let F

:

Rn ÷ R n

be piecewise linear and continuously differentiable on this subdivision, that is, PC I with affine on each piece of Z.

Since Z contains a finite number of pieces,

outside some compact region, points of R n lie in some unbounded piece in Z.

Let

these unbounded pieces be numbered al,~2,...,Uk and let F(x) l~" = Aix - a i for i some n x n matrices Ai, and n vectors a i.

Theorem 2 :

Then we have the following:

Suppose that the Jacobian matrix of each piece of linearity of F has

a positive determinant.

Also, let there exist

a matrix B such that (l-t)B+ tA i

is non-singular for each t E [0,i] and i = 1,2,...,k.

Then F is a homeomorphism

of R n onto R n.

Remark :

Let z s Rn,

containing z. for each

such that det JF (z) is positive

Then, there exists s > 0

6 s (0,~).

Proof of theorem 2 :

(negative) for every

such that deg (F,B~(z), F(z)) ~ i (~ - i)

For a ~roof of this remark see [32].

Let y s R n

be arbitrary.

Now consider the homotopy

H(x,t) : (l-t)Bx + t(F(x)-y), t s [0,i]. We will first prove that H-I(0) has no unbounded component.

Assume the contrary

this means that for some ai we can find a sequence (xm,tm) s ~--i(0), m such that x m s ~i and llxmll ÷ ~. t -~ t*, t* E [0,i] and x* # 0.

1,2,...

Also on some subsequence xm/llxmll ÷ x*, ~ence we have

m

(l-t*)Bx* + t* A.x* = 0 ]

95

which is a contradiction to our hypothesis that (l-t)B + tA. is non-singular for 1

each t s [0,i].

Since H-I(0) is bounded for each y, and det B > 0, from the

homotopy invariance theorem the degree of F(x)-y is + i for all y. remark it follows that F is one-one and onto. Now we will give two examples. are satisfied.

From the above

This terminates the proof of theorem 2.

In the first example conditions of theorem 2

In the second example one of the conditions of theorem 2 will be

violated but the map will be a homeomorphism.

Example I :

Consider F:R 2 + R 2

where

=f(x-y,y)

if

x~ 0

if

x < 0

F (x,y) ~(2x-y,y) In this example A I = [0

i ]' A2 = [0

i ] and let B = [

check that (l-t)B + tA. is n o n - s i n ~ l a r is a homeomo~hism

].

One can easily

for each t c [0,i] s~_d i = 1,2.

~nce

F

of mR2 onto R 2.

We need the following lemma for examole 2.

Lemma :

-i Let A I = [-2

0 1 ]'

-i A2 = [ 2

0 i ]'

i A3 = [0

there does not exist any matrix B such that tB + t s [O,i]

Proof :

-2 -i ]

and

i A 4 = [0

2 -i ]"

Then

(l-t)A i is non-singular for every

and i = 1,2,3,4.

One can possibly give a direct proof but our proof depends on theorem 2.

Consider the following diagram, which represents a piecewise linear function F in R 2.

~+ For example

F(x,y)

=

(-x,-2x+y)

when

y>x>0

=

(x+ 2y, -y)

when

0~-y~x.

96

Clearly the given AI, A2, A3, A 4

are the Jacobians of F in different regions.

Also note that F(½, I) = F(½,-I) = (-½, 0) and consequently the given homeomorphism.

F

is not a

Hence from theorem 2 we can conclude that there cannot exist a

matrix B with tB + (l-t)A i nonsingular for every t s [0,I] and i = 1,2,3,4.

This

terminates the proof of the lemma. We will use this lemma in our next example to show that the condition given in theorem 2 is not a necessary condition for an F to be a homeomorphism.

Example 2 :

See the diagram below.

In this example R 2 is divided into 6 regions

and in each region we have given the Jacobian of the function F.

Clearly F is a

homeomorphism but this F does not satisfy the condition imposed in theorem 2.

In

other words this example demonstrates that the condition "tB+ (l~t)A i is non-singular for every t s [0,i] and i = 1,2,...,k for some matrix B" is not a necessary condition for homeomorphism.

I [0

-2 _l ]

-i 0 [21 ] Because of the growing i~oortance of PL-functions it would be nice to give univalence conditions that are both necessary and sufficient.

Recently Schramm in

fact has given in [70] necessary and sufficient conditions for PL-functions to be a homeomorphism.

Among other results he proves that when the determinants

associated with a PL-funetion F:Rn ÷ R n have the same sign, F is an open mapping. Also he is able to give bounds for the number of solutions of certain PL-equations. For details see [69].

¢~le-Nikaido's results have also been extended by Kojima-

Saigal for piecewise continuously differentiable, pcl-fumctions [32].

On a global univalent result when the Jacobian vanishes :

In this section we will

give a global tunivalent result due to Chua and Lam [9] when the Jacobian is allowed

97

to vanish on a set of isolated points. thereby extending

Gale-Nikaido's

Kojima-Saigal

fundamental

have shown a similar result

theorem.

For the proof of Chua and

Lam we need the following results on local homeomorphism where the Jacobian may be allowed to vanish. (i)

Let F be C I map from an open set U e - R n + R n and n > 3.

t s U\{t

}

where t

O

t .

is an isolated point.

Then F is a local homeomorphism

(ii)

[ii].

Let F be a C (n) map from R n to R n where n ~ 3.

zeros of the Jacobian.

If Rn_ I is zero-dimensional

This result is due to Church

Remarks

:

Let Rn_ I denote the set of then F is a local homeomorphism.

[i0].

It is interesting to note that both the results are valid when n > 3.

For n = 2, the analytic function f(z) = z 2 is an obvious counter example. of

at

O

This result is due to Church and Hemmingsen

O

Suppose J(t) # 0,

(i) and (ii) depend on the fact that the set of branch points

points at which F fails to be a local homeomorphism)

Proofs

(= collection of

forms a oerfect set when n ~ 3.

We are ready to state the following

Theorem 3 :

(Chua and Lam)

and J(x) is singular)

Let F:R n ÷ R n

cient for F to be a homeomorphism

Proof

(i)

det J(x) > 0

(2)

F is norm-coercive.

:

be a C I mao where n # 2.

and I c = (x:x ~ I).

are suffi-

of R n onto Rn:

for all x s I c and I is at most a set of isolated points.

First we will prove theorem 3 when n = i.

Suppose F:R I + R I

This means we can find x I # x 2 with F(x I) = F(x2). exists an interval

Let l = { x : x s R n

Then the following conditions

Condition

is not one-one

(i) implies that there

(c,d) with x I ~ c < d ~ x 2 such that F'(x) > 0 on (c,d).

Since

F is C I, we have x2

F(x 2)-F(x l)

d

= / F'(x)dx~ x1

which is a contradiction. Since F is norm coercive,

c

Fence F is one-one and consequently F is a homeomorphism. F is onto R

and this terminates

We will now consider the case when n > 3. an open set N u about u in R n with N u ¢~ I = (u}.

quoted above that F is a local homeomorphism

the proof when n = i.

Let u be a point in I.

There exists

Since det J(x) ~ 0 on N u except

at the isolated point u where the Jacobian vanishes,

non-singular

/ F'(x)dx>O

it follows from the results

on N u for each u s I.

Since J(x) is

for every x s I c, it follows that F is a local homeomorphism

for all n ~ 3.

Hence we can conclude that F is a homeomorohism

norm-coercive and a local homeomorphism).

on R n

of R n (as F is

98

Proof of the theorem will be complete if we show that F is onto R n • Suppose F(R n) is a proper subset of R n.

Observe F(R n) is open as R n is open.

be a boundary point of F (Rn) and let ~ •

.

n

Since F is a finlte covering mapping on R , F number of components.

N~ are open and connected. onto M b.

-I

(Mb) has a finite and a non-zero

Let N b be a component of

Let N~ = F(Rn)g] F-I(Mb).

Let b s R n

an open connected neighbourhood of F (b).

F-I(Mb ) that contains the point b.

Since F is continuous F -I is open.

Fence, both N b and

Also observe that F maps both N b and N~ topologically

Clearly Nbf~ F(Rn)# ~

as the point b belongs to both N b and F(Rn).

follows that N b C ~ N ~ ~ ~, for otherwise there will be at least one point

xI

It in

N b (] F(R n) and a point x 2 in N~ such that F(x I) = F(x 2) s M b and F restricted to F(R n) will not be one-one, which is a contradiction. we have N b = N~. R n.

As both N b and N~ are connected,

Hence it follows that F(R n) cannot be an open proper subset of

That is F(Rn), is closed in R n.

and it is non-empty.

We have F(R n) which is both open and closed

Therefore we can conclude that F(R n) = R n.

This completes

the proof of theorem 3.

Remark i :

We have used theorem 3 in chaoter V to prove theorem 2".

We have given

an example there which will serve as a counter example to theorem 3 in R 2.

For

extensions of Gale-Nikaido's univalent results and other results along these lines we refer the readers to [32], [9].

Another result which is closely related to

theorem 3 is the following theorem due to Cronin and

Theorem 4 :

McAuley [14].

Let A and B be the interiors of two unit balls in R n and F a continuously

differentiable light and open map from A to B. of A and B respectively].

Suppose F(A\A)

[Here A and B stand for the closure

= B\B

and F I(A\A) is a homeomorphism.

Let J(x) stand for the Jacobian evaluated at x s A as usual, and let =

{x s A

:

J(x) = O}

=

(x s A

:

J(x) > O)

J

=

{x s A

:

J(x) < O}

J1

=

~+~

J O

J

+

~ g~ A

If dim Jl < (n-l), then F is a homeomorphism.

Proof :

See pp. 406-408

Remark i :

in Cronin and

McAuley [14].

This result enables us to study subsets of the set of points Jo where the

Jacobian is zero instead of dealing directly with the branching or singular set which is a subset of J . O

99

Remark 2 :

The following example shows that there are many homeomorphie onto

mappings with their Jacobian vanishing on an (n-l) dimensional set.

The following

simple example is a case in point. Let F:R 3 ÷ R 3

This function

F

be defined by Yl

=

fl (x)

=

x~

Y2

=

f2 (x)

=

x~

Y3

=

f3(x) = x I + x 2 + x 3.

has a global inverse on R3; namely :

:

:

Hence, F : R 3 + R 3

is a homeomorphic onto mapping.

2 2 But det J(x) = 9x I x 2 vanishes

on two 2-dimensional hyperplanes : one defined by x I = 0 and the other defined by x 2 = O.

This shows that the condition "dim Jl < n-l" imposed in theorem 4 is not

necessary.

The set J

Remark 3 :

contains the singular set or branch set - see

Ma~uley [37].

O

But it x s J

the map F can be locally one-to-one at x as the example of the map O

Yl

=

x~

Y2

=

x~

in a neighbourhood of (0,O) shows.

Remark 4 :

It may be possible to strengthen theorem 3 as well as theorem 4 with the

help of theorem 6 (as well as theorem 7) of chapter IV but we have not made any attempt to do so.

Injectivity of quasi-isometric mappings : Let ~ be an open subset of X.

Let X,Y stand for real Banach spaces.

Following John, a mapping F:~ ÷ Y is said to be

(m,M)-isometric if it is a local homeomorohism one-one)

for which M > D+F(x)

respectively,

(that is continuous, open and locally

and 0 < m < D-F(x) where D+F(x) and D-F(x) are

the upper and lower limits l lF(y)-F(x)II/IIy-xll

as

y + x.

Less

precisely, F is called quasi-isometric if it is (m,M) - isometric for some m, M. We have then the following theorem due to

Gevirtz.

100 Theorem 5 : Let ~ C - X

Let X and Y be Banach spaces [In particular we can take X = Y = R n]"

be an open ball and let F:~ ÷ Y be an (m,M) isometric mapping.

Then

F is one-one if any one of the following conditions is satisfied.

(a)

M/m < ~o x

=

[x+

where

U o is the unique real root of the equation

(25x2-8x)½]/2(3x2-x).

(b)

X is a Hilbertspace

and M/m < ~

.

(c)

X and Y are Hilbertspaces and M/m < (i + ~)½.

For a proof of this result see J.Cevirtz [Proc. Amer. Math. Soc. 85 (1982), 345-349]. For related results see also F.John [On quasi-isometric mappings I, II Communications to Pure and Appl. Math. 1968, 1969

21,22 pp. 77-110, 265-278].

We will close this chapter by mentioning an

old conjecture due to Jacobi.

[This conjecture was brought to my attention by Professor Garcia]. be a polynomial map.

That is

n variables Zl,Z2,...,z n.

F = (fl,f2,...,fn)

Let F : ~ n + (~n

where each fi is a polynomial in

If F has a polynomial inverse

G = (gl,g2,...,gn),

9f. 1 then the determinant o~ the Jacobian matrix ( ~7. ) is a non-zero constant. J follows from the chain rule. Zi

= gi(fl,f2,...,fn), 6ij

-

This

Since F o G is the identity, we have

so ~ gi(fl'f2'''''fn ) - L

~zj

~gi t=l-~Zt

~ft (fl'

"" "'fn )" 9zj

This shows that the product ~f (~gi "~z. (fl' "'" 'fn ) )" (~zl.) J J is the identity matrix.

Thus the Jacobian determinant of F is a non-vanishing

polynomial, hence a constant.

The Jacobian conjecture :

Now we are ready to state

Let F: ~ n ÷ ~ n

determinant is a non-zero constant.

be a polynomial map such that the Jacobian

Then F has a polynomial inverse.

For more

information on this conjecture see the following articles: D.Wright (1981), On the Jacobian conjecture, ILL.Jour. Math. 25, 423-439. H.Bass, E.H. Connell and D. Wright (1982), The Jacobian conjecture - reduction of degree and formal expansion of the inverse, Bull. Amer. Math. Soc. 7, 287-330.

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I~DEX Browder

29

Local univalent theorem

Caceioppoli

92

Local inverse function theorem

Ch~nder

79

Invariance of domain theorem

Chua-Lam

48,96

Inverse isotone

2 2,91 3 71

Clarke

90

Jacobian

Condition (L)

28

John

99

Complete with resoect to arc length

30

Kojima-Saigal

94

Complimentarity problem

81

Leontief type Jaeobians

68

Kestelman

1,17

Covering soace

3

Critical points

59

Leontief type matrix

Cronin

35

Light mao

37

Line liftin~ property

28 92

Differentiable

I

Diffeomorphism

30

LipschitzianFadamard theorem

Diriehlet distribution

81

Local homeomorohism

Fund~mental global univalence theorem

4,20

Cele-Nikaido-lnada

4,26

Garcia C~rcia-Zangwill Ceneralized Jacobian

6,68

3

Local k-extension

33

Local univalence

2

Mas-Colell

41,78

P-matrices

6

N-matrices

6

N-matrices first kind

6

N-matrices second kind

6

i00 41 90,91

Gevirtz

99

Globally asymptotically stable

59

Globally uniquely solvable

81

Cordon

86

PL functions Positive quasidefinite matrix Positive dominant diagonal matrix Hadamard

26,27

93 6,21 ll

30,86 Merely positive definite matrix

~artman

59

Inada

17

Infinite divisibility theorem

87

7

NVL-Matrix

52

Uniformly dominant diagonal matrix

69

106

Megiddo-Kojima McAuley

83 36,98

Samuelson

77

Samuelson-Thrall-Wesler

83

M-function

71

Scarf

4

Monotone function

92

Schramm

4

Strictly monotone function

92

(Positive) Stable

Uniformly monotone function

92

Stolper-Samuelson

Markus-Yamabe

59

Subdivided polyhedron of dimension n

More-Rheinboldt

15 (condition)

Transformation of class q

34,74

}{ultiplieation operation

85

Nikaido

54

I0 94 i

Univalent result when Jacobian vanishes

96

Univalent (results in R 2 and R 3)

49

Vidossich

65

Non-linear complimentarity theory 81 Norm-coercive mapping

41

Norm-coerciveness theorem

41

Noshira

21

Off-diagonally antitone

yon NeumarLn value

7

Weak P-matrix

48

71

Weakly positive quasi-definite matrix

36

Olech

59

x-simple domain

51

Open map

37

x-simple curve

51

Ortega-Rheinboldt

45

P-function

20

Paranjape

87

Parthasarathy

4

Plastock

4,29

Pourciau

92

Proper mapping

28

Quasi-isometric mappings

99

Radulescu

33

Raghavan

15

Rao

88

Rectangular region

17

Reverse the sign of a vector

i0

Rothe

35