Implicit Function Theorems for Multi-Valued Mappings

Working Paper

IMPLICIT FUNCTION THEORE3IS FOR MULTI-VALUED MAPPINGS

B.N.

Pshenichn y

September 1986 WP-86-45

International Institute for Applied Systems Analysis

A-2361 Laxenburg, Austria

NOT FOR QUOTATION WITHOUT THE PERMISSION OF THE AUTHOR

IKPLICIT FUNCl'ION THEOREXS MIR MULTI-VALUED MAPPINGS

B.N. Pshenichn y

September 1986 WP-86-45

Working Papers a r e interim r e p o r t s on work of t h e International Institute f o r Applied Systems Analysis and have received only limited review. Views o r opinions e x p r e s s e d h e r e i n d o not necessarily r e p r e s e n t those of t h e Institute o r of i t s National Member Organizations.

INTERNATIONAL INSTITUTE FOR APPLIED SYSTElMS ANALYSIS A-2361 Laxenburg, Austria

FOREWORD

This p a p e r d e a l s with t h e proof of implicit function t h e o r e m s f o r c e r t a i n c l a s s e s of set-valued functions. The techniques a p p l i e d h e r e are mainly b a s e d o n t h e duality t h e o r y of convex analysis. The p a p e r was written d u r i n g a visit of P r o - f e s s o r Pshenichny t o t h e System and Decision S c i e n c e s P r o g r a m of IIASA.

Alexander B. Kurzhanski Chairman System a n d Decision S c i e n c e s P r o g r a m

AUTHOR

P r o f e s s o r B.N. P s h e n i c h n y i s a n A s s o c i a t e Member of t h e Academy of S c i e n c e s of t h e U k r a i n e a n S S R a n d now i s t h e h e a d of t h e d e p a r t m e n t in V. Glushkov Insti- t u t e of C y b e r n e t i c s , Kiev, USSR.

CONTENTS

1 Implicit Function Theorems f o r Convex Mappings 2 Locally Smooth Maps

R e f e r e n c e s

-

vii

-

W L I C I T

FLTNCTION THEOREXS FOR MULTI-VALUED MAPPINGS

B.N.

Pshenichn y

V. Glushkov Institute of Cybernetics Kiev, USSR

Let us consider t h r e e Banach s p a c e s X , Y, Z and o p e r a t o r F : X x Y -+ Z. We are i n t e r e s t e d in t h e solution of t h e following equation:

Suppose t h a t t h e points (zo, yo) satisfy t h i s equation. Then implicit function theorems yield c e r t a i n sufficient conditions f o r solvability of t h e equation (1) with r e s p e c t t o y f o r all

z

f r o m t h e c e r t a i n neighborhood of zo.

Let u s somewhat reformulate t h e problem now. Define

Then implicit function theorems give conditions f o r a (z) #

4

in t h e neighborhood of zo. W e shall b e concerned mostly with t h i s reformulation.

Let us introduce some notations. Take Z

=

X

x

Y. Dual s p a c e s of continuous l i n e a r functionals will b e denoted by X* , Y* , Z*

.

P a i r of points from X and Y will b e defined (z, y ) while C c , z*

>

is r e s e r v e d f o r t h e value of t h e functional

z*

at

* * *

t h e point

z .

Taking

z =

(z , y ),

z =

(z , y ) w e obtain

Multivalued mapping t r a n s f o r m s e a c h point z E X into set a(z) G Y. The s e t a(z) may b e empty.

Some more notations:

aom a

= [z

:a(z) # @ {

Mapping a is called convex if gf a is a convex set and closed if gf a is closed. Some more notations will b e introduced in due c o u r s e . Terminology is close t o [I].

I. mPLIcrr

F'JNCTION THEOREMS FOR CONVEX MAPPINGS Let us s t a r t with some definitions. For any convex set M e x i s t

which is convex cone associated with t h e set M . For convex mapping a l e t us de- fine

Suppose t h a t z E gf a . Then

a , ( Z )

= I G : ( Z , G )

M a ( z ) j

* * *

a,

(v

)

=

^{) I} : ( - z * , v * ) E K ; ( Z ) { , where Ka ( z

*

) is cone dual t o Ka ( z ). Thus,

z

*

E a:(Y * ) if and only if

Let u s p r o v e two auxiliary lemmas:

* *

LEMMA 1 a, ( 0 )

= -

(dom a, )

.

PROOF The mapping a, i s a positively homogenious convex mapping and t h e r e - f o r e dom a, is convex cone and dual cone t o t h i s cone exists. According t o defini-

* *

tion of a, ( 0 ) i t contains t h o s e and only t h o s e elements z which satisfy t h e follow- ing condition:

which is equivalent t o

ti,

- z

*

> 2 O , Z E d o m a ,

The l a s t inequality means t h a t

*

*

- z E (dom a, ) The proof i s completed.

LEMMA 2

S u p p o s e t h a t

K i s

cone

in X , int K # 0

a n d

K*

= to{. T h e n

K

=

X . I f X

=

Rn

t h e n r e q u i r e m e n t

i n t k # d

c a n be d r o p p e d .

PROOF I t follows from t h e well-known t h e o r e m s of convex analysis t h a t K**

=

c l K , w h e r e symbol c l defines c l o s u r e of t h e s e t . The f a c t t h a t K*

=

101 im- p l i e s c l K

=

K

* * =

X. T h e r e f o r e cone X i s d e n s e e v e r y w h e r e in X a n d i t i s l e f t t o p r o v e t h a t i t coincides with X. L e t u s assume t h e o p p o s i t e , namely s u p p o s e t h a t ex- i s t s

z o

s u c h t h a t

z o

K . T a k e

z l

E i n t K a n d s e l e c t

2 = 2 z 0 - z l .

The c o n e K i s d e n s e e v e r y w h e r e in X, t h u s e x i s t s e q u e n c e

zk

^E K such t h a t

zk

⁴

^2.

Let u s t a k e

z l k = 2 z 0 - z k . -

I t i s c l e a r t h a t

z l k

⁴ X a n d f o r l a r g e k w e h a v e

z l k

E K . A c -

1 1

-

c o r d i n g t o definition

z 0 = zzk ^{+ 2 z k ,}

a n d d u e t o convexity of K w e o b t a i n

z o

^E K.

The proof i s completed b e c a u s e in t h e finite-dimensional case assumption i n t K # q5 i s fulfilled automatically, which c a n b e v e r i f i e d in t h e s t a n d a r d way. These two lemmas l e a d t o t h e following r e s u l t :

THEOREM 1 S u p p o s e t h a t a i s a c o n v e z m a p p i n g , z

= ( z O ,

y o ) E gf

a a n d t h e following c o n d i t i o n s a r e s a t i s f i e d :

1 int dom

a

# I0

T h e n f o r a n y element z e z i s t s u c h n u m b e r

^S

>

0 ,

t h a t a ( z o +

h z ) # q5

f o r

all h E [O, dl. X i s

of f i n i t e d i m e n s i o n t h e n t h e f i r s t r e q u i r e m e n t c a n be d r o p p e d a n d in a d d i t i o n , e z i s t s u c h a member

S

>

⁰

t h a t a ( z o + z)

^{# q5}

for

all

Z , 11; 11 r

^S.

B e f o r e s t a r t i n g t h e proof l e t u s make o n e comment. W e h a v e in t h e s t a t e m e n t of t h e t h e o r e m c e r t a i n point y o E

a ( z O )

a n d i t is n o t defined how t o s e l e c t it. I t might seem t h e r e f o r e t h a t t h e r e s u l t of t h e t h e o r e m d e p e n d s on a p p r o p r i a t e s e l e c - tion of t h i s point. L e t u s show t h a t t h i s is n o t t h e case. F i r s t l y w e s h a l l i n t r o d u c e t h e following notations [I]:

Notice t h a t Wa

( z

^{, y * )}

= + -

^if

^{a ( z ) = 6}

a n d t h e function Wa

( z

, y * ) i s convex with r e s p e c t t o

z .

Let us t a k e

a ( z , y ) ( ~ * ) * = 1

^{if y}

F a ( z ,

y * )

a,

W a ( z , Y * ) if Y E

a ( z ,

Y * )

If y

* =

0 t h e n

I

T h e r e f o r e

ZZ

Wa ( x , 0 )

= a

6 ( x (dom a )

= -

[con(dom a

-

^{z ) ]}

* .

Comparison of t h e s e s t a t e m e n t s l e a d s t o t h e conclusion t h a t

f o r any point y E a ( z ) . T h e r e f o r e w e c a n t a k e a n a r b i t r a r y point in t h e s t a t e m e n t of t h e t h e o r e m 1 . Let u s start t h e proof now. If assumption 1 i s s a t i s f i e d t h e n i t i s e a s y t o g e t t h a t doma,

=

X. L e t u s s e l e c t a r b i t r a r y

5 .

We h a v e

z

^E doma, a n d t h e r e f o r e e x i s t v e c t o r

y

- =

7 ( y - y o ) , 7

>

0 , ( z , y ) E gf a Taking now 6

=

7 - 1 w e o b t a i n

( z o

+ X Z ,

y o

+

A ; )

=

( ( 1 - X 7 ) z o

+

X 7 z , ( 1

-

X 7 ) y o

+

X 7 y ) E g f a

f o r X E [0, dl i.e. a ( z o

+ Xz)

^{# 0 .} In t h e c a s e of a f i n i t e dimension t h e f i r s t as- sumption i s n o t n e c e s s a r y . F u r t h e r m o r e , if X

= R n

t h e n i t i s possible t o find s u c h

- -

v e c t o r s

z i ,

ⁱ

=

1 , .

. .

, n

+

1 t h a t simplex S

=

I X l z l

+ .

_{+ An} Z , + I : X i ² 0,

z r 2

^{X i}

⁼

^lj c o n t a i n s 0 as i n n e r point. If w e r e d u c e t h e length of t h e v e c t o r s

Zi

a p p r o p r i a t e l y w e c a n o b t a i n t h a t all sets a ( z o

+ z i )

^are not empty. T h e r e f o r e a n y point from t h e c e r t a i n neighborhood of z e r o c a n b e r e p r e s e n t e d as follows:

This t o g e t h e r with p r o p e r t i e s of t h e convex maps implies

The proof i s completed.

This proof i s f a i r l y simple, b u t t h e r e s u l t i s q u i t e i n t e r e s t i n g . Let u s i l l u s t r a t e t h i s with some examples:

EXAMPLE 1 Take X

= R ~

Y

, = R ~

A

,

a n d B are m a t r i c e s r x n a n d r x m .

-

^{0. Then Ka (0)}

⁼

Define g f a

=

) ( z , y ) : A z - B y = 0 ] a n d s e l e c t z o =O, y o -

I(Z, y)

: f i - B ;

=

01,

* *

T h e r e f o r e a. ( y )

=

)A* u

=

B* u , u E Rr j . Condition a : ( 0 )

=

)O j means now t h a t B * U

=

0 which implies A * u

=

0. Thus, f o r solvability of t h e system of equa- tion Ax - B y

=

0 with r e s p e c t t o y f o r a l l x , i t is sufficient t h a t Kern B*

c

Kern A

* .

H e r e , a s usual, Kern C

=

! v : C v = O j

EXAMPLE 2 Take t h e same assumption as in t h e previous example and consid- e r g f a

=

) ( x , y ) : h - B y SO, y 2 0 1 . Nowwe have

where Rf, and R y a r e positive o r t h a n t s . This gives

Hence condition a: ( y * )

=

101 implies in this c a s e t h a t from inequalities B * u 5 0 , u 2 0 t h e equality A * u

=

0 follows. T h e r e f o r e f o r solvability of t h e system Ax - B y 5 0 , y r u f o r all z i t i s sufficient t h a t t h e following inclusion is satis- fied:

ern

^A*

>=

Iu 2 0 : B * u

s

01.

EXAMPLE 3 Suppose now t h a t X, Y, W a r e Banach s p a c e s , Z

=

X x Y , F - convex multivalued map from Z into W , i.e. F ( x , y ) C W , M-convex s u b s e t of W.

Define a ( x )

=

^{) y} : F ( x , y )

n

^M

+ $1.

I t is c l e a r t h a t points ( x , y ) E g f a if and only if exist such point w t h a t ( x , y , w ) E gf F , w E M . Let us s e l e c t t h e point s o such t h a t a ( x o )

+ 4,

y o E a ( x O ) i.e. ( s o , y o ) E gf a . Suppose t h a t w o i s such a point from M t h a t ( x o , y o , w o ) E g f F . Let us denote z o

=

( s o , y o ) and define

~ : ~ ( y * ) a s follows. According t o definition x * E az* ( y * ) if and only if

-

^<z

-

^{x O .}

0

z *

> +

^<y

-

^{y o , y}

* >

2 0 , ( x , y ) E gf a . Taking into account t h e description of gf a given above, w e a r r i v e a t t h e conclusion t h a t t h e l a s t inequality i s equivalent t o t h e following:

f o r a l l ( x , y , w ) E g f F , w E M , where 0: i s z e r o of t h e s p a c e

w * .

But t h e last inequality is equivalent t o t h e following:

Suppose now t h a t e x i s t t h e point ( x , y , y ) E gf F such t h a t (2) u E int M . This im- plies g f F

5

int (X X Y X M)

+

0. I t is consequence of t h e well-known r e s u l t s of convex analysis t h a t u n d e r t h i s assumption t h e cone in t h e r i g h t hand side of (2) is equal t o t h e sum of convex cones dual t o t h e i n t e r s e c t e d ones. Taking into account easily established r e l a t i o n s

* * * *

(con (X x Y x M

-

( z O , Y 0 , ~ 0 ) ) )

=

(Ox ^I O y , (con (M - w o ) ) ) w e obtain t h a t (2) is equivalent t o t h e following statement:

* *

o r , in o t h e r words, e x i s t functional w E (con (M

-

wo)) such t h a t t h e following inclusion holds:

And, finally using a definition of t h e conjugate mapping introduced b e f o r e w e ob-

* * *

tain t h a t x E G* ( y

*

⁾ only if f o r some w E (con (M

-

w 0)) t h e following inclu-

2 0

sion holds:

Hence a,

*

,(O)

=

10{ only if t h e following inclusion

* *

implies equality x

=

0, , i.e.

THEOREM 2 Suppose t h a t X, Y, W a r e Banach s p a c e s , M i s a convex set w h i c h belongs to W , F i s c o n v e z m u l t i v a l u e d m a p p i n g from X X Y to W a n d

s u p p o s e t h a t

y o E

a ( x

O )

a n d w

^E M

a r e p o i n t s s u c h t h a t ( x o , y w o )

E gf F .

T h e n t h e following c o n d i t i o n s a r e s u m c i e n t for e x i s t e n c e of n u m b e r

d

>

0

for a n y s u c h t h a t a ( x o + h z ) + 4 , h

E [O,

63:

1 int dom

a +

^@;

2 g f F

9

( X XY X i n t M )

+

^@;

* * * *

F ( z ~ w ~ ) (- (con (M

- w o > > > ^{= lo,,}

^0,

I .

I f X

i s

o f f i n i t e d i m e n s i o n t h e n i t

i s

n o t n e c e s s a r y to check t h e f i r s t c o n d i t i o n . The proof follows d i r e c t l y f r o m theorem

1

a n d t h e above a r g u m e n t .

EXAMPLE

4 Suppose t h a t

b

and

c

a r e convex multivalued mappings from X t o Y and

a ( x ) = b ( x ) r.~ c ( x ) .

THEOREM

3

S u p p o s e t h a t

y E

a ( x o ) a n d t h e following c o n d i t i o n s a r e s a - t i s f i e d :

1 int d o m a

+ ^4;

2 gf

b n

^{int gf}

c + 4;

3

for a n y y * e i t h e r one of t h e s e t s b:o(y

*),

cze0(-

^{* )} i s

e m p t y , o r e x i s t s

* * *

s u c h f u n c t i o n a l x * t h a t bz*o(y *

⁾

= Ix 1, c

( -

y

⁾

=

)

- x * 1. T h e n f o r a n y

Z E X

e x i s t

b

> 0 s u c h t h a t a ( x o + h z ) + 4 f o r h

^E

[O,

61.

I f X

i s

o f f i n - i t e d i m e n s i o n t h e n i t

i s

n o t n e c e s s a r y to check c o n d i t i o n

1.

The proof of t h e theorem follows from theorem 1, t h e f a c t t h a t gf

a =

gf

b n

gf

c

and d i r e c t calculation of

a z O ( O ) *

by using convex analysis tech- niques.

Let us show how t o use this theorem by t h e following example. Take

b ( x ) = Ix : A x

- B y 5

01, c ( x )

= M where M i s a fixed convex s e t in

R m , _A

and

B

a r e matrices of dimension

r

xn and

r

X

m

,

X = R n

, Y

= R m .

Suppose t h a t 0 E M and e x i s t points

x l ,

y such t h a t Axl

- B y

⁵

0 ,

y ^E int M . A straightforward a r - gument shows t h a t

Kc

(0)

= X

x con M

W e can obtain more from t h e following r e s u l t by applying theorem 3. In o r d e r t h a t t h e system of inequalities Ax - B y 4 0, y E M h a s a solution with r e s p e c t t o y f o r a l l z from some neighborhood of z e r o i t i s sufficient t h a t conditions

* *

imply equality A* u

=

0. In f a c t b o ( y )

=

@ if y

*

c a n not b e r e p r e s e n t e d in t h e

* *

form y = B * u , u 2 0 . In case if y = B * u , u 2 0 , and - B * U Z ( c o n ~ ) * . If, however,

* * *

t h e n f o r f *

=

B u w e h a v e c o ( - y )

=

101. T h e r e f o r e according t o t h e theorem

* *

3 t h e set b o ( y ) a l s o should contain only one point i.e. z e r o . I t follows from

* * *

r e p r e s e n t a t i o n of t h e set b o ( y ) t h a t A u

=

0.

If M

= ^Ry

i.e. M consists of nonnegative v e c t o r s t h e n t h i s r e s u l t coincides with one obtained from example 2.

2. LOCALLY SMOOTH MAF'S

Let K b e a convex cone in a Banach s p a c e X. I t is obvious t h a t LinK = K

- K

is t h e minimal l i n e a r manifold containing K. If M i s a convex set, t h e n LinM =con(M - 2 ) -con(M - 2 ) ,

where z i s a n a r b i t r a r y point of M . I t is not difficult t o show t h a t LinM does not depend upon z E M

.

W e s h a l l s a y t h a t t h e point x belongs t o t h e r e l a t i v e i n t e r i o r of M (denote i t r i M ) , if f o r some E

>

0

x

+

(EB)

n

^{Lin M}

^c

^{M ,}

wnere B i s t h e unit ball of t h e s p a c e X with t h e c e n t e r in t h e origin.

If X

= ^R",

t h e n i t i s well known [Z], [3], t h a t r i M

+ 4.

In g e n e r a l t h i s r e s u l t i s n o t t r u e .

L e t M b e a n a r b i t r a r y set of X.

DEFINITION Call t h e set K m a r q u e e for M

at

t h e p o i n t x € M, if'K i s t h e con- v e x cone a n d for e a c h € K e x i s t v a l u e s E

>

0, d

>

0 a n d c o n t i n u o u s l y d i f - ferentiable

in

t h e neighborhood of t h e o r i g i n f u n c t i o n

+:

^{X -4 X} e x i s t s u c h t h a t

1 +(O)

=

0, +'(O)

=

I ( u n i t operator);

2 z

+ +(y)

^E M for a l l

This definition i s b a s e d on i d e a s of V.G. Boltyanski, h e developed them f o r

n-

dimensional s p a c e . But t h e s e i d e a s c a n n o t b e generalized o n infinite dimensional s p a c e without c h a n g e s . F o r t h i s r e a s o n t h e i n t r o d u c e d definition d i f f e r s f r o m V.G.

Boltyanski's definition [3].

F u r t h e r i t i s convenient t o s u p p o s e without r e d u c i n g g e n e r a l i t y t h a t point z coincide with t h e origin.

THEOREM 4 Let M o i s

a

s e t , 0 ^€ M o . KO t h e m a r q u e e for M o

at

p o i n t x

=

0 a n d f u n c t i o n s f i ( z ) , i

=

1,

. . .

^, K satisf'g t h e c o n d i t i o n s :

1 f i ( 0 ) = 0 , i = ,

. . . .

^{K ;}

2

yi

a r e c o n t i n u o u s l y Freshet-dinerentiable

in

t h e neighborhood of zero a n d d e r i v a t i v e s f i ( 0 ) a r e l i n e a r l y i n d e p e n d e n t o n L o

=

Lin KO.

T h e n

i s t h e Local m a r q u e e a t t h e p o i n t x

=

0 for t h e set

PROOF Since f i ( 0 ) are linearly independent on l i n e a r manifold L o , t h e n vec- t o r s ej E L o e x i s t [ I ] , such t h a t

Let E K. Then

and smooth in t h e neighborhood of z e r o . T h e r e f o r e e x i s t s function q0 such t h a t q O ( o )

=

0 ,

q&o) =

I ,

f o r a l l

i7

^E [ c o n ( z

+

( E B )

n

^{L O ) ]}

n

^{( 6 B ) ,}

f o r E

>

0 , 6

>

0 . Let

k

where z i s t h e v e c t o r with components z j , j

=

1 ,

. . .

, k . Let u s consider t h e system of equations

g i ( y , z ) = 0 , i = I , .

. .

, k with r e s p e c t t o z

.

I t i s evident

Since gi are compositions of smooth functions, t h e n they a r e smooth functions. U s - ing r u l e s f o r differentiation of t h e complex function i t i s not difficult t o g e t

gi& ( 0 , 0 )

= P i

( O ) q i , ( 0 )

- P i

( 0 )

=

0 , Because 4'6 ( 0 )

=

I . F u r t h e r

Thus, t h e matrix with elements g i , , (0, 0) i s non-degenerate a n d a c c o r d i n g t o t h e

1

implicit f u n c t i o n s t h e o r e m [4] t h e solution z (y') of t h e system (7) e x i s t s , s u c h t h a t z (0)

=

0. F u r t h e r a c c o r d i n g t o t h e same t h e o r e m z

(c)

i s continuously d i f f e r e n t i - a b l e in t h e neighborhood of t h e origin of c o o r d i n a t e s a n d t a k i n g i n t o a c c o u n t (8)

Let now

Then Q(0)

=

0 , a n d

a c c o r d i n g t o (10). Denote

Let u s c h o o s e el

>

0, dl

>

0 sufficiently small s o t h a t t h e inclusion

y' ^E [con (5 + (clB)

fl

LO)]

n

( ~ I B ) (12)

e n s u r e t h a t

c

s a t i s f i e s t h e inclusion (5). I t is possible, b e c a u s e ( l o ) , b u t if

c

^sa-

t i s f i e s (5) t h e n

Denote

L1

= ) Z :

f i ( 0 ) Z = O f i

=

1 , .

. .

, K {

.

I t i s e a s y t o see t h a t LinK G L o r-?L1

.

Let now

y

^E [con

(5 +

(elB) r\ LinK)]

n

(dlB)

.

(13)

The set from t h e r i g h t p a r t of (13) i s t h e s u b s e t of set from t h e r i g h t p a r t of (12).

F o r t h i s r e a s o n (13) implies t h e inclusion Q(G) E Mo. F u r t h e r , s i n c e E L1, t h e n

E L 1 . T h e r e f o r e due to (7) and (6)

Thus (13) assumes

'k(y)

^E M and consequently K is t h e local marquee f o r M .

Since t h e p r o o f s of t h e following theorems r e p e a t t h e similar p a r t s of t h e theorem 4 proof t h e details will b e omited.

THEOREM 5 Let M 1 a n d M 2 be t w o s e t s , M

=

M 1

n

^{M 2 , 0 E} M a n d K1, K2 be r e s p e c t i v e l y m a r q u e e s f o r M 1 a n d M 2 .

V

t h e Linear m a n i f o l d s e x i s t s u c h t h a t : 1 L 1 c L i n K 1 , L 2 ~ L i n K 2 ;

2 L 1

+

L 2 = X ;

3 for a l l x l E L l , x2 E L 2

t h e n K

=

K1

n

^{K2 i s} t h e m a r q u e e for M a t t h e p o i n t 0.

PROOF In a c c o r d a n c e t o t h e condition 2 a r b i t r a r y x E X can b e r e p r e s e n t e d as follows

That r e p r e s e n t a t i o n i s unique, because if d i f f e r e n t r e p r e s e n t a t i o n e x i s t s

then

i.e. x l = x i , x 2 = x i .

Consequently o p e r a t o r s P 1 and P 2 are defined such t h a t

Due t o t h e condition 3

i t i s e a s y t o see t h a t o p e r a t o r s Pi are l i n e a r a n d consequently t h e y a r e l i n e a r a n d continuous o p e r a t o r s .

Let now 5 EK1

n

K 2 a n d E , d i , \ki c o r r e s p o n d t o

5

in t h e m a r q u e e Ki. Consid- e r t h e equation

with r e s p e c t t o z . I t i s e a s y t o see

In a c c o r d a n c e t o t h e implicit functions t h e o r e m t h e smooth function z ( c ) i s de- fined in t h e neighborhood of z e r o a n d z ( 0 )

=

0 , z ' ( 0 )

=

0 .

Suppose

Taking into a c c o u n t z ' ( 0 )

=

0 , i.e.

z

(-1

+ _II

_Y

_I1

^{0 , f o r}

^{I I ~ O --,o}

i t i s n o t difficult t o p r o v e t h a t \k i s t h e d e s i r e d function f o r

E .

This p r o o f i s b a s e d o n t h e f a c t t h a t f o r

c

c l o s e t o t h e d i r e c t i o n 'kl(ii + p l z

( 5 ) )

^E M 1

.

\k2(Y + P ~ Z

(GI)

^{E M 2 I}

a n d consequently d u e t o ( 1 5 )

\k(y)

^E M l

n

M 2 . In t h i s connection if

y

^E LinKi, i

=

1 , 2 , t h e n

5 ⁺

^{P l z}

(c)

^{E L l ,}

^-

^{P 2 z}

^{( 5 )}

^E L 2 a n d taking i n t o a c c o u n t t h e smallness of z

( y )

all conditions r e l a t e d with t h e c h o i c e of E , 6 c a n b e satisfied. Be- s i d e s

b e c a u s e \k; ( 3 )

=

I , z ' ( 0 )

=

0 . The proof i s completed.

REMARK Conditions 1-3 f o r X

=

Rn c a n b e r e p l a c e d by t h e condition

In a g e n e r a l c a s e t h a t condition is not sufficient b e c a u s e condition 2 means t h a t t h e s e t s M1 a n d M 2 h a v e sufficiently l a r g e dimensions. The condition 3 means t h a t l i n e a r manifolds i n t e r s e c t at some angle. Indeed, if X i s a H i l b e r t s p a c e with i n n e r p r o d u c t [ x , y ] t h e n condition 3 i s equivalent t o condition

Let u s c o n s i d e r now t h e t h e o r e m a b o u t implicit functions f o r nonconvex multivalued maps.

THEOREM 6 Let a be a m u l t i v a l u e d m a p , K be a m a r q u e e f o r ^g~' a L X X Y at t h e p o i n t z

=

( x o , y o ) ,

If t h e following conditions are s a t i s f i e d 1 i n t dom a,, # @ ;

2 e x i s t a l i n e a r r e s t r i c t e d o p e r a t o r P : X -+ Y s u c h t h a t ( z , P z ) E Lin K ; a,O

*

(0)

=

I O j , t h e n f o r a n y e x i s t s d

>

0 s u c h t h a t

a ( x o + X z ' ) #

4

f o r X E [ O , d l

.

If X and I' h a v e f i n i t e dimensions t h e n t h e f i r s t two conditions are satisfied and be- s i d e s value d

>

0 e x i s t s s u c h t h a t a ( x o

+

z ) # @ f o r a l l

5

^E (dB).

PROOF Without l o s s of t h e g e n e r a l i t y w e c o n s i d e r x

=

0 , y o

=

0.

Lemma 2 implies

i.e. f o r any ji EX t h e v e c t o r e x i s t s s u c h t h a t

(z, c)

^{E K.} Consequently f o r any

z

^E X t h e v e c t o r

c

EY e x i s t s s u c h t h a t ( z , c ) E Lin K.

Thus, l e t Z o b e a v e c t o r f r o m X and z o

=

( Z o , g o ) E K . S i n c e K i s t h e m a r q u e e t h e n E

>

0 , 6

>

0 and e x i s t s smooth function 9 ( 0 )

=

0 , 9'(O)

=

I,, w h e r e I, i s t h e unit o p e r a t o r in Z a n d 9 ( z ) E gf a f o r a l l

w h e r e B, i s t h e unit ball in Z. S i n c e

Z =

X x Y, t h e n

a n d condition 9 ' ( 0 )

=

I, c a n b e r e w r i t t e n as follows

i.e. 91;-(0, 0 ) =I,, 9 ' ( 0 , 0 ) = 0 , ' k k ( 0 , 0 ) = 0 , 'k&(O, 0 ) = I y . lfi

Consider t h e system of e q u a t i o n s

w h e r e A E

R',

z E X. Taking i n t o a c c o u n t p r e v i o u s r e l a t i o n s w e g e t

Thus d u e to t h e t h e o r e m a b o u t implicit f u n c t i o n s t h e system (17) h a s t h e solution r ( A ) a n d a l s o

L e t u s c o n s i d e r now t h e point

; ( A )

=

(Ago

+

r ( A ) ,

Ago +

Pr ( A ) )

.

Taking i n t o a c c o u n t r ' ( 0 )

=

0 , r (A)

=

o ( A ) f o r sufficiently s m a l l ^A

>

0 inclusion (16) i s s a t i s f i e d , b e c a u s e by definition of t h e o p e r a t o r P w e h a v e ;(A) E LinK.

Consequently

. k ( z ( A ) ) E gf a f o r small A. But

h e r e t h e condition (17) w a s used. Thus

i.e. a ( A s o ) # ^t$ f o r small A. Thus t h e f i r s t p a r t of t h e t h e o r e m i s p r o v e d . L e t u s c o n s i d e r t h e case with t h e f i n i t e dimension. Condition 1 c a n b e omitted in a c c o r - d a n c e t o lemma 1 a n d 2. W e will show t h a t condition 2 c a n b e omitted too.

Indeed b e c a u s e LinK G R n X R m t h e n e x i s t t h e m a t r i c e s A a n d B with dimen- sions r x n a n d r X m r e s p e c t i v e l y s u c h t h a t t h e points ( x , y ) E LinK a n d only t h e y s a t i s f y t h e equations

w h e r e rows of t h e matrix (A, - B ) with t h e dimension r x ( n

+

m ) are l i n e a r l y in- dependent. This follows f r o m t h e f a c t t h a t in t h e f i n i t e dimensional s p a c e l i n e a r manifold c a n b e d e s c r i b e d as set of solution of some l i n e a r equations system.

S i n c e t h e rows of t h e m a t r i x (A, - B ) are l i n e a r l y independent t h e n e x i s t non-degenerate submatrix of t h i s matrix with dimension r x r . Consequently system

(18) h a s solution y f o r a r b i t r a r y x

.

F o r t h i s r e a s o n ( s e e example 1

5

1)

K e r n B * c_ K e r n A * (19)

* *

However K e r n B *

=

^{f O}

1.

Indeed l e t y E R ~ , y ^# 0 a n d B* y

* =

0, t h e n d u e t o (19) A * y

* =

0. This means t h a t e x i s t t h e non-zero v e c t o r y

*

o r t h o g o n a l t o a l l columns of t h e matrix (A, - B ) . The l a s t s t a t e m e n t c o n t r a d i c t s t h e e x i s t e n c e of non-degenerated r X r s u b m a t r i x of t h e m a t r i x ( A ,

-

B ) . Thus K e r n B

* =

fOj, i.e.

columns of t h e matrix B* are l i n e a r l y independent. F o r t h i s r e a s o n B* (and conse- quently B ) c o n t a i n s non-degenerated submatrix B1 with r a n k r X r . L e t B

=

(B1, B2). Consider t h e v e c t o r s y

w h e r e y E

R~

a n d 0,

-,

i s t h e non-zero v e c t o r with dimension m

-

r . I t i s c l e a r t h a t By

=

Bly l. If y

=

B1-lAx, t h e n v e c t o r

s a t i s f i e s (18). I t i s obvious t h a t t h e l i n e a r o p e r a t o r

s a t i s f i e s condition 2 of t h e t h e o r e m . Q.E.D.

REMARK Let u s c o n s i d e r t h e condition 2 in g e n e r a l . Let L ^C X XY i s a l i n e a r manifold. Denote

I t i s n o t difficult t o see t h a t 1 1 ( x ) i s a n a f f i n e manifold;

2 1 ( x l

+

x z )

=

1 ( x l )

+

1 ( x z ) ; 3 L ( X x ) = X l ( z ) f o r X + O .

Let in a c c o r d a n c e t o t h e t h e o r e m 6 dom 1

=

X. Thus t h e map 1 from t h e s p a c e X t o t h e set of a f f i n e manifolds of y is l i n e a r . Condition 2 of t h e o r e m 6 implies t h a t ex- i s t a l i n e a r continuous map P such t h a t Pz € 1 ( x ) . S i n c e 1 is a l i n e a r map t h e n ex- i s t e n c e of P i s a n a t u r a l condition. A s was shown e a r l i e r in a f i n i t e dimensional s p a c e , t h i s o p e r a t o r e x i s t s . I t would b e i n t e r e s t i n g in g e n e r a l t o formulate a n addi- tional condition f o r L g u a r a n t e e i n g e x i s t e n c e of a continuous l i n e a r selector of P.

Let u s c o n s i d e r now t h e questions connected with t h e c o n s t r u c t i o n of a mar- q u e e f o r a convex set.

THEOREM 7 Let M be a convex s e t i n B a n a c h s p a c e X, 0 E M and r i M

+ 4.

Then t h e c o n e K

=

c o n ( r i M )

is t h e m a r q u e e f o r s e t M a t t h e p o i n t z

=

0.

PROOF I t i s e v i d e n t t h a t r i M

c

M , f o r t h i s r e a s o n Lin ( r i M)

c

LinM a n d LinL L LinM.

Let

5

^{E K} i.e. z

- =

y x , y > 0 , z E r i M . If x f O t h e n f o r some E > O x

+

(EB)

n

LinM

c

M in a c c o r d a n c e t o t h e definition of t h e r e l a t i v e i n t e r i o r of a set M.

Let

where y i s a n a r b i t r a r y point of Lin

K , 11

^y

1)

^g 1. If

t h e n y' E M d u e t o t h e definition ri M . Since 5

+

^{0 , t h e n}

F o r t h i s r e a s o n if

< (Ig (1

t h e n

Thus if

>

^{0 , 6}

>

⁰ are chosen s o t h a t

t h e n f o r

y

satisfying condition

y

^{E [con}

( g + ( c l B ) n

LinK)]

n

( d B )

the, following inciusion is t r u e :

c

^{E M .} Thus, in t h i s c a s e

9(5) ⁼ c

^{c a n be taken.}

If 5

=

⁰ t h e n in a c c o r d a n c e with t h e definition of t h e r e l a t i v e i n t e r i o r e x i s t s such ^E

>

0 t h a t

( E B )

n

Lin K L M

i.e. any point of LinK L K with norm less t h a n ^E belongs t o M. lt is c l e a r t h a t in t h i s c a s e

9 ( y ) = 5

too. Q.E.D.

Let u s now consider some applications of t h e s e r e s u l t s . In p a r t i c u l a r it is in- t e r e s t i n g f o r u s t o generalize t h e implicit functions theorem in c a s e s when solu- tions belong t o some set M. I t is formulated below.

THEOREM 8

Let t h e f u n c t i o n s f c ( z ) ,

i

=

1,

. . .

, K

be d e f i n e d o n t h e s p a c e

Z

=

X

x Y, t h e s e m n c t i o n s be smooth in t h e neighborhood of t h e o r i g i n of coor- d i n a t e s ,

M

be

a

c o n v e x s e t c o n t a i n i n g

0.

Let in a d d i t i o n :

1

g r a d i e n t s f i ( z o ) a r e L i n e a r l y i n d e p e n d e n t o n s u b s p a c e

LinM;

2

e x i s t p o i n t z s u c h that

3

for a n y vector

u E R k

t h e set

i s

e m p t y or c o n s i s t s from t h e u n i q u e vector

f ;

* u .

Then for a n y vector z, [I 2 1) <

d e x i s t s vector

s u c h t h a t

PROOF Define

a ( x ) = l y : f i ( x , y ) = O , i =I. . . . ,

k ,

( x , y )

E M J

.

In a c c o r d a n c e with t h e theorems 4 and 7 t h e cone

is t h e marquee f o r gf

a

at t h e point

z o = ( x o , y o ) .

Taking into account assumptions and well known theorems of convex analysis w e g e t

where

f ; ( z O )

is t h e F r e s h e t derivative of t h e map f : R n + m -+ R k , i.e. matrix with rows

pi, ( z o )

^E R n ⁺ m. Condition 3 of t h e theorem 6 means t h a t r e l a t i o n s

( X I * , y * )

^E (con(M

- z o ) ) *

* *

Y + f $ ( z 0 b

= D assume t h e equality

* *

x +p3; ( z 0 ) u = o .

The l a s t condition i s equivalent t o condition 3 of t h e theorem.

THEOREM 9

Let

Z

=

R n X R m ,

Pi ( z ) ,

i

=

1,

. . .

^{, k}

be a smooth f u n c t i o n a n d

U

be a convex set

i n R n

.

Ip

( x o , y

O ) i s

a point s u c h t h a t

t h e n for t h e existence of t h e v a l u e d >

0

s u c h t h a t for a n y

E R n

e x i s t vector

5

^{E R m}

s a t i s j 5 i n g

i t i s su.tj%cient

1 the vectors f f ; ( z o ) a r e LinearLy independent;

2 e x i s t vector

( z l , y l )

s u c h t h a t

3 The set

c o n t a i n s onLy zero.

The proof follows d i r e c t l y from t h e p r e v i o u s t h e o r e m , t a k i n g i n t o a c c o u n t t h e e q u i v a l e n c e of e q u a l i t i e s f g ( z o ) u

=

0 a n d u

=

0 which, in t u r n follows f r o m l i n e a r independence of v e c t o r s f

t;

( z O ) , i

=

1,

. . .

, k

L e t u s c o n s i d e r now t h e solvability of t h e system of i n e q u a l i t i e s

for a n y x from vicinity of some point x o . S u p p o s e t h a t t h e point ( x o , y o ) i s o n e of t h e solutions of t h i s system.

This p r o b l e m c a n b e r e d u c e d to t h e p r e v i o u s o n e by introducing a u x i l i a r y v a r i a b l e s w i , i

=

1,

. . .

, k a n d c o n s i d e r i n g t h e following system:

The t h e o r e m 8 c a n b e a p p l i e d now. To d o t h i s let u s t a k e X

=

Rn a n d t h e s p a c e Y from t h i s t h e o r e m will b e t h e s p a c e of p a i r s ( y , w ) E Rm X

R ~ .

T h e set M i s now t h e set ( R n , R m , R : ) . T h e r e f o r e LinM

=

( R n . R m , R k ) . Let us n o t e t h a t in t h e conditions ( 2 1 ) e a c h new v a r i a b l e c o r r e s p o n d s to s e p a r a t e equality, t h e r e f o r e condition 1 of t h e o r e m 8 i s t r u e . F u r t h e r m o r e , w e c a n assume without l o s s of gen- e r a l i t y t h a t

This assumption will considerably simplify t h e argument. What is needed now f o r fulfillment of t h e second condition of t h e theorem is existence of t h e v e c t o r z l

- =

(5 l, GI) such t h a t

Due t o t h e f a c t t h a t M

=

( R n , R m , Rk+) w e have [con (M

-

Zo)l

* =

(On, O m , Rk,)

.

The third condition of theorem 8 easily follows now from t h e assumption t h a t condi-

* *

tions u r 0 ,

f $

( z 0 ) u

=

0 i m p l y

f $

( z o ) u

=

0. O r in o t h e r words Kern f; ( z O ) 2 (Kern

* fi*

⁽²⁰⁾⁾

^f7

R*

Thus, we have obtained t h e following r e s u l t :

THEOREM 1 0 S u p p o s e t h a t x E R n , y E k m , j'unctions

Pi

( z ) , i

=

1,

. . .

, k a r e smooth f o r z

=

( x , y ) a n d t h e p o i n t z o

=

(xo, y o ) is s u c h t h a t

Let US t a k e i n a d d i t i o n t h e following a s s u m p t i o n s : 1 E x i s t s vector

zl ⁼

( z l , GI) s u c h t h a t

2 _Kern

I; *

( z O ) 2 (Kern f $ ( z O ) )

n

Rk+

Then e z i s t s d

>

0 s u c h t h a t f o r a n y

z,

^llz'

11 <

d e z i s t s s u c h t h a t

REFERENCES

1 Pshenichny, B.N.: Convex analysis and extremal problems. Moscow, Nauka, 1980 (in Russian).

2 Rockafellar, T. : Convex analysis. Princeton Univ. P r e s s , Princeton NJ, 1970.

3 Boltiansky, B.G.: The method of marquees in t h e t h e o r y of extremal problems.

Uspekhi motematicheskih nauk, 1975, v. 30, No. 3 , pp. 1-55 (in Russian).