SURJECTIVE TOEPLITZ OPERATORS 1. Introduction We ... - NTNU

4 ANDREAS HARTMANN, DONALD SARASON, AND KRISTIAN SEIPThe full theorem of Hayashi provides a recipe for constructing the most general nontrivialkernel of a Toeplitz operator. One starts with functions a, b 0 , I such that (i) a and b 0 arein H ∞ , a is outer, a(0) > 0, and |a| 2 + |b 0 | 2 = 1 on T; (ii) g 2 0 is rigid, where g 0 = a/(1 − b 0 );(iii) the measure having Poisson integral (1 − |b 0 | 2 )/|1 − b 0 | 2 is absolutely continuous; (iv)I is inner, with I(0) = 0. The function g = a/(1 − b), where b = Ib 0 , is then the extremalfunction for the nearly invariant subspace g(H 2 ⊖ IH 2 ), and that subspace is the kernel ofT ϕ , where ϕ = Ig/g. Hayashi’s recipe underlies the examples we shall give at the ends ofSections 3 and 4.Retaining the preceding notations, we can now state our main result.Theorem 1. The operator T ϕ is surjective if and only if |g 0 | 2 is a Helson–Szegő weight.As we shall show below, the surjectivity of T ϕ does not imply that |g| 2 itself is a Helson–Szegő weight. If T ϕ is surjective and |g| 2 is not a Helson–Szegő weight, the inner functionI associated with ker T ϕ cannot serve as one of the inner functions J appearing in thesurjectivity criterion discussed earlier coming from Nakazi’s observation.A proof of Theorem 1 is given in Section 3; it involves de Branges–Rovnyak spaces.Those spaces were the basis of the analysis in [Sa88], and also in [Sa94b], which containsan alternative approach to Hayashi’s characterization of kernels of Toeplitz operators. Theneeded properties of de Branges–Rovnyak spaces are reviewed in Section 2. In Section 4we make some explicit computations of inverse images under T ϕ of reproducing kernelfunctions. This will enable us, in case T ϕ is surjective, to express in terms of Toeplitzoperators the right inverse of T ϕ whose range is orthogonal to ker T ϕ . It will also lead toan alternative proof of Theorem 1.To close this introduction we mention that the projection operator P + can be appliedto L 1 -functions as well as to L 2 -functions. For f in L 1 we interpret P + f as the Cauchyintegral of f, the function in D given by∫(P + f)(z) =Tf(ζ)1 − ζz dm(ζ).Of course, P + is unbounded as an operator on L 1 , but it is continuous relative to thenorm topology in L 1 and the topology of locally uniform convergence in H(D), the spaceof holomorphic functions in D.A Toeplitz operator with an unbounded symbol in L 2 can be regarded, according toconvenience, either as an unbounded operator on H 2 (with domain containing H ∞ ) or,via the preceding interpretation of P + , as a continuous operator from H 2 to H(D). Belowwe shall have occasion to form products of such operators. This will present no technicalproblems worth mentioning, and basic properties of bounded Toeplitz operators, such asthe product lemma, are easily seen to carry over. It is not uncommon for the productof two unbounded Toeplitz operators to be bounded. For example, if our operator T ϕ isinvertible, and if G is as in part (iv) of the Devinatz–Widom theorem, then Tϕ −1 = T G T 1/G .(Note that 1/G is in L 2 because |G| 2 is in this case a Helson–Szegő weight.)

SURJECTIVE TOEPLITZ OPERATORS 52. de Branges–Rovnyak spacesWe introduce just those properties of de Branges–Rovnyak spaces relevant for presentpurposes. The spaces were first studied in [dBrRo66]; a more recent account is [Sa94a].We assume, as in the second half of Section 1, that ϕ is a unimodular function such thatT ϕ has a nontrivial kernel. We retain the notations g, I, a, b, b 0 , g 0 introduced earlier. Thespace H(b) is by definition the range of the operator (1−T b T b) 1/2 with the range norm, i.e.,the norm that makes (1 − T b T b) 1/2 a coisometry of H 2 onto H(b). The norm on H(b) willbe denoted by ‖ · ‖ b . The spaces H(b 0 ) and H(I) are defined analogously. In particular,H(I) is just the range of the projection 1 − T I T I , in other words, it is the S ∗ -invariantsubspace H 2 ⊖ IH 2 . Thus ker T ϕ = T g H(I).The space M(a) is by definition the range of the operator T a , with the range norm. Wenote the inclusion M(a) ⊂ H(b) (see [Sa94a, p. 24]). The norm in M(a) will not play arole here.Proposition A. (See [Sa94a, p. 10].) The space H(b) is the orthogonal direct sum of H(I)and T I H(b 0 ). The operator T I acts as an isometry of H(b 0 ) into H(b).Proposition B. (See [Sa94a, p. 30].) The operator T 1−b T g is an isometry of H 2 ontoH(b), and the operator T 1−b0 T g0 is an isometry of H 2 onto H(b 0 ).Proposition C. (See [Sa94b, p. 161].) The operator T 1−b T g maps ker T ϕ onto H(I). Itacts on ker T ϕ as division by g.By a corona pair is meant a pair of functions in H ∞ that satisfies the hypothesis of thecorona theorem, namely, the maximum of the moduli of the functions is bounded awayfrom 0 in D.Proposition D. (See [Sa94a, p. 66].) The following conditions are equivalent.(i) H(b) = M(a)(ii) T g/g is invertible(iii) T a/a is invertible, and the functions a and b form a corona pair.3. Proof of Theorem 1As before, we assume ϕ is a unimodular function such that T ϕ has a nontrivial kernel.In addition to the extremal function g and the associated functions I, a, b, b 0 , g 0 introducedearlier, we let ϕ 0 = g 0 /g 0 . Theorem 1 follows quickly once the following lemma has beenestablished.Lemma 1. T ϕ0 T ϕ0 = T ϕ T ϕ .Proof. It will suffice to show that ‖T ϕ0 f‖ = ‖T ϕ f‖ for all f in H 2 . By Proposition B,Applying the product lemma, we get‖T ϕ f‖ = ‖T 1−b T g T ϕ f‖ b .T 1−b T g T ϕ = T 1−b T g T Ig/g = T 1−b T Ig = T Ia .

6 ANDREAS HARTMANN, DONALD SARASON, AND KRISTIAN SEIPThus ‖T ϕ f‖ = ‖Iaf‖ b , which by Proposition A equals ‖af‖ b0 . Exactly the same reasoning,with g 0 and b 0 in place of g and b, gives ‖T ϕ0 f‖ = ‖af‖ b0 .Proof of Theorem 1. After the elaborate preparations for Theorem 1, its proof seems almostan afterthought. By Lemma 1, the operators T ϕ and T ϕ0 are surjective or not surjectivetogether. But, by Hayashi’s theorem, the function g0 2 is rigid, so that T g0 /g 0= T ϕ0 hasa trivial kernel. Hence T ϕ0 is surjective if and only if it is invertible, in other words, if andonly if |g 0 | 2 is a Helson–Szegő weight.If |g| 2 is a Helson–Szegő weight, then T g/g is invertible, and it follows that T ϕ = T I T g/gis surjective. The next result shows that the reverse implication fails.Theorem 2. Let T ϕ be surjective. Then |g| 2 is a Helson–Szegő weight if and only if thefunctions a and I form a corona pair.Proof. By Theorem 1 and Proposition D, the functions a and b 0 form a corona pair, whilea and b do if and only if |g| 2 is a Helson–Szegő weight. But a and b form a corona pair ifand only if a and I do.To obtain an example in which T ϕ is surjective but |g| 2 is not a Helson–Szegő weight,we start with a normalized outer function a such that ‖a‖ ∞ < 1, |a| 2 is a Helson–Szegőweight, but a is not invertible in H ∞ . Let b 0 be the normalized outer function such that|b 0 | 2 = 1 − |a| 2 on T, and let g 0 = a/(1 − b 0 ). The function b 0 is invertible in H ∞ , so aand b 0 form a corona pair in a trivial way. Since |a|/2 ≤ |g 0 | ≤ 2/|a|, the functions g 0and 1/g 0 are both in H 2 , hence g 2 0 is rigid. Moreover, (1 + b 0 )/(1 − b 0 ) = (1 + b 0 )g 0 /a isin H 1 , guaranteeing that its real part, (1 − |b 0 | 2 )/|1 − b 0 | 2 , is the Poisson integral of itsboundary function, i.e., of an absolutely continuous measure. Because a is not invertiblein H ∞ , it is easy to find an inner function I with I(0) = 0 such that a and I do not forma corona pair. We have all the ingredients needed in Hayashi’s recipe. Letting b = Ib 0 ,g = a/(1 − b), and ϕ = Ig/g, we obtain by Hayashi’s theorem that g is the extremalfunction for ker T ϕ = T g H(I). Because of the properties we arranged for a and b 0 topossess, it follows from Proposition D, the Devinatz–Widom theorem, and Theorem 1 thatT ϕ is surjective. Finally, we conclude by Theorem 2 that |g| 2 is not a Helson–Szegő weight.4. Inverse imagesWe continue to let ϕ be a unimodular function such that T ϕ has a nontrivial kernel, andwe retain the notations g, I, a, b, b 0 , g 0 , ϕ 0 introduced earlier. Also, for λ in D, we let k λdenote the kernel function in H 2 for the functional of evaluation at λ: k λ (z) = (1 − λz) −1 .Our first objective in this section is to determine the inverse image of k λ under T ϕ thatis orthogonal to ker T ϕ . This will enable us, in case T ϕ is surjective, to determine theright inverse of T ϕ whose range is orthogonal to ker T ϕ . It will also form the basis for analternative proof of Theorem 1.It is easy to find at least one function in Tϕ−1 (k λ ). Formally, disregarding concerns aboutboundedness, we can factor T ϕ as T ϕ = T Ig T 1/g . Hence, formally, the operator T Ig T 1/g

SURJECTIVE TOEPLITZ OPERATORS 7is a right inverse of T ϕ , suggesting that the function T Ig T 1/g k λ = Igk λ /g(λ) should be in(k λ ). That this is indeed the case can be verified directly:T −1ϕT ϕ (Igk λ ) = P + (gk λ ) = g(λ)k λ .To find the function in Tϕ−1 (k λ ) orthogonal to ker T ϕ , we need to project the functionIgk λ /g(λ) onto (ker T ϕ ) ⊥ . The first part of the following lemma will help.Lemma 2. (i) P + (|g| 2 Ik λ ) = Ik λ1 − b + b 0(λ)k λ1 − b(λ) .(ii) P + (|g| 2 k λ ) =k λ1 − b + b(λ)k λ1 − b(λ) .Proof. We shall prove (i); the proof of (ii) is entirely analogous. On T we haveSince b = Ib 0 , we obtain, again on T,|g| 2 = 1 − |b|2|1 − b| 2 = 11 − b + b1 − b .|g| 2 Ik λ = Ik λ1 − b + b 0k λ1 − b .Now suppose ‖b‖ ∞ < 1. Then both summands on the right side of the preceding equalityare in L 2 , and application of P + givesP + (|g| 2 Ik λ ) = Ik λ1 − b + b 0(λ)k λ1 − b(λ) ,the desired equality. To pass to the case ‖b‖ ∞ = 1, it suffices to apply what was justdone with rb in place of b (0 < r < 1) and let r → 1. Indeed, since the L 1 norm of(1 − |rb| 2 )/|1 − rb| 2 coincides with that of |g| 2 and (1 − |rb| 2 )/|1 − rb| 2 tends to |g| 2 almosteverywhere, it follows that (1 − |rb| 2 )/|1 − rb| 2 tends to |g| 2 in L 1 . Thus we obtain thedesired conclusion by using the continuity of P + as a map from L 1 to H(D). We can now identify the function in T −1ϕ (k λ ) belonging to (ker T ϕ ) ⊥ .Lemma 3. The function d λ = g(I − b 0 (λ))k λ /a(λ) lies in (ker T ϕ ) ⊥ and is mapped by T ϕto k λ .Proof. We first obtain an expression for the orthogonal projection in H 2 with range ker T ϕ .We let P I = 1−T I T I , the orthogonal projection in H 2 with range H(I). By Propositions Band C, the operator T 1−b T g is an isometry of H 2 onto H(b), and it sends ker T ϕ onto H(I),acting on ker T ϕ as division by g. By Proposition A, the orthogonal projection in H(b)with range H(I) coincides with the restriction of P I to H(b). The orthogonal projectionin H 2 with range ker T ϕ therefore equals T g P I T 1−b T g .

8 ANDREAS HARTMANN, DONALD SARASON, AND KRISTIAN SEIPWe apply the preceding operator to the function Igk λ and use Lemma 2 (i):[]T g P I T 1−b T g (Igk λ ) = T g P I T 1−b P + (I|g| 2 Ik λk λ ) = T g P I T 1−b1 − b + b 0(λ)k λ1 − b(λ)= T g P I (Ik λ ) + b 0(λ)1 − b(λ) T gP I T 1−b k λ .The first summand on the right side is 0. Moreover, because I divides b, we have P I T b = 0,so P I T 1−b = P I . Since P I k λ = (1 − I(λ)I)k λ , we obtainT g P I T 1−b T g (Igk λ ) =b 0(λ)1 − b(λ) (1 − I(λ)I)gk λ.Finally, we divide the last function by g(λ) and subtract the result from Igk λ /g(λ) to getthe function in (ker T ϕ ) ⊥ sent by T ϕ to k λ . After a little algebra one finds that the desiredfunction is d λ .We are now able to identify the canonical right inverse of T ϕ , assuming T ϕ is surjective.Theorem 3. Let T ϕ be surjective. Then the right inverse of T ϕ whose range is orthogonalto ker T ϕ equals T g T I−b0T 1/a .Proof. Let A be the right inverse in question, and let B = T g T I−b0T 1/a . We first note that,by Theorem 1, the operator T ϕ0 is invertible. An application of Proposition D, with g 0replacing g, then shows that the operator T a/a is invertible, implying that |a| 2 is a Helson–Szegő weight. In particular, 1/a is in L 2 , so, as noted at the end of Section 1, the operatorT 1/a is continuous as an operator from H 2 to H(D). The same is true of T b0 /a . Combiningthese observations, one sees that B is continuous as an operator from H 2 to H(D).By Lemma 3, we must have Ak λ = d λ . A direct computation shows that Bk λ = d λ (infact, B was selected with this in mind). The operators A and B thus agree on a densesubset of H 2 , namely, the linear span of the kernel functions. Continuity now guaranteesthat A = B.Corollary. If T ϕ is surjective, then the left inverse of T ϕ with the same kernel as T ϕ equalsT 1/a T I−b0 T g .The next lemma will be used in the alternative proof of Theorem 1.Lemma 4. If f is in H 2 and T ϕ f = d λ , then f is the function e λ = (1−b 0 (λ)b 0 )k λ /(a(λ)a).Conversely, if e λ is in H 2 , then T ϕ e λ = d λ .Proof. Assuming T ϕ f = d λ , we apply T g to the equality. Since T g T ϕ = T Ig , we get()g(I − b 0 (λ))k λIgf = T g = 1a(λ) a(λ) P ( )+ |g| 2 (I − b 0 (λ))k λ .

SURJECTIVE TOEPLITZ OPERATORS 9Applying Lemma 2 to the right side, one finds after some algebra that the equality can berewritten as( )Igf =k λ I − b 0 (λ)b.a(λ) 1 − bDividing through by Ig, one obtains the desired equality. A reversal of the reasoning givesthe converse.Lemmas 3 and 4 have versions with ϕ 0 in place of ϕ.Lemma 5. (i) The function d 0 λ = g 0k λ /g 0 (λ) is mapped by T ϕ0 to k λ .(ii) If f is in H 2 and T ϕ0 f = d 0 λ , then f = e λ. Conversely, if e λ is in H 2 , thenT ϕ0 e λ = d 0 λ .We omit the details of the proof. Part (i) can be verified directly. The proof of (ii) usesthe same idea as the proof of Lemma 4, but is less complicated. We note that the functionse λ lie in H 2 if and only if the function 1/a does.Alternative proof of Theorem 1. The proof will use Lemmas 3–5 and Hitt’s theorem. Therigidity of g0, 2 which is given by Hayashi’s theorem, will emerge from other considerationsin the argument to follow.Assume T ϕ is surjective. We prove that |g 0 | 2 is then a Helson–Szegő weight. Since T ϕ T ϕis invertible, there is a function f in H 2 such that T ϕ T ϕ f = k λ . By Lemmas 3 and 4, wehave f = e λ and so e λ is in H 2 . In fact, again by the invertibility of T ϕ T ϕ , the functionse λ (λ ∈ D) span H 2 because the kernel functions k λ (λ ∈ D) do. Since the functions e λare in H 2 , so is 1/a, and hence so is 1/g 0 = (1 − b 0 )/a. Now it is a simple matter to showthat if f is an outer function in H 2 , then the operator T f/fhas a trivial kernel. Applyingthis with f = 1/g 0 , we conclude that T ϕ0 has a trivial kernel. Also, we can then infer byLemma 5 that T ϕ0 T ϕ0 e λ = k λ . Since T ϕ T ϕ and T ϕ0 T ϕ0 coincide on a dense subset of H 2 ,we conclude that T ϕ0 T ϕ0 = T ϕ T ϕ , and therefore T ϕ0 is surjective, hence invertible. TheDevinatz–Widom theorem now tells us that |g 0 | 2 is a Helson–Szegő weight, as desired.Assume, conversely, that |g 0 | 2 is a Helson–Szegő weight. We prove that T ϕ is surjective.By the Devinatz–Widom theorem, T ϕ0 is invertible. Using Lemma 5, we infer that thefunctions e λ are in H 2 and that T ϕ0 T ϕ0 e λ = k λ . From Lemmas 3 and 4 we get T ϕ T ϕ e λ = k λ ,so arguing as above T ϕ T ϕ = T ϕ0 T ϕ0 , giving the surjectivity of T ϕ .We close with a few comments. First, the inclusion of 1/a in H 2 is not automatic. Infact, as shown in [Sa89, pp. 491–492], if a(z) = (1 + z)/2 and b 0 (z) = z(1 − z)J(z)/2,where J is a Blaschke product with zeros on (−1, 0) and tending to −1, then the measurehaving Poisson integral (1 − |b 0 | 2 )/|1 − b 0 | 2 is absolutely continuous and g 2 0 = (a/(1 − b 0 )) 2is rigid. Let I be any inner function vanishing at 0, b = b 0 I, g = a/(1 − b), and ϕ = Ig/g.By Hayashi’s theorem, g is the extremal function for ker T ϕ = T g H(I). For this particularϕ, then, the corresponding function 1/a fails to be in H 2 .Second, even if 1/a is in H 2 , the operator T ϕ need not be surjective. In fact, take anormalized outer function a such that ‖a‖ ∞ < 1 and 1/a is in H 2 but not in H 2+ε forany ε > 0. Let b 0 be the outer function whose modulus on T equals (1 − |a| 2 ) 1/2 , and let

10 ANDREAS HARTMANN, DONALD SARASON, AND KRISTIAN SEIPg 0 = a/(1 − b 0 ). As in the example at the end of Section 3, the measure having Poissonintegral (1 − |b 0 | 2 )/|1 − b 0 | 2 is then absolutely continuous because (1 + b 0 )/(1 − b 0 ) is inH 1 . Since 1/g 0 is in H 2 , the function g 2 0 is rigid. As above, take any inner function Ivanishing at 0 and let b = b 0 I, g = a/(1 − b), ϕ = Ig/g. By Hayashi’s theorem, g is theextremal function for ker T ϕ = T g H(I). However, the function |a| 2 is not a Helson–Szegőweight, because a Helson–Szegő weight has the property that its reciprocal is in L 1+ε forsufficiently small ε. By Theorem 1 and Proposition D, the operator T ϕ is not surjective.Finally, we wish to remark that, while we believe our characterization of surjectivity isinteresting from a theoretical perspective, it is unclear how useful it might be in analyzingspecific Toeplitz operators or classes of Toeplitz operators. One obstacle to its use, possibly,is that it seems extremely rare to be able to describe concretely, in terms of the symbol ϕ,the extremal function g of ker T ϕ and its associated inner function I. This matter awaitsfurther investigation.References[BöSi90] A. Böttcher & B. Silbermann, Analysis of Toeplitz Operators. Springer-Verlag, Berlin, 1990.[dBrRo66] L. de Branges & J. Rovnyak, Square Summable Power Series, Holt, Rinehart and Winston,New York, 1966.[Ha85] E. Hayashi, The solution sets of extremal problems in H 1 , Proc. Amer. Math. Soc. 93 (1985),690–696.[Ha86] E. Hayashi, The kernel of a Toeplitz operator, Integral Equations Operator Theory 9 (1986),588–591.[Ha90] E. Hayashi, Classification of nearly invariant subspaces of the backward shift, Proc. Amer.Math. Soc. 110 (1990), 441–448.[HeSz60] H. Helson & G. Szegő, A problem in prediction theory, Ann. Mat. Pura Appl. 51 (1960),107–138.[Hi88] D. Hitt, Invariant subspaces of H 2 of an annulus, Pacific J. Math. 134 (1988), 101–120.[Na93] T. Nakazi, Toeplitz operators and weighted norm inequalities, Acta Sci. Math. (Szeged) 58(1993), 443–452.[Ni02] N. Nikolski, Operators, Functions, and Systems: An Easy Reading. Vol. 1, Hardy, Hankel, andToeplitz; Vol.2, Model Operators and Systems, Math. Surveys Monographs 92 and 93, Amer.Math. Soc., Providence, RI, 2002.[Sa88] D. Sarason, Nearly invariant subspaces of the backward shift, Contributions to Operator Theoryand Its Applications (Mesa, AZ, 1987), 481–493, Oper. Theory Adv. Appl., 35, Birkhäuser,Basel, 1988.[Sa89] D. Sarason, Exposed points in H 1 , I, The Gohberg Anniversary Collection, Vol. II (Calgary,AB, 1988) 485–496, Oper. Theory Adv. Appl., 41, Birkhäuser, Basel, 1989.[Sa94a] D. Sarason, Sub-Hardy Hilbert Spaces in the Unit Disk, University of Arkansas Lecture Notesin the Mathematical Sciences 10, John Wiley & Sons Inc., New York, 1994.[Sa94b] D. Sarason, Kernels of Toeplitz operators, Toeplitz Operators and Related Topics (Santa Cruz,CA, 1992), 153–164, Oper. Theory Adv. Appl., 71, Birkhäuser, Basel, 1994.

SURJECTIVE TOEPLITZ OPERATORS 11Laboratoire de Mathématiques Pures de Bordeaux, Université Bordeaux I, 351 coursde la Libération, 33405 Talence, FranceE-mail address: hartmann@math.u-bordeaux.frDepartment of Mathematics, University of California, Berkeley, CA 94720, USAE-mail address: sarason@math.berkeley.eduDepartment of Mathematical Sciences, Norwegian University of Science and Technology(NTNU), NO-7491 Trondheim, NorwayE-mail address: seip@math.ntnu.no