A NEW RESULT ON REVERSE ORDER LAWS FOR { 1 , 2 , 3 }-INVERSE OF A TWO-OPERATOR PRODUCT

In this note, reverse order laws for {1,2,3}-inverse of a two-operator product is mainly investigated by making full use of block-operator matrix technique. First, an example is given, which demonstrates there is a gap in the main result in [X. J. Liu, S. X. Wu, D. S. Cvetković-Ilić. New results on reverse order law for {1,2,3}and {1,2,4}-inverses of bounded operators. Mathematics of Computation, 2013, 82(283): 1597-1607]. Next, The new necessary and sufficient conditions for B{1,2, i}A{1,2, i} ⊆ (AB){1,2, i}(i ∈ {3,4}) are presented respectively, when all ranges R(A), R(B) and R(AB) are closed. Which will fill up the gap in the above paper.


Introduction
Throughout this paper, let H , K and L be separable Hilbert spaces and B(K , H ) be the set of all bounded linear operators from K into H and abbreviate B(K , H ) to B(H ) if K = H .If A ∈ B(H , K ), write N(A) and R(A) for the null space and the range of A, respectively.
For an operator A ∈ B(H , K ), a generalized inverse of A is an operator G ∈ B(K , H ) which satisfies some of the following four equations, which is said to be the Moore-Penrose conditions: (1)AGA = A, (2)GAG = G, (3)(AG) * = AG, (4) (GA) * = GA.
Let A{i, j, • • • , l} denote the set of operators G ∈ B(K , H ) which satisfy equation (i), ( j), • • • , (l) from among the above equations.An operator G ∈ A{i, j, • • • , l} is called an {i, j, • • • , l}-inverse of A, and also denoted by A (i j•••l) .The unique {1, 2, 3, 4}-inverse of A is denoted by A + , which is called the Moore-Penrose inverse of A. As is well known, A is Moore-Penrose invertible if and only if R(A) is closed.
The main result in [18] could fill up the gap in Theorem 1.1.In this paper, we shall give a new result about the reverse order law for {1, 2, 3}-and {1, 2, 4}-reverses by the relationship of the range conclusion.In section 2, we shall give some preliminaries.Some necessary and sufficient conditions for an operator G ∈ B(K , H ) to be in A{1, 2, 3} and A{1, 2, 4} are pointed.In section 3, we will derive a new sufficient and necessary conditions for B{1, 2, i}A{1, 2, i} ⊆ (AB){1, 2, i}(i ∈ {3, 4}) respectively, when R(A), R(B), R(AB) are closed.And also our result will fill up the gap in Theorem 1.1.

Preliminaries
In this section, we mainly introduce some notations and lemmas.Let A ∈ B(H , K ) with closed range.Then under the orthogonal decompositions H = R(A * ) ⊕ N(A) and K = R(A) ⊕ N(A * ) respectively, A has the matrix form where A 1 ∈ B(R(A * ), R(A)) is invertible.The Moore-Penrose inverse A + of A has the matrix form as follows The {1, 3}, {1, 2, 3}-inverses also have similarly matrix forms.
Lemma 2.1.([12]) Let A ∈ B(H , K ) with closed range and the matrix form (2.1).Then A (13) and A (123) have the matrix form and under the orthogonal decompositions respectively, such that A 1 is invertible and A 3 is surjective, then there are some operators G ji ∈ such that A (123) has the matrix form In [10], the authors have given the necessary and sufficient conditions for G ∈ A{1, 2, 3} and G ∈ A{1, 2, 4} for any matrix A. Now, we generalize these results to an operator on an infinite dimensional Hilbert space. ( It is sufficient to show one of the two statements holds.We next show the statement (1) holds for A with closed range.Since R(A) is closed, A has the matrix form as the formula (2.1).So , then G has the matrix form as the formula (2.3) by Lemma 2.1.Thus and and then Because R(G * ) = R(A), by a simple calculation G 4 = 0 and the Moore-Penrose condition (2) holds.Therefore G ∈ A{1, 2, 3}.The proof is complete.
The proof of Theorem 2.4 implies the following result.
Xiong and Zheng [17] obtained the equivalent condition for B{1, 2, i}A{1, 2, i} ⊆ (AB){1, 2, i}(i ∈ {3, 4}).And another equivalent conditions of above inclusions were given under conditions of operators A, B, AB and A − ABB + are regular in [4], which equivalent to the rang of A, B, AB and A − ABB + are closed since A is regular if and only if A + exists.Here, the sufficient and necessary conditions for B{1, 2, i}A{1, 2, i} ⊆ (AB){1, 2, i}(i ∈ {3, 4}) will be presented respectively, when R(A), R(B) and R(AB) are closed.And the range of A − ABB + not necessarily closed.Suppose that A = 0 and B = 0, then A and B have the matrix forms as follows, By Lemma2.1, we have the {1, 2, 3}-inverses of A and B have the matrix forms, where Hence From B{1, 2, 3}A{1, 2, 3} ⊆ (AB){1, 2, 3}, it is easy to get B −1 21 G 21 = 0, so G 21 = 0 since B = 0.But G 21 is arbitrary by Lemma2.1, then A = 0.It is a contradiction with the assumption.Hence A = 0 or B = 0 in this case.It is natural to get that the result holds.
We firstly claim that FG ∈ (AB){1, 2, 3} if and only if Therefore, This means that It follows that from Corollary 2.5.On the other hand, Thus, It follows that FG ∈ AB{1, 2, 3} if and only if from Lemma 2.4 and formulae (3.3) and (3.4).
Moreover, if we set and respectively, then it is known that In particular, it is elementary that A is of the matrix form such that A 11 is invertible and A 22 is surjective.Then there are some operators such that G has the matrix form from Lemma 2.3.We note that all of G 31 , G 32 , G 41 and G 42 are arbitrary.From the matrix forms (3.6) and (3.8), we have and If K 2 = {0}, then R(A) = R(AB) and A 22 = 0.In this case, it is immediate that from the formulae (3.9) and (3.10).Since B{1, 2, 3}A{1, 2, 3} ⊆ AB{1, 2, 3}, FG ∈ (AB){1, 2, 3}. Hence The proof is complete.Proof if AB = 0, by the discussion for AB = 0 in the proof of Theorem 3.1, we can get the result holds.So assume that AB = 0 and denote H i (i = 1, 2, 3, 4), K j ( j = 1, 2, 3) as in (3.5) Let J 2 = B + H 3 and J 1 = R(B * ) J 2 .B has the following matrix form, which B 11 and B 22 invertible.According to Lemma 2.1, {1, 2, 3}-inverse A (123) and B (123) of A and B has the matrix forms, , which G 31 , G 41 , F 31 , F 32 are arbitrary.This follows that Combining formulae (3.11) and (3.12), we have , Using Lemma 2.1 again, we get that Hence A has the matrix form with respect to the orthogonal decompositions H = H 1 ⊕ H 2 ⊕ H 4 and K = K 1 ⊕ K 2 , respectively, such that A 11 and A 22 are invertible.B has the matrix form  Comparing the formula (3.17 (2) R(A * AB) = R(B) or R(A * ) ⊆ R(B).
{0}, A 12 = 0 and A 22 = 0. Then A has the matrix form as follows,