Doubly-invariant subspaces for the shift on the vector

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and Operator Theory. Doubly-invariant subspaces for the shift on the vector-valued Sobolev spaces of the disc and an- nulus. I. Chalendar and J. R. Partington.
Integr. equ. oper. theory 99 (9999), 1–10 0378-620X/99000-0, DOI 10.1007/s00020-003-0000 c 2008 Birkh¨

auser Verlag Basel/Switzerland

Integral Equations and Operator Theory

Doubly-invariant subspaces for the shift on the vector-valued Sobolev spaces of the disc and annulus I. Chalendar and J. R. Partington Abstract. This paper characterizes the closed doubly shift-invariant subspaces of the Sobolev space of vector-valued functions defined on the unit circle. Partial results are obtained on the analogous problem for the Sobolev space on the boundary of the annulus.

Mathematics Subject Classification (2000): Primary: 46E20, 47B38. Secondary: 30D55, 47A15.

1. Introduction The purpose of this paper is to study the shift operator S (multiplication by the independent variable) on certain Sobolev spaces, of which formal definitions will be given later in the paper. These spaces consist of vector-valued functions on the boundary of the annulus A = {z ∈ C : r0 < |z| < 1}, where r0 is a positive real number less than unity. In particular, we are interested in the invariant subspaces of S. We consider also the Hardy–Sobolev spaces of vector-valued functions on the annulus. In the case of scalar-valued Hardy spaces on the annulus there has been much significant work [14, 6, 12, 15, 1], and some extensions to the vector-valued case were recently obtained in [2]. However, as a preliminary step, we need also to study certain spaces of functions defined on the unit disc in C. The function space W 1,2 (T) denotes the set of functions f defined on T = {z ∈ C P : |z| = 1} for which there exists a sequence (an )n∈Z in C satisfying P f (eit ) = n∈Z an eint and kf k2W 1,2 (T) := |a0 |2 + n∈Z n2 |an |2 < ∞. The functions f in W 1,2 (T) have their derivatives with respect to t, denoted by ft , in L2 (T) and

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W 1,2 (T) is a Hilbert space endowed with the scalar product defined by Z 2π 1 hf, giW 1,2 (T) = g (0). ft (eit )gt (eit ) dt + fb(0)b 2π 0

The Hardy–Sobolev space H 1,2 (D) is the subspace of W 1,2 (T) consisting of functions whose Poisson extension to the disc D = {z ∈ C : |z| < 1} P is analytic; ∞ n equivalently, it may be regarded as the space of Taylor series f : z → 7 n=0 an z P∞ 2 2 2 2 for which the norm given by kf k = |a0 | + n=1 n |an | is finite. For 0 < r0 < 1 we b \ r0 D), consisting of functions with an shall also be interested in the space H 1,2 (C P b analytic extension to C \r0 D, represented by a Laurent series f : z 7→ 0n=−∞ an z n with finite norm given by kf k2 = |a0 |2 +

−1 X

n2 r02n |an |2 .

n=−∞

Denote by S the operator of multiplication by z on W 1,2 (T). Clearly S is invertible and S −1 f (eit ) = e−it f (eit ) for f ∈ W 1,2 (T). A closed subspace M in W 1,2 (T) is said to be invariant for S if SM ⊂ M, doubly invariant for S if M is both invariant for S and S −1 and reducing for S if M is both invariant for S and S ∗ . For f ∈ W 1,2 (T), 1. IS [f ] will denote the smallest closed subspace M in W 1,2 (T) containing f and invariant for S. 2. DS [f ] will denote the smallest closed subspace M in W 1,2 (T) containing f and doubly invariant for S. 3. RS [f ] will denote the smallest closed subspace M in W 1,2 (T) containing f and reducing for S. Let E be a closed subset of T. Denote by MW 1,2 (T) (E) the closed doubly-invariant subspace of W 1,2 (T) defined by MW 1,2 (T) (E) = {f ∈ W 1,2 (T) : f|E = 0}. Such subspaces are called spectral. It can happen that spectral subspaces completely characterize the doubly-invariant subspaces, as in the Wiener theorem asserting that a every doubly-invariant subspace of L2 (T) has the form χE L2 (T) for some measurable set E ⊂ T (see [5, 10, 11]). We employ an analogous notation for subspaces of W 1,2 (T, Cm ); in general we use lower case letters for scalar functions and capital letters for vector-valued functions. In particular,     (f1 )t f1     for F =  ...  in W 1,2 (∂A, Cm ), (F )t denotes  ... . fm

(fm )t

The structure of this paper is as follows. In Section 2 we characterize the doubly-invariant subspaces of W 1,2 (T, Cm ), first in the known case m = 1, and

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then for general m. The case when m = 2 and the subspace is the graph of an operator is of particular importance, as this enables a characterization of unbounded shift-invariant operators. The Sobolev space (and Hardy–Sobolev space) of the annulus are much harder to analyse: in Section 3 we show that under some circumstances the doubly-invariant subspaces are spectral as in the case of the disc (i.e., determined by the set on which the functions in the subspace vanish), although in general this is not the case.

2. Doubly-invariant subspaces in W 1,2 (T, Cm ) 2.1. Scalar case The next result was announced in [8], having been obtained by a different method based on the approximation of the functions in the Sobolev space defined on the real axis. We give a direct proof, which we later apply to the vectorial case. Some related results, in the context of spectral synthesis, have been given by Netrusov (see [9, 4]). Proposition 2.1. Let K be a closed doubly-invariant subspace in W 1,2 (T). Then there exists a function φ ∈ W 1,2 (T) such that K = DS (φ) and moreover K = {f ∈ W 1,2 (T) : f|Z(φ) = 0}, where Z(φ) = {eit ∈ T : φ(eit ) = 0}. In other words the closed doubly-invariant subspace in W 1,2 (T) are spectral. Proof. Let PK denote the orthogonal projection from W 1,2 (T) onto K, and take φ = PK 1, where 1 is the function identically equal to 1 on T. Obviously, if φ = 1, then K = W 1,2 (T). Suppose now that φ 6= 1. Then it follows that kφkW 1,2 (T) 6= 1 and thus b 6= 0. 1 − φ(0) (2.1)

Since φ = PK 1, φ ∈ K and 1 − φ is orthogonal to K. Since K is doubly-invariant it follows that (2.2) hS n φ, 1 − φiW 1,2 (T) = 0, n ∈ Z. Now, (2.2) means that Z 2π Z int it int it it (ine φ(e ) + e φt (e ))(−φt (e )) dt + 0

0

for all n ∈ Z, and then we get: Z

0



−ine

int

it

φ(e )φt

(eit ) dt

=

Z



b dt = 0, φ(eit )eint (1 − φ(0))



e

int

it

φt (e )φt

(eit ) dt−

0

Z

0



b dt φ(eit )eint (1− φ(0))

for all n ∈ Z. It follows that φφt is absolutely continuous and b (φφt )t = φt φt − φ(1 − φ(0).

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Since (φt )t = −φtt , we then obtain, b −φφtt = −φ(1 − φ(0)).

(2.3)

b 6= 0 on T \ Z(φ). Combining (2.3) and (2.1) we have φtt = 1 − φ(0)

Thus the circle can be written as the union of a closed set on which φ = 0 and a countable union of intervals, on each of which φ has constant second derivative. In order to check that K = DS (φ), take f ∈ K orthogonal to DS (φ). Then hS n φ, f iW 1,2 (T) = 0, n ∈ Z.

(2.4)

Moreover, since f ∈ K, it follows that hS n f, 1 − φiW 1,2 (T) = 0, n ∈ Z.

(2.5)

In the sequel we will show that if f ∈ W 1,2 (T) satisfies (2.4) and (2.5), then f is identically equal to 0. Using (2.4) we get: Z 2π Z 2π int it int it it (ine φ(e ) + e φt (e ))ft (e ) dt + φ(eit )eint fb(0) dt = 0, n ∈ Z, 0

0

and then, for all n ∈ Z, we get: Z Z 2π −ineint φ(eit )ft (eit ) dt = 0

0



eint (φt (eit )ft (eit ) + φ(eit )fb(0)) dt.

It follows that φft is absolutely continuous and (φft )t = φt ft + φfb(0). We then obtain, φftt = φfb(0), which means that ftt = fb(0) on T \ Z(φ).

(2.6)

b f (φtt − (1 − φ(0))) = 0.

(2.7)

Using a similar calculation, (2.5) implies that f φt is absolutely continuous and

We claim now that f = 0 on Z(φ). Recall that on each “component interval”, b = (a, b), forming a component of the open set T \Z(φ), we may write φtt = 1− φ(0) c 6= 0, say (a.e.), and so, integrating and recalling that φ vanishes at the end points, we find that c for a ≤ t ≤ b. φ(eit ) = (t − a)(t − b) 2 Note that φt (a+) = −c(b − a)/2 and φt (b−) = c(b − a)/2. Take z ∈ Z(φ). If z ∈ int Zφ , then φtt = 0 on an interval including z and so f (z) = 0 by continuity and (2.7). Otherwise, suppose now that z ∈ ∂Z(φ) and f (z) 6= 0, so f (eit ) 6= 0 on some nontrivial interval (z − δ, z + δ). Now f φt is known to be absolutely continuous, and the condition on f implies that φt must be continuous on (z −δ, z +δ). But this is absurd, since there is either a left-hand endpoint a of a component interval (a, b) lying in [z, z + δ) or a right-hand endpoint b lying in

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(z − δ, z]. Suppose the first case (the second is similar): then φt (a+) = −c(b − a)/2; on the other hand, either there is an interval (a − η, a] ⊆ Z(φ) (so that φt (a−) = 0 and φt is discontinuous at a) or there is a sequence of component intervals (an , bn ) with bn → a− (possibly bn = a for each n). In this case φt (bn −) = c(bn − an )/2, which cannot converge to −c(b − a)/2. Moreover, by (2.6), ftt equals a unique constant on T \Z(φ), and since φtt has the same property, there exists λ ∈ C such that f = λφ. Finally hφ, λφiW 1,2 (T) = 0 so λ = 0. Therefore f is identically equal to 0 and thus K = DS (φ). It remains to check that K = M(Z(φ))(= {f ∈ W 1,2 (T) : f|Z(φ) = 0}). Obviously, K is a subset of M(Z(φ)). In order to prove the converse inclusion, take f ∈ M(Z(φ)) satisfying (2.4). Then f|Z(φ) = 0 and ftt = fb(0) on T \ Z(φ) (see (2.6). It follows that there exists λ ∈ C such that f = λφ and as previously hφ, λφiW 1,2 (T) = 0 implies that λ = 0. Therefore f is identically equal to 0 and thus K = M(DS (φ)).  2.2. Vectorial case Let m be a positive integer. Theorem 2.2. Let K be a closed doubly-invariant subspace in W 1,2 (T, Cm ). Then, for almost all eiw ∈ T, there exists a subspace I(eiw ) of Cm such that K = {F ∈ W 1,2 (T, Cm ) : F (eiw ) ∈ I(eiw )}. Proof. Let PK : W 1,2 (T, Cm ) → K denote the orthogonal projection onto K. Let {e1 , . . . , em } denote the canonical basis of Cm . For r ∈ Z and 1 ≤ k ≤ m, let Ψr,k denote the vector-valued function defined on T by z 7−→ PK z r ek . For each w ∈ T, let I(eiw ) denote the linear span of the set of vectors {Ψr,k (eiw ) : r ∈ Z, 1 ≤ k ≤ m}. Note that I(eiw ) is closed as a subspace of finite dimension. Define P (eiw ) to be the orthogonal projection from Cm onto I(eiw ). It is clear that K ⊂ {F ∈ W 1,2 (T, Cm ) : F (eiw ) ∈ I(eiw )} and that {F ∈ W 1,2 (T, Cm ) : F (eiw ) ∈ I(eiw )} is both closed and doublyinvariant. If the two spaces were unequal, then there would exist F ∈ W 1,2 (T, Cm ) such that F (eiw ) ∈ I(eiw ) and with F orthogonal to S n Ψr,k for all n, r ∈ Z and k ∈ {1, . . . , m}. Thus we get, for all n, r ∈ Z and k ∈ {1, . . . , m}: Z 2π Z 2π int it it m h(eint Ψr,k (eit )), Fb(0)iCm dt = 0. h(e Ψr,k (e ))t , Ft (e )iC dt + 0

0

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It follows that for all n, r ∈ Z and k ∈ {1, . . . , m}: Z 2π eint (hinΨr,k (eit ), Ft (eit )iCm + h(Ψr,k (eit ))t , Ft (eit )iCm 0

So hΨr,k , Ft i is absolutely continuous and

+ hΨr,k (eit ), Fb(0)iCm ) dt = 0.

(hΨr,k , Ft i)t = h(Ψr,k )t , Ft i + hΨr,k , Fb (0)i.

Or, writing G = Ft − tFb (0), we have

hΨr,k , Git = h(Ψr,k )t , Gi,

implying that the the directional derivative hΨr,k , Gt i exists and is zero almost everywhere; hence hF, Gt i also exists and is zero a.e. So Z 2π hF (eit ), Gt (eit )i dt = 0. 0

That is, Z

0



hF, Ftt − Fb (0)i dt

= =

[hF, Ft i]2π 0 −

Z

0





Z

2π 0

|Ft |2 dt − 2π|Fb (0)|2

|Ft |2 dt − 2π|Fb (0)|2 = 0.

Thus F is constant and Fb (0) = 0. Hence, F is identically equal to 0 and K = {F ∈ W 1,2 (T, Cm ) : F (eiw ) ∈ I(eiw )}.

 Remark 2.3. If we define P (eiw ) to be the orthogonal projection from Cm onto I(eiw ), P is a measurable function of eiw but it does not even need to be continuous. Indeed, let m ≥ 2 and consider I(eiw ) to be the linear span of the set of vectors e1 and e1 + (eiw − 1)e2 , where {e1 , . . . , em } denote the canonical basis of Cm . Then clearly P (eiw ) is the identity map whenever eiw 6= 1 and P (eiw ) is the projection onto e1 otherwise. Note that eiw 7−→ e1 and eiw 7−→ e1 + (eiw − 1)e2 belong to W 1,2 (T, Cm ). One application of these results is to the graphs of operators. We recall that, for Banach spaces X and Y the graph of a (possibly unbounded) operator T : D(T ) → Y , with domain D(T ) ⊂ X , is the subspace G(T ) = {(x, y) : x ∈ D(T ), y = T x} ⊂ X × Y. Theorem 2.2 can be used to obtain information on the structure of doublyinvariant graphs; however, this does not seem to give the full analogue of the known result for L2 (T, C2 ), which may be found in [7] and [11, Sec. 3.2].

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3. Doubly-invariant subspaces in the Sobolev space in the annulus Let r0 be a positive real number less than unity and let A be the annulus defined by A = {z ∈ C : r0 < |z| < 1}. The boundary ∂A of A consists of two circles T and r0 T. W 1,2 (rT) as the set of functions f defined on rT satisfying f (reit ) = P Define n int and n∈Z bn r e X n2 r2n |bn |2 < ∞. kf k2W 1,2 (rT) := |b0 |2 + n∈Z Define W 1,2 (∂A) as the set of functions f on ∂A of the form f = f1 ⊕ f0 with f1 ∈ W 1,2 (T) and f0 ∈ W 1,2 (r0 T). The subspace H 1,2 (∂A) set of functions f0 in W 1,2 (T) ⊕ P is theint Pf = f1 n⊕int 1,2 it it W (rT) where f1 (e ) = n∈Z an e , f0 (r0 e ) = n∈Z an r0 e and X n2 (1 + r02n )|an |2 < ∞. kf k2H 1,2 (∂A) := 2|a0 |2 + n∈Z Note that H 1,2 (∂A) can be isometrically with the set H 1,2 (A) of funcP identified n tions analytic in A of the form f (z) = n∈Z an z with X n2 (1 + r02n )|an |2 < ∞. 2|a0 |2 + n∈Z We shall start with a factorization theorem for functions in H 1,2 (A). Similar formulae have been found useful in H 2 (A) (see for example [6], [3], [1, Thm. 2.1]), and our method of proof is similar to that of [3]. Theorem 3.1. Let f ∈ H 1,2 (A). Then there exist m ∈ Z,f1 ∈ H 1,2 (D) and f0 ∈ b \ r0 D) such that f (z) = z m f0 (z)f1 (z). Moreover, the zeros of f0 and f1 H 1,2 (C can be taken to lie in A. b \ r0 D) be the Blaschke product associated Proof. Let r1 ∈ (r0 , 1) and B0 ∈ H ∞ (C with the zeros of f in {z ∈ C : r0 < |z| ≤ r1 } and B1 ∈ H ∞ (D) be the Blaschke product associated with the zeros of f in {z ∈ C : r1 < |z| < 1}. Then B1fB0 ∈ H 2 (A) and B1fB0 has no zero.   Now, there exists m ∈ Z such that h := log B1 Bf0 zm is single-valued. Note R h(z) P+∞ 1 n that h ∈ H 1 (A), and h(z) = n=−∞ an z , where an = 2iπ Γ z n+1 dz for any circle Γ lying in A. Moreover, there exists C > 0 such that |an | ≤ C forP n ≥ 0 and |an | ≤ rCn for n ≤ 0. It follows that h(z) = h1 (z)+h0 (z) where h1 (z) := n≥0 an z n 0 P b \ r0 D. an z n is holomorphic in C is holomorphic in D, and h0 (z) := n 0 such that kF (ξ)kCm ≥ δ almost everywhere on ∂A. Proposition 3.3. Suppose that F ∈ W 1,2 (∂A, Cm ) satisfies kFt k∞ < ∞ and F is essentially bounded below. Then DS (F ) = H 1,2 (∂A)F . Proof. We need to show that H 1,2 (∂A)F is closed, in which case it will be DS (F ), PN being limits of expressions of the form −N an z n F (z).

Suppose that {gn F } is a Cauchy sequence in W 1,2 (∂A, Cm ), where gn ∈ H (∂A). Now (gn F )t = (gn )t F + gn Ft , in the sense that gn F is absolutely continuous and is the integral of the L2 (∂A, Cm ) function (gn )t F + gn Ft . 1,2

Since {gn F } converges in L2 (∂A, Cm ) and F is bounded below, we get convergence of {gn } in L2 (∂A, Cm ). Hence we have L2 (∂A, Cm ) convergence of {gn Ft } as Ft is in L∞ (∂A, Cm ). By subtraction, we have L2 (∂A, Cm ) convergence of {(gn )t F }, and hence of {(gn )t }. Hence {gn } is Cauchy in H 1,2 (∂A), which is enough to show that H 1,2 (∂A)F is closed.  One circumstance under which a subspace DS (f ) will be spectral is given by the following result. Theorem 3.4. Take f = f1 ⊕f0 in W 1,2 (∂A), and suppose that f1 = 0 or f0 = 0 on an interval of positive measure. Then DS (f ) = DS (f1 ) ⊕ DS (f0 ) and thus DS (f ) is spectral. Proof. Clearly DS (f ) ⊂ DS (f1 ) ⊕ DS (f0 ), so we need to show that if g = g1 ⊕ g0 ∈ DS (f1 ) ⊕ DS (f0 ) then it is in DS (f ).

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Suppose to fix ideas that f = 0 on an arc Z = {r0 eit : π − α ≤ t ≤ π + α} with 0 < α < π. It will be clear how the proof can be adapted in general. We have a sequence of pairs of trigonometric polynomials {pn ⊕ qn } with kpn f1 − g1 kW 1,2 (T) → 0, and kqn f2 − g2 kW 1,2 (r0 T) → 0. If we can find a sequence of polynomials in z and 1/z, say {rn (z)} with krn − pn kW 1,∞ (T) → 0 and krn − qn kW 1,∞ (r0 T\Z) → 0, then we shall also have krn f − gkW 1,2 (∂A) → 0, and the result follows. We are grateful to the referee for the following conclusion to the proof, which is somewhat simpler than our original argument using Mergelyan’s theorem. We apply Runge’s theorem [13, Thm. 13.6] to a small neighbourhood U of ∂A \ Z with two components, not containing 0. This gives us a sequence {rn } such that rn − pn → 0 uniformly on an open subset containing T and rn − qn → 0 uniformly on an open subset containing r0 T \ Z. By an easy estimate involving Cauchy’s integral formula, this implies also that rn′ − p′n → 0 uniformly on T and rn′ − qn′ → 0 uniformly on r0 T \ Z.  Same arguments imply the following result. Corollary 3.5. Take F = F1 ⊕ F0 ∈ W 1,2 (∂A, Cm ) and suppose that F1 = 0 or F0 = 0 on an interval of positive measure. Then DS (F ) = DS (F1 ) ⊕ DS (F0 ) and thus DS (F ) is spectral. We conclude with the following question: in [14], it was shown that if f = f1 ⊕ f0 ∈ L2 (∂A) satisfies f1 is not log-integrable on T or f0 is not log-integrable on r0 T, then DS (f ) = DS (f1 ) ⊕ DS (f0 ). Is a similar property still true in analytic Sobolev spaces? Acknowledgements. The authors are grateful to the EPSRC for financial support, and they thank Janyne and Claude Charlier for providing such a pleasant environment in which to do research. They also wish to express their gratitude to the referee for reading the manuscript very carefully and helping them to remove a number of errors.

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