APPROXIMATION OF MOORE-PENROSE INVERSE OF A CLOSED OPERATOR BY A SEQUENCE OF FINITE RANK OUTER INVERSES

APPROXIMATION OF MOORE-PENROSE INVERSE OF A CLOSED OPERATOR BY A SEQUENCE OF FINITE RANK OUTER INVERSES S. H. KULKARNI AND G. RAMESH Abstract. Let T be a densely defined closed linear operator between complex Hilbert spaces H 1 and H 2 with domain D(T ) H 1 and separable range R(T ). In this note we approximate the Moore- Penrose inverse T of T by its finite rank bounded outer inverses. We also illustrate this method with an example. 1. Introduction Suppose T is a densely defined closed linear operator between Hilbert spaces H 1 and H 2 with domain D(T ) H 1 and has a separable range. Let T denote the Moore-Penrose inverse of T. In this note we prove the following results: For each n N, there exists a bounded finite rank outer inverse T n # of T such that D(T ) = {y H 2 : lim T n # y exists} and T y = lim T # n y for all y D(T ). Such methods were studied by Huang et al., [9] for the case of bounded operators with separable range. Earlier these results were proved for the bounded operators with closed and separable range by J. Ma and Z. Ma [14]. The proofs mentioned in these two articles are not applicable when the operators under consideration are unbounded. We overcome this difficulty by constructing new operators. The important point here is to note that a large number of the operators which arise naturally in applications e.g Mathematical Physics, Quantum Mechanics and Partial differential Equations are all unbounded (See [17, 18] for more details). In fact, many of these unbounded operators have compact inverse. To solve operator equations involving such unbounded operators it is necessary to generalize the existing results to the case of unbounded operators. The theory of generalized inverses has many applications in diverse mathematical fields like Optimization [8], Statistics, Economics, Games, Programming and Networks, Science and Engineering [1]. In this note we have made an attempt to generalize the existing results to the case of unbounded operators. A point worth noting is that while dealing with unbounded operators one has to be more careful with the domains of the operators as those are Date: July 16, 2008. 2000 Mathematics Subject Classification. 47A05, 47A50. Key words and phrases. Outer inverse, generalized inverse, closed operator, weakly bounded operator. The second author is greatly indebted to the Council of Scientific and Industrial Research for the financial support in the form of SRF (No: 9/84 (408) 07-EMR-I). 1

2 S. H. KULKARNI AND G. RAMESH proper subspaces of the whole space. Hence, in most cases the techniques of the bounded operators do not work. The same is true for the Moore-Penrose inverse if the operator under consideration does not have closed range. This paper is a sequel to an earlier paper [11], in which we have discussed projection methods for the inverse of an unbounded operator. The paper is organized in the following manner: In the second section we set up notations and state some of the definitions and results which will be frequently used throughout the remaining part of the paper. The third section contains the main results and an example to illustrate these results. 2. Notations and Basic results Throughout the paper we consider the complex Hilbert spaces which will be denoted by H, H 1, H 2 etc. The inner product and the induced norm are denoted respectively by, and.. If T : H 1 H 2 is a linear operator with domain D(T ) H 1, then it is denoted by T L(H 1, H 2 ). The null space and range space of T are denoted by N(T ) and R(T ) respectively. The graph of T L(H 1, H 2 ) is defined by G(T ) := {(x, T x) : x D(T )} H 1 H 2. If G(T ) is closed, then T is called a closed operator. The set of all closed operators is denoted by C(H 1, H 2 ). By the closed graph Theorem [6, Page 281], an everywhere defined closed operator is bounded. The set of all bounded operators is denoted by B(H 1, H 2 ). If H 1 = H 2 = H, then B(H 1, H 2 ) and C(H 1, H 2 ) are denoted by B(H) and C(H) respectively. If S and T are two linear operators such that D(T ) D(S) and T x = Sx for all x D(T ), then T is called a restriction of S and S is called an extension of T. We denote this fact by T S. If M is a closed subspace of a Hilbert space H, then P M is the orthogonal projection onto M and M is the orthogonal complement of M in H. For closed subspaces M 1 and M 2 of H, the direct sum and the orthogonal direct sum are denoted by M 1 M 2 and M 1 M 2 respectively. Definition 2.1. [1, Definition 1.12, Page 13] Let T C(H 1, H 2 ). If there exists an operator T # L(H 2, H 1 ) such that T # T T # = T #, then T # is called an outer inverse of T (This is called {2} inverse in [4]). Definition 2.2. [4] Let T L(H 1, H 2 ). If D(T ) = H 1, then T is called densely defined. The subspace C(T ) := D(T ) N(T ) is called the carrier of T. Note 2.3. If T C(H 1, H 2 ), then D(T ) = N(T ) C(T ) [4, page 340]. Definition 2.4. [Moore-Penrose Inverse] [4] Let T C(H 1, H 2 ) be densely defined. Then there exists a unique densely defined operator T C(H 2, H 1 ) with domain D(T ) = R(T ) R(T ) and has the following properties; (1) T T y = P R(T ) y, for all y D(T ). (2) T T x = P N(T ) x, for all x D(T ). (3) N(T ) = R(T ). This operator T is called the Moore-Penrose inverse of T. The following property of T is also well known. For every y D(T ), let L(y) := {x D(T ) : T x y T u y u D(T )}.

FINITE RANK APPROXIMATION OF MOORE-PENROSE INVERSE 3 Here any u L(y) is called a least square solution of the operator equation T x = y. The vector x = T y L(y) and satisfies, T y x x L(y) and is called the least square solution of minimal norm. A different treatment of T is described in [4, Pages 336, 339, 341], where the authors call it the Maximal Tseng generalized Inverse. Theorem 2.5. [3, 5] Let {H k }, k = 1, 2, 3,... be closed subspaces of H and let P k = P Hk. Suppose {P k } is a monotone(h k H K+1 or H k+1 H k ) sequence of orthogonal projections. Then the strong limit P = lim P H k exists and P is k the projection onto k H k in case P k is non-increasing and onto k H k if {P k } is non-decreasing. Proposition 2.6. [4] Let T C(H 1, H 2 ) be a densely defined operator. Then (1) N(T ) = R(T ) (2) N(T ) = R(T ) (3) N(T T ) = N(T ) and (4) R(T T ) = R(T ). Proposition 2.7. [7, 16] Let T C(H 1, H 2 ) be densely defined. Then (1) (I + T T ) 1 B(H 1 ), (I + T T ) 1 B(H 2 ). (2) (I + T T ) 1 T T (I + T T ) 1 and T (I + T T ) 1 1 (3) (I + T T ) 1 T T (I + T T ) 1 and T (I + T T ) 1 1. 3. Main Results In this section, first we prove a lemma which is helpful in proving the main theorem. Lemma 3.1. Let T C(H 1, H 2 ) be densely defined. Let Y n R(T ) be such that (a) Y n Y n+1 for each n N (b) dim Y n = n (c) n=1 Y n = R(T ). Let Z n := (I + T T ) 1 Y n and X n := T Z n = T (I + T T ) 1 Y n. Then (1) X 1 X 2 X n X n+1 R(T ) = N(T ), dim X n = n and (2) n=1 Z n = R(T ) (3) n=1 X n = R(T ) (4) n=1 T X n = R(T ). Proof. By the definition of X n, X n C(T ) N(T ) = R(T ) for all n and X n X n+1. Since the operator T (I + T T ) 1 R(T ) is injective dim X n = n = dim Y n. For a proof of (2), we make use of the following observation: (I + T T ) 1 (R(T )) = R(T ). It can be proved easily that (I + T T ) 1 (N(T T )) = N(T T ). By the Projection Theorem [12, 21.1, Page 420 ], H 2 = N(T T ) N(T T ). That is H 2 = N(T T ) R(T T ). But (I + T T ) 1 (H 2 ) = D(T T ) = N(T T ) C(T T ).

4 S. H. KULKARNI AND G. RAMESH Hence (I + T T ) 1 H 2 = (I + T T ) 1 (N(T T ) R(T T )) = N(T T ) (I + T T ) 1 (R(T T )). From this we can conclude that (I +T T ) 1 (R(T T )) = C(T T ) and as C(T T ) = N(T T ), we have (I + T T ) 1 (R(T T )) = R(T T ). Hence (I + T T ) 1 (R(T )) = R(T ), by Proposition (2.6). Thus R(T ) = (I + T T ) 1 (R(T )) = (I + T T ) 1 ( n=1 Y n) = n=1 (I + T T ) 1 Y n = n=1 Z n. This proves (2). It is clear that n=1 X n R(T ) = N(T ). Suppose n=1 X n N(T ). Then there exists a 0 z 0 N(T ) such that z 0 ( n=1 X n). That is z 0, T (I + T T ) 1 y = 0 for all y R(T ) By the continuity of T (I + T T ) 1, this holds for all y R(T ). We claim that this holds for all y H 2. Let y H 2. Then y = u + v for some u R(T ) and v R(T ) = N(T ) D(T ). Hence by Proposition 2.7, T (I + T T ) 1 v = (I + T T ) 1 T v = 0. Hence z 0, T (I + T T ) 1 y = z 0, T (I + T T ) 1 u = 0. This proves the claim. Next, since C(T ) = N(T ) [10], there exists a sequence {z n } C(T ) such that z n z 0. Hence for all y H 2, 0 = z 0, T (I + T T ) 1 y = lim z n, T (I + T T ) 1 y = lim T z n, (I + T T ) 1 y = lim (I + T T ) 1 T z n, y = lim T (I + T T ) 1 z n, y. This shows that T (I +T T ) 1 w z n 0 (weakly), but since T (I +T T ) 1 is bounded, we have T (I + T T ) 1 z 0 = 0. That is (I + T T ) 1 z 0 N(T ). Let y = (I + T T ) 1 z 0. Then T y = 0. Hence z 0 = (I + T T )y = y N(T ). Thus z 0 N(T ) N(T ) = {0}. Hence z 0 = 0, a contradiction to our assumption. This proves (3). Using a similar argument we can prove (4). Remark 3.2. We may note that Lemma 3.1 implies that if R(T ) is separable, then R(T ) is separable. This generalizes an analogous result for bounded operators proved in [2, Page 362].

FINITE RANK APPROXIMATION OF MOORE-PENROSE INVERSE 5 Theorem 3.3 (Compare Theorem 2.1 of [9]). Let T C(H 1, H 2 ) be a densely defined operator with separable range R(T ). Then for each n N, there exists a bounded outer inverse T n # of T of rank n such that and T y = lim T # n y for all y D(T ). D(T ) = {y H 2 : lim T # n y exists} Proof. Assume that R(T ) is infinite dimensional. Since R(T ) is separable, we can find a sequence of subspaces of Y n of R(T ) with the following properties: (1) Y n Y n+1 and dimy n = n for all n N. (2) n=1 Y n = R(T ). ( For example if {φ1, φ 2,...,} is an orthonormal set that spans R(T ), then define Y n := span({φ 1, φ 2,..., φ n }) ). Let Z n and X n be as in Lemma 3.1. Then Z n Z n+1 and dim Z n = n. Similar results hold for X n. Let P n : H 2 H 2 and Q n : H 1 H 1 be sequences of orthogonal projections with R(P n ) = Z n and R(Q n ) = X n. Let T n := P n T. Here D(T n ) = D(T ) and T n x T x for all x D(T ). Next we claim that R(T n ) = R(P n ) = Z n. It is clear that R(T n ) R(P n ) = Z n. To show the other way inclusion, it is enough to show N(Tn) N(P n ). Now let z N(Tn). Then T P n z = 0. Hence P n z N(T ) = R(T ). But, P n z R(T ). Hence P n z = 0. Thus z N(P n ). Note that Tn = T P n = T R(Pn) = T Zn. Hence R(Tn) = T Z n = X n = N(T n ). That is N(T n ) = Xn. R(T n ) = N(T n ) = Zn. That is R(T n ) = Z n. So T n Xn : X n Z n is a bijective operator. Hence dim X n = dim Z n = n. Construction of outer inverses: Define T n # : H 2 H 1 by { T n # (T n Xn ) 1 y, if y Z n, y := 0, if y Zn. Here T n # = T n and T n # is bounded since R(T n ) is closed. It is also true that T n # is an outer inverse of T n. Here N(T n # ) = Zn and R(T n # ) = X n. Next we claim that T n # is also an outer inverse of T. For this we make use of the following observation: T n # y = T n P n y, for all y H 2. To see this let y H 2. Then y = u + v for some u Z n and v Zn. Hence T # n y = T # n (u + v) = T # n u ( T # n (v) = 0, because v Z n ) = T # n P n y. Since T # n is an outer inverse of T n, We can now prove that T # n T T # n y = T # n P n T T # n y = T # n T n T # n y = T # n y. D(T ) = {y H 2 : lim T # n y exists}.

6 S. H. KULKARNI AND G. RAMESH Let y D(T ). Then T y C(T ). Since Q n x x for all x C(T ) N(T ) = R(T ), it is clear that Q n T y T y. Next we show that Q n T y = T # n y, for all y D(T ). From the facts Q n T y X n, (Q n I)T y N(T n ) and Theorem 2.5, Q n T y = T # n T n Q n T y = T # n T n Q n T y + T # n P n y T # P n y = T # n (T n Q n T y P n y) + T # P n y = T # n (T n Q n T y P n T T y) + T # n P n y = T # n (T n Q n P n T )T y + T # n P n y = T # n T n (Q n I)T y + T # n P n y = T # n P n y = T # n y. As Q n T y T y for all y D(T ), and by the above argument lim T n # y exists and equals T y. This shows that D(T ) {y H 2 : lim T n # y exists}. Next, we prove that if y H 2 such that lim T n # y exists, then y D(T ). Let x 0 = lim T n # y. We can easily verify that T T n # is a projection with R(T T n # ) = T X n and N(T T n # ) = (T X n ). Since by Lemma 3.1 it follows that n=1 T X n = R(T ), note that T T n # y P R(T ) y. As T is closed, x 0 D(T ) and T x 0 = P R(T ) y. Since x 0 C(T ) = R(T ), there exists u D(T ) such that x 0 = T u. Now, P R(T ) y = T x 0 = T T u = P R(T ) u. Hence y u R(T ) = N(T ). That is y = y u + u D(T ). Thus x 0 = T u = T y. This completes the proof. Theorem 3.4 (Compare Corollary 2.1 of [9]). Let T C(H 1, H 2 ) be a densely defined operator. Then the following statements are equivalent: (1) R(T ) is closed. (2) T is bounded. (3) D(T ) = H 2. (4) 0 is not an accumulation point of σ(t T ). If, in addition R(T ) is separable and T n # are as in Theorem 3.3, then each of the above statements is also equivalent to each of the following; (5) lim T n # y exists for all y H 2. (6) T n # is uniformly bounded. Proof. The equivalence of (1), (2) and (3) is well known and can be found in ([4]). The equivalence of (1) and (4) is proved in ([13, Theorem 3.3]). The equivalence of (3) and (5) follows from Theorem 3.3. The implication (5) (6) follows from the Uniform boundedness principle. (6) (5): By Theorem 3.3, lim T n # y exists for every y D(T ). Since D(T ) = H 2 [4, Theorem 2, Page 320], the conclusion follows by [15, Theorem 6.4, Page 220]. Using similar arguments as in Theorem 3.3, we can prove the following result. Theorem 3.5. Let T C(H 1, H 2 ) be a densely defined closed operator. If there exists a sequence of increasing orthogonal projections P n on H 2 onto subspace of

FINITE RANK APPROXIMATION OF MOORE-PENROSE INVERSE 7 R(T ) with the property that P n y P R(T ) y for all y H 2 and R(P n T ) is closed, then for each n, there exists an outer inverse T # n and Example 3.6. Let T : l 2 l 2 be with Define such that D(T ) := {y H 2 : lim T # n y exists} T y = lim T # n y, for all y D(T ). D(T ) := {(x 1, x 2,... ) l 2 : (0, 2x 2, 0, 4x 4,... ) l 2 }. T (x 1, x 2,... ) = (0, 2x 2, 0, 4x 4,... ) for all (x 1, x 2,... ) D(T ). It can be shown that T = T and R(T ) is closed. Let {e n } n=1 be the standard orthogonal basis for l 2. Here R(T ) = span(e 2, e 4,..., e 2n,... ). Let Y n := span{e 2, e 4,..., e 2n }. Then Y n Y n+1, dim(y n ) = n and n=1 Y n = R(T ). Since T = T, we have I + T T = I + T 2. For any x = (x 1, x 2,... ) D(T 2 ), (I + T 2 )x = (x 1, 5x 2, x 3, 17x 4,..., x 2n 1, (1 + 4n 2 )x 2n,... ). For any y = (y 1, y 2,... ) l 2, (I + T 2 ) 1 y = (y 1, y 2 5, y 3, 4 17 y y 2n 4,..., y 2n 1, 1 + 4n 2,... ), y = (y 1, y 2,..., ) l 2. In particular, (I + T 2 ) 1 (e 2n ) = e 2n 1 + 4n 2. Hence Z n = (I + T 2 ) 1 Y n = Y n. Also X n = T Z n = Y n. Then X n = Y n = Z n. Hence P n = Q n. That is P n x = x 2 e 2 + x 4 e 4 + + x 2n e 2n for all x = (x 1, x 2,..., x n,..., ) l 2. T n = P n T. There fore T n x = 2x 2 e 2 + 4x 4 e 4 + + 2nx 2n e 2n. Hence T # n (y) = { y2 2 e 2 + y 4 4 e 4 +... y 2n 2n e 2n if y Y n 0 if y Yn. Hence by Theorem 3.3, D(T ) = {y l 2 : lim T # n T y = lim T # n exists} = l 2 and = lim ((0, 1 2 y 2, 0, 1 4 y 4)) for all y = (y 1, y 2,... ) l 2. Concluding remarks: For an arbitrary densely defined closed operator T, computing Z n, X n may be difficult. A procedure to compute (I + T T ) 1 is indicated in [7]. We hope to apply this procedure to some concrete densely defined closed operators in future. References 1. Generalized inverses and applications, Academic Press [Harcourt Brace Jovanovich Publishers], New York, 1976, Edited by M. Zuhair Nashed, University of Wisconsin, Mathematics Research Center, Publication No. 32. MR MR0451661 (56 #9943) 2. Y. A. Abramovich and C. D. Aliprantis, Problems in operator theory, Graduate Studies in Mathematics, vol. 51, American Mathematical Society, Providence, RI, 2002. MR MR1921783 (2003h:47073) 3. N. I. Akhiezer and I. M. Glazman, Theory of linear operators in Hilbert space, Dover Publications Inc., New York, 1993, Translated from the Russian and with a preface by Merlynd Nestell, Reprint of the 1961 and 1963 translations, Two volumes bound as one. MR MR1255973 (94i:47001)

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