Spectral decomposition of normal absolutely minimum attaining operators
aa r X i v : . [ m a t h . F A ] M a y SPECTRAL DECOMPOSITION OF NORMAL ABSOLUTELYMINIMUM ATTAINING OPERATORS
NEERU BALA AND G. RAMESH
Abstract.
Let T : H Ñ H be a bounded linear operator defined be-tween complex Hilbert spaces H and H . We say T to be minimum at-taining if there exists a unit vector x P H such that } T x } “ m p T q , where m p T q : “ inf t} T x } : x P H , } x } “ u is the minimum modulus of T . We say T to be absolutely minimum attaining ( AM -operators in short), if for anyclosed subspace M of H the restriction operator T | M : M Ñ H is minimumattaining. In this paper, we give a new characterization of positive absolutelyminimum attaining operators ( AM -operators, in short), in terms of its essen-tial spectrum. Using this we obtain a sufficient condition under which theadjoint of an AM -operator is AM . We show that a paranormal absolutelyminimum attaining operator is hyponormal. Finally, we establish a spectraldecomposition of normal absolutely minimum attaining operators. In provingall these results we prove several spectral results for paranormal operators. Weillustrate our main result with an example. Introduction
The class of minimum attaining operators on Hilbert spaces is first introducedby Carvajal and Neves [6]. An important property of this class of operators is thatit is dense in B p H , H q , the space of all bounded linear operators between H and H with respect to the operator norm [18]. For densely defined closed operators inHilbert spaces, this class of operators is studied in [26].An important subclass of the minimum attaining operators is the class of abso-lutely minimum attaining operators, which was introduced by Carvajal and Nevesin [6]. We say T is an absolutely minimum attaining operator or AM -operator,if for every closed subspace M of H the restriction T ˇˇ M : M Ñ H is minimumattaining.The minimum attaining operators and the absolutely minimum attaining oper-ators are defined analogous to those of norm attaining and the absolutely normattaining operators, respectively. Recall that T P B p H , H q is norm attaining op-erator if there exists a x P H with } x } “ } T x } “ } T } and absolutelynorm attaining or AN -operator, if for every closed subspace N of H the operator T ˇˇ N : N Ñ H is norm attaining.There is a close connection between the minimum attaining operators and normattaining operators. Similarly, absolutely minimum attaining operators and theabsolutely norm attaining operators are related with each other. Date : 03:55; Friday 18 th May, 2018.2010
Mathematics Subject Classification.
Key words and phrases.
Minimum modulus, absolutely minimum attaining operator, essentialspectrum, generalized inverse, paranormal operator.
A partial characterization of positive, absolutely norm attaining operators wasgiven in [6], which was further improved in [20, 22]. Another characterization ofpositive, absolutely norm attaining operators is discussed in [23].In the present article we give a characterization for positive absolutely minimumattaining operators. Similar to absolutely norm attaining operators, adjoint of anabsolutely minimum attaining operator need not be absolutely minimum attaining.Here we will give a condition similar to the one given in [23], for the adjoint of anabsolutely minimum attaining operator to be absolutely minimum attaining.Next we will study the class of paranormal, AM -operators. In particular, wewill show that the class t T P AM p H q : T is paranormal with N p T q “ N p T ˚ qu issame as the class of hyponormal AM -operators. We prove a spectral decompositiontheorem for normal AM -operators. Our results can be compared to those of para-normal AN -operators given in [23]. Specifically, we will show that if T is normal AM -operator then there exists pairs p H β , U β q , where H β is a reducing subspacefor T , U β P B p H β q is a unitary such that,(1) H “ ‘ β P σ p| T |q H β ,(2) T “ ‘ β P σ p| T |q βU β .There are four sections in this article. In the second section we will provide thebasic definitions and results which we will be using through out the article.In the third section we will give a characterization for positive AM -operatorsand a sufficient condition for the adjoint of an AM -operator to be AM . Fourthsection consists of spectral decomposition of normal AM -operators.2. Preliminaries
Throughout this article we consider complex Hilbert spaces which are denotedby
H, H , H etc. Mostly we assume that these spaces are infinite dimensional.We denote the space of all bounded linear operators from H to H by B p H , H q and in case if H “ H “ H , we denote this by B p H q . By an operator we meana linear operator all the time. If T P B p H , H q , the adjoint of T is denoted by T ˚ P B p H , H q . An operator T P B p H q is called normal if T T ˚ “ T ˚ T , unitary if T T ˚ “ I “ T ˚ T , self-adjoint if T “ T ˚ and positive if T is self-adjoint and x T x, x y ě x P H . If S and T are two self adjoint operators in B p H q then S ď T if and only if x Sx, x y ď x
T x, x y for all x P H . Let A Ď B p H q , then wedenote the set of all self-adjoint elements (operators) of A by A sa and the set of allpositive elements (operators) of A by A ` . The set of all positive bounded linearoperators on H is denoted by B p H q ` .Let H , H be two Hilbert spaces. Then H ‘ H “ tp h , h q : h P H , h P H u is a Hilbert space with the inner product x ., . y , given by xp h , h q , p k , k qy “ x h , k y H ` x h , k y H for h , k P H and h , k P H .Let T i P B p H i q , i “ ,
2. Define T ‘ T : H ‘ H Ñ H ‘ H by p T ‘ T qp x , x q “ p T x , T x q , for all x i P H i , i “ , . This can be represented by a 2 ˆ T ‘ T “ ˆ T T ˙ . PECTRAL DECOMPOSITION OF NORMAL AM -OPERATORS 3 For T P B p H , H q , N p T q and R p T q denote the null space and range space of T , respectively. If M is a closed subspace of a Hilbert space H , then M K is theorthogonal complement of M , T ˇˇ M denote the restriction of the operator T to M and the orthogonal projection onto M in H , is denoted by P M . The unit sphere of M is S M “ t x P M : } x } “ u .An operator T P B p H , H q is said to be finite-rank , if R p T q is finite dimensional.The space of all finite-rank operators in B p H , H q is denoted by F p H , H q and F p H, H q “ F p H q . If T P B p H , H q , then T is said to be compact, if for everybounded set A Ď H , the set T p A q Ď H is pre-compact. The space of all compactoperators in B p H , H q , is denoted by K p H , H q .If V P B p H q , then V is called an isometry if } V x } “ } x } for all x P H . We say V to be a co-isometry if V ˚ is an isometry. Equivalently, V is an isometry if andonly if V ˚ V “ I and co-isometry if and only if V V ˚ “ I .For T P B p H q , the set σ p T q “ t λ P C : T ´ λI is not invertible in B p H qu iscalled the spectrum of T . Note that σ p T q is a non-empty compact subset of C .The spectrum of T P B p H q decomposes as the disjoint union of the point spectrum , σ p p T q , the continuous spectrum , σ c p T q and the residual spectrum σ r p T q , where σ p p T q “t λ P C : T ´ λI is not injective u ,σ r p T q “t λ P C : T ´ λI is injective but R p T ´ λI q is not dense in H u ,σ c p T q “ σ p T qz p σ p p T q Y σ r p T qq . If T is normal, then σ r p T q is empty. For a self-adjoint operator T P B p H q , the spec-trum can be divided into disjoint union of the discrete spectrum and the essentialspectrum. We summarize these details here. Theorem 2.1. [25, Theorem 7.10,7.11, Page 236]
Let T “ T ˚ P B p H q , the spec-trum σ p T q of T decomposes as the disjoint union of the essential spectrum, σ ess p T q and the discrete spectrum of T , σ d p T q , where σ d p T q , is the set of isolated finitemultiplicity eigenvalues of T . Furthermore, λ P σ ess p T q if and only if one of thefollowing holds:(1) λ is an eigenvalue of infinite multiplicity.(2) λ is a limit point of σ p p T q .(3) λ P σ c p T q . More details about essential spectrum and discrete spectrum can be found in[25, Page 235, 236]. Now we will quote two results from [27], namely [Theorem5.2, Page 288] and [Theore 5.4, Page 289]. These results are proved for unboundedoperators in [27], but they are true for bounded operators as well. A boundedversion of these results is given below.
Theorem 2.2.
Let T i P B p H i q , i “ , and T “ T ‘ T . Then(1) N p T q “ N p T q ‘ N p T q . (2) R p T q “ R p T q ‘ R p T q . (3) T ´ exists if and only if T ´ and T ´ exists.(4) R p T q “ H ‘ H if and only if R p T q “ H and R p T q “ H . Theorem 2.3.
Let T be as defined in Theorem (2.2). Then(1) σ p T q “ σ p T q Y σ p T q .(2) σ p p T q “ σ p p T q Y σ p p T q . If it is further assumed that σ p T q and σ p T q have no point in common, it follows that NEERU BALA AND G. RAMESH (3) σ c p T q “ σ c p T q Y σ c p T q .(4) σ r p T q “ σ r p T q Y σ r p T q . Definition 2.4. [4] Let T P B p H , H q . Then(1) The minimum modulus of T is m p T q : “ inf t} T x } : x P S H u .(2) The essential minimum modulus of T is m e p T q : “ inf t| λ | : λ P σ ess p T qu .It is to be noted that m p T q ą R p T q is closed and T is one-to-one.In particular, if H “ H “ H and T is normal, then m p T q ą T ´ exists and T ´ P B p H q . Definition 2.5. [6, Definition 1.1] Let T P B p H , H q . Then(1) T is norm attaining, if there exist x P S H such that } T } “ } T x } , where } T } “ sup t} T x } : x P H , } x } “ u .(2) T is absolutely norm attaining operator, if for every closed subspace M Ď H , T ˇˇ M : M Ñ H is norm attaining operator.We denote the classes of norm attaining and absolutely norm attaining operatorsin B p H , H q by N p H , H q and AN p H , H q , respectively. In particular, N p H, H q and AN p H, H q are denoted by N p H q and AN p H q , respectively. Definition 2.6. [6] Let T P B p H , H q . Then(1) T is minimum attaining, if there exist x P S H such that m p T q “ } T x } .(2) T is absolutely minimum attaining operator, if for every closed subspace M Ď H , T ˇˇ M is minimum attaining operator.We denote the classes of minimum attaining and absolutely minimum attainingoperators in B p H , H q by M p H , H q and AM p H , H q , respectively. In particu-lar, M p H, H q and AM p H, H q are denoted by M p H q and AM p H q , respectively. Theorem 2.7. [8, Theorem 5.9]
Let H be a complex Hilbert space of arbitrarydimension and let P be a positive operator on H . Then P is an AM -operator ifand only if P is of the form P “ βI ´ K ` F , where β ě , K P K p H q ` with } K } ď β and F P F p H q ` , satisfying KF “ F K “ . Let T P B p H , H q . For y P H , we say that u P H is a least square solutionof the equation T x “ y , if } T u ´ y } ď } T x ´ y } for all x P H . Moreover, if u is a least square solution of T x “ y and } u } ď } u } for any least square solution u of T x “ y , then u is called the least square solution of minimal norm . Such asolution is unique. Definition 2.8. [10, Page 223] Suppose T P B p H , H q and let D p T : q “ R p T q ‘ R p T q K , where D p T : q is domain of T : . The generalized inverse or the Moore-Penroseinverse of T is the operator T : : D p T : q Ñ H which assigns to each b P D p T : q theunique least square solution of minimal norm of the equation T x “ b .If T has closed range, then T : is the unique operator in B p H , H q satisfying T T : “ P R p T q and T : T “ P R p T : q . Here we list out some of the properties of the Moore-Penrose inverse, which weneed to prove our results. The following theorem is true for densely defined closedoperators in a Hilbert space, which is also true for bounded operators. We state itfor bounded operators.
Theorem 2.9. [3, Theorem 2, Page 341]
Let T P B p H , H q . Then PECTRAL DECOMPOSITION OF NORMAL AM -OPERATORS 5 (1) R p T : q “ N p T q K .(2) N p T : q “ R p T q K “ N p T ˚ q (3) T : is continuous if and only if R p T q is closed.(4) p T : q : “ T .(5) p T ˚ q : “ p T : q ˚ .(6) N pp T ˚ q : q “ N p T q .(7) T : p T ˚ q : is positive and p T ˚ T q : “ T : p T ˚ q : .(8) p T ˚ q : T : is positive and p T T ˚ q : “ p T ˚ q : T : . AM -operators In this section we give a new characterization of positive AM -operators in termsof the essential spectrum. Using this we give a sufficient condition under which theadjoint of an AM -operator is again an AM -operator.First, we have the following observation. Remark . Let T P AM p H q ` . By Theorem (2.7), T “ βI ´ K ` F , where β ě K P K p H q ` , F P F p H q ` , KF “ “ F K and } K } ď β . Then we have thefollowing:(1) β “ T is finite-rank.(2) σ ess p T q “ t β u .(3) β “ N p T q is infinite dimensional.Here we will give a relation between the essential spectrum of a self-adjointinvertible operator and its inverse. Lemma 3.2.
Let T P B p H q be self-adjoint and T ´ P B p H q . Then λ P σ ess p T q ifand only if λ ´ P σ ess p T ´ q .Proof. It is enough to prove that λ P σ ess p T q implies λ ´ P σ ess p T ´ q , as p T ´ q ´ “ T . We know that σ p T ´ q “ t λ ´ : λ P σ p T qu and σ p p T ´ q “ t λ ´ : λ P σ p p T qu . If λ P σ ess p T q then either λ is an eigenvalue of T with infinite multiplicity or λ is alimit point of σ p p T q or λ P σ c p T q .If λ is an eigenvalue of T with infinite multiplicity, then 1 { λ is an eigenvalue of T ´ of infinite multiplicity with same eigenvectors. Hence λ ´ P σ ess p T ´ q .If λ is a limit point of σ p p T q , then there exist a sequence p λ n q Ď σ p p T q such that λ n converges to λ . Since 0 R σ p T q , so λ n , λ ‰ n P N and hence 1 { λ n converges to 1 { λ. So 1 { λ P σ ess p T ´ q as 1 { λ n P σ p p T ´ q .If λ P σ c p T q i.e. λ R σ p p T q , so 1 { λ R σ p p T ´ q and hence 1 { λ P σ c p T ´ q , because σ r p T ´ q is empty. Hence 1 { λ P σ ess p T ´ q . (cid:4) Remark . Let T P B p H q . If N p T q be a reducing subspace for T , then σ p T q Ď σ p T q Ď σ p T q Y t u , where T “ T ˇˇ N p T q K . Proof.
This is clear from (1) of Theorem (2.3). (cid:4)
Lemma 3.4.
Let T P B p H q be self adjoint and T “ T ˇˇ N p T q K . Then(1) σ ess p T q Ď σ ess p T q Ď σ ess p T q Y t u .(2) σ d p T q Ď σ d p T q Ď σ d p T q Y t u . NEERU BALA AND G. RAMESH
Proof.
Since N p T q is reducing, we have T “ „ T . (1) Note that as T is self-adjoint, N p T q reduces T . If λ P R , then T ´ λI “ „ ´ λI N p T q T ´ λI N p T q K . Using Theorem (2.2), we have(3.1) N p T ´ λI q “ N ` ´ λI N p T q ˘ ‘ N ` T ´ λI N p T q K ˘ . and R p T ´ λI q “ N p T q Y R p T ´ λI N p T q K q . Thus we can conclude that σ p p T q Ď σ p p T q and σ c p T q Ď σ c p T q .Let λ P σ ess p T q . By Theorem (2.1), one of the following hold(a) λ P σ c p T q .(b) λ is an eigenvalue of T of infinite multiplicity.(c) λ is a limit point of σ p p T q .Using above argument, we get λ P σ ess p T q . Second containment followsfrom Theorem (2.3), which says that σ p p T q “ σ p p T q Y t u and σ c p T q “ σ c p T q Y t u .(2) From Equation (3.1), we get σ d p T q Ď σ d p T q Y t u . Other containmentfollows from Remark (3.3) and Equation (3.1). (cid:4) Next we will prove that Lemma (3.2) can be generalized to T : . Lemma 3.5.
Let T P B p H q . Suppose T is a self-adjoint operator with closed range.Let ‰ λ P R . Then(1) λ P σ ess p T q if and only if { λ P σ ess p T : q .(2) P σ ess p T q if and only if P σ ess p T : q .Proof. Let T “ T ˇˇ N p T qK . It is enough to show one way implication in both thecases, as p T : q : “ T .(1) As λ ‰
0, by Lemma (3.4), λ P σ ess p T q if and only if λ P σ ess p T q . UsingLemma (3.2), we get 1 { λ P σ ess p T ´ q . Hence 1 { λ P σ ess p T : q , by p q ofLemma (3.4).(2) Let 0 P σ ess p T q . As T is self-adjoint and R p T q is closed so 0 R σ c p T q and by[16, Theorem 4.4], 0 is not a limit point of σ p p T q . So 0 is an eigenvalue of T with infinite multiplicity. As T is self-adjoint, we have N p T q “ N p T : q . So0 is an eigenvalue of T : with infinite multiplicity. Hence 0 P σ ess p T : q . (cid:4) Remark . Let T be as in Lemma (3.5). Define λ : “ λ ´ if λ ‰ , λ “ . (1) By Lemma (3.5), we can conclude that λ P σ ess p T q if and only if λ : P σ ess p T : q .(2) Since σ p T q “ σ d p T q Y σ ess p T q , we can also conclude that λ P σ d p T q if andonly if λ : P σ d p T : q . Proposition 3.7.
Let T P AM p H q ` and H be a finite dimensional Hilbert space.If S P B p H q ` , then S ‘ T P AM p H ‘ H q ` . PECTRAL DECOMPOSITION OF NORMAL AM -OPERATORS 7 Proof.
Let T P AM p H q ` . By Theorem (2.7), T “ βI H ´ K ` F , where β ě K P K p H q ` with } K } ď β and F P F p H q ` satisfying KF “ F K “
0. Then S ‘ T “ „ S βI H ´ K ` F “ βI ´ „ K ` „ S ´ βI H F “ βI ´ ˜ K ` ˜ F , where ˜ K “ „ K and ˜ F “ „ S ´ βI H F . Again by using Theorem (2.7), we get S ‘ T P AM p H ‘ H q ` . (cid:4) By similar arguments as above, we can prove the following Remark for AN -operators. Remark . Let T P AN p H q ` and H be a finite dimensional Hilbert space. If S P B p H q ` , then S ‘ T P AN p H ‘ H q ` .Now we will generalize the result of [17, Theorem 5.1], to any bounded linearoperator, by dropping the injectivity condition. Theorem 3.9.
Let T P B p H q be a positive operator. Then T P AM p H q if andonly if R p T q is closed and T : P AN p H q .Proof. Let T P AM p H q and T “ T ˇˇ N p T q K . Then by [9, Proposition 3.3], R p T q is closed and by Theorem (2.7), T “ βI ´ K ` F where β ě K P K p H q ` , F P F p H q ` satisfying KF “ “ F K and } K } ď β . We consider the following twocases which exhaust all possibilities.Case (1): Let N p T q be infinite dimensional. We get β “ T is finite-rankoperator. By [15, Theorem 3.2], T : is finite-rank and hence T : P AN p H q .Case (2): Let N p T q be finite dimensional. Since T is AM -operator, so T P AM p N p T q K q . By [17, Theorem 5.1], T ´ P AN p N p T q K q . Using Proposition 3.7,we get T : “ „ T ´ P AN p H q . Conversely, assume that R p T q is closed and T : P AN p H q . By [20, Theorem 5.1],we get T : “ αI ` K ` F , where α ě , K P K p H q ` and F P F p H q is a self-adjointoperator.Case (1): Let N p T q be infinite dimensional. Then 0 is an eigenvalue of T withinfinite multiplicity. So, α “ T : is compact. Since restriction of a compactoperator to a closed subspace is compact, T ´ is compact. We know that a compactoperator is invertible if its domain is finite dimensional, so R p T q is finite dimensionaland hence T is finite-rank operator. Thus T P AM p H q .Case (2): Let N p T q be finite dimensional. Since T : P AN p H q , T ´ P AN p N p T q K q .By [17, Theorem 5.1], T P AM p N p T q K q . Since N p T q is finite dimensional, byProposition 3.7, T P AM p H q . (cid:4) Lemma 3.10.
Let T P B p H q be a positive operator. If σ ess p T q is singleton and p m e p T q , } T }s contains only finitely many eigenvalues of T , then R p T q is closed. NEERU BALA AND G. RAMESH
Proof.
To prove R p T q is closed, it is enough to show that 0 is not a limit pointof σ p T q , by [16, Theorem 4.4]. On the contrary assume that 0 is a limit point of σ p T q . Since σ ess p T q is singleton, 0 must be a limit point of σ d p T q . In fact 0 is alimit point of a decreasing sequence in σ p p T q , this implies σ ess p T q “ t u . Hencethe sequence which is converging to 0 is the zero sequence, which is a contradiction.So 0 is not a limit point of σ p T q and hence R p T q is closed. (cid:4) Here we will give a new characterization for positive AM -operators, which issimilar to [23, Theorem 2.4]. Theorem 3.11.
Let T P B p H q be a positive operator. Then T P AM p H q if andonly if σ ess p T q is singleton and p m e p T q , } T }s contains only finitely many eigenvaluesof T .Proof. Let σ ess p T q “ t β u and λ , λ , . . . λ m be the eigenvalues of T contained in p m e p T q , } T }s . By Lemma (3.10), R p T q is closed.Case (1): Let β “
0. By Lemma (3.5), σ ess p T : q “ t u and r m p T : q , m e p T : qs “t u . So, by [23, Theorem 2.4], T : P AN p H q . Hence by Theorem (3.9), T P AM p H q .Case (2): Let β ą
0. This implies that σ ess p T : q “ t { β u and r m p T : q , { β q contains only finitely many eigenvalues of T : , namely either 0 , { λ , { λ , . . . { λ m or 1 { λ , { λ , . . . { λ m . By [23, Theorem 2.4], T : P AN p H q . Hence by Theorem(3.9), T P AM p H q .Conversely, let T P AM p H q and T “ T ˇˇ N p T q K . By [23, Theorem 2.4], it isenough to show that T : P AN p H q . Since T P AM p H q , T P AM p N p T q K q and by[17, Theorem 5.1], T ´ is AN -operator. Firstly if N p T q is finite dimensional, then N p T : q is finite dimensional and hence T : is an AN -operator, by Proposition (3.7).Secondly, if N p T q is infinite dimensional, then T is finite-rank operator. By [15,Theorem 3.2], T : is finite-rank and hence AN -operator. (cid:4) In general, if T P AM p H q , then T ˚ need not be an AM -operator (see [7, 9]for more details). The same is true for AN -operators. A sufficient condition tohold this result for AN -operators is given in [23], which depends on the essentialspectrum. A similar condition works for AM -operators too. The details are givenbelow. Theorem 3.12.
Let T P B p H q and σ ess p T ˚ T q “ σ ess p T T ˚ q . Then(1) T T ˚ P AM p H q if and only if T ˚ T P AM p H q .(2) T P AM p H q if and only if T ˚ P AM p H q .Proof. By Lemma (3.5), σ ess pp T : q ˚ T : q “ σ ess p T : p T : q ˚ q .(1) By Theorem (3.9), T T ˚ P AM p H q if and only if R p T T ˚ q is closed and p T T ˚ q : P AN p H q , i.e. p T : q ˚ T : P AN p H q . Since σ ess pp T : q ˚ T : q “ σ ess p T : p T : q ˚ q ,by [23, Theorem 2.7], we get T : p T : q ˚ P AN p H q . Again by applying Theo-rem (3.9), we get T ˚ T P AM p H q .(2) This follows by Case (1) and by [9, Corollary 4.9], that T P AM p H q if andonly if T ˚ T P AM p H q . (cid:4) It is well known that for T P B p H q , σ p T ˚ T qzt u “ σ p T T ˚ qzt u . We can askwhether the same is true if the spectrum is replaced by the essential spectrum. Weanswer this question affirmatively. Lemma 3.13.
Let T P B p H q . Then σ ess p T ˚ T qzt u “ σ ess p T T ˚ qzt u . PECTRAL DECOMPOSITION OF NORMAL AM -OPERATORS 9 Proof.
To prove the result it is enough to show that σ ess p T ˚ T qzt u Ď σ ess p T T ˚ qzt u .Let α P σ ess p T ˚ T qzt u . By Theorem (2.1), either α is an eigenvalue of T ˚ T withinfinite multiplicity or α P σ c p T ˚ T q or α is a limit point of σ p p T ˚ T q .Case (1): Let α be an eigenvalue of T ˚ T with infinite multiplicity. This impliesthere exist t x δ P H : δ P Λ u such that T ˚ T x δ “ αx δ , where Λ is an indexing set.So we get T T ˚ p T x δ q “ α p T x δ q . We have T x δ ‰ T x ¯ δ if δ ‰ ¯ δ and δ, ¯ δ P Λ. Becauseif
T x δ “ T x ¯ δ then T ˚ T x δ “ T ˚ T x ¯ δ , thus x δ “ x ¯ δ . Hence α is an eigenvalue of T T ˚ of infinite multiplicity.Case (2): Let α P σ c p T ˚ T q . This implies T ˚ T ´ αI is one-one with R p T ˚ T ´ αI q “ H but T ˚ T ´ αI does not have a bounded inverse. We know that σ r p T T ˚ q isempty and σ p T T ˚ qzt u “ σ p T ˚ T qzt u . Thus we get that either α P σ p p T T ˚ q or α P σ c p T T ˚ q . Now we will show that α R σ p p T T ˚ q . Let p T T ˚ ´ αI q x “ ‰ x P H . Applying on both sides T ˚ , we get T ˚ T T ˚ x “ αT ˚ x , thus T ˚ x P N p T ˚ T ´ αI q . As T ˚ T ´ αI is one-one, so T ˚ x “ x “
0. Thuswe get α P σ c p T T ˚ q .Case (3): Let α be a limit point of σ p p T ˚ T q . So there exist a sequence p λ n q Ď σ p p T ˚ T qzt u such that p λ n q converges to α . Also p λ n q Ď σ p p T T ˚ q , so α is a limitpoint of σ p p T T ˚ q . Hence α P σ ess p T T ˚ q .In all the three cases we get α P σ ess p T T ˚ q . Hence σ ess p T ˚ T qzt u Ď σ ess p T T ˚ qzt u . (cid:4) Theorem 3.14.
Suppose T P B p H q , such that N p T q “ N p T ˚ q . Then we have thefollowing:(1) σ p p T ˚ T q “ σ p p T T ˚ q .(2) σ c p T ˚ T q “ σ c p T T ˚ q .(3) σ p T ˚ T q “ σ p T T ˚ q .(4) σ ess p T ˚ T q “ σ ess p T T ˚ q .(5) σ d p T ˚ T q “ σ d p T T ˚ q .Proof. First note that as N p T ˚ T q “ N p T q and N p T T ˚ q “ N p T ˚ q , by the assump-tion it follows that N p T ˚ T q “ N p T T ˚ q .Proof of (1): Let λ P σ p p T ˚ T q . First, assume that λ “
0. Then t u ‰ N p T ˚ T q “ N p T T ˚ q , we can conclude that 0 P σ p p T T ˚ q . The other implication follows in thesimilar lines. Next, assume that λ ‰
0. Let 0 ‰ x P H be such that T ˚ T x “ λx .Then p T T ˚ q T x “ λT x . Since x P R p T ˚ T q Ď N p T q K , T x ‰
0. This means that λ P σ p p T T ˚ q . Similarly, the other way implication can be proved.Proof of (2): Let λ P σ c p T ˚ T q . Then λ R σ p p T ˚ T q and R p T ˚ T ´ λI q is not closed.By (1), λ R σ p p T T ˚ q . By [16, Theorem 4.4], it follows that λ is an accumulationpoint of σ p T ˚ T q . As σ p T ˚ T qzt u “ σ p T T ˚ qzt u , λ is an accumulation point of σ p T T ˚ q . Hence by [16, Theorem 4.4], R p T T ˚ ´ λI q is not closed, concluding λ P σ c p T T ˚ q . The other implication can be proved with similar arguments.Proof of (3): Since for a self-adjoint operator the residual spectrum is empty andthe spectrum is disjoint union of the point spectrum, continuous spectrum and theresidual spectrum, by (1) and (2), the conclusion follows.Proof of (4): In view of Lemma (3.13), it is enough to show that 0 P σ ess p T ˚ T q if and only if 0 P σ ess p T T ˚ q . But this follows by the definition of the essentialspectrum and (1) and (2) proved above.Proof of (5): For a self-adjoint operator A P B p H q , σ p A q “ σ ess p A q Y σ d p A q , theconclusion follows by (3) and (4) above. (cid:4) Corollary 3.15.
Let T P B p H q be such that N p T q “ N p T ˚ q . Then (1) m p T q “ m p T ˚ q .(2) m e p T q “ m e p T ˚ q .Proof. Since σ p T ˚ T q “ σ p T T ˚ q by (3) of Theorem (3.14), by the spectral mappingtheorem, σ p| T |q “ σ p| T ˚ |q . Now, m p T q “ inf t λ : λ P σ p| T |qu“ inf t λ : λ P σ p| T ˚ |qu“ m p T ˚ q . Next, by [5] and by (4) of Theorem (3.14), we have m e p T q “ inf t λ : λ P σ ess p| T |qu“ inf t λ : λ P σ ess p| T ˚ |qu“ m e p T ˚ q . (cid:4) Corollary 3.16.
Let T P B p H q be such that N p T q “ N p T ˚ q . Then T P M p H q ifand only if T ˚ P M p H q .Proof. First, note that m p T q “ m p T ˚ q by (1) of Corollary (3.15). It is enough toshow that T P M p H q implies T ˚ P M p H q . Assume that T P M p H q . Thus m p T q P σ p p T ˚ T q . But by Theorem (3.14), m p T ˚ q P σ p p T T ˚ q , concluding T T ˚ P M p H q and hence T ˚ P M p H q . (cid:4) Normal AM -operators In this section we describe a spectral decomposition of normal AM -operators. Definition 4.1.
Let T P B p H q .Then(1) T is hyponormal if T T ˚ ď T ˚ T . Equivalently, T is hyponormal if } T ˚ x } ď} T x } for all x P H .(2) T is paranormal if } T x } ď } T x } for all x P S H . Equivalently, T isparanormal if } T x } ď } T x }} x } for all x P H .It is easy to see that every hyponormal operator is paranormal [14]. More detailsabout hyponormal and paranormal operators can be found in [2, 4, 13, 14, 12]. Remark . Let T P B p H q .(1) If T is paranormal, then N p T q “ N p T q .(2) If T is hyponormal, then N p T q Ă N p T ˚ q . This inclusion is strict, forexample, the right shift operator R on l p N q is hyponormal such that N p R q Ĺ N p R ˚ q . Also note that R ˚ is not paranormal. Theorem 4.3.
Let T P B p H q , with N p T q “ N p T ˚ q . Then(1) T is paranormal if and only if T : is paranormal.(2) T P AM p H q if and only if R p T q is closed and T : P AN p H q .Proof. Let T “ T | N p T q K : N p T q K Ñ N p T q K .Proof of (1): Let T be paranormal. By [23, Lemma 3.8], T is paranormal. Since N p T q “ N p T ˚ q , we have T ´ is paranormal, by [13, Theorem 1]. As T : x “ x P N p T q ,T ´ x if x P N p T q K . PECTRAL DECOMPOSITION OF NORMAL AM -OPERATORS 11 Hence T : is paranormal. Reverse implication is clear, as p T : q : “ T .Proof of (2): By Theorem (3.14) σ ess p T T ˚ q “ σ ess p T ˚ T q . Thus using Theorem(3.12) T ˚ T P AM p H q if and only if T T ˚ P AM p H q . Now using all these argumentsand Theorem (3.9) we conclude the following. T P AM p H q ðñ T ˚ T P AM p H qðñ T T ˚ P AM p H qðñ R p T T ˚ q is closed and p T T ˚ q : P AN p H qðñ R p T q is closed and p T : q ˚ T : P AN p H qðñ R p T q is closed and T : P AN p H q . (cid:4) The following result is not used in the article, but it is of independent interest.
Theorem 4.4.
Let T P B p H q be such that R p T q “ R p T q . Also assume that R p T q is closed. Then T is paranormal implies T : is paranormal.Proof. Assume that T is paranormal. Then } T x } ď } T x }} x } , @ x P H. Since T : u “ u P R p T q K . It suffices to show that } T : y } ď } T : y }} y } , @ y P R p T q . Let y P R p T q “ R p T q . Then y “ T v for some v P N p T q K “ N p T q K (see (1) ofRemark (4.2)). Now T : y “ T : T v “ P N p T q K T v “ T v , as R p T q Ď N p T q K . Also p T : q y “ T : T v “ P R p T : q v “ v . Thus we have } T : y } “ } T v } ď } v }} y } “ } T : y }} y } . Thus T : is paranormal. (cid:4) Now we will show that the classes of paranormal AM -operators and hyponormal AM -operators are the same, under some assumption. Theorem 4.5.
Let T P B p H q with N p T q “ N p T ˚ q . If T P AM p H q is paranormalthen T is hyponormal.Proof. Since T P AM p H q , T ˚ T P AM p H q . Hence by Theorem (2.7), T ˚ T “ βI ´ K ` F , where K P K p H q ` with } K } ď β and F P F p H q ` satisfies KF “ “ F K .Case (1): Let β “
0. As a consequence of Remark (3.1), we conclude that T ˚ T is a finite-rank operator and so is T . Also we know that a paranormal compactoperator is normal [21], hence T is normal.Case (2): Let β ą
0. By Theorem (3.14), we have σ ess p T ˚ T q “ t β u “ σ ess p T T ˚ q .Now using Theorem (3.12), we get T ˚ P AM p H q . So H has a basis, say t y α : α P Λ u , consisting of eigenvectors of T T ˚ . Let T T ˚ y α “ λ α y α where λ α P r , .If λ α ‰ α , then } T ˚ y α } “ |x T ˚ y α , T ˚ y α y| “ |x T T ˚ y α , y α y| ď } T T ˚ y α } } y α } . As T is paranormal, by the above inequality, we have } T ˚ y α } ď } T T ˚ y α } } y α } ď } T T ˚ y α }} T ˚ y α }} y α } “ λ α } T y α }} T ˚ y α }} y α } “ } T y α }} T ˚ y α }x T T ˚ y α , y α y“ } T y α }} T ˚ y α }} T ˚ y α } . This implies } T ˚ y α } ď } T y α } .If for some α, λ α “
0, then } T ˚ y α } “ “ } T y α } , since N p T q “ N p T ˚ q . Thisimplies for each y P H , } T ˚ y } ď } T y } . Hence T is hyponormal. (cid:4) By a similar argument as in Theorem (4.5), we can prove the following;
Theorem 4.6.
Let T P AN p H q be paranormal operator with N p T q “ N p T ˚ q .Then T is hyponormal. Now we will give a characterization of normal AM -operators, which is similarto the result for paranormal AN -operators, given in [23]. Theorem 4.7.
Let T P B p H q be normal. Suppose T P AM p H q with Λ “ σ p| T |q ,where | T | “ ? T ˚ T . Then there exist p H β , U β q β P Λ , where H β is a reducing subspacefor T , U β P B p H β q is a unitary such that,(1) H “ ‘ β P Λ H β ,(2) T “ ‘ β P Λ βU β .Proof. As T P AM p H q is normal, T : is normal AN -operator, by Theorem (2.9).Using [23, Theorem 3.9], there exist p G α , V α q α P Γ , where G α is a reducing subspacefor T : , V α P B p G α q is an unitary such that,(i) H “ ‘ α P Γ G α ,(ii) T : “ ‘ α P Γ αV α ,where Γ “ σ p| T : |q . Now we claim that G α is a reducing subspace for T as well. If α “ G α “ N p T ˚ q “ N p T q . Clearly G α is a reducing subspace for T . Onthe other hand if α ‰
0, then G α Ď N p T q K .Let x P G α . Assume that T x “ a ` b , where a P G α X N p T q K and b P G K α X N p T q K .Now T : b “ T : T x ´ T a “ P R p T : q x ´ T a “ P N p T q K x ´ T a “ x ´ T a P G K α . Since G α is a reducing subspace for T : , we get T : b P G K α , thus T : b P G α X G K α . So T : b “
0, but b P N p T q K , so b “ T x “ a P G α . We get G α is invariantunder T .To show that G K α is invariant under T , let y P G K α . If T y “ u ` v , where u P G α and v P G K α , then T : T y “ P N p T q K y P N p T q K X G K α . Now by similar argumentas above, we get T : u P G α X G K α , so T : u “
0. This implies u P N p T q , but u P G α Ď N p T q K , so u “ T y “ v P G K α . By (i) and (ii), we get PECTRAL DECOMPOSITION OF NORMAL AM -OPERATORS 13 (1) T “ ‘ α P Γ α : V ˚ α .where α : is defined as in Remark (3.6). By using [17, Proposition 3.15], we get α : P σ p| T : | : q “ σ p| T ˚ |q . But by Theorem (3.14), we have that σ p| T |q “ σ p| T ˚ |q , as N p T q “ N p T ˚ q . Thus we get the result, by taking β “ α : , H β “ G α and U β “ V ˚ β .As V β is an unitary, it is clear that U β is a unitary. (cid:4) Below we illustrate Theorem (4.7) with an example.
Example 4.8.
Let p X, Σ , µ q be a σ -finite measure space. For f P L p X q definethe multiplication operator M f : L p X q Ñ L p X q by M f p g q “ f g, for all g P L p X q . Then(1) m p M f q “ ess inf p f q , where ess inf p f q “ sup t α P R : µ pt x P X : | f p x q| ă α uq “ u .(2) M f is minimum attaining if and only if there exist A P Σ with µ p A q ą | f p t q| “ ess inf p f q for all t P A .(3) M f P AM p L p X qq if and only if there exist a sequence p A i q Ď Σ with µ p A i q ą | f p t q| “ ess inf p f i q for all t P A i and X “ X Y ´ Y i “ A i ¯ ,where X P Σ with µ p X q “ f i “ f with domain X z ` Y i ´ j “ A j ˘ . Proof. (1) Let g P L p X q . Then } M f p g q} “ ż X | f p t q| | g p t q| dµ p t qěp ess inf p f qq } g } . Thus we get m p M f q ě ess inf p f q . Now for every n P N , define E n “ t t P X : | f p t q| ď ess inf p f q ` { n u . By the definition of ess inf p f q , it is clear that µ p E n q ą
0. As p X, Σ , µ q is a σ -finite measure space, choose a measurable set F n Ď E n such that0 ă µ p F n q ă 8 , for every n P N . Let g n “ χ Fn ? µ p F n q , then } M f p g n q} “ ż X | f p t q| χ F n µ p F n q dµ “ µ p F n q ż F n | f p t q| dµ p t qďp ess inf p f q ` { n q , @ n P N . Thus we get m p M f q ď ess inf p f q . Hence } M f } “ ess inf p f q .(2) To prove this, we use a similar technique that is used in [1, Lemma 2.6].First we assume that M f is a minimum attaining operator. Then thereexists a g P S L p X q , such that } M f p g q} “ m p M f q “ ess inf p f q . Now we have ż X p ess inf p f qq | g p t q| dµ p t q “p ess inf p f qq “} M f p g q} “ ż X | f p t q| | g p t q| dµ p t q . This implies ess inf p f q| g p t q| “ | f p t q|| g p t q| a.e. Let C “ t t P X : g p t q ‰ u . Then C is a measurable set with µ p C q ą
0. There exist a subset Z of C such that µ p Z q “
0. Thus | f p t q| “ ess inf p f q for all t P A : “ C z Z .Conversely, let A P Σ satisfying the given property. Choose a measurablesubset B of A such that 0 ă µ p B q ă 8 . Define g “ χ B ? µ p B q . It is easy tosee that g P S L p X q . Now } M f p g q} “ ż X | f g | dµ “ ż B | f g | dµ “ p ess inf p f qq µ p B q ż B dµ “p ess inf p f qq . Hence M f is a minimum attaining operator.(3) First assume that M f P AM p L p X qq . As M f is minimum attaining, weget A P Σ such that µ p A q ą | f p t q| “ ess inf p f q , for all t P A .Let X “ X z A . If µ p X q “ M | f | “ M ess inf p f q . Otherwise, it iseasy to see that L p X q “ t p g | X : p g P L p X q , p g “ A u . DefineΣ “ t X X A : A P Σ u and µ “ µ | Σ . Let f “ f | X P L p X q and define M f : L p X q Ñ L p X q as M f g “ f g , for g P L p X q .Let G “ t g P L p X q : g “ A u . We claim that G is a closedsubspace of L p X q . To prove our claim, let p p g n q Ď G and p g n convergesto some h P L p X q . As p p g n q is a Cauchy sequence in G , p g n q is a Cauchysequence in L p X q , where g n “ ˆ g n | X for every n P N . Using the complete-ness of L p X q , we can conclude that p g n q converges to some g P L p X q .Hence h “ p g P G , where ˆ g p t q “ g p t q for t P X and ˆ g p t q “ m p M f | G q “ inf t}| M f | G p p g q} : p g P S G u“ inf t} M f g } : g P L p X qu“ m p M f q . As M f P AM p L p X qq , there exist ˆ g P S G such that } M f | G ˆ g } “ m p M f | G q .Thus we get } M f ˆ g | X } “ m p M f q . Hence M f is minimum attaining.Again by (2), we get A P Σ with µ p A q ą | f p t q| “ ess inf p f q for all t P A , also A X A “ H .Continuing this way, we get a sequence of sets p A i q Ď Σ with m p A i q ą A i X A j “ H if i ‰ j and | f p t q| “ ess inf p f i q for all t P A i , where f i “ f with domain X z ` Y i ´ j “ A j ˘ . PECTRAL DECOMPOSITION OF NORMAL AM -OPERATORS 15 Case(1): There exist n P N such that µ ` X z ` Y n j “ A j ˘˘ “ X “ X z ` Y n j “ A j ˘ and M | f | “ n ‘ i “ M ess inf p f i q .Case(2): We have the infinite sequence p A i q Ď Σ. Since p ess inf p f i qq isa monotonically increasing sequence and bounded above by } f } , so it isconvergent. Let p ess inf p f n qq converges to α . Now set X “ X z pY i “ A i q . If µ p X q “ X and again get a sequence of sets p B j q with µ p B j q ą | f p t q| “ ess inf p g j q for all t P B j , where g j “ f with domain X z ˆ j ´ Y B kk “ ˙ . As in case (1),there exists a stage m such that µ p X zp m Y k “ B k qq “ . Otherwise σ ess p M | f | q will have more than one element, which is a contradiction to the hypothesisthat M f P AM p L p X qq . By taking X “ X z "´ Y i “ A i ¯ Y ˆ m Y j “ B j ˙* , we get the result.Conversely, let p A i q Ď Σ be a sequence satisfying the given condition.First we will show that M f is minimum attaining. We have µ p A q ą | f p t q| “ ess inf p f q for all t P A . Choose a subset B of A such that0 ă µ p B q ă 8 . Consider g “ χ B ? µ p B q , we get } M f g } “ ess inf p f q .Now to show that M f P AM p L p X qq , consider a non-trivial closed sub-space E Ď L p X q . Let F “ t B P Σ : g “ a.e. on B, @ g P E u and define arelation „ on F as, A „ B if A Ď B , for A, B P F . Now p F , „q is a partiallyordered set. Using Zorn’s Lemma F has a maximal element, say B . Then µ p X z B q ą
0, otherwise E will be the trivial space. Let i be the smallestnatural number for which µ pp X z B q X A i q ą
0. So there exist g i P E suchthat g i ‰ p X z B q X A i . Existence of such a g i is guaranteed,as if for every g P E, g “ C : “ p X z B q X A i , then B Y C P F which contradict the maximality of B in F . Now choose a measurablesubset D of C such that 0 ă µ p D q ă 8 and consider h p t q “ χ D p t q a µ p D q . g i p t q| g i p t q| . Then } M f h } “ ż D | f p t q| dµ “ ess inf p f i q “ m p M f | E q Thus M f | E is minimum attaining. Hence M f P AM p L p X qq . (cid:4) Using similar arguments as above, we can prove the following;
Theorem 4.9.
Let M f be defined as in Example 4.8. Then we have the following;(1) M f P N p L p X qq if and only if there exists a set A P Σ with µ p A q ą suchthat | f p t q| “ ess sup p f q , for all t P A , where ess sup p f q “ inf t α P R : µ pt x P X : | f p x q| ą α uq “ u .(2) M f P AN p L p X qq if and only if there exist a sequence p A i q Ď Σ with µ p A i q ą such that | f p t q| “ ess sup p f i q for all t P A i and X “ X Y ´ Y i A i ¯ ,where X P Σ with µ p X q “ and f i “ f with domain X z ` Y i ´ j “ A j ˘ . We close this section with the following question.
Question 4.10.
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Department of Mathematics, Indian Institute of Technology - Hyderabad, Kandi,Sangareddy, Telangana, India 502 285.
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