J. Ana. Num. Theor. 1, No. 1, 1-5 (2013) 1 Journal of Analysis & Number Theory An International Journal

°c 2013 NSP Natural Sciences Publishing Cor.

Multiplication Operators with Closed Range in Operator Algebras

P. Sam Johnson Department of Mathematical and Computational Sciences, National Institute of Technology Karnataka, Mangalore, India

Received: 16 Jul. 2012; Revised 22 Oct. 2012; Accepted 23 Oct. 2012 Published online: 1 Jan. 2013

Abstract: Let B(H) denote the C∗-algebra of all bounded linear operators from a H into itself. Let T ∈ B(H). Define LT : B(H) → B(H) by LT (S) = TS and define RT : B(H) → B(H) by RT (S) = ST . Consider the following three statements: (i) T has closed range in H, (ii) LT has closed range in B(H), (iii) RT has closed range in B(H). It is proved that all these three statements are equivalent. Some possibilities of extending this result to Banach spaces have been discussed.

Keywords: Closed range operator ; Hahn-Banach extension property.

1. Introduction

Many of the concrete applications of mathematics in science and engineering, eventually result in a problem involving operator equations. This problem can be usually represented as an operator equation T x = y (1) where T : X → Y is a linear or nonlinear operator (between certain function spaces or Euclidean spaces) such as differential operator or integral operator or a matrix. The spaces X and Y are linear spaces endowed with certain norms on them. Solving linear equations with infinitely many variables is a problem of , while solving equations with finitely many variables is one of the main themes of linear algebra. The problem of solving the equation (1) is well-posed if it asserts existence and uniqueness of a solution of (1) and the continuous dependence of the solution on the data y. It is well-known that the problem of solving the operator equation (1) is essentially well-posed if the range of T is closed. The study of operators with closed range plays a vital role in perturbation theory. The composition of closed range continuous linear operators between Frechet´ spaces does not have closed range, in general. We have given necessary and sufficient conditions for the composition operator TS to have closed range in [3, 8]. Bouldin [1,2] has given a geometric characterization in terms of the angle between two linear subspaces for operators on Hilbert spaces and for the operators on Banach spaces. The closedness of range of a linear operator T is helpful in establishing the Hyers-Ulam stability of T between Frechet´ spaces [7]. The range and null spaces of a linear operator T are denoted by R(T ) and N(T ) respectively. In this paper, we first discuss the closedness of multiplication operators on Hilbert spaces. Kulkarni and Nair [5] proved that for a bounded linear operator T from a Hilbert space H into a Hilbert space K, T has closed range in K if and only if σ(T ∗T ) ⊆ {0} ∪ [γ, kT k2], for some γ > 0. In this paper it is proved that ∗ 2 σ(x x) ⊆ {0} ∪ [γ, kxk ], for some γ > 0 if and only if Lx : A → A defined by Lx(y) = xy has closed range in A where x is any element of a commutative C∗-algebra A with identity. Moreover, the multiplication operators with closed range are found on some classical Banach algebras. Throughout the paper, the following characterization for closed range operators on Banach spaces is used. Theorem 1. [3] Let T : X → Y be a bounded linear operator from a X into a Banach space Y . Then T has closed range in Y if and only if there is a constant c > 0 such that for given x ∈ X, there is an element y ∈ X such that T x = T y and kyk ≤ ckT xk.

∗ Corresponding author: e-mail: [email protected]

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We shall apply this theorem to multiplication operators on Banach algebras. For a given A and a given element x in A, Lx and Rx will denote the left multiplication operator and the right multiplication operator, respectively; that is; Lx(y) = xy and Rx(y) = yx, for every y ∈ A. (1) If A is a Banach algebra with a multiplicative identity element e, then L0 and Le have closed ranges. (2) If x is invertible, then Lx has closed range ; this result follows from the next proposition. (3) If A is a commutative Banach algebra with an identity element e and with a non trivial idempotent element e0, then 0 0 0 0 0 Le0 has closed range (because e xn → x implies that e xn = e e xn → e x = x). Proposition 2 Let A be a Banach algebra with multiplicative identity e. Then x ∈ A is right invertible in A if and only if Lx(A) = A. −1 −1 −1 Proof. Suppose x is right invertible in A with right inverse x . Then for y ∈ A, Lx(x y) = xx y = ey = y. Thus Lx(A) = A. Conversely, assume that Lx(A) = A. Then for e ∈ A, there is a y ∈ A such that e = Lx(y) = xy so that x is right invertible with the right inverse y. We now find multiplication operators with closed ranges in some classical Banach algebras.

Proposition 3 For 1 ≤ p < ∞, the collection of all x ∈ `p such that Lx has closed range is the space ∞ c00 = {x = (xn)n=1 : xn = 0 for all except for finitely many n} . ∞ Proof.If x ∈ c00, then the range of Lx is finite dimensional and hence it is closed in `p. If x = (xn)n=1 ∈/ c00, then there ∞ ∞ 1 (m) is a subsequence (xnm )m=1 of (xn)n=1 such that |xnm | < m and xnm 6= 0 for every m. For each nm, define x ∈ `p by ½ (m) m if i = n x = m i 0 elsewhere.

(m) (m) (m) (m) (m) (m) If, for given m, there is a z ∈ `p such that Lxz = Lxx , then znm = m so that kz kp ≥ m and kLxx kp ≤ (m) (m) 1. In this case, there is no positive constant c such that kz kp ≤ ckLxx kp, for every m. Therefore Lx does not have closed range in `p; by theorem 1.

Proposition 4 For a given element x ∈ `∞, Lx has closed range in `∞ if and only if x = (0, 0, 0,...), or, infn {|xn| : xn 6= 0} > 0. This proposition follows from the next proposition. The argument used in the previous proof shall be extended to the following proposition.

Proposition 5 Let (X, M, µ) be a (non-negative) positive measure space. For a given element f ∈ L∞(X), Lf has closed range in L∞(X) if and only if f = 0 almost everywhere on X or ess. inf. {|f(x)| : x ∈ X, f(x) 6= 0} > 0 on {x ∈ X : f(x) 6= 0}.

Proof.If f = 0, then Lf (L∞(X)) = {0} so that it has closed range in L∞(X). Suppose f 6= 0 almost everywhere on X and assume that 0 < ess. inf. {|f(x)| : x ∈ X, f(x) 6= 0} = c, say, on {x ∈ X : f(x) 6= 0}. Then for each g ∈ L∞(X), define h such that ½ g(x) if f(x) 6= 0 h(x) = 0 if f(x) = 0 then ckhk∞ ≤ kfgk∞ = kLf (g)k∞ and Lf (g) = Lf (h). Thus, by theorem 1, Lf has closed range in L∞(X). Suppose f 6= 0 almost everywhere on X and ess. inf. {|f(x)| : x ∈ X, f(x) 6= 0} = 0

∞ on {x ∈ X : f(x) 6= 0}. Then we can find a sequence (En)n=1 of disjoint measurable subsets of X with positive measures 1 such that |f(x)| ≤ n , for all x ∈ En, and f(x) 6= 0 almost everywhere on En. For each integer n, define gn ∈ L∞(X) by ½ n if x ∈ E g (x) = n n 0 elsewhere.

If, for each n, there is a hn ∈ L∞(X) such that Lf (hn) = Lf (gn), then hn(x) ≥ n on En almost everywhere, khnk∞ ≥ n and kLf (gn)k ≤ 1. In this case, there is no positive constant c > 0 such that ckhnk∞ ≤ kLf (gn)k∞, for every n. Therefore Lf does not have closed range in L∞(X), by theorem 1.

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It is possible to find multiplication operators with closed range to each individual Banach algebra. The following theorem gives a characterization of closed range multiplication operators on commutative C∗-algebras with multiplicative identity elements. Theorem 6. Let A be a commutative C∗-algebra over C with a multiplicative identity element. Let x ∈ A. Then the following three statements are equivalent: (1) Lx has closed range in A. (2) 0 is not a limit point of σ(x). (3) σ(x∗x) ⊆ {0} ∪ [γ, kxk2], for some γ > 0.

Proof. By the Gelfand-Naimark theorem [10], let us assume that A = C(∆), the algebra of all complex valued continuous functions on the maximal ideal space ∆ of A, where ∆ is a compact Hausdorff space under the Gelfand-topology. Fix x ∈ A. Note that Lx has closed range in A if and only if there is a positive constant c > 0 such that for each y ∈ A there is a z ∈ A such that kzk ≤ ckxyk and xy = xz; that is, sup |z(m)| ≤ c sup |x(m)y(m)| and xy = xz. (2) m∈∆ m∈∆ 1 If {m ∈ ∆ : x(m) = 0} is open in ∆, then we can take c = supm∈∆ |x(m)| and ½ y(m) if x(m) 6= 0 z(m) = 0 if x(m) = 0 so that the previous condition (2) is satisfied. Suppose {m ∈ ∆ : x(m) = 0} is not open in ∆. Then, for each n > 0, find yn ∈ A such that yn(m) ∈ [0, 1], for all 1 1 m ∈ ∆, yn(m) = 0 for |x(m)| ≥ n and yn(m) = 1 for |x(m)| ≤ 2n . This is possible by the Urysohn’s lemma. 1 If there is an zn ∈ A such that xyn = xzn, then zn(m) = yn(m) for some m with 0 < |x(m)| ≤ 2n and hence 1 supm∈∆ |zn(m)| ≥ 1 = supm∈∆ |yn(m)| but supm∈∆ |x(m)yn(m)| ≤ n and hence there is no positive constant c such that supm∈∆ |zn(m)| ≤ c supm∈∆ |x(m)yn(m)|, for each n = 1, 2,.... Thus the condition (2) fails to be true in this case. Therefore, we conclude that Lx has closed range in A if and only if {m ∈ ∆ : x(m) = 0} is open in ∆. That is, Lx has closed range in A if and only if 0 is not a limit point of σ(x). Thus (6) and (6) are equivalent. It is clear that (6) and (6) are equivalent.

We now discuss the closedness of range of multiplication operators on Hilbert spaces. Let B(H) denote the C∗- algebra of all bounded linear operators from a Hilbert space H into itself. For given T ∈ B(H), let LT : B(H) → B(H) and RT : B(H) → B(H) be the left and right multiplication operators defined by LT (S) = TS and RT (S) = ST , respectively. The following theorem gives a relation between the closedness of range of T and the range of left/right multiplication operators.

Theorem 7. Let H be a Hilbert space and T ∈ B(H). Then the following statements are equivalent : (1) T has closed range in H. (2) LT has closed range in B(H). (3) RT has closed range in B(H).

Proof. Suppose that the range R(T ) is closed in H. Suppose RT (Tn) = TnT converges to S in B(H), as n → ∞, where Tn ∈ B(H). If x ∈ N(T ), then S(x) = 0 = limn→∞ Tn(T (x)). For each x ∈ H, define Λ(T (x)) = Sx. If T x = T y, then x − y ∈ N(T ) and hence S(x − y) = 0 so that Sx = Sy. Thus Λ is well defined on R(T ). The linearity of T and S imply the linearity of Λ on R(T ). Since R(T ) is closed in H, there is a positive constant c > 0 such that for given x ∈ H, there is a y ∈ H satisfying kyk ≤ ckT xk and T x = T y. For such elements x and y we have

kΛ(T x)k = kΛ(T y)k = kSyk ≤ kSk kyk ≤ kSk c kT xk.

This proves that Λ is continuous on R(T ). 0 0 ⊥ 0 Extend Λ to Λ : H → H by Λ (y1 + y2) = Λ(y1) for y1 ∈ R(T ) and y2 ∈ R(T ) . Then Λ ∈ B(H). Also 0 0 S = ΛT = Λ T = RT (Λ ). Thus S ∈ RT (B(H)), the range of RT . This proves that the range of RT is closed in B(H). Therefore (7) implies (7). Now, assume that R(T ) is not closed in H. Then there is a sequence (xn) in H and element y0 ∈ H such that T xn → y0 as n → ∞ and y0 ∈/ R(T ). Fix a non-zero continuous linear finctional f on H. To each n, define Tn : H → H

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by Tn(x) = f(x)xn. Then each Tn ∈ B(H) and TTn ∈ B(H). Also TTn converges to S defined by S(x) = f(x)y0, in 0 0 B(H). Therefore S is in the closure of LT (B(H)). If there is some T ∈ B(H) such that S = TT , then y0 ∈ R(S) ⊆ R(T ), which is impossible. Therefore S/∈ LT (B(H)). Thus LT (B(H)) is not closed in B(H). Therefore, if LT has closed range in B(H), then R(T ) is closed in H. Thus, (7) is proved. Assume that (7) is true. Therefore {ST : S ∈ B(H)} is closed in B(H). Then {(ST )∗ : S ∈ B(H)} is closed in B(H). ∗ ∗ ∗ By Banach’s closed range theorem [10], {T S : S ∈ B(H)} is closed in B(H). That is, LT ∗ (B(H)) is closed in B(H). Therefore, the previous part of the proof implies that T ∗ has closed range in H. The first part of the proof implies that {S∗T ∗ : S∗ ∈ B(H)} is closed in B(H). That is, {(TS)∗ : S ∈ B(H)} is closed in B(H). Therefore, {TS : S ∈ B(H)} is closed in B(H). Thus (7) is proved.

The author is not able to extend this theorem to Banach spaces. However some partial extensions are obtained. It is proved in the next section that any RT has closed range in B(X), if X has Hahn-Banach extension property and T has closed range in X. It is also observed that T has closed range in any Banach space X, if LT or RT has closed range in B(X).

2. Partial Extensions

Definition 1. Let Z be a normed space. Z is said to have Hahn-Banach extension property if for any linear subspace Y of a normed space and for any continuous linear operator T : Y → Z, there is a continuous linear extension Te : X → Z to T such that kT k = kTek.

There are many characterizations for normed spaces with Hahn-Banach property in [4] and [9].

Theorem 8. Let X be a Banach space with Hahn-Banach extension property. Let T ∈ B(X) has closed range in X. Then LT : B(X) → B(X) has closed range.

Proof. Suppose TnT converges to some S in B(X), for a sequence (Tn) in B(X). If x ∈ N(T ), then Sx = 0 = limn→∞ TnT x. For each x ∈ X, define Λ(T x) = Sx. If T x = T y, then x − y ∈ N(T ), S(x − y) = 0 and Sx = Sy. Thus Λ is well defined on R(T ). The linearity of T and S imply the linearity of Λ on R(T ). Since R(T ) is closed in X, by theorem 1, there is a positive constant c > 0 such that for given x ∈ X, there is an element y ∈ X satisfying kyk ≤ ckT xk and T x = T y. For such elements x and y we have kΛ(T x)k = kΛ(T y)k = kSyk ≤ kSk kyk ≤ ckSk kT xk. This proves that Λ is continuous on R(T ). Extend Λ to a continuous linear operator Λ0 : X → X, which is possible because X has Hahn-Banach extension 0 0 property. Then S = ΛT = Λ T = RT (Λ ) and hence S ∈ RT (B(X)), the range of RT . This proves that the range of RT is closed in B(X).

Theorem 9. Let X be a Banach space. Let T ∈ B(X). Suppose LT : B(X) → B(X) has closed range. Then T has closed range in X.

∞ Proof. On the contrary, assume that R(T ) is not closed in X. Then there is a sequence (xn)n=1 in X and an element y0 ∈ X such that T xn → y0 as n → ∞, and y0 ∈ R(T ). Fix a non zero continuous linear functional f on X. To each n, define Tn : X → X by Tn(x) = f(x)xn. Then each Tn ∈ B(X) and TTn ∈ B(X). Also TTn converges to S defined 0 0 by S(x) = f(x)y0 in B(X). Therefore S is in the closure of LT (B(X)). If there is some T ∈ B(X) such that S = TT then y0 ∈ R(S) ⊆ R(T ), which is impossible. Therefore S ∈ LT (B(X)). Thus LT (B(X)) is not closed in B(X), which is a contradiction. This proves the result.

Theorem 10. Let X be a Banach space. Let T ∈ B(X). Suppose RT : B(X) → B(X) has closed range. Then T has closed range in X.

Proof. Since kSk = kS∗k and (ST )∗ = T ∗S∗, for all S ∈ B(X), and since {ST : S ∈ B(X)} is closed in B(X), { T ∗S∗ : S∗ ∈ B(X∗) } is also closed in B(X∗), where X∗ is the dual of X and T ∗ : X∗ → X∗ is the adjoint of T . That ∗ ∗ ∗ ∗ is, LT ∗ : B(X ) → B(X ) has closed range. So, by theorem 9, T has closed range in X . This implies that T has closed range in X, because it is known in [10] that S ∈ B(X) has closed range in X if and only if S∗ ∈ B(X∗) has closed range in X∗.

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Further Remarks. If X = H is a Hilbert space and T ∗ is the Hilbert adjoint of T ∈ B(H), then {ST : S ∈ B(H)} ∗ ∗ ∗ is closed in B(H) if and only if {T S : S ∈ B(H)} is closed in B(H). So, LT : B(H) → B(H) has closed range if 0 and only if RT : B(H) → B(H) has closed range. In the proof of the theorem 7, Λ can be extended to Λ by making Λ0(R(T )⊥) = {0}, if X is a Hilbert space. This type of extension is possible only in Hilbert spaces because every closed subspace is complemented in a Banach space if and only if the Banach space is a Hilbert space [6]. Some new arguments are expected to extend the theorem 7 to Banach spaces. Acknowledgements The present work of the author was partially supported by SERC, DST, Government of India, through the Fast Track Scheme for Young Scientists (D.O. No. SR/FTP/MS-050/2011).

References

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