CHAPTER III

NONLINEAR MAPS ON P(X)

In this short chapter an attempt is made

to study certain type of nonlinear maps on 13(X)

where X is a complex Banach space.

Let 1: 13(X) --'^ (3(X) be a transformation.

The problem is to find conditions under which there

exist bounded linear operators A and B in R(X) such

that, -^(T) = AT2B for all T in p(X).

PROPOSITION 3.1.1.

Let 1: {3(X) >1( X) be a map such that

(1) I(Tl+T2) + q(Tl -T2)2 _ ^(Tl) + T(T2) for all

T1,T2 in R(X).

(2) Rank T(T) < 1 whenever rank T=1 and

Rank T<l, a(T) = 0 implies ^(T) = 0

(3) J(aT) = a2 J(T) for all T E 13(X) and for

all a ' (3 . Then either,

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(a) J(Lx ) c Ly(x) for every x in X, or

(b) J(Lx) c Rf(x) for every x in X

PROOF

Let Mx be the vector space generated X ) .

Case 1

Dimension Mx=l. If dim (Mx) = 0, there is

nothing to be proved. If the dimension is 1, then

Mx = [a J( x ® f0) : a E (!I for some f0 in X

=[a (y00 g0):a

Since I(x (Df0) is of rank 1. Hence

E CTJ for some y0 in X and

g0 in X*

(0

Case 2

dim (Mx) > 2.

Let if possible, there exist an x0 in X and

fl,f2 in X* linearly independent, such that

I(x0®fl) = xl® 91 0

^(x0 ®f2) = x2 ®g2 0

where xl,x21 and Lg1,g21 are linearly independent

sets in X and X* respectively.

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Now by (1) we-have,

But

x0® fl+ f2) + J(x0 0 f1-f2)2 = T(x0(& fI )+ 7(x0® f2 )

J(x0f1+f2) = Yo ®go

(x0fl-f2) = zo®h0

for some y0,z0 in X and g0,h0 in X*; by condition (3).

Thus,

I(x0®f1+f2) = Yo ®g0

= 2x1 ®gl + 2x2 0g2 - zo®go

By'multiplying f2 by a suitable scalar if necessary,

we may assume that fl(x0) = f2(xo) so that

6(x0®f1-f2) = [03

Therefore I (x0(D fl-f2) = 0 = z0 0 h0

Thus we get,

^(x0(D f1+f2) = 2x1®gl + 2x2® 92

This would imply that ^(x0(@f1+f2) is a rank 2

operator, since {x1,x2., Lgl,g22 are linearly

independent. This is contradictory to the assumption (3).

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Hence ^ ( L)c.L for some y in X

and ( Lx) c R 9 for some g in X*

Now we show that either J (Lx) C Ly( x) for every x EX.

or (Lx) c RfX) for every x £ X.

Let M ={x E XI ^(Lx) c Ly(x)l and

N =fx E XI J(Lx) c Rf(x)1J

We found that M U N = X and M n N = 0. So assuming

that M A 0, it is enough to establish that N = 0.

Let if possible xl E N and let x0 be in M.

Also put,

.(xo® f ) = Yo ®g0 and

(xl ®f) = Y10 91

We choose an f in X* so that y 0 and yl are linearly

independent . Also by multiplying x 0 with a suitable

scalar , if necessary , we may assume that f(x0)=f(x1) .

Now, as before, we get

(^(xo+xl ® f ) + (I(xo-xl) ® f

= 2 I(xo(Z f) + 2 T(xl(& f)

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Thus, since 1((xo-xl) ®f) = 0, we get

(xo+xl)® f = 2 ^(x0 Qf) + 21(x10 f)

But left side is of rank 1 and right is of rank 2

which is impossible. Hence N = 0 if M A 0.

Similarly we can show that if N ^ 0, then M = 0.