Uniform Boundedness And Convergence Of Generalized Inverses, Moore-Penrose Inverses And Group Inverses

Let X and Y be Banach spaces.Tn and T?B(X,Y)satisfy Tn?T.If Tn is invertible and sup n?N||Tn-1||<+?,then T is invertible and Tn-1?T-1.It can be claimed that in the case of invertible operators,the uniformly boundedness of ||Tn-1|| can imply the invertibility of T and the convergence Tn-1?T-1.Particularly,does this conclusion hold for the case of Moore-Penrose inverse and group inverse or generalized inverse?J.Koliha,J.Benitez?D.Cvetkovic-Ilic and X.Liu gave the conclusion:the uniformly boundedness of ||Tn(?)|| and ||Tn#||can imply the convergence Tn(?)?T(?)and Tn#?T#This paper firstly gives examples to illustrate that in the case of matrics,Tn+ is uniformly bounded but not convergent,and that is to say supn?N |Tn+|<+? cannot imply Ta+?T+ although Rank Tn = Rank T.This shows that the case of generalized inverse is totally different from that of Moore-Penrose inverse and group inverse.Using the stable perturbation for generalized inverse and the gap between closed linear subspace,we prove the equivalence of the uniform boundedness and convergence for generalized inverse.At last,these results are used to consider the case for the Moore-Penrose inverses and group inverses.Some new and concise expressions and convergence theorems are provided.The obtained results extend and improve known results in operator theory and matrix theory.Theorem:Let Tn,T ? B(X,Y)be generalized invertible and Tn?T.If the generalized inverse Tn+:satisfies sup n?N ||Tn+||<+?,then for any generalized inverse T+,there exists a generalized inverse Tn(?)of Tn,such that Tn(?)? T+Theorem:Let X,Y be Hilbert spaces and Tn,T?(X,Y)with Tn?T,If Tn is Moore-Penrose invertible and sup n?N||Tn(?)||<+?,then T is Moore-Penrose invertible,Tn(?)?T(?)and for any sufficiently large n,Tn(?)=[I-BnTn-(BnTn)*]-1Bn[I-TnBn-(TnBn)*]-1,where Bn=T(?)[I+(Tn-T)T(?)]-1.Theorem:Let X bea Banach space and Tn,T?B(X)with Tn?T.If the group inverses Tn#exist and sup n?N||Tn#||<+?,then T has the group inverses T#satisfying,Tn#?T#and for any sufficiently large n,and for any sufficiently large n,where Bn =T#[I+(Tn-T)T#]-1=[I + T#(Tn-T)]-'T#and Wn = BnTn+TnBn-I.