Compressions of k th --order slant Toeplitz operators to model spaces
aa r X i v : . [ m a t h . F A ] N ov COMPRESSIONS OF k th –ORDER SLANT TOEPLITZ OPERATORS TOMODEL SPACES BARTOSZ LANUCHA, MA LGORZATA MICHALSKA
Abstract.
In this paper we consider compressions of k th –order slant Toeplitz operatorsto the backward shift invariant subspaces of the classical Hardy space H . In particular,we characterize these operators using compressed shifts and finite rank operators ofspecial kind. Introduction
Let T = { z : | z | = 1 } and denote by L = L ( T ) the space of all Lebesgue measurablefunctions on T with square integrable modulus, and by L ∞ = L ∞ ( T ) – the space ofall Lebesgue measurable and essentially bounded functions. Moreover, let H be theclassical Hardy space in the disk D = { z : | z | < } . As usual, we will view H as a spaceof functions analytic in D or (via radial limits) as a closed subspace of L (see [6, 8] fordetails). The orthogonal projection from L onto H will be denoted by P .For ϕ ∈ L ∞ the classical Toeplitz operator T ϕ is defined on H by T ϕ f = P M ϕ f, f ∈ H , where M ϕ : L → L is the multiplication operator f ϕf . For ϕ ∈ L the operators T ϕ and M ϕ can be defined on a dense subset of H and L , respectively (namely, on boundedfunctions).Classical Toeplitz operators, as compressions of multiplication operators to H , areamong the most studied linear operators on the Hardy space. Their study goes back to thebeginning of the 20th century (for details and references see, e.g., [23]). In recent years,compressions of multiplication operators to model spaces were widely studied. Modelspaces are the non–trivial subspaces of H which are invariant for the backward shiftoperator S ∗ = T ¯ z . Each of these subspaces is of the form K α = H ⊖ αH , where α is aninner function: a function analytic and bounded in D ( α ∈ H ∞ ) such that its boundaryvalues are of modulus one a.e. on T .For an inner function α and for ϕ ∈ L let A αϕ f = P α ( ϕf ) , f ∈ K ∞ α = K α ∩ H ∞ , Mathematics Subject Classification.
Key words and phrases. model space, compressed shift, Toeplitz operator, slant Toeplitz operator,generalized slant Toeplitz operator, truncated Toeplitz operator. here P α is the orthogonal projection from L onto K α . Note that A αϕ is densely definedsince K ∞ α is a dense subset of K α . These operators, called truncated Toeplitz operators,gained attention in 2007 with D. Sarason’s paper [24]. Ever since, truncated Toeplitzoperators have been under constant intensive study, which led to many interesting resultsand applications (see [10] and references therein). More recently, authors in [4, 5] and [3]introduced the so–called asymmetric truncated Toeplitz operators: for two inner functions α , β and for ϕ ∈ L an asymmetric truncated Toeplitz operator A α,βϕ is defined by A α,βϕ f = P β ( ϕf ) , f ∈ K ∞ α . As above, A α,βϕ is densely defined. Moreover, A α,αϕ = A αϕ . These operators were thenstudied in [12, 17, 18, 21, 22].Recall that if ϕ ( z ) = ∞ X j = −∞ a j z j ∈ L ∞ , then for each n ∈ Z , M ϕ ( z n ) = ∞ X j = −∞ a j − n z j andthe matrix of M ϕ with respect to the standard basis { z n : n ∈ Z } is given by . . . ... ... ... ... . . . a a − a − a − . . .. . . a a a − a − . . .. . . a a a a − . . .. . . a a a a . . . ... ... ... ... . . . . Similarly, for each n ∈ N , T ϕ ( z n ) = ∞ X j =0 a j − n z j and T ϕ is the operator on H representedby the matrix (with respect to { z n : n ∈ N } ) a a − a − a − . . .a a a − a − . . .a a a a − . . .a a a a . . . ... ... ... ... . . . . In [1, 2], for k ∈ N , k ≥
2, and ϕ ( z ) = ∞ X j = −∞ a j z j ∈ L ∞ , the authors define the k th –orderslant Toeplitz operator U ϕ on L by U ϕ ( z n ) = ∞ X j = −∞ a kj − n z j for n ∈ Z , that is, as theoperator on L represented by the matrix . . . ... ... ... ... . . . a a − a − a − . . .. . . a k a k − a k − a k − . . .. . . a k a k − a k − a k − . . .. . . a k a k − a k − a k − . . . ... ... ... ... . . . . quivalently, U ϕ : L → L , ϕ ∈ L ∞ , is given by U ϕ f = W k M ϕ f, f ∈ L , where W k ( z n ) = ( z m if nk = m ∈ Z , nk Z . The authors also consider V ϕ , the compression of U ϕ to H , defined by V ϕ f = P U ϕ f, f ∈ H , and represented by the matrix a a − a − a − . . .a k a k − a k − a k − . . .a k a k − a k − a k − . . .a k a k − a k − a k − . . . ... ... ... ... . . . . Note that if ϕ ∈ L , then U ϕ and V ϕ are densely defined. For k = 2, operators U ϕ (called slant Toeplitz operators) and their compressions to H were first studied in [13,26](see also [14–16]). These operators have connections with wavelet theory and dynamicalsystems (see, e.g., [11, 14, 25]). In [19, 20] the authors investigate commutativity of k th –order slant Toeplitz operators. Observe that if we consider k = 1 in the above definitions,then U ϕ = M ϕ and V ϕ = T ϕ .Here we study compressions of k th –order slant Toeplitz operators to model spaces. Fora fixed k ∈ N , two inner functions α , β and for ϕ ∈ L define U α,βϕ f = P β U ϕ f = P β W k ( ϕf ) , f ∈ K ∞ α , and denote by S k ( α, β ) the set of all the compressions U α,βϕ , ϕ ∈ L , which can beboundedly extended to K α .In Section 2 we investigate conditions on ϕ which imply that U α,βϕ = 0. We show that,in contrast with the case of U ϕ and V ϕ [1, 2], U α,βϕ can be equal to the zero operator for ϕ not equal to zero.In Section 3 we characterize operators from S k ( α, β ) using the compressed shifts S α = A αz and S β = A βz . It is well known that a bounded linear operator T : H → H is aToeplitz operator if and only if T − S ∗ T S = 0, where S is the shift operator S = T z .A similar characterization was given in [1, 2] for V ϕ . Namely, T is a compression of a k th –order slant Toeplitz operator to H if and only if T − S ∗ T S k = 0. We show, forexample, that a bounded linear operator U from K α into K β belongs to S k ( α, β ) if andonly if U − S β U ( S ∗ α ) k is a special kind of operator of rank at most k + 1. This is done inthe spirit of D. Sarason’s characterization of truncated Toeplitz operators given in [24], here, among other results, he shows that U is a truncated Toeplitz operator if and onlyif U − S α U S ∗ α is of rank two and special kind (see also [3, 12, 21] for the asymmetric case).2. Operators from S k ( α, β ) equal to the zero operator In this section we investigate for which ϕ ∈ L , U α,βϕ = 0.We start with some basic properties of the operator W k and its adjoint W ∗ k . Some ofthese properties can be found for example in [13] (for k = 2) and in [1, 2]. Lemma 2.1.
Let k ∈ N . Then (a) W ∗ k f ( z ) = f ( z k ) , | z | = 1 , f ∈ L , (b) W ∗ k ( f · g ) = W ∗ k f · W ∗ k g for all f, g ∈ L such that f · g ∈ L , (c) W k W ∗ k = I L and W ∗ k W k is the orthogonal projection from L onto the closed linearspan of { z km : m ∈ Z } , (d) W k f = W k f and W ∗ k f = W ∗ k f , f ∈ L , (e) P reduces W k , that is, P W k = W k P and P W ∗ k = W ∗ k P , (f) M ϕ W k = W k M W ∗ k ϕ for every ϕ ∈ L ∞ , (g) for every inner function α the function W ∗ k α is also inner and P α W k = W k P W ∗ k α .Proof. The proofs of (a)–(e) are straightforward. We only prove (f) and (g).To show (f) fix ϕ ∈ L ∞ and take any f, g ∈ L . Then by (b) and (d), h M ϕ W k f, g i = h f, W ∗ k ( ϕg ) i = (cid:10) f, W ∗ k ϕ · W ∗ k g (cid:11) = (cid:10) W k M W ∗ k ϕ f, g (cid:11) . If α is an inner function, then W ∗ k α is also an inner function by (a). Since P α = I L − M α P M α , using (e) and (f) we obtain P α W k = W k − M α P W k M W ∗ k α = W k − M α W k P M W ∗ k α = W k ( I L − M W ∗ k α P M W ∗ k α ) = W k P W ∗ k α . That is, (g) holds. (cid:3)
For the reminder of this section fix k ∈ N . Proposition 2.2.
Let α and β be two inner functions and let ϕ ∈ L . If ϕ ∈ αH +( W ∗ k β ) H , then U α,βϕ = 0 .Proof. Assume that ϕ = αh + W ∗ k β · h for some h , h ∈ H . Then for all f ∈ K ∞ α and g ∈ K ∞ β we have (by Lemma 2.1(g)), (cid:10) U α,βϕ f, g (cid:11) = h ϕf, W ∗ k g i = (cid:10) αh f, W ∗ k g (cid:11) + h W ∗ k β · h f, W ∗ k g i = h f, αh · W ∗ k g i + h P β W k ( W ∗ k β · h f ) , g i = (cid:10) W k P W ∗ k β ( W ∗ k β · h f ) , g (cid:11) = 0 . Corollary 2.3. If U = U α,βϕ , then ϕ can be chosen from K α + K W ∗ k β . Hence U α,βϕ is not uniquely determined by its symbol. Moreover, boundedness of thesymbol is not necessary for the boundedness of U α,βϕ .In general, the implication in Proposition 2.2 cannot be reversed as the following ex-ample shows. Example 2.4.
Let k = 2, α ( z ) = z and β ( z ) = z . In that case, for ϕ = P ∞ n = −∞ a n z n ∈ L , the operator U α,βϕ is represented by the matrix a a − a − a − a a a a − a a a a . Hence, if for example ϕ ( z ) = z , then U α,βϕ = 0. But here z / ∈ αH + ( W ∗ k β ) H = z H + z H . Observe that U α,βϕ can be seen as a composition of W k and an asymmetric truncatedToeplitz operator. Indeed, using Lemma 2.1(g), for f ∈ K ∞ α we get U α,βϕ f = P β W k ( ϕf ) = W k P W ∗ k β ( ϕf ) = W k A α,W ∗ k βϕ f. Now Proposition 2.2 follows from the fact that A α,βϕ = 0 if and only if ϕ ∈ αH + βH (see [3, 17]).Let α be an inner function. Then the operator C α : L → L , defined by the formula C α f ( z ) = α ( z ) zf ( z ) , | z | = 1 , is a conjugation on L (an antilinear, isometric involution). Moreover, C α ( K α ) = K α (see [9] and [8, Chapter 8]). It is known that all truncated Toeplitz operators are C α –symmetric, that is, C α A αϕ C α = ( A αϕ ) ∗ = A αϕ [24]. It was observed in [21] that C β A α,βϕ C α = A α,βαϕβ (see also [12]). Here we have Proposition 2.5.
Let α , β be two inner functions and let ϕ ∈ L . Then C β U α,βϕ C α = U α,βψ with ψ = z k − αϕW ∗ k β .Proof. Let ϕ ∈ L . Then for f ∈ K ∞ α , g ∈ K ∞ β we have, by Lemma 2.1, (cid:10) C β U α,βϕ C α f, g (cid:11) = h C β P β W k M ϕ C α f, g i = h C β g, P β W k ( ϕC α f ) i = h C β g, W k ( ϕC α f ) i = (cid:10) zβg, W k ( ϕzαf ) (cid:11) = (cid:10) W ∗ k ( zβg ) , ϕzαf (cid:11) = D z k − αϕW ∗ k β · f, W ∗ k g E = D P β W k M z k − αϕW ∗ k β f, g E = D U α,βz k − αϕW ∗ k β f, g E . As a corollary to Lemma 2.5 we get the following.
Corollary 2.6.
Let α , β be two inner functions and let U be a bounded linear operatorfrom K α into K β . Then U ∈ S k ( α, β ) if and only if C β U C α ∈ S k ( α, β ) . Proposition 2.7.
Let α , β be two inner functions and let ϕ ∈ L . If ϕ ∈ αH + z k − ( W ∗ k β ) H , then U α,βϕ = 0 .Proof. Let ϕ = αh + z k − W ∗ k β · h with h , h ∈ H . By Proposition 2.2 we have U α,βαh = 0and, consequently, U α,βϕ = U α,βz k − W ∗ k β · h . Now, using Lemma 2.5, we obtain C β U α,βϕ C α = C β U α,βz k − W ∗ k β · h C α = U α,βz k − αz k − W ∗ k β · h · W ∗ k β = U α,βαh = 0and it follows that U α,βϕ = 0. (cid:3) Corollary 2.8. If U = U α,βϕ , then ϕ can be chosen from K α + z k − K W ∗ k β . Let us note that if an inner function β is such that β (0) = 0, then z k divides W ∗ k β = β ( z k ) and z k − W ∗ k β is also an inner function. Moreover, αH + ( W ∗ k β ) H = αH + z k − ( W ∗ k β ) z k − H ⊂ αH + z k − ( W ∗ k β ) H . Characterizations of operators from S k ( α, β )Let α and β be two inner functions. In this section we will characterize operators from S k ( α, β ), k ∈ N , using compressed shifts and finite rank operators of special kind. Recallthat the compressed shift S α is defined as S α = A αz = P α S | K α . As K α is S ∗ –invariant, wehave S ∗ α = A αz = S ∗| K α .For each n ∈ N and w ∈ D the functional f f ( n ) ( w ) is bounded on H and sothere exists k w,n ∈ H such that f ( n ) ( w ) = h f, k w,n i . It is not difficult to verify that k w,n ( z ) = n ! z n (1 − wz ) n +1 and, in particular, k ,n ( z ) = n ! z n . Denote k w = k w, . Lemma 3.1.
Let k ∈ N . For f ∈ H we have (a) ( S ∗ ) k f ( z ) = z k f ( z ) − k − X j =0 h f, k ,j i j ! z k − j , | z | = 1 , (b) W ∗ k f − z k W ∗ k S ∗ f = h f, k i = f (0) . roof. Let k ∈ N and f ∈ H . Then ( S ∗ ) k f = ( T z ) k f = P ( z k f ). Since f ( z ) = P ∞ j =0 h f,k ,j i j ! z j , we have, for | z | = 1,( S ∗ ) k f ( z ) = P ∞ X j =0 h f, k ,j i j ! z j − k ! = ∞ X j = k h f, k ,j i j ! z j − k = z k f ( z ) − k − X j =0 h f, k ,j i j ! z k − j . Now, observe that W ∗ k f ( z ) − z k W ∗ k S ∗ f ( z ) = f ( z k ) − z k W ∗ k ( zf ( z ) − zf (0)) = f ( z k ) − (cid:0) f ( z k ) − f (0) (cid:1) = f (0) , which proves (b). (cid:3) As K α is a closed subspace of H , f f ( n ) ( w ) is also bounded on K α for each n ∈ N and w ∈ D . Then f ( n ) ( w ) = h f, k αw,n i for all f ∈ K α and k αw,n = P α k w,n (see, e.g., [7, p.204]). Denote e k αw,n = C α k αw,n . In particular, k αw ( z ) = P α k w ( z ) = 1 − α ( w ) α ( z )1 − wz and e k αw ( z ) = α ( z ) − α ( w ) z − w . Lemma 3.2.
Let k ∈ N and ϕ ∈ L . Then U α,βϕ − S β U α,βϕ ( S ∗ α ) k = k β ⊗ χ + k − X j =0 ψ j ⊗ k α ,j , where the equality holds on K ∞ α with χ = P α ( ϕ ) and ψ j = j ! S β P β W k ( ϕz k − j ) , ≤ j ≤ k − . Proof.
Let f ∈ K ∞ α and g ∈ K ∞ β . By Lemma 3.1(a) we have (cid:10)(cid:0) U α,βϕ − S β U α,βϕ ( S ∗ α ) k (cid:1) f, g (cid:11) = h ϕf, W ∗ k g i − (cid:10) ϕ ( S ∗ α ) k f, W ∗ k S ∗ β g (cid:11) = h ϕf, W ∗ k g i − (cid:10) ϕ ( S ∗ ) k f, W ∗ k S ∗ g (cid:11) = h ϕf, W ∗ k g i − (cid:10) ϕz k f, W ∗ k S ∗ g (cid:11) + k − X j =0 h f, k ,j i j ! (cid:10) ϕz k − j , W ∗ k S ∗ g (cid:11) = (cid:10) ϕf, W ∗ k g − z k W ∗ k S ∗ g (cid:11) + k − X j =0 j ! (cid:10) f, k α ,j (cid:11) (cid:10) S β P β W k ( ϕz k − j ) , g (cid:11) . By Lemma 3.1(b), (cid:10) ϕf, W ∗ k g − z k W ∗ k S ∗ g (cid:11) = h ϕf, h g, k β ii = hh f, P α ( ϕ ) i k β , g i . (cid:3) For k = 2 the following was partly noted in [13]. Lemma 3.3.
Let k ∈ N and m ∈ N , | m | < k . Then W k M z m W ∗ k = ( I L if m = 0 , if < | m | < k. roof. Let f ∈ L , f = ∞ P j = −∞ a j z j . Then W ∗ k f ( z ) = ∞ X j = −∞ a j z kj and M z m W ∗ k f ( z ) = ∞ X j = −∞ a j z kj + m . If 0 < | m | < k , then kj + m is not divisible by k and so W k M z m W ∗ k f = 0. On the otherhand, if m = 0, then W k W ∗ k f = f (Lemma 2.1(c)). (cid:3) Lemma 3.4.
Let k ∈ N , χ ∈ K α and ψ , . . . , ψ k − ∈ K β be such that ψ (0) = . . . = ψ k − (0) = 0 . Then for ϕ = χ + k X j =1 ( k − j )! (cid:0) W ∗ k S ∗ β ψ k − j (cid:1) · z j ∈ L we have (3.1) U α,βϕ − S β U α,βϕ ( S ∗ α ) k = k β ⊗ χ + k − X j =0 ψ j ⊗ k α ,j . Proof.
Note that for each j ∈ { , . . . , k } we have (cid:0) W ∗ k S ∗ β ψ k − j (cid:1) · z j = ψ k − j ( z k ) z k z j ∈ H since ψ k − j (0) = 0. This and the fact that P α = P α P give P α ( ϕ ) = P α P ( ϕ ) = P α χ = χ. Moreover, since P β = P β P and P W k = W k P , for 1 ≤ l ≤ k we have P β W k ( ϕz l ) = P β W k P ( ϕz l ) = P β W k k X j =1 ( k − j )! (cid:0) W ∗ k S ∗ β ψ k − j (cid:1) · z j − l ! = P β k X j =1 ( k − j )! W k M z j − l W ∗ k S ∗ β ψ k − j ! = ( k − l )! S ∗ β ψ k − l , where the last equality follows from Lemma 3.3 and the fact that 0 < | j − l | < k − j ∈ { , . . . , k } \ { l } . Hence, for 0 ≤ j ≤ k − j ! S β P β W k ( ϕz k − j ) = 1 j ! S β ( j ! S ∗ β ψ j ) = S β S ∗ β ψ j = ψ j ( ψ j (0) = 0) , and (3.1) follows from Lemma 3.2. (cid:3) Theorem 3.5.
Let U be a bounded linear operator from K α into K β . Then U ∈ S k ( α, β ) , k ∈ N , if and only if there exist functions χ ∈ K α and ψ , . . . , ψ k − ∈ K β such that U − S β U ( S ∗ α ) k = k β ⊗ χ + k − X j =0 ψ j ⊗ k α ,j . (3.2) roof. If U ∈ S k ( α, β ), then U = U α,βϕ for some ϕ ∈ L and it satisfies (3.2) by Lemma3.2.Assume that U satisfies (3.2) for some χ ∈ K α and ψ , . . . , ψ k − ∈ K β . Without anyloss of generality we can additionally assume that ψ (0) = . . . = ψ k − (0) = 0 (otherwisewe would replace ψ j and χ with ψ j − ψ j (0) k k β k k β and χ + k − X j =0 ψ j (0) k k β k k α ,j , respectively). By(3.2), for every l ∈ N , S lβ U ( S ∗ α ) kl − S l +1 β U ( S ∗ α ) ( l +1) k = S lβ k β ⊗ S klα χ + k − X j =0 S lβ ψ j ⊗ S klα k α ,j . It follows that for any n ∈ N , U = n X l =0 S lβ k β ⊗ S klα χ + k − X j =0 S lβ ψ j ⊗ S klα k α ,j ! + S n +1 β U ( S ∗ α ) ( n +1) k . (3.3)Since for every f ∈ K α , S n +1 β U ( S ∗ α ) ( n +1) k f → n → ∞ , we have U f = ∞ X l =0 S lβ k β ⊗ S klα χ + k − X j =0 S lβ ψ j ⊗ S klα k α ,j ! f. (3.4)Let now ϕ = χ + ψ with ψ = k X j =1 ( k − j )!( W ∗ k S ∗ β ψ k − j ) · z j ∈ H . By Lemma 3.4, U α,βϕ − S β U α,βϕ ( S ∗ α ) k = k β ⊗ χ + k − X j =0 ψ j ⊗ k α ,j on K ∞ α , and as above (3.3) holds with U α,βϕ in place of U (on K ∞ α ). For f ∈ K ∞ α and g ∈ K ∞ β wehave (cid:10) S n +1 β U α,βϕ ( S ∗ α ) ( n +1) k f, g (cid:11) = (cid:10) ϕ ( S ∗ α ) k ( n +1) f, W ∗ k ( S ∗ β ) n +1 g (cid:11) → n → ∞ . (3.5)Indeed, since (cid:10) ϕ ( S ∗ α ) k ( n +1) f, W ∗ k ( S ∗ β ) n +1 g (cid:11) = (cid:10) χT z k ( n +1) f, W ∗ k ( S ∗ ) n +1 g (cid:11) + (cid:10) ( S ∗ ) k ( n +1) f, ψW ∗ k T z n +1 g (cid:11) = D P (cid:16) χz k ( n +1) f (cid:17) , W ∗ k ( S ∗ ) n +1 g E + (cid:10) ( S ∗ ) k ( n +1) f, ψW ∗ k P ( z n +1 g ) (cid:11) = D P (cid:16) χz k ( n +1) f (cid:17) , W ∗ k ( S ∗ ) n +1 g E + (cid:10) ( S ∗ ) k ( n +1) f, P (cid:0) z k ( n +1) ψW ∗ k g (cid:1)(cid:11) , (cid:12)(cid:12)(cid:12)D P (cid:16) χz k ( n +1) f (cid:17) , W ∗ k ( S ∗ ) n +1 g E(cid:12)(cid:12)(cid:12) ≤ k χf k · k ( S ∗ ) n +1 g k → n → ∞ , and (cid:12)(cid:12)(cid:10) ( S ∗ ) k ( n +1) f, P (cid:0) z k ( n +1) ψW ∗ k g (cid:1)(cid:11)(cid:12)(cid:12) ≤ k ( S ∗ ) k ( n +1) f k · k ψ · W ∗ k g k → n → ∞ . This means that for f ∈ K ∞ α and g ∈ K ∞ β , (cid:10) U α,βϕ f, g (cid:11) = * ∞ X l =0 S lβ k β ⊗ S klα χ + k − X j =0 S lβ ψ j ⊗ S klα k α ,j ! f, g + , and by (3.4), (cid:10) U α,βϕ f, g (cid:11) = h U f, g i . It follows that U = U α,βϕ ∈ S k ( α, β ). (cid:3) Corollary 3.6.
If a bounded linear operator U : K α → K β satisfies (3.2) , then U = U α,βϕ with ϕ = χ + k − X j =0 ψ j ( z k ) j ! z j . Proof.
Assume that U satisfies (3.2). If ψ (0) = . . . = ψ k − (0) = 0, then following theproof of Theorem 3.5 we get that U = U α,βϕ with ϕ = χ + k − X j =0 j !( W ∗ k S ∗ β ψ j ) · z k − j = χ + k − X j =0 z k ψ j ( z k ) j ! z k − j = χ + k − X j =0 ψ j ( z k ) j ! z j . Otherwise, as explained at the beginning of the proof of Theorem 3.5, replace ψ j and χ with ψ j − ψ j (0) k k β k k β and χ + k − X j =0 ψ j (0) k k β k k α ,j , respectively. Then, by the first part of theproof, U = U α,βϕ with ϕ = χ + k − X j =0 ψ j (0) k k β k k α ,j ! + k − X j =0 (cid:18) ψ j ( z k ) − ψ j (0) k k β k k β ( z k ) (cid:19) j ! z j . However, in that case ϕ − ϕ = k − X j =0 ψ j (0) k k β k k α ,j − k − X j =0 ψ j (0) k k β k (1 − β (0) β ( z k )) j ! z j = k − X j =0 ψ j (0) k k β k P α ( j ! z j ) − k − X j =0 ψ j (0) k k β k j ! z j + β (0) β ( z k ) k − X j =0 ψ j (0) k k β k j ! z j = P α k − X j =0 ψ j (0) k k β k j ! z j ! − k − X j =0 ψ j (0) k k β k j ! z j ! + z k − W ∗ k β · β (0) k − X j =0 ψ j (0) k k β k j ! z k − − j ∈ αH + z k − ( W ∗ k β ) H nd so, by Proposition 2.7, U = U α,βϕ = U α,βϕ . (cid:3) Note that if ψ (0) = . . . = ψ k − (0) = 0, then the pairwise orthogonal functions ψ j ( z k ) j ! z j , 0 ≤ j ≤ k −
1, all belong to H . Corollary 3.7. If U ∈ S k ( α, β ) , k ∈ N , then there exist functions χ ∈ K α and ψ , . . . , ψ k − ∈ K β such that U = U α,βϕ with ϕ = χ + ψ (cid:0) z k (cid:1) + 1 z ψ (cid:0) z k (cid:1) + 1 z ψ (cid:0) z k (cid:1) + . . . + 1 z k − ψ k − (cid:0) z k (cid:1) and ψ (0) = . . . = ψ k − (0) = 0 . Moreover, the above decomposition is orthogonal. Corollary 3.8.
Let U be a bounded linear operator from K α into K β . Then U ∈ S k ( α, β ) , k ∈ N , if and only if there exist functions χ ∈ K α and ψ , . . . , ψ k − ∈ K β such that U − S ∗ β U S kα = e k β ⊗ χ + k − X j =0 ψ j ⊗ e k α ,j . (3.6) Proof.
By Corollary 2.6, U ∈ S k ( α, β ) if and only if C β U C α ∈ S k ( α, β ). By Theorem 3.5,the latter happens if and only if(3.7) C β U C α − S β ( C β U C α )( S ∗ α ) k = k β ⊗ µ + k − X j =0 ν j ⊗ k α ,j for some functions µ ∈ K α and ν , . . . , ν k − ∈ K β . Since C α is an involution, (3.7) isequivalent to C β U C α − C β S β C β U C α ( S ∗ α ) k C α = C β ( k β ⊗ µ ) C α + k − X j =0 C β ( ν j ⊗ k α ,j ) C α . By C α -symmetry of the compressed shift, the above can be written as U − S ∗ β U S kα = e k β ⊗ χ + k − X j =0 ψ j ⊗ e k α ,j , with χ = C α µ ∈ K α and ψ j = C β ν j ∈ K β for j = 0 , , . . . , k − (cid:3) Corollary 3.9. If U satisfies (3.6) , then U = U α,βϕ with ϕ = β ( z k ) z k χ + α k − X j =0 ψ j ( z k ) j ! z j +1 . Proof.
Assume that U satisfies (3.6). A reasoning similar to the one given in the proofof Corollary 3.8 shows that C β U C α satisfies (3.7) with µ = C α χ and ν j = C β ψ j . ByCorollary 3.6, C β U C α = U α,βψ with ψ = µ + k − X j =0 ν j ( z k ) j ! z j nd by Proposition 2.5, U = C β U α,βψ C α = U α,βϕ with ϕ = z k − αψW ∗ k β = W ∗ k β z k − αµ + z k − α k − X j =0 ν j ( z k ) j ! z j ! = β ( z k ) z k − ααzχ + k − X j =0 z k − αβ ( z k ) z k ψ j ( z k ) j ! z j ! = β ( z k ) z k χ + k − X j =0 αψ j ( z k ) j ! z j +1 . (cid:3) Using Theorem 3.5 and Corollary 3.8 one can prove the following.
Corollary 3.10.
Let U be a bounded linear operator from K α into K β . Then U ∈S k ( α, β ) , k ∈ N , if and only if there exist functions χ ∈ K α and ψ , . . . , ψ k − ∈ K β (possibly different for different conditions) such that one (and all) of the following condi-tions holds: (a) S ∗ β U − U ( S ∗ α ) k = e k β ⊗ χ + k − X j =0 ψ j ⊗ k α ,j , (b) S β U − U S kα = k β ⊗ χ + k − X j =0 ψ j ⊗ e k α ,j .Proof. Use reasoning analogous to the one presented in the proof of [12, Cor. 2.4]. (cid:3)
Note that if dim K β = 1, then by Theorem 3.5 all bounded linear operators from K α into K β belong to S k ( α, β ) for each k ∈ N . Corollary 3.11.
Let α and β be two inner functions and assume that dim K α = m < + ∞ .If k ≥ m , then every bounded linear operator from K α into K β belongs to S k ( α, β ) .Proof. Let U be a bounded linear operator from K α into K β . If dim K α = m < + ∞ , then U − S β U ( S ∗ α ) k has rank at most m , which by assumption is less or equal to k . Hence,there exist functions g , . . . , g k − ∈ K α and f , . . . , f k − ∈ K β (some of these functionspossibly equal to zero), such that U − S β U ( S ∗ α ) k = k − X j =0 f j ⊗ g j . Since here the kernels k α , , k α , , . . . , k α ,m − are linearly independent and span K α , each g j can be written as a linear combination of these kernels and so U satisfies (3.2). (cid:3) xample 3.12. As in Example 2.4 consider α ( z ) = z and β ( z ) = z . Then, for k = 5and ϕ = P ∞ n = −∞ a n z n ∈ L , the operator U α,βϕ is represented by the matrix a a − a − a − a a a a a a a a . It follows easily that every bounded linear operator from K α into K β belongs to S ( α, β ).Note that here U α,βz = 0 , but z / ∈ αH + z k − ( W ∗ k β ) H = z H + z H . Finally, we describe some rank–one operators from S k ( α, β ). Proposition 3.13.
Let α , β be two inner functions and let k ∈ N . Then for each l ∈ { , , . . . , k − } the rank–one operators e k β ⊗ k α ,l and k β ⊗ e k α ,l belong to S k ( α, β ) .Proof. Let l ∈ { , , . . . , k − } . Since S β e k β = − β (0) k β , we have e k β ⊗ k α ,l − S β ( e k β ⊗ k α ,l )( S ∗ α ) k = e k β ⊗ k α ,l − ( S β e k β ) ⊗ ( S kα k α ,l )= e k β ⊗ k α ,l + ( β (0) k β ) ⊗ ( S kα k α ,l )= k β ⊗ ( β (0) S kα k α ,l ) + e k β ⊗ k α ,l and by Theorem 3.5, e k β ⊗ k α ,l ∈ S k ( α, β ) since it satisfies (3.2) with χ = β (0)( S kα k α ,l ), ψ l = e k β and ψ j = 0 for j = l . It follows that, by Proposition 2.5, k β ⊗ e k α ,l = C β ( e k β ⊗ k α ,l ) C α ∈ S k ( α, β ) . (cid:3) Corollary 3.14.
For each l ∈ { , , . . . , k − } , (a) e k β ⊗ k α ,l = U α,βϕ with ϕ = W ∗ k β · l ! z l + k , (b) k β ⊗ e k α ,l = U α,βψ with ψ = α · l ! z l +1 .Proof. By Corollary 3.6 and the proof of Proposition 3.13, e k β ⊗ k α ,l = U α,βϕ with ϕ = β (0) (cid:0) S kα k α ,l (cid:1) + e k β ( z k ) l ! z l . Since for each g ∈ K α , h S kα k α ,l , g i = h P α ( l ! z l ) , ( S ∗ α ) k g i = h l ! z l , T z k g i = h l ! z l + k , g i , we have S kα k α ,l = P α ( l ! z l + k ) and ϕ = β (0) P α ( l ! z l + k ) + ( β ( z k ) − β (0)) l ! z l + k = W ∗ k β · l ! z l + k − β (0)( l ! z l + k − P α ( l ! z l + k )) . Since ϕ − ϕ ∈ αH we get e k β ⊗ k α ,l = U α,βϕ = U α,βϕ , that is, (a) holds.Now (b) follows from Proposition 2.5. (cid:3) eferences [1] S. C. Arora, R. Batra, On generalized slant Toeplitz operators , Indian J. Math. (2003), no. 2,121–134.[2] S. C. Arora, R. Batra, Generalized slant Toeplitz operators on H , Math. Nachr. (2005), no. 4,347–355.[3] C. Cˆamara, K. Kli´s-Garlicka, J. Jurasik, M. Ptak, Characterizations of asymmetric truncated Toeplitzoperators , Banach J. Math. Anal. (2017), no. 4, 899–922.[4] M. C. Cˆamara, J. R. Partington, Asymmetric truncated Toeplitz operators and Toeplitz operatorswith matrix symbol , J. Operator Theory (2017), no. 2, 455–479 .[5] M. C. Cˆamara, J. R. Partington, Spectral properties of truncated Toeplitz operators by equivalenceafter extension , J. Math. Anal. Appl. (2016), no. 2, 762–784.[6] P. L. Duren,
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E-mail address : [email protected]@poczta.umcs.lublin.pl