GeneralizedInverseAnalysisontheDomain ?(A,A+)in B(E,F)

Zhaofeng Ma

Department of Management and Informations,Zhejiang Institute of Communications,Hangzhou 311112,China

Abstract.Let B(E,F)be the set of all bounded linear operators from a Banach space E into another Banach space F,B+(E,F)the set of all double splitting operators in B(E,F)and GI(A)the set of generalized inverses of A∈B+(E,F).In this paper we introduce an unbounded domain ?(A,A+)in B(E,F)for A∈B+(E,F)and A+∈GI(A),and provide a necessary and sufficient condition for T∈?(A,A+).Then several conditions equivalent to the following property are proved:B=A+(IF+(T?A)A+)?1is the generalizedinverseof T with R(B)=R(A+)and N(B)=N(A+),for T∈?(A,A+),where IF is the identity on F.Also we obtain the smooth(C∞)diffeomorphism MA(A+,T)from?(A,A+)onto itself with the fixed point A.Let S={T∈?(A,A+):R(T)∩N(A+)={0}},M(X)={T∈B(E,F):TN(X)?R(X)}for X∈B(E,F)},and F={M(X):?X∈B(E,F)}.Using the diffeomorphism MA(A+,T)we prove the following theorem:S is a smooth submanifold in B(E,F)and tangent to M(X)at any X∈S.The theorem expands the smooth integrability of F at A from a local neighborhoold at A to the global unbounded domain ?(A,A+).It seems to be useful for developing global analysis and geomatrical method in differential equations.

Key Words:Generalized inverse analysis,smooth diffeomorphism,smooth submanifold.

1 Introduction

Let E,F be two Banach spaces,B(E,F)the set of all linear bounded operators from E into F,B+(E,F)that of all double splitting operators in B(E,F),and GI(A)that of all generalized inverses of A for A∈B+(E,F).Write V(A,A+)={T∈B(E,F):kT?Ak

2 The domain ?(A,A+)in B(E,F)

Let B(F)=B(F,F)and BX(F)be the set of all invertible operators in B(F).Write CA(A+,T)=IF+(T?A)A?1∈B(F)for A∈B+(E,F)and A+∈GI(A),where IFdenotes the identity on F.We define

for A∈B+(E,F)and A+∈GI(A).For abbreviation,writeandrespectively,in the sequel.Then CA(A+,T)=We have

Theorem 2.1.T belongs to ?(A,A+)if and only if the following conditions hold:

and

Proof.Assume T∈?(A,A+).We claim N(T)∩R(A+)={0}.Let x∈N(T)satisfy x=A+y for some y∈R(A),then

so that x=A+y=0 because of T∈?(A,A+).This says N(T)∩R(A+)={0}.Next go to prove that the equality(2.1)holds.We claim that N(A+)∩R(TA+)={0}and R(TA+)is closed.Let y1∈N(A+)satisfy y1=TA+y0for some y0∈R(A),then

and so,y1=0=y0because of T∈ ?(A,A+). So N(A+)∩R(TA+)={0}.Note CA(A+,T)R(A)=TA+R(A)=R(TA+).It follows that R(TA+)is closed for T∈?(A,A+)and R(A)is closed.Now go to show the equality(2.1).Obviously,it is enough to show F?R(TA+)⊕N(A+).Let z belong to F,and z=CA(A+,T)y for some y∈F.Then

Thus the necessity of the theorem is proved.

Assume that the conclusions of the theorem hold.We claim N(CA(A+,T))={0}.Obviously

By(2.1),PN(A+)y=TA+PR(A)=0.By the assumption N(T)∩R(TA+)={0},PR(A)y=0,so y=0.This says N(CA(A+,T))={0}.Next go to show that CA(A+,T)is surjective.By(2.1)there exist y0∈N(A+)and y1∈R(A)such that

Therefore

The proof of the theorem ends.

Corollary 2.1.If T∈?(A,A+),then so does xT,?λ6=0.

Note R(λTA+)=R(TA+)and N(λT)=N(T),?λ 6=0.The corollary follows from Theorem 2.1.

Corollary 2.2.?(A,A+)is an unbounded domain in B(E,F).

Proof.By Corollary 2.1,?(A,A+)is unbounded.Next go to show that ?(A,A+)is an open set in B(E,F).Let T0be arbitrary one of ?(A,A+)and T?=CA(A+,T0)∈ BX(F).Note that BX(F)is an open set in B(F)and CA(A+,T):B(E,F)→B(F)is continuous.It is obvious that T0is an inner point in ?(A,A+).Thus ?(A,A+)is an unbounded domain in B(E,F).

The following equalities will be often used in the sequel:for T∈?(A,A+),

These equalities are immediate from CA(A+,T)PN(A+)=PN(A+)and CA(A+,T)PR(A)=TA+PR(A)for T∈?(A,A+).

3 A basic theorem in generalized inverse analysis in B+(E,F)

In this section we will prove a basic theorem in B+(E,F)generalized inverse analysis,which expands some inportent results in Theorem 1.1 in[6]from V(A,A+)to the unbounded domain ?(A,A+).

Theorem 3.1.The following conditions for T∈?(A,A+)are equivalent:

(4)R(TA+)=R(T);

(5)F=R(T)⊕N(A+);

(6)R(T)∩N(A+)={0}.

Proof.To show(1)?(2).Obviously,R(B)=R(A+)and

because of(2.2).While by(2.2)

and

Then one concludes(1)?(2).

To show(2)?(3).By(2.2)for any x∈E.Hence(2)?(3).

To show(3)? (4).Assume that(3)holds.Write(A+,T)TPN(A)x=y,?x∈E,then y∈R(A).By(3.1)

and so,one can conclude R(T)=R(TA+).

Assume R(T)=R(TA+).Then for any x∈E there exists y∈R(A)such that Tx=TA+y and so(A+,T)Tx=y because of(2.2).Thus,by(3.1),

Hence(4)holds.This shows(3)?(4).

To show(4)? (5).Assume(4)holds.By(2.1),F=R(T)⊕N(A+),i.e.,(5)holds.Assume(5)holds.Also by(2.1),for any y∈R(T)there exists y0∈R(TA+)and y1∈N(A+)such that y?y0=y1∈N(A+);while R(T)∩N(A+)={0}because of(5)and so,y∈R(TA+).This shows R(T)=R(TA+).

To show(5)?(6).Obviously(5)? (6).Assume R(T)∩N(A+)={0}.By(2.1),for any y∈R(T)there exists y0∈R(TA+)and y1∈N(A+)such that y?y0=y1∈N(A+).Then by the assumption R(T)=R(TA+)one concluds that(5)holds.

4 Some applications

In this section we consder the following two maps:

and

Obviously

by(2.2)

So

for any T∈?(A+,T).Moreover,we have

Theorem 4.1.MA(A+,T)is a smooth diffeomorphism from ?(A,A+)onto ilself with a fixed point A.

Proof.Obviously,MA(A+,T)is smooth(i.e.,of C∞)in ?(A,A+)and MA(A+,A)=A.By(4.1)and(2.2)

Similarly

The proof ends.

Let S={T∈?(A,A+):R(T)∩N(A+)={0}}.By Theorem 3.1,X∈S has the generalized inverse X+with R(X+)=R(A+)and N(X+)=N(A+).For X∈S consider

As indicated in the proof of Theorem 4.1,

is its inverse.

By direct computing

Note that JX(X+,X)=X and MX(X+,JY(X+,T))=T.It follows

Let M(X)={T∈B(E,F):TN(X)?R(X)}for X∈B(E,F).Recall that S is said to be tangent to M(X)at X∈S provided

Evidently

and

Write M?={PN(A+)TPN(A):?T∈B(E,F)}.Obviously B(E,F)=M(A)⊕M?,i.e.,M(A)is splitting in B(E,F).We now are in the position to prove the following Theorem.

Theorem 4.2.S is a smooth submanifold in B(E,F)and tangent to M(X)at any X∈S.

Proof.By Theorem 4.1,(MA(A+,T),?(A,A+))is a smooth coordinate chart of B(E,F)at any X∈S.In view of the equivalence of conditions(2)and(6)in Theorem 3.1,

and so,MA(A+,S)is an open set in M(A).This says that S is a smooth submanifold in B(E,F).Now we are going to show that S is tangent to M(X)at any X∈S.

Note N(X+)=N(A+)and X∈?(A,A+)∩?(X,X+),?X∈S.One can conclude that there exists a neighborhood UXat X satisfying UX??(A,A+)∩?(X,X+),and so

Let C(t)?S∩UYbe any C1curve with C(0)=X,then by the equivalence of conditions(2)and(6),D(t)=MX(X+,C(t))is a C1curve contained in M(X)with D(0)=X.By(4.3),(0)=(0)∈M(X).This shows M(X)?{(0):for any C1curve C(t)?S with C(0)=X}.

Let X+UT=D(T)?M(X)∩UXfor any T∈M(X),then C(t)=JX(X+,D(t))?S is a C1curve with C(0)=X.By(4.2),˙C(0)=(0)=T∈M(X).Thus S is tangent to M(X)at any X∈S.Then proof ends.

The theorem expands Theorem 4.1 in[6]from the case of a local neighborhood at A to that of global domain ?(A,A+).Also we have

Corollary 4.1.1.If T∈S,then λT∈S for λ0;

2.MA(A+,λA)=λA for λ0.

Proof.The part 1 of the corollary is immedaite from Corollary 2.1. Obviously JA(A+,λA)=λA.Notefor λ 60,then

The proof ends.

Our results seem to be useful for developing global analysis,differential topology and geometrical method in differential equations.