Partially isometric Toeplitz operators on the polydisc
aa r X i v : . [ m a t h . F A ] F e b PARTIALLY ISOMETRIC TOEPLITZ OPERATORS ON THEPOLYDISC
DEEPAK K. D., DEEPAK PRADHAN, AND JAYDEB SARKAR
Abstract.
A Toeplitz operator T ϕ , ϕ ∈ L ∞ ( T n ), is a partial isometry if and only ifthere exist inner functions ϕ , ϕ ∈ H ∞ ( D n ) such that ϕ and ϕ depends on differentvariables and ϕ = ¯ ϕ ϕ . In particular, for n = 1, along with new proof, this recovers aclassical theorem of Brown and Douglas.We also prove that a partially isometric Toeplitz operator is hyponormal if and only ifthe corresponding symbol is an inner function in H ∞ ( D n ). Moreover, partially isometricToeplitz operators are always power partial isometry (following Halmos and Wallen), andhence, up to unitary equivalence, a partially isometric Toeplitz operator with symbol in L ∞ ( T n ), n >
1, is either a shift, or a co-shift, or a direct sum of truncated shifts. Alongthe way, we prove that T ϕ is a shift whenever ϕ is inner in H ∞ ( D n ). Introduction
Toeplitz operators are one of the most useful and prevalent objects in matrix theory,operator theory, operator algebras, and its related fields. For instance, Toeplitz operatorsprovide some of the most important links between index theory, C ∗ -algebras, functiontheory, and non-commutative geometry. See the monograph by Higson and Roe [14] for athorough presentation of these connections, and consult the paper by Axler [2] for a rapidintroduction to Toeplitz operators.Evidently, a lot of work has been done in the development of one variable Toeplitzoperators, and it is still a subject of very active research, with an ever-increasing list ofconnections and applications. But on the other hand, many questions remain to be settledin the several variables case, and more specifically in the open unit polydisc case (however,see [7, 8, 11, 16, 21]). The difficulty lies in the obvious fact that the standard (and classical)single variable tools are either unavailable or not well developed in the setting of polydisc.Evidently, advances in Toeplitz operators on the polydisc have frequently resulted in anumber of new tools and techniques in operator theory, operator algebras, and relatedfields.Our objective of this paper is to address the following basic question: Characterizepartially isometric Toeplitz operators on H ( D n ), where H ( D n ) denotes the Hardy spaceover the unit polydisc D n . Recall that a partial isometry [12] is a bounded linear operatorwhose restriction to the orthogonal complement of its null space is an isometry. Mathematics Subject Classification.
Key words and phrases.
Toeplitz and Laurent operators, Hardy space over polydisc, partial isometries,hyponormal operators, inner functions, power partial isometries.
Before we answer the above question, we first recall that H ( D n ) is the Hilbert spaceof all analytic functions f on D n such that k f k := (cid:16) sup ≤ r< Z T n | f ( rz , . . . , rz n ) | d m ( z ) (cid:17) < ∞ , where d m ( z ) is the normalized Lebesgue measure on the n -torus T n , and z = ( z , . . . , z n ).We denote by L ( T n ) the Hilbert space L ( T n , d m ( z )). From the radial limits of squaresummable analytic functions point of view [19], one can identify H ( D n ) with a closedsubspace H ( T n ) of L ( T n ). Let L ∞ ( T n ) denote the standard C ∗ -algebra of C -valuedessentially bounded Lebesgue measurable functions on T n . The Toeplitz operator T ϕ withsymbol ϕ ∈ L ∞ ( T n ) is defined by T ϕ f = P H ( D n ) ( ϕf ) ( f ∈ H ( D n )) , where P H ( D n ) denotes the orthogonal projection from L ( T n ) onto H ( D n ). Also recallthat H ∞ ( D n ) = L ∞ ( T n ) ∩ H ( D n ) , where H ∞ ( D n ) denotes the Banach algebra of all bounded analytic functions on D n . Afunction ϕ ∈ H ∞ ( D n ) is called inner if ϕ is unimodular on T n .The answer to the above question is contained in the following theorem: Theorem 1.1.
Let ϕ be a nonzero function in L ∞ ( T n ) . Then T ϕ is a partial isometry ifand only if there exist inner functions ϕ , ϕ ∈ H ∞ ( D n ) such that ϕ and ϕ depends ondifferent variables and T ϕ = T ∗ ϕ T ϕ . In particular, if n = 1, then the only nonzero Toeplitz operators that are partial isome-tries are those of the form T ϕ and T ∗ ϕ , where ϕ ∈ H ∞ ( D ) is an inner function. This wasproved by Brown and Douglas in [5].A key ingredient in the proof of the Brown and Douglas theorem is the classical Beurlingtheorem [3]. Recall that the Beurling theorem connects inner functions in H ∞ ( D ) withshift invariant subspaces of H ( D ). However, in the present case of higher dimensions,this approach does not work, as is well known, Beurling type classification does not holdfor shift invariant subspaces of H ( D n ), n > H ( D n ) and L ( T n ) to achievethe main goal. Section 3 contains the proof of the above theorem.Along the way to the proof of Theorem 1.1, in Section 2 we prove some basic propertiesof Toeplitz operators on the polydisc. Some of these observations are perhaps known (ifnot readily available in the literature) to experts, but they are necessary for our purposeshere. We also remark that the proof of k T ϕ k = k ϕ k ∞ , ϕ ∈ L ∞ ( T n ), in Proposition 2.2seems to be different even in the case of n = 1, as it avoids the standard techniquesof the spectral radius formula (see Brown and Halmos [6, page 99] and the monographs[9, 17, 18]).Moreover, in Section 4, we prove the following result, which connects inner functionswith shift operators, and is also of independent interest: If ϕ ∈ H ∞ ( D n ) is a nonconstantinner function, then M ϕ is a shift. ARTIALLY ISOMETRIC TOEPLITZ OPERATORS 3
Here, and in what follows, M ϕ denotes the analytic Toeplitz operator T ϕ whenever ϕ ∈ H ∞ ( D n ). In this case, M ϕ is simply the standard multiplication operator on H ( D n ), thatis, M ϕ f = ϕf for all f ∈ H ( D n ).In Section 5, as a first application to Theorem 1.1, we classify partially isometric hy-ponormal Toeplitz operators. Recall that a bounded linear operator T on some Hilbertspace is called hyponormal if T ∗ T − T T ∗ ≥
0. In Corollary 5.1, we prove the following: If T ϕ , ϕ ∈ L ∞ ( T n ), is a partial isometry, then T ϕ is hyponormal if and only if ϕ is an innerfunction in H ∞ ( D n ).Secondly, following the Halmos and Wallen [13] notion of power partial isometries (alsosee an Huef, Raeburn and Tolich [1]), in Corollary 5.2 we prove that partially isometricToeplitz operators are always power partial isometry. In Theorem 5.3, we further exploitthe Halmos and Wallen models of power partial isometries, and obtain a connection be-tween partially isometric Toeplitz operators, shifts, co-shifts, and direct sums of truncatedshifts.Finally, collecting all these results together, from an operator theoretic point of view,we obtain the following refinement of Theorem 1.1: Suppose T ϕ , ϕ ∈ L ∞ ( T n ) , is partially isometric. Then, up to unitary equivalence, T ϕ iseither a shift, or a co-shift, or a direct sum of truncated shifts. We stress that the latter possibility is only restricted to the n >
Preparatory results
In this section, we develop the necessary tools leading to the proof of Theorem 1.1.In this respect, we again remark that in what follows, we will often identify (via radiallimits) H ( D n ) with H ( T n ) without further explanation. Given ϕ ∈ L ∞ ( T n ), we denoteby L ϕ the Laurent operator on L ( T n ), that is, L ϕ f = ϕf for all f ∈ L ( T n ). Note that k L ϕ k B ( L ( T n )) = k ϕ k ∞ , where k ϕ k ∞ denotes the essential supremum norm of ϕ . The Toeplitz operator T ϕ withsymbol ϕ ∈ L ∞ ( T n ) is given by T ϕ = P H ( D n ) L ϕ | H ( D n ) . Clearly, T ϕ ∈ B ( H ( D n )). Also note that a function f = P k ∈ Z n a k z k ∈ L ( T n ) is in H ( D n )if and only if a k = 0 whenever at least one of the k j , j = 1 , . . . , n , in k = ( k , . . . , k n ) isnegative.We start with a several variables analogue of brother Riesz theorem (also see [22] forthe same for functions in H ∞ ( D n )). We denote the set of zeros of a scalar-valued function f by Z ( f ). Lemma 2.1. If f ∈ H ( D n ) is nonzero, then m ( Z ( f )) = 0 .Proof. Let m denote the normalized Lebesgue measure on T . Suppose f is a nonzerofunction in H ( D ). For each fixed w and w in T , we define the slice functions f w and f w by f w ( z ) = f ( w , z ) and f w ( z ) = ( z, w ) for all z ∈ T . Set Z = { w ∈ T : f w ≡ } . DEEPAK, PRADHAN, AND SARKAR
Note that
Z ⊆ Z ( f w ) for all w ∈ T . If m ( Z ) >
0, then the classical brother Riesztheorem implies that f is identically zero. Therefore, m ( Z ) = 0. Evidently m ( Z ( f w )) = ( w ∈ Z w ∈ Z c , and hence w m ( Z ( f w )) is a measurable function. By the Tonelli and Fubini theorem,we see that ( m × m )( Z ( f )) = Z T m ( Z ( f z )) dm ( z )= Z Z m ( Z ( f z )) dm ( z ) + Z Z c m ( Z ( f z )) dm ( z )= 0 . The rest of the proof now follows easily by the induction on n . (cid:3) We now prove that k T ϕ k B ( H ( D n )) = k ϕ k ∞ . As we have pointed out already in theintroductory section above, this may be known to experts. However, even when n = 1, ourproof seems to be direct as it avoids the standard techniques of the spectral radius formula.For instance, see the classic monograph [9, Corollary 7.8] and the recent monograph [17,Corollary 3.3.2]. Proposition 2.2. k T ϕ k = k ϕ k ∞ for all ϕ ∈ L ∞ ( T n ) .Proof. Without loss of generality assume that k T ϕ k = 1. Then, k ϕ k ∞ ≥
1, as k T ϕ k = k P H ( D n ) L ϕ | H ( D n ) k ≤ k ϕ k ∞ . Assume that k ϕ k ∞ > c for some c >
1. Then there exists f ∈ L ( T n ) such that k f k = 1and k ϕf k > c . Let f = P k ∈ Z n a k z k so that k f k = P k ∈ Z n | a k | < ∞ . We rewrite k f k = ∞ X m =0 X | k | = m | a k | , where | k | = n P i =1 | k i | for all k = ( k , . . . , k n ) ∈ Z n . There exists m ∈ N such that (cid:16) ∞ X m = m X | k | = m | a k | (cid:17) < k ϕ k ∞ ( c − . Let f = f ⊕ f , where f = ∞ X m = m X | k | = m a k z k . Observe that f is a polynomial in { z i , ¯ z i : 1 ≤ i ≤ n } , and k f k ≤
1. In particular, thereexists l ∈ Z n + such that g := z l f ∈ H ( D n ). Therefore k g k H ( D n ) = k z l f k H ( D n ) ≤ . Moreover, k ϕf k L ( T n ) >
1, since otherwise k ϕf k ≤ k ϕf k + k ϕf k ≤ k ϕ k ∞ k f k < c −
1) = c, ARTIALLY ISOMETRIC TOEPLITZ OPERATORS 5 contradicts the fact that k ϕf k > c . Since ϕz l f = z l ϕf in L ( T n ), we have k ϕg k L ( T n ) > . Now suppose ϕg = P k ∈ Z n b k z k . Then there exists p ∈ N such that (cid:16) p X m =0 X | k | = m | b k | (cid:17) > . Then k ˜ g k >
1, where ˜ g = p X m =0 X | k | = m b k z k . We write ϕg = X m>p X | k | = m b k z k ⊕ ˜ g. There exist a monomial z k , for some k ∈ Z n + , and a function h ∈ L ( T n ) such that z k ˜ g ∈ H ( D n ), and z k ϕg = h ⊕ z k ˜ g , which implies that P H ( D n ) ( z k ϕg ) = z k ˜ g ⊕ g forsome g ∈ H ( D n ). Finally, k T ϕ ( z k g ) k = k P H ( D n ) ( z k ϕg ) k ≥ k z k ˜ g k = k ˜ g k > , along with k z k g k H ( D n ) = k g k H ( D n ) ≤ k T ϕ k >
1. This is a contradiction.Thus we have k ϕ k ∞ = 1, which completes the proof. (cid:3) We also wish to point out that the above proof can be further simplified if we use thefollowing classification of Toeplitz operators [16, Theorem 3.1]: Suppose T ∈ B ( H ( D n ).Then T is a Toeplitz operator if and only if T ∗ z i T T z i = T for all i = 1 , . . . , n . In thiscase, the proof goes along the same lines as in the proof of [18, page 244, Theorem 4.1.4].We also believe that the above technique can be used to solve the problem of computingnorms of Toeplitz operators in other settings.Before proceeding to the proof of the main theorem, we conclude this section with aresult concerning unimodular functions in L ∞ ( T n ). Corollary 2.3.
Suppose ϕ is a nonzero function in L ∞ ( T n ) . If k T ϕ f k = k ϕ k ∞ k f k forsome nonzero f ∈ H ( D n ) , then k ϕ k ∞ ϕ is unimodular in L ∞ ( T n ) .Proof. In view of Proposition 2.2, without loss of generality we may assume that k T ϕ k = 1.Then Z T n | ϕ ( z ) | | f ( z ) | d m ( z ) = Z T n | f ( z ) | d m ( z ) . By Lemma 2.1, | ϕ ( z ) | = 1 for all z ∈ T n a.e. and the result follows. (cid:3) In particular, if T ϕ , ϕ ∈ L ∞ ( T n ), is a partial isometry, then ϕ is unimodular. DEEPAK, PRADHAN, AND SARKAR Proof of Theorem 1.1
In this section, without explicitly mentioning it in each instance, we always assumethat T ϕ , ϕ ∈ L ∞ ( T n ), is partially isometric.For simplicity we denote by R ( T ) the range of a bounded linear operator T . Clearly, R ( T ϕ ) is a closed subspace of H ( D n ). Lemma 3.1. R ( T ϕ ) is invariant under M z i , i = 1 , . . . , n .Proof. Note that, since k T ϕ k = 1, we have k ϕ k ∞ = 1. Suppose f ∈ R ( T ϕ ). By Corollary2.3, it follows that ϕ is unimodular, and hence k L ¯ ϕ f k = k f k . Since T ∗ ϕ is an isometry on R ( T ϕ ), we have k f k = k T ∗ ϕ f k ≤ k L ¯ ϕ f k = k ¯ ϕf k = k f k . Therefore, k P H ( D n ) ( ¯ ϕf ) k = k ¯ ϕf k , that is, P H ( D n ) ( ¯ ϕf ) = ¯ ϕf . This implies that¯ ϕf ∈ H ( D n ) , (3.1)and hence z i ¯ ϕf ∈ H ( D n ) for all i = 1 , . . . , n . Then T ϕ T ∗ ϕ ( z i f ) = T ϕ ( ¯ ϕz i f ) = P H ( D n ) ( | ϕ | z i f ) = P H ( D n ) ( z i f ) = z i f, implies that z i f ∈ R ( T ϕ ) for all i = 1 , . . . , n . This completes the proof. (cid:3) In what follows, if i ∈ { , . . . , n } and k i is a negative integer, then we write z k i i = ¯ z − k i i . Lemma 3.2.
For each i = 1 , . . . , n , the function ϕ cannot depend on both z i and ¯ z i variables at a time.Proof. We shall prove this by contradiction. Assume without loss of generality that ϕ depends on both z and ¯ z . Then ϕ = ∞ X k =1 ¯ z k ϕ − k ⊕ ∞ X k =0 z k ϕ k , and ϕ − k = 0 for some k = 0. Here ϕ k ∈ L ( T n − ), k ∈ Z , is a function of { z i , ¯ z j : i, j = 2 , . . . , n } . There exist non-negative integers k , . . . , k n , and l , . . . , l n such that thecoefficient of ¯ z k · · · ¯ z k n n z l · · · z l n n in the expansion of the Fourier series of ϕ − k is nonzero.Set Z kl := z k · · · z k n n z l · · · z l n n , and f := T ϕ ( z k Z kl ) − z T ϕ ( z k − Z kl ) . Note that f is a nonzero function in H ( D n ), and f does not depend on z . Since T ϕ ( z k − Z kl ) ∈ R ( T ϕ ), Lemma 3.1 implies that f ∈ R ( T ϕ ). In particular, by (3.1),¯ ϕf ∈ H ( D n ). On the other hand, since¯ ϕf = ∞ X k =1 z k ( f ¯ ϕ − k ) ⊕ ∞ X k =0 ¯ z k ( f ¯ ϕ k ) , it follows that f ¯ ϕ k = 0 for all k >
0. Since m ( { z ∈ T n : f ( z ) = 0 } ) = 0, we have ¯ ϕ k = 0for all k >
0. This yields ϕ = ∞ X k =0 ¯ z k ϕ − k , ARTIALLY ISOMETRIC TOEPLITZ OPERATORS 7 and hence ϕ depends on ¯ z and does not depend on z . This is a contradiction. (cid:3) We are now ready for the proof of Theorem 1.1.
Proof of Theorem 1.1.
Suppose T ϕ is a partial isometry. In view of Lemma 3.2, thereexists a (possibly empty) subset C of { z , . . . , z n } such that ϕ is analytic in z i for all z i ∈ A := C c , and co-analytic in z j for all z j ∈ C . Let A = { z i , . . . , z i p } and C = { z j , . . . , z j q } . Then p + q = n , and ϕ = X k ∈ Z q + ¯ z kC ϕ A,k , where ϕ A,k ∈ H ( D p ) is a function of { z i , . . . , z i p } , ¯ z kC = ¯ z k j · · · ¯ z k q j q , and k = ( k , . . . , k q ) ∈ Z q + . Note that ϕ A,l ∈ R ( T ϕ ) ( l ∈ Z q + ) . Indeed, ϕ A, = T ϕ ∈ R ( T ϕ ). Moreover, for each l ∈ Z q + \ { } , we have T ϕ z l = P H ( D n ) (cid:16) X k ∈ Z q + z l − kC ϕ A,k (cid:17) , that is T ϕ z l = X l − k ≥ z l − kC ϕ A,k . Here l − k ≥ l i − k i ≥ i = 1 , . . . , q . Thus the claim follows byinduction. By (3.1), we have ¯ ϕϕ A,l ∈ H ( D n ), l ∈ Z q + . Therefore¯ ϕϕ A,l = X k ∈ Z q + z kC ϕ A,k ϕ A,l ∈ H ( D n ) ( l ∈ Z q + ) . Consequently, ϕ A,k ϕ A,l ∈ H ( D p ) for all k and l , and hence, in particular, we have ϕ A,l ϕ A,l ∈ H ( D p ) ( l ∈ Z q + ) . This immediately implies that ϕ A,l ϕ A,l is a constant function, and hence ϕ A,l = α l ψ l forsome inner function ψ l ∈ H ∞ ( D p ) and scalar α l such that | α l | ≤ l ∈ Z q + . Assumewithout loss of generality that ϕ A, = 0. Now by the fact that ϕ A, ϕ A,k and ϕ A,k ϕ A, arein H ( D p ), we have ϕ A,k = β k ψ , k ∈ Z q + . Therefore ϕ = (cid:16) X k ∈ Z q + β k ¯ z kC (cid:17) ψ = ¯ ϕ ϕ , where ϕ = P k ∈ Z q + ¯ β k z kC and ϕ = ψ . We now turn to the converse part. First we haveclearly T ϕ T ϕ = T ϕ T ϕ . (3.2)We also claim that T ϕ T ∗ ϕ = T ∗ ϕ T ϕ . (3.3)This holds trivially when one of the functions ϕ or ϕ is constant. We continue with theabove notation, and assume that both A and C are nonempty subsets of { z , . . . , z n } . First DEEPAK, PRADHAN, AND SARKAR we observe that ϕ and ϕ depends only on { z i , . . . , z i p } and { z j , . . . , z j q } , respectively.Consider a monomial z k ∈ C [ z , . . . , z n ]. Suppose k = ( k , . . . , k n ), and write z k = z k c C z k a A , where k c = ( k j , . . . , k j q ) ∈ Z q + , and k a ∈ Z p + is the ordered p tuple made out of { k i } ni =1 \{ k j t } qt =1 . Since the analytic function ϕ depends only on z j s ∈ C , s = 1 , . . . , p , it is clearthat ¯ ϕ z k c C = ϕ a + ϕ c , where ϕ a depends only on { z j s } ps =1 (and hence it is an analytic function) and ϕ c ∈ L ( T q ) ⊖ H ( D q ) is a function of { z j t , ¯ z j t } qt =1 . Note that the latter property ensures that ϕ c (0) = 0.Then, on one hand, we have T ∗ ϕ T ϕ z k = P H ( D n ) ( ¯ ϕ ϕ z k ) = P H ( D n ) (cid:16) ( ϕ a + ϕ c ) ϕ z k a A (cid:17) = ϕ a ϕ z k a A , and on the other hand that T ϕ T ∗ ϕ z k = ϕ P H ( D n ) ( ¯ ϕ z k ) = ϕ P H ( D n ) (cid:16) ( ϕ a + ϕ c ) z k a A (cid:17) = ϕ ϕ a z k a A . Consequently, T ∗ ϕ T ϕ z k = T ϕ T ∗ ϕ z k for all k ∈ Z n + , which proves our claim. Now supposethat T ϕ = T ∗ ϕ T ϕ , where ϕ and ϕ depends on different variables. Using (3.2) and (3.3),we obtain T ϕ T ∗ ϕ = T ∗ ϕ T ϕ T ∗ ϕ T ϕ = ( T ∗ ϕ T ϕ )( T ϕ T ∗ ϕ ) = P R ( T ϕ ) , (3.4)which implies that T ϕ is a partial isometry. (cid:3) We remark that the commutativity and doubly commutativity of T ϕ and T ϕ in (3.2)and (3.3) will be useful in the particular applications to Theorem 1.1 in the final section.4. Inner functions and shifts
In this short section, we pause to prove an auxiliary result that is both a necessary toolfor our final refinement of partial isometric Toeplitz operators and a subject of independentinterest with its own applications.Let ϕ ∈ H ∞ ( D n ), and suppose the multiplication operator M ϕ is an isometry on H ( D n ). Then k ϕ k ∞ = k M ϕ k B ( H ( D n )) = 1 , and hence Corollary 2.3 implies that ϕ is a unimodular function in H ∞ ( D n ), that is, ϕ isan inner function. Now we prove that a nonconstant inner function always defines a shift(and not only isometry). Recall that an operator V ∈ B ( H ) is said to be a shift if V isan isometry and V ∗ m → m → ∞ in the strong operator topology.Recall that a closed subspace S ⊆ H ( D n ) is of Beurling type if there exists an innerfunction θ ∈ H ∞ ( D n ) such that S = θH ( D n ). It is also known that (cf. [15, Corollary6.3]) a closed subspace S ⊆ H ( D n ), n >
1, is of Beurling type if and only if R ∗ i R j = R j R ∗ i for all 1 ≤ i < j ≤ n , where R p = M z p | S ∈ B ( S ) is the restriction operator and p = 1 , . . . , n . Note that R ∗ i R j = P S M ∗ z i M z j | S and R j R ∗ i = M z j P S M ∗ z i | S , (4.1)for all i, j = 1 , . . . , n . Theorem 4.1. If ϕ ∈ H ∞ ( D n ) is a nonconstant inner function, then M ϕ is a shift. ARTIALLY ISOMETRIC TOEPLITZ OPERATORS 9
Proof.
It is well known (as well as easy to see) that M ϕ is an isometry. Following theclassical von Neumann and Wold decomposition for isometries, we only need to prove that H u := ∞ \ m =0 ϕ m H ( D n ) = { } . Assuming the contrary, suppose that H u = { } . We claim that H u is of Beurling type.Since the n = 1 case is obvious, we assume that n >
1. As ϕ p H ( D n ) ⊆ ϕ q H ( D n ) for all p ≥ q , we have P H u = SOT − lim m →∞ P ϕ m H ( D n ) . Since ϕ m H ( D n ), m ≥
1, is a Beurling type invariant subspace, in view of (4.1), it followsthat P H u M ∗ z i M z j h = M z j P H u M ∗ z i h, for all h ∈ H u . Then (4.1) again implies that H u is of Beurling type. Therefore, thereexists an inner function θ ∈ H ∞ ( D n ) such that H u = θH ( D n ) (note that the n = 1 casedirectly follows from Beurling). Then, for each m ≥
1, there exists an inner function ψ m ∈ H ∞ ( D n ) such that θ = ϕ m ψ m (for instance, see (5.1)). Since ϕ is a nonconstantinner function, by the maximum modulus principle [20, §
2, Theorem 6], we have | ϕ ( z ) | < z ∈ D n . For each fixed z ∈ D n , it follows that | θ ( z ) | = | ϕ ( z ) | m | ψ m ( z ) | ≤ | ϕ ( z ) | m → m → ∞ , and hence θ ≡
0. This contradiction shows that H u = { } . (cid:3) In fact, the above argument yields something more: Suppose {S m } m ≥ be a sequenceof Beurling type invariant subspaces of H ( D n ). Then T ∞ m =1 S m is also a Beurling typeinvariant subspace. Indeed, we let H m = T mi =1 S m . Then {H m } m ≥ forms a decreasingsequence of Beurling type invariant subspaces, and hence P T ∞ m =1 S m = P T ∞ m =1 H m = SOT − lim m →∞ P H m . The rest of the proof is then much as before.We also wish to point out that Theorem 4.1 can be proved by using (analytic) reproduc-ing kernel Hilbert space techniques. We believe that the algebraic tools described abovemight be useful in other settings.5.
Applications and further refinements
We begin with partially isometric Toeplitz operators that are hyponormal. A boundedlinear operator T acting on a Hilbert space is called hyponormal if [ T ∗ , T ] ≥
0, where[ T ∗ , T ] = T ∗ T − T T ∗ , is the self commutator of T .Now suppose T ϕ , ϕ ∈ L ∞ ( T n ), is a partial isometry. If ϕ ∈ H ∞ ( D n ) is inner, then T ϕ is an isometry and hence is hyponormal. For the converse direction, we note by Theorem1.1 that T ϕ = T ∗ ϕ T ϕ for some inner functions ϕ and ϕ in H ∞ ( D n ) which depends ondifferent variables. If ϕ is a constant function, then T ϕ = T ϕ = M ϕ is an isometry, and hence T ϕ is hyponormal. If ϕ is a constant function, then T ϕ = T ∗ ϕ = M ∗ ϕ is a co-isometry, and hence T ϕ cannot be hyponormal. Suppose both ϕ and ϕ are nonconstantfunctions. Now (3.2) and (3.3) implies that T ∗ ϕ T ϕ = T ∗ ϕ T ϕ T ∗ ϕ T ϕ = ( T ∗ ϕ T ϕ )( T ϕ T ∗ ϕ ) = T ϕ T ∗ ϕ . Then, by (3.4) we see that [ T ∗ ϕ , T ϕ ] ≥ T ϕ T ∗ ϕ ≤ T ϕ T ∗ ϕ . By noting that ϕ and ϕ are analytic functions, we see M ϕ M ∗ ϕ ≤ M ϕ M ∗ ϕ , which, by the Douglas range inclusion theorem, is equivalent to M ϕ = M ϕ X for some X ∈ B ( H ( D n )). Observe that M ϕ M z i X = M z i M ϕ X = M z i M ϕ = M ϕ M z i = M ϕ XM z i , (5.1)implies that M z i X = XM z i for all i = 1 , . . . , n , and hence X = M ψ for some ψ ∈ H ∞ ( D n ).Hence, we conclude that ϕ = ϕ ψ . Since ϕ and ϕ are inner functions, ψ ∈ H ∞ ( D n ) isinner. Moreover, since ϕ and ϕ depends on different variables, that ϕ = ϕ ψ is possibleif and only if ψ is a unimodular constant. Suppose ϕ = αϕ , where | α | = 1. Then T ϕ = T ∗ ϕ T ϕ = αT ∗ ϕ T ϕ = αT | ϕ | = αI H ( D n ) , as | ϕ | = 1 on T n , that is, T ϕ is a unimodular constant times the identity operator. Wehave therefore shown the following result: Corollary 5.1.
Let T ϕ , ϕ ∈ L ∞ ( T n ) , be a partial isometry. Then T ϕ is hyponormal ifand only if ϕ is an inner function in H ∞ ( D n ) . Therefore, in view of Theorem 4.1, T ϕ is hyponormal if and only if (up to unitaryequivalence) T ϕ is a shift.We recall [13, Halmos and Wallen] that a bounded linear operator T acting on someHilbert space is called a power partial isometry if T m is partially isometric for all m ≥ Corollary 5.2.
Partially isometric Toeplitz operators are power partial isometry.
We also recall from Halmos and Wallen [13] (also see [1]) that every power partialisometry is a direct sum whose summands are unitary operators, shifts, co-shifts, andtruncated shifts. Recall that a truncated shift S of index p , p ∈ N , on some Hilbert space H is an operator of the form S = · · · I H · · · I H · · · · · · · · · I H p × p , where H is a Hilbert space, and H = H ⊕ · · · ⊕ H | {z } p .We prove that, up to unitary equivalence, a partial isometric T ϕ is simply direct sumof truncated shifts, or a shift, or a co-shift (that is, adjoint of a shift). The proof isessentially contained in Theorem 4.1 and the Halmos and Wallen models of power partialisometries. ARTIALLY ISOMETRIC TOEPLITZ OPERATORS 11
Theorem 5.3.
Up to unitary equivalence, a partially isometric Toeplitz operator is eithera shift, or a co-shift, or a direct sum of truncated shifts.Proof.
Suppose T ϕ , ϕ ∈ L ∞ ( T n ), is a partial isometry. By Theorem 1.1, T ϕ = T ∗ ϕ T ϕ ,where ϕ and ϕ are inner functions in H ∞ ( D n ) and depends on different variables. More-over, by Corollary 5.2, T ϕ is a power partial isometry. If ϕ is a constant function, then T ϕ is a shift, and if ϕ is a constant function, then T ϕ is a co-shift. Now let both ϕ and ϕ are nonconstant functions. Following the construction of Halmos and Wallen [13, page660] (also see [1]), we set E m = T ∗ mϕ T mϕ and F m = T mϕ T ∗ mϕ for the initial and final projec-tions of the partial isometry T mϕ , m ≥
1. By (3.2) and (3.3) it follows that E m = T mϕ T ∗ mϕ and F m = T mϕ T ∗ mϕ , and hence R ( E m ) = ϕ m H ( D n ) and R ( F m ) = ϕ m H ( D n ) , for all m ≥
1. Then, by Theorem 4.1, we have \ m ≥ R ( E m ) = \ m ≥ ϕ m H ( D n ) = { } , and similarly T m ≥ R ( F m ) = { } . Therefore, the unitary part, the shift part, and the co-shift part of the Halmos and Wallen model of T ϕ are trivial (see [13, page 661] or [1]).Hence in this case, T ϕ is a direct sum of truncated shifts. (cid:3) Clearly, Corollary 5.1 immediately follows from the above result as well. Also, notethat the Halmos and Wallen models of power partial isometries played an important rolein the proof of the above theorem. Thus, in view of Theorem 5.3, we are motivated topose the following natural problem:
Problem 5.4.
Characterize all power partial isometric Toeplitz operators on H ( D n ).We refer [1, 4, 10] for a more recent view point of power partial isometries. Finally,summarizing our results from an operator theoretic point of view, we conclude the follow-ing: Let T ϕ , ϕ ∈ L ∞ ( T n ), be a partially isometric Toeplitz operator. Then the followinghold:(1) If n = 1, then T ϕ is either a shift, or a co-shift. This is due to Brown and Douglas.(2) If n >
1, then, up to unitary equivalence, T ϕ is either a shift, or a co-shift, or adirect sum of truncated shifts. Acknowledgement:
The research of the second named author is supported by NBHM(National Board of Higher Mathematics, India) post-doctoral fellowship no: 0204/27-/2019/R&D-II/12966. The third author is supported in part by the Mathematical Re-search Impact Centric Support, MATRICS (MTR/2017/000522), and Core ResearchGrant (CRG/2019/000908), by SERB, Department of Science & Technology (DST), andNBHM (NBHM/R.P.64/2014), Government of India.
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