Ronalda Benjamin. Definition A normed space X is complete if every Cauchy sequence in X converges in X.

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1 Group inverses in a Banach algebra Ronalda Benjamin Talk given in mathematics postgraduate seminar at Stellenbosch University on 27th February 2012 Abstract Let A be a Banach algebra. An element a A is said to be group invertible if there exists an element b A such that ab = ba, b = bab and a = aba. It should be noted that this concept makes sense even if we only had an algebra with identity. For the purpose of the talk I wanted to work in one algebraic structure, namely a Banach algebra. Observe from the definition of a group inverse that if a A 1, then b = a 1 trivially satisfies the three conditions; hence group invertibility is a generalization of invertibility. In this talk we discussed basic properties like the existence and uniqueness of a group inverse and the spectrum of a group invertible element. The main part of the talk, however, was to present continuity results for group inversion in a Banach algebra. Preliminaries Definition A normed space is a vector space X with a norm defined on it. Here, a norm on a complex vector space X is a real-valued function on X whose value is denoted by x and which has the properties (a) x 0 (b) x = 0 x = 0 (c) αx = α x (d) x + y x + y for all x, y in X and α C. Definition A normed space X is complete if every Cauchy sequence in X converges in X. Definition A complete normed space is called a Banach space. 1

2 Definition An algebra is a vector space A over a field K such that for each ordered pair of elements x, y A, a unique product xy A is defined satisfying the properties (a) x(yz) = (xy)z (b) (x + y)z = xz + yz (c) x(y + z) = xy + xz (d) λ(xy) = (λx)y = x(λy) for all x, y, z in A and λ C. If K = R then A is said to be a real algebra, whereas if K = C then A is called a complex algebra. If the multiplication is commutative, that is, for all x, y A xy = yx, then A is called a commutative algebra. Definition If A is an algebra over C which is also a Banach space and the algebra multiplication and Banach space norm satisfies the norm inequality xy x y for all x, y A and 1 = 1 where 1 is the identity of A, then we say that A is a Banach algebra. Examples of Banach algebras: Let K be a compact set of C. Then C(K), the vector space of all complex continuous functions on K is a Banach algebra with: Multiplication: (xy)(t) = x(t)y(t) Norm: x = sup{ x(t) : t K} Identity: 1(t) = 1. Let X be a Banach space. Then L(X), the vector space of all bounded linear operators on X, is a Banach algebra with: Multiplication: (ST )x = S(T x) T (x) Norm: T = sup{ : 0 x X} x Identity: 1(x) = x. Definition Let A be a Banach algebra. An element a A is nilpotent if there exists a positive integer n such that a n = 0. Definition Let A be a Banach algebra. An element a A is an idempotent if it satisfies p 2 = p. 2

3 Theorem 1 Let A be a Banach algebra. If x A satisfies x < 1, then 1 x A 1 and we have that (1 x) 1 = x k. k=0 Definition Let A be a Banach algebra. The spectrum σ A (a) of an element a A is defined as follows: σ A (a) := {λ C : λ1 a / A 1 } Examples of the spectrum of a Banach algebra element include: Consider the Banach algebra C[a, b]. Then x C[a, b] has σ C[a,b] (x) = x[a, b]. Consider the Banach algebra M n (C). Then a M n (C) has σ Mn(C)(a) = {λ : λ is an eigenvalue of a} A very important property of σ A (a) is that it is non-empty and compact, that is, σ A (a) is a closed and bounded subset of C. The algebra of all complex valued functions f, defined and holomorphic on an open set Ω C, is denoted by H(Ω). Theorem 2 Let A be a Banach algebra and a A. Suppose that Ω is an open set containing σ A (a) and that Γ is a smooth contour included in Ω and surounding σ A (a). Then the mapping from H(Ω) into A, defined by f f(a) = 1 f(λ)(λ1 2πi Γ a) 1 dλ, has the following properties: (1) (f 1 + f 2 )(a) = f 1 (a) + f 2 (a), (2) (f 1.f 2 )(a) = f 1 (a).f 2 (a), (3) 1(a) = 1 and I(a) = a, where I is the identity function on C, (4) If (f n ) converges to f uniformly on all compact subsets of Ω, then f(a) = lim n f n (a) and (5) σ(f(a)) = f(σ(a)). Number (5) above is called the spectral mapping theorem. 3

4 Theorem 3 Let A be a Banach algebra. Suppose that a A has a disconnected spectrum. Let U 0 and U 1 be two disjoint open sets such that σ A (a) U 0 U 1, σ A (a) U 1 and σ A (a) U 0. Then there exists a non-trivial idempotent p commuting with a, such that σ A (pa) = (σ A (a) U 1 ) {0} and σ A (a pa) = (σ A (a) U 0 ) {0}. Moreover, p = f(a), where { 0 if λ U0 f(λ) = 1 if λ U 1. We call p in Theorem 3 the spectral idempotent of a. Uniqueness and existence of group inverses Definition Let A be a Banach algebra. An element a A is said to be group invertible if there exists an element b A such that: ab = ba b = bab a = aba We call b a group inverse of a. It is obvious that if b is a group inverse of a, then a is a group inverse of b, that is, group inverses are symmetric. It is also clear from the definition of a group inverse that every idempotent is group invertible with itself as a group inverse. Let A g denote the set of group invertible elements of a Banach algebra A. It is easy to verify that the equations in the Definition above hold when a is invertible (b = a 1 ); hence, A 1 A g. The inclusion is strict since 0 A g, but 0 / A 1. Lemma 4 Let A be a Banach algebra and a, b A. (1) If a is group invertible with group inverse b, then a k is group invertible 4

5 with group inverse b k. (2) If a is group invertible with group inverse b, then ab is an idempotent. (1) Since a and b commute, we have that a k b k = (ab) k = (ba) k = b k a k, b k a k b k = (bab) k = b k and a k b k a k = (aba) k = a k. Hence b k is a group inverse of a k. (2) Using the fact that bab = b, we have that (ab) 2 = a(bab) = ab. Since group invertibility is a generalization of invertibility, it is natural to ask whether certain results available for invertibility also hold for group invertibility. It is well known that inverses are unique provided they exist. In the next result we show that this fact is also true for group inverses. Proposition 5 Let A be a Banach algebra. An element in A g can have at most one group inverse. Let a be group invertible with group inverses b and c. Then, using Lemma 4(2) and the fact that both b and c commute with a, we have that and b = bab = b 2 a = b 2 aca = b 2 a(ca) 2 = b 2 ac 2 a 2 = b 2 a 3 c 2 c = cac = ac 2 = abac 2 = (ab) 2 ac 2 = a 2 b 2 ac 2 = b 2 a 3 c 2. Hence b = c. The group inverse of a Banach algebra element a will be denoted by a g. Our next concern is the existence of a group inverse of a Banach algebra element. We observe that, for an arbitrary Banach algebra A, we do not have that A = A g, that is, not every element is group invertible. For instance, if we consider the Banach algebra C[0, 1] and a C[0, 1], defined by a(x) = x for x [0, 1], then this fact can be easily seen from a result we will prove later on (Lemma 9). This lemma describes the spectrum of a group invertible element. In our next result we address the problem of developing conditions for the existence of a group inverse in a Banach algebra. Lemma 6 Let A be a Banach algebra. An element a A is group invertible 5

6 if and only if there exists an idempotent p A such that ap = pa, a + p is invertible and ap = 0. (0.1) If the conditions in (0.1) are satisfied, then a g = (a + p) 1 (1 p). Suppose that a A is group invertible with a g = b. Let p = 1 ba. Then by Lemma 4(2), p is an idempotent. We also have that ap = pa, pb = bp and ap = 0 = pb. So it follows that (a + p)(b + p) = ab + ap + pb + p 2 = ab + p = 1. Hence a + p is invertible with (a + p) 1 = b + p. Conversely, suppose that the conditions in (0.1) hold. We show that b = (a + p) 1 (1 p) is the group inverse of a : Since ap = pa we have that ba = (a + p) 1 (1 p)a = a(a + p) 1 (1 p) = ab. Using the fact that p(1 p) = 0, it follows that (a + p)(1 p) = a(1 p). Hence and bab = [(a + p) 1 (1 p)]a[(a + p) 1 (1 p)] = (a + p) 1 (1 p)(a + p) 1 a(1 p) = (a + p) 2 (1 p)a = [(a + p) 2 (1 p)](a + p) = (a + p) 1 (1 p) = b aba = a[(a + p) 1 (1 p)]a = (a + p)[(a + p) 1 (1 p)]a = (1 p)a = a. Corollary 7 Let A be a Banach algebra and a A g. The idempotent p satisfying (0.1) is unique. Suppose that a A g and that p 1 and p 2 satisfies (0.1). By Lemma 6, both (a + p 1 ) 1 (1 p 1 ) and (a + p 2 ) 1 (1 p 2 ) are group inverses of a. It then follows from Proposition 5 that (a + p 1 ) 1 (1 p 1 ) = (a + p 2 ) 1 (1 p 2 ), and hence (1 p 1 )(a + p 2 ) = (a + p 1 )(1 p 2 ). This gives a + p 2 ap 1 p 1 p 2 = a ap 2 + p 1 p 1 p 2. Using the fact that ap 1 = 0 = ap 2, we have that p 1 = p 2, 6

7 confirming the uniqueness. We call p in Lemma 6 the group idempotent of a. Corollary 8 Let A be a Banach algebra. If a A g, then the group idempotent is given by 1 a g a. Spectrum of a group invertible element We denote by iso σ(a) the set of all isolated spectral points of a. Lemma 9 Let A be a Banach algebra and a A g. Then either 0 / σ A (a) (i.e. a is invertible) or 0 iso σ A (a). If a = 0, then σ A (a) = {0}, and hence 0 is an isolated point of the spectrum. If a 0 is group invertible with a g = b, then b 0. Let λ C be such that 0 < λ < 1 ; that is λb < 1. By Theorem 1 we have that 1 λb is b invertible. Using simple algebraic arguments and the fact that b is the group inverse of a, we obtain that [(1 λb) 1 b 1λ ] (1 ba) (a λ1) = 1, so that a λ1 is invertible. Since λ / σ A (a) for all λ satisfying 0 < λ < 1, we have that σ b A(a) {0} {λ : λ 1 }. The result then follows. b A natural question would be whether the converse of Lemma 9 also holds, that is, if a Banach algebra element a is such that 0 iso σ A (a), can we conclude that a is group invertible? The answer is no in general and we provide the following example to illustrate this. Consider the Banach algebra M 2 (C). Let a = ( ). Then σm2 (C)(a) = {0}; hence 0 iso σ M2 (C)(a). But a is a nonzero nilpotent element; hence it is not group invertible. 7

8 Definition Let A be a Banach algebra. An element a A is quasinilpotent if σ A (a) = {0}. The set of all quasinilpotent elements in A will be denoted by QN(A). The radical of a Banach algebra A, denoted by Rad(A), is defined by Rad(A) = {x A : zx QN(A), for all z A}. If Rad(A) = {0}, then A is called semisimple. If a Banach algebra A is commutative, then Rad A = QN(A). The following proposition shows that in the case of semisimple commutative Banach algebras, the spectral condition in Lemma 9 is sufficient. The proof of the result requires knowledge on the holomorphic functional calculus (Theorems 2 and 4) which I did not discuss during my talk. Therefore, only for interest sake I included the proof. Proposition 10 Let A be a semisimple commutative Banach algebra. An element a A is group invertible if and only if 0 / σ A (a) (i.e. a is invertible) or 0 iso σ A (a). By Lemma 9, if a is group invertible, then 0 / σ A (a) or 0 iso σ A (a). Since a A 1 implies that a is group invertible, we are only left to show that 0 iso σ A (a) implies that a is group invertible. Let 0 iso σ A (a) and U 1 and U 0 be disjoint neighbourhoods containing {0} and σ A (a)\{0} respectively. Then U = U 0 U 1 is a neighbourhood containing σ A (a). Define f : U C as follows: { 0 if λ U0 f(λ) = 1 if λ U 1. By Theorem 3, p = f(a) is the spectral idempotent of a corresponding to 0. Moreover, p 0 and commutes with a. Let g(λ) = f(λ) + λ. Then { λ if λ U 0 g(λ) = 1 + λ if λ U 1. Moreover, g H(U) and g(σ A (a)) = {g(λ) : λ σ A (a)\{0}} {g(λ) : λ = 0} = σ A (a)\{0} {1}. 8

9 Hence 0 / g(σ A (a)) and by the spectral mapping theorem (Theorem 2(5)) it follows that 0 / σ A (g(a)), so that p + a = f(a) + a = g(a) A 1. Let h(λ) = λf(λ). Then { 0 if λ U0 h(λ) = λ if λ U 1. Moreover, h H(U) and h(a) = af(a) = ap. Since h(σ A (a)) = h[(σ A (a) U 1 ) (σ A (a) U 0 )] = h({0}) {0} = {0}, we have again by the spectral mapping theorem that σ A (ap) = σ A (h(a)) = h(σ A (a)) = {0}. Hence ap QN(A). Since A is semisimple and commutative, QN(A) = Rad(A) = {0}, that is ap = 0. By Lemma 6 we have that a is group invertible. Continuity of group inversion Group inversion refers to the map defined on A g, which maps a group invertible element to its group inverse. It is well known that inversion is continuous on A 1, so a natural question would be whether group inversion is continuous on A g. The answer is no in general and we provide the following example to illustrate this. Consider the Banach algebra M 2 (C). For n N, let ( 1 1 a n = + 1 ) n 1 0 n and a = ( ) Since det a n 0, we have that all a n are invertible and hence group invertible with ( ) 1 (1 + n) a g n = a 1 n =. 0 n 9

10 We also have that, since a is idempotent, a is group invertible with group inverse itself. It is clear that a n a as n, but (a g n) does not converge to a g, infact (a g n) diverges. We continue by providing the following criteria to obtain continuity of group inversion. Lemma 11 Let A be a Banach algebra and (a n ) a convergent sequence in A with limit a such that every a n and a are group invertible. Then the group inverses of a n converge to the group inverse of a if and only if the group idempotents of a n converge to the group idempotent of a. Suppose that b n and b are the group inverses of a n and a respectively and that b n b as n. By using Lemma 6 and Corollaries 7 and 8, we have that p n = 1 b n a n and p = 1 ba are the group idempotents of a n and a respectively. It now follows that lim p n = lim (1 b n a n ) = 1 ba = p. n n Conversely, suppose that p n and p are the group idempotents of a n and a respectively and that p n p as n. Using Lemma 6 again, we have that b n = (a n + p n ) 1 (1 p n ) and b = (a + p) 1 (1 p) are the corresponding group inverses of a n and a. The result then follows from the fact that a n a, p n p and the continuity of inversion. We introduce the next concept and use it to state another characterization for continuity of group inversion. Definition Let A be a Banach algebra and p and q idempotents in A. If there exists an invertible element c A such that q = c 1 pc, then p and q are similar and we write p q. Lemma 12 Let A be a Banach algebra and p and q idempotents in A such that p q < 2p 1 1. Then p q. Suppose that p and q are idempotents in A satisfying p q < 2p 1 1. Observe that the element 2p 1 is invertible with inverse itself. By assumption we have that (2p 1) 1 (q p) (2p 1) 1 q p < 2p 1 2p 1 1 = 1, and hence 1 + (2p 1) 1 (q p) is invertible by Theorem. Let c := p + q 1. 10

11 It then follows that c = (2p 1) + (q p) = (2p 1)[1 + (2p 1) 1 (q p)] is also invertible. Now, pc = p(p + q 1) = p + pq p = pq = pq + q q = (p + q 1)q = cq, that is, q = c 1 pc; hence p q. The algebra pap is defined by pap = {pap : a A}. We state the following result without proof. Theorem 13 Let A be a Banach algebra, (a n ) A a convergent sequence with limit a A, and suppose all a n and a are group invertible with corresponding group idempotents p n and p. Also assume that pap is a finite dimensional algebra. Then the following statements are equivalent: (a) the group inverses of a n converge to the group inverse of a (b) the group idempotents p n of a n are similar to the group idempotent p of a. Group inverses has various applications, especially the group inverse of a square matrix. An important application of the group inverse of a sqaure matrix is in the study of finite Markov chains. References [1] S. Roch and B. Silbermann: Continuity of generalized inverses in Banach algebras. Studia Math. 136 (1999),