An inverse problem for a class of lacunary canonical systems with diagonal Hamiltonian
aa r X i v : . [ m a t h . F A ] J u l AN INVERSE PROBLEM FOR A CLASS OF DIAGONALHAMILTONIANS
MASATOSHI SUZUKI
Abstract.
Hamiltonians are 2-by-2 positive semidefinite real symmetric matrix val-ued functions satisfying certain conditions. In this paper, we solve the inverse problemfor which recovers a Hamiltonian from the solution of a first-order system consistingof ordinary differential equations parametrized by complex numbers attached to agiven Hamiltonian, under certain conditions for the solutions. This inverse problemis a generalization of the inverse problem for a class of two-dimensional Hamiltoniansystems. Introduction
In this paper, we generalize the theory on the inverse problem for a class of two-dimensional Hamiltonian systems in [15] together with some simplifications of argument.A 2 × H defined on an interval I = [ t , t )( −∞ < t < t ≤ ∞ ) is called a Hamiltonian if H ( t ) is positive semidefinite for almostall t ∈ I , H is not identically zero on any subset of I with positive Lebesgue measure,and H belongs to L ([ t , c ) , R × ) for any t < c < t . An open subinterval J of I iscalled H -indivisible , if the equality H ( t ) = h ( t ) (cid:18) cos θ cos θ sin θ cos θ sin θ sin θ (cid:19) holds on J for some positive valued function h on J and 0 ≤ θ < π . A point t ∈ I iscalled regular if it does not belong to any H -indivisible interval, otherwise t is called singular . The first-order system − ddt (cid:20) A ( t, z ) B ( t, z ) (cid:21) = z (cid:20) −
11 0 (cid:21) H ( t ) (cid:20) A ( t, z ) B ( t, z ) (cid:21) (1.1)associated with a Hamiltonian H on an interval I parametrized by all z ∈ C is called a canonical system on I . In (1.1), the sign is different from the usual definition. This isbecause, we want to regard the value at the right end t of I as the initial value for ourconvenience. A typical source of Hamiltonians is entire functions of the Hermite–Biehlerclass , which is the set HB of all entire functions satisfying | E ♯ ( z ) | < | E ( z ) | for all z ∈ C + , (1.2)and the subset HB of HB consisting of E such that E ( z ) = 0 for any z ∈ R , where C + = { z | ℑ ( z ) > } is the upper half-plane and F ♯ ( z ) := F (¯ z ), the notation is oftenused in this paper. Every E ∈ HB generates a de Branges space H ( E ) which is areproducing kernel Hilbert space of entire functions. Every de Branges space H ( E )has a unique maximal chain of de Branges subspaces H ( E t ) parametrized by t in aninterval I such that H ( E t ) contained isometrically in H ( E ) for almost all t ∈ I . For thegenerating functions E t , A t = ( E t + E ♯t ) / B t = i ( E t − E ♯t ) / I associated with some Hamiltonian H . Such Hamiltonian iscalled the structure Hamiltonian of H ( E ). Recently, a complete characterization ofstructure Hamiltonians of de Branges spaces is obtained by Romanov–Woracek [13].The inverse problem for the recovering the structure Hamiltonian from given E ∈ HB Mathematics Subject Classification. has been studied by many authors; see Winkler [18], Remling [11], Romanov [12], Suzuki[16], and references therein, for example. However, if E does not necessarily belong to theHermite–Biehler class, nothing can be said about whether Hamiltonian can be obtainedfrom E , in general.In this paper, we prove that if we suppose several conditions on a function E , aHamiltonian is obtained from E by an explicit way of the construction, even though E does not necessarily belong to the Hermite–Biehler class. Those conditions describedbelow may look artificial, but they naturally arise from number theory; see the finalpart of the introduction. The first condition for a function E is the following.(K1) There exists c >
6∈ Z (which is possibility empty orinfinite) of the horizontal strip − c ≤ ℑ z ≤ c such that it is closed under thecomplex conjugation z ¯ z and the negation z
7→ − z and E is analytic in C \ Z and that E satisfies E ♯ ( z ) = E ( − z ) for z ∈ C \ Z .Note that this condition (K1) is more general than that in [15]. In particular, it isallowed that E has an essential singularity in Z . We defineΘ( z ) = Θ E ( z ) := E ♯ ( z ) E ( z )under (K1). Then, Θ(0) = 1,Θ( z )Θ( − z ) = 1 for z ∈ C \ Z (1.3)and | Θ( u ) | = 1 for u ∈ R \ Z (1.4)by definition. We denote by( F f )( z ) = Z ∞−∞ f ( x ) e ixz dx, ( F − g )( z ) = 12 π Z ∞−∞ g ( u ) e − ixu du (1.5)the Fourier integral and inverse Fourier integral, respectively. Then, additional condi-tions are as follows.(K2) There exists a real-valued continuous function K defined on the real line suchthat | K ( x ) | ≪ exp( c | x | ) as | x | → ∞ and that Θ( z ) = ( F K )( z ) holds for ℑ ( z ) > c ,where c is the constant in (K1).(K3) K vanishes on the negative real line ( −∞ , K is continuously differentiable outside a discrete subset Λ ⊂ R and the deriva-tive K ′ belongs to L ( R ).These three conditions are the same as [15]. Under conditions (K2) and (K3), the map K [ t ] : f ( x ) ≤ t ( x ) Z t −∞ K ( x + y ) f ( y ) dy defines a Hilbert–Schmidt operator on L ( −∞ , t ) for every t ∈ R , where ≤ t stands forthe characteristic function of ( −∞ , t ]. In fact, the Hilbert-Schmidt norm of K [ t ] is finite: Z t −∞ Z t −∞ | K ( x + y ) | dxdy ≤ Z t − t dy Z t − t | K ( x ) | dx = 2 t Z t | K ( x ) | dx < ∞ . For t ≤
0, we understand K [ t ] = 0 by (K3). Moreover, K [ t ] is self-adjoint, because thekernel K ( x + y ) is real-valued and symmetric. Therefore, the spectrum of K [ t ] consistsonly of real eigenvalues of finite multiplicity and 0. Finally, we consider the followingconditions.(K5) There exists 0 < τ ≤ ∞ such that both ± K [ t ] for every t < τ .(K6) Θ can not be expressed as a ratio of two entire functions of exponential type. N INVERSE PROBLEM FOR A CLASS OF DIAGONAL HAMILTONIANS 3
Condition (K5) is the same as [15], but condition (K6) is added in this paper, though itis rarely used. The requirement for eigenvalues of K [ t ] in (K5) is trivial for t ≤
0, since K [ t ] = 0. The set of functions satisfying (K1) ∼ (K6) is not empty but a large; see thefinal part of the introduction.Now we assume that E satisfies (K1) ∼ (K5) and define m ( t ) := det(1 + K [ t ])det(1 − K [ t ]) , (1.6) H ( t ) := (cid:20) /γ ( t ) 00 γ ( t ) (cid:21) , γ ( t ) := m ( t ) , (1.7)where we understand that m ( t ) = 1 and H ( t ) is the identity matrix if t ≤
0. Then γ ( t ) is a continuous positive real-valued function on I τ = ( −∞ , τ ) (Theorem 1.4 andProposition 2.1). Therefore H is a Hamiltonian on I τ consisting of continuous functionssuch that it has no H -indivisible intervals, that is, all points of I τ are regular. Thesolution of the associated first-order system (1.1) on I τ recovers the original E as followsas well as the case of the inverse problem for entire functions of the Hermite–Biehlerclass in [15]. The solution of the first-order system is explicitly described by using theunique solutions of the integral equationsΦ( t, x ) + Z t −∞ K ( x + y )Φ( t, y ) dy = 1 , (1.8)Ψ( t, x ) − Z t −∞ K ( x + y )Ψ( t, y ) dy = 1 . (1.9) Theorem 1.1.
Let E be a function satisfying (K1) ∼ (K5), and define A and B by A ( z ) := 12 ( E ( z ) + E ♯ ( z )) and B ( z ) := i E ( z ) − E ♯ ( z )) . (1.10) Let H be the Hamiltonian on I τ = ( −∞ , τ ) defined by (1.6) and (1.7) . Let Φ( t, x ) and Ψ( t, x ) be the unique solutions of (1.8) and (1.9) , respectively. Define A ( t, z ) and B ( t, z ) by A ( t, z ) = − iz E ( z ) Z ∞ t Ψ( t, x ) e izx dx, − iB ( t, z ) = − iz E ( z ) Z ∞ t Φ( t, x ) e izx dx. (1.11) Then, (1) for each t ∈ I τ , A ( t, z ) and B ( t, z ) are well-defined for ℑ ( z ) > c and extend toanalytic functions on C \ Z satisfying A ( t, − z ) = A ( t, z ) , B ( t, − z ) = − B ( t, z ) , A ( t, z ) = A ( t, ¯ z ) , and B ( t, z ) = B ( t, ¯ z ) , (2) for each z ∈ C \Z , A ( t, z ) and B ( t, z ) are continuously differentiable with respectto t , (3) A ( t, z ) and B ( t, z ) solves the first-order system (1.1) associated with H on I τ for every z ∈ C \ Z (4) A ( z ) = A (0 , z ) , B ( z ) = B (0 , z ) and E ( z ) = A (0 , z ) − iB (0 , z ) . For H ( t ) ≡ I , the identity matrix, the functions A ( t, z ) = A ( z ) cos( tz ) + B ( z ) sin( tz )and B ( t, z ) = − A ( z ) sin( tz )+ B ( z ) cos( tz ) satisfy (1) ∼ (4) of Theorem 1.1 for any intervalof t containing 0, and this is the case for the subinterval ( −∞ ,
0] of I τ in the theorem.In this trivial case, H has no information about the original E . Different from thesecases, we will describe below that H in Theorem 1.1 has nontrivial information about E on (0 , τ ).Let H ∞ = H ∞ ( C + ) be the space of all bounded analytic functions in C + . A function θ ∈ H ∞ is called an inner function in C + if lim y → | θ ( x + iy ) | = 1 for almost all x ∈ R with respect to the Lebesgue measure. M. SUZUKI
Theorem 1.2.
Assume that E satisfies (K1) ∼ (K3), and (K5) with τ = ∞ . Then Θ = E ♯ /E is an inner function in C + . Remark 1.1.
Compare this with [15, Theorem 2.4] , where some additional conditionsare assumed to conclude that Θ is an inner function in C + . If E ∈ HB , Θ = E ♯ /E is an inner function in C + . Theorem 1.2 shows that (K5) playsthe role of the condition E ∈ HB for entire functions E ; Z = ∅ . On the other hand, it isknown that if θ is an inner function and meromorphic in C + , there exists E ∈ HB suchthat θ = E ♯ /E ([8, Sections 2.3 and 2.4]). However, the existence of τ > E ♯ /E is inner in C + . Therefore, for the converse ofTheorem 1.2, we require (K6). Theorem 1.3.
Assume that E satisfies (K1) ∼ (K3), and (K6). In addition, assumethat Θ is an inner function in C + . Then (K5) holds for τ = ∞ . Theorems 1.1, 1.2, and 1.3 emphasize the importance of the function m ( t ). Thefollowing simple formula is interesting from both theoretical and computational aspects. Theorem 1.4.
Assume that E satisfies (K2) ∼ (K5). Then, m ( t ) = 1Φ( t, t ) = Ψ( t, t ) (1.12) holds for every t ∈ R . See Propositions 2.2 and 2.5 for other formulas of m ( t ). If E satisfies (K1) ∼ (K5) andΘ is an inner function in C + , K = lim t →∞ K [ t ] defines a bounded operator on L ( R )(Lemma 4.1), and the Fourier transform F ( V t ) of the space V t = L ( t, ∞ ) ∩ K L ( t, ∞ )forms a reproducing kernel Hilbert space for each 0 ≤ t < τ (Section 6.1). Theorem 1.5.
The following statements hold. (1)
Assume that E satisfies (K1) ∼ (K5) and that Θ is an inner function in C + . Let A ( t, z ) and B ( t, z ) be as in Theorem 1.1, and let j ( t ; z, w ) be the reproducingkernel of F ( V t ) for ≤ t < τ . Then, j ( t ; z, w ) = 1 E ( z ) E ( w ) · A ( t, z ) B ( t, w ) − A ( t, w ) B ( t, z ) π ( w − ¯ z ) , (1.13) and j ( t ; z, z ) as a function of z ∈ C + for any ≤ t < τ . (2) Assume that E satisfies (K1) ∼ (K5) with τ = ∞ . Then, ( Θ is an inner functionin C + and) lim t →∞ j ( t ; z, w ) = 0 for every z, w ∈ C + . Theorem 1.5 shows that H of (1.7) provides the genuine structure Hamiltonian of thede Branges space H ( E ) if E ∈ HB satisfies (K1) ∼ (K3) and (K6) by [2, Theorem 40].The basic idea for achieving the above results originates from the work of J.-F. Burnol[6] (and [3, 4, 5]) as well as [14, 15], and [17]. However, in this paper, the method usedin [6] for Γ(1 − s ) / Γ( s ) standing on the theory of Hankel transforms is axiomatized,reorganized and generalized, and some arguments are simplified.Before closing the introduction, we mention a few examples of functions E satisfyingconditions (K1) ∼ (K6). Let ζ ( s ) be the Riemann zeta function, and let ξ ( s ) = s ( s − π − s/ Γ( s/ ζ ( s ). Then ξ ( s ) is an entire function taking real-values on the critical line ℜ ( s ) = 1 / ζ ( s ).We put E n ( z ) = ξ (cid:18)
12 + 2 n − iz (cid:19) n , E ⋊⋉ ( z ) = exp (cid:18) ξ ′ ξ (cid:18) − iz (cid:19)(cid:19) (1.14)for n ∈ Z > . Then, E ⋊⋉ ( z ) = lim n →∞ E n ( z ) /ξ (1 / − iz ) n , and it is proved that E n (resp. E ⋊⋉ ) satisfies the conditions (K1) ∼ (K6) in [15, Proposition 4.1 and Lemma 4.3] N INVERSE PROBLEM FOR A CLASS OF DIAGONAL HAMILTONIANS 5 (resp. [17, Theorem 2]), where τ > τ = ∞ if theRiemann hypothesis is assumed; all zeros of ξ ( s ) lie on the critical line. In [15], manyexamples of E satisfying the conditions (K1) ∼ (K6) are made from L -functions in theSelberg class.The function ξ ( s ) has no zeros in ℜ ( s ) > + N if and only if E n belongs to HB foreach n ≥ N ([15, Theorem 9.1]). In particular, the de Branges space H ( E n ) is definedfor each n ∈ Z > under the Riemann hypothesis, and its structure Hamiltonian H n is constructed in [15]. Therefore, it is natural to ask about the limit behavior of H n as n → ∞ . However, lim n →∞ E n does not make sense, and E ⋊⋉ is no longer an entirefunction, because E ⋊⋉ has an essential singularity at a zero of ξ (1 / − iz ). Therefore, themethod constructing H n in [15] can not be applied to E ⋊⋉ . This is the main reason whywe generalized condition (K1) as above in this paper.The paper is organized as follows. In Section 2, we study basic properties of solutionsΦ( t, x ) and Ψ( t, x ) of integral equations (1.8) and (1.9) in preparation for the proof ofTheorem 1.1, and prove Theorem 1.4. In Section 3, we prove Theorem 1.1 by usingresults in Section 2. In Section 4, we prove Theorems 1.2 and 1.3 by studying thebehavior of the operator norm of K [ t ] when t varies. In Section 5, we review the theoryof model subspaces and de Branges spaces in preparation for the proof of Theorem1.5. In Section 6, we prove Theorem 1.5 by studying the vectors representing the pointevaluation maps in a model subspace. In Section 7, we comment on a relation betweenour inverse problem, the Cauchy problem for certian hyperbolic systems, and dampedwave equations. Acknowledgments
This work was supported by JSPS KAKENHI Grant NumberJP17K05163. 2.
Solutions of related integral equations
We suppose that E satisfies (K2) ∼ (K5) throughout this section. In particular, weunderstand that c and τ are constants in (K2) and (K5), respectively. Let L p ( I ) be the L p -space on an interval I with respect to the Lebesgue measure. If J ⊂ I , we regard L p ( J ) as a subspace of L p ( I ) by the extension by zero.2.1. Properties of Φ( t, x ) and Ψ( t, x ) .Proposition 2.1. The integral equations (1.8) and (1.9) for ( x, t ) ∈ R × ( −∞ , τ ) haveunique solutions Φ( t, x ) and Ψ( t, x ) , respectively. Moreover, (1) Φ( t, x ) and Ψ( t, x ) are real-valued, (2) Φ( t, x ) and Ψ( t, x ) are continuously differentiable functions of x ∈ R , (3) Φ( t, x ) = Ψ( t, x ) = 1 for x < − t and Φ( t, x ) ≪ exp( c ′ x ) and Ψ( t, x ) ≪ exp( c ′ x ) as x → + ∞ for any t < τ and c ′ > c , where implied constants depend on t , (4) Φ( t, t ) = 0 and Ψ( t, t ) = 0 for every t < τ , (5) if t ≤ , Φ( t, t ) = Ψ( t, t ) = 1 and Φ( t, x ) = 1 − Z x + t K ( y ) dy, Ψ( t, x ) = 1 + Z x + t K ( y ) dy. (2.1) Proof.
We prove only in the case of Φ( t, x ), because the case of Ψ( t, x ) is proved in thesimilar argument. First, we prove the uniqueness of Φ( t, x ). If Φ ( t, x ) and Φ ( t, x ) solve(1.8), the difference f ( t, x ) = Φ ( t, x ) − Φ ( t, x ) satisfies f ( t, x )+ R t −∞ K ( x + y ) f ( t, y ) dy =0. This shows that ≤ t ( x ) f ( t, x ) = 0, since (1 + K [ t ]) is invertible, and hence f ( t, x ) = 0.Then (1) is obvious, since the kernel K is real-valued by (K2). To prove other statements,we first suppose that t > L ( − t, t ), we find that [ − t,t ] ( x )Φ( t, x ) is a continuous function of x on [ − t, t ] by the continuity of K and M. SUZUKI [ − t,t ] ( x ), where A stands for the characteristic function of A ⊂ R . On the other hand,Φ( t, x ) = 1 for x < − t , since the integral in (1.8) is zero for x < − t by (K3). Therefore,Φ( t, x ) = 1 − Z − t − x K ( x + y ) dy − Z t − t K ( x + y )Φ( t, y ) dy (2.2)by (1.8), where the middle integral is understood as zero if x < t . This equality and(K2) shows that Φ( t, x ) is a continuous function of x ∈ R and satisfies the estimate in(3). Moreover, differentiating (1.8) with respect to x , ∂∂x Φ( t, x ) + Z t −∞ K ′ ( x + y )Φ( t, y ) dy = 0 . (2.3)This shows that Φ( t, x ) is differentiable for x and ( ∂/∂x )Φ( t, x ) is a continuous functionof x by (K4).We prove (4) by contradiction. Differentiating (1.8) with respect to x , and thenapplying integration by parts, ∂∂x Φ( t, x ) + K ( x + t )Φ( t, t ) − Z t −∞ K ( x + y ) ∂∂y Φ( t, y ) dy = 0 . (2.4)Therefore, if we suppose that Φ( t, t ) = 0, ∂∂x Φ( t, x ) − Z t −∞ K ( x + y ) ∂∂y Φ( t, y ) dy = 0 . (2.5)This asserts that the restriction [ − t,t ] ( x )( ∂/∂x )Φ( t, x ) is a solution of the homogeneousequation (1 − K [ t ]) f = 0 on L ( − t, t ), and thus ( ∂/∂x )Φ( t, x ) = 0 by (2.5) and theequality Φ( t, x ) = 1 for x < − t , since K ( x + y ) = 0 if x < − t and y < t . Hence, we have c (cid:16) R x + t K ( y ) dy (cid:17) = 1 for arbitrary x if Φ( t, x ) = c . This implies that K ≡ R ,and therefore, Φ( t, x ) = 1 for all x ∈ R by (1.8). This is a contradiction.Finally, we prove (5). If t ≤ K ( x + y ) = 0 for x < t and y < t . Thus Φ( t, x ) = 1 for x < t , the first equality of (2.1) holds, and Φ( t, t ) = 1. Hence Φ( t, x ) is a continuouslydifferentiable function of x on R satisfying the desired estimate. (cid:3) For convenience of studying the solutions Φ( t, x ) and Ψ( t, x ), we consider the solutionsof integral equations φ + ( t, x ) + Z t −∞ K ( x + y ) φ + ( t, y ) dy = K ( x + t ) , (2.6) φ − ( t, x ) − Z t −∞ K ( x + y ) φ − ( t, y ) dy = K ( x + t ) . (2.7)The usefulness of solutions φ + ( t, x ) and φ − ( t, x ) comes from relationships with theresolvent kernels R + ( t ; x, y ) and R − ( t ; x, y ) of (1 + K [ t ]) and (1 − K [ t ]), respectively: ≤ t ( x ) φ + ( t, x ) = R + ( t ; x, t ), ≤ t ( x ) φ − ( t, x ) = R − ( t ; x, t ) ([15, Section 3]). Proposition 2.2.
The integral equations (2.6) and (2.7) for ( x, t ) ∈ R × ( −∞ , τ ) haveunique solutions φ + ( t, x ) and φ − ( t, x ) , respectively. Moreover, (1) φ + ( t, x ) and φ − ( t, x ) are continuous on R and continuously differentiable on R \ { λ − t | λ ∈ Λ } as a function of x , where Λ is the set in (K4). (2) φ + ( t, x ) and φ − ( t, x ) are continuous on [0 , τ ) and continuously differentiable on (0 , τ ) except for points in { λ − x | λ ∈ Λ } as a function of t , (3) for fixed t ∈ [0 , τ ) , φ ± ( t, x ) = 0 for x < − t and φ ± ( t, x ) ≪ e cx as x → + ∞ for c > in (K2), where the implied constants depend on t , (4) if t ≤ , φ + ( t, t ) = φ − ( t, t ) = 0 and φ + ( t, x ) = φ − ( t, x ) = K ( x + t ) , N INVERSE PROBLEM FOR A CLASS OF DIAGONAL HAMILTONIANS 7 (5) the following equations hold φ ± ( t, t ) = ± ddt log det(1 ± K [ t ]) , (2.8)(6) the following equation holds m ( t ) = exp (cid:18)Z t µ ( s ) ds (cid:19) , µ ( t ) = φ + ( t, t ) + φ − ( t, t ) , (2.9)(7) m ( t ) is a continuous positive real-valued function on ( −∞ , τ ) , m (0) = 1 , andcontinuously differentiable outside a discrete subset.Proof. The proof is the same as that in [15, Section 3], because conditions (K2) ∼ (K5)are exactly the same. (cid:3) Proposition 2.3.
Let Φ( t, x ) and Ψ( t, x ) be unique solutions of (1.8) and (1.9) , re-spectively. Let φ + ( t, x ) and φ − ( t, x ) be unique solutions of (2.6) and (2.7) , respectively.Then, Φ( t, x ) and Ψ( t, x ) are continuously differentiable with respect to t on ( −∞ , τ ) and equalities φ + ( t, x ) = − t, t ) ∂∂t Φ( t, x ) = 1Ψ( t, t ) ∂∂x Ψ( t, x ) , (2.10) φ − ( t, x ) = − t, t ) ∂∂x Φ( t, x ) = 1Ψ( t, t ) ∂∂t Ψ( t, x ) (2.11) hold.Proof. Applying integration by parts to (2.3), ∂∂x Φ( t, x ) + K ( x + t )Φ( t, t ) − Z t −∞ K ( x + y ) ∂∂y Φ( t, y ) dy = 0 . (2.12)This shows that − ( ∂/∂x )Φ( t, x ) / Φ( t, t ) solves (2.7) by Proposition 2.1 (4). Hence theuniqueness of the solution of (2.6) concludes the first equality of (2.11).On the other hand, we find that [ − t,t ] ( x )Φ( t, x ) is also continuous in t , because theresolvent kernel R + ( t ; x, y ) of (1 + K [ t ]) is continuous in all variables ([15, Section 3]).Therefore, Φ( t, x ) is continuous in t by (2.2). By differentiating (1.8) with respect to t (in the sense of weak derivative), we find that − ( ∂/∂t )Φ( t, x ) / Φ( t, t ) solves (2.6) byProposition 2.1 (4). Hence the uniqueness of the solution concludes the first equalityof (2.10). Moreover, the first equality of (2.10) shows that ( ∂/∂t )Φ( t, x ) is continuouswith respect to t by Proposition 2.2 (2), (3), and (4). Hence Φ( t, x ) is differentiable withrespect to t in the usual sense, and the derivative with respect to t is continuous in t .The differentiability of Ψ( t, x ) with respect to t and the second equalities of (2.10) and(2.11) are proved by the similar argument. (cid:3) Proof of Theorem 1.4.
Taking x = t in equation (1.8) and then differentiatingit with respect to t ,0 = ddt (Φ( t, t )) + 2 K (2 t )Φ( t, t ) − Z t −∞ K ( t + y ) ∂∂y Φ( t, y ) dy + Z t −∞ K ( t + y ) ∂∂t Φ( t, y ) dy. Using the first equalities of (2.10) and (2.11) on the right-hand side, ddt (Φ( t, t ))+ 2 K (2 t )Φ( t, t ) − Φ( t, t ) Z t −∞ K ( t + x )( φ + ( t, x ) − φ − ( t, x )) dx = 0 . (2.13)On the other hand, by the proof of [15, Theorem 6.1], we have12 ( φ + ( t, x ) + φ − ( t, x )) = K ( x + t ) − Z t −∞ K ( x + y ) 12 ( φ + ( t, y ) − φ − ( t, y )) dy. M. SUZUKI
Substituting this into (2.13) after taking x = t , we get ddt (Φ( t, t )) + Φ( t, t )( φ + ( t, t ) + φ − ( t, t ))) = 0 . (2.14)Therefore, Φ( t, t ) = C exp (cid:16) − R t ( φ + ( τ, τ ) + φ − ( τ, τ )) , dτ (cid:17) = Cm ( t ) − by (2.9). Todetermine C , we take x = t = 0 in equation (1.8). Then Φ(0 ,
0) = 1, since the integralon the left-hand side is zero because K ( x ) = 0 for x <
0, and thus C = 1 by m (0) = 1.Hence we obtain the first equality of (1.12). The second equality of (1.12) is proved bythe same way. (cid:3) From Theorem 1.4 and Proposition 2.1, we find that H of (1.7) is a Hamiltonian on( −∞ , τ ) such that it consists of continuous functions and has no H -indivisible intervals.These properties also obtained from Proposition 2.2.2.3. Corollaries of Proposition 2.3.
Here we state a few results that easily obtainedfrom Proposition 2.3, but note that these are of their own interest and are not used toprove the main results.
Proposition 2.4.
The solutions of (1.8) and (1.9) are related each other as follows Ψ( t, x ) = 1 − t, t ) Z x − t ∂∂t Φ( t, y ) dy, (2.15)Φ( t, x ) = 1 − t, t ) Z x − t ∂∂t Ψ( t, x ) dy. (2.16) Proof.
Integrating the second equalities of (2.10) and using (1.12), Proposition 2.1 (3)and Proposition 2.2 (6), Ψ( t, x ) = 1 + m ( t ) Z x − t φ + ( t, y ) dy. (2.17)Substitute the first equality of (2.10) and (1.12) into (2.17), we obtain (2.15). (2.16) isalso proved in a similar way. (cid:3) Proposition 2.5.
We have m ( t ) = 1 − Z t − t φ + ( t, y ) dy, m ( t ) = 1 + Z t − t φ − ( t, y ) dy. (2.18) Proof.
The first equality is obtained by taking x = t in (2.17) and noting (1.12). Thesecond equality is proved in a similar way. (cid:3) Proposition 2.6.
The following partial differential equations hold ∂∂t φ + t ( x ) − ∂∂x φ − t ( x ) = − µ ( t ) φ − t ( x ) , ∂∂t φ − t ( x ) − ∂∂x φ + t ( x ) = µ ( t ) φ − t ( x ) . (2.19) Proof.
We have ∂∂t φ + ( t, x ) = − Ψ t ( t, t ) + Ψ x ( t, t )Ψ( t, t ) φ + ( t, x ) + Ψ xt ( t, x )Ψ( t, t ) , ∂∂x φ − ( t, x ) = Ψ tx ( t, x )Ψ( t, t )by (2.10) and (2.11). On the other hand,Ψ t ( t, t ) + Ψ x ( t, t )Ψ( t, t ) = ddt log m ( t ) = µ ( t ) . by (1.12) and Proposition 2.2 (6). Hence we obtain the first equation. The secondequation is proved in a similar way. (cid:3) N INVERSE PROBLEM FOR A CLASS OF DIAGONAL HAMILTONIANS 9 Proof of Theorem 1.1
We suppose that E satisfies (K1) ∼ (K5) throughout this section. We use the samenotation (1.5) for the Fourier transforms on L ( R ) and L ( R ) if no confusion arises. If weunderstand the right-hand sides of (1.5) in L -sense, they provide isometries on L ( R )up to a constant multiple: k F f k L ( R ) = 2 π k f k L ( R ) , k F − f k L ( R ) = (2 π ) − k f k L ( R ) .3.1. Proof of (1).
The right-hand side of (1.11) is defined for ℑ ( z ) > c by Proposition2.1 (3). Therefore, Ω( t, z ) = Z ∞ t (Φ( t, x ) − Ψ( t, x )) e izx dx is defined for ℑ ( z ) > c . Subtracting (1.9) from (1.8),(Φ( t, x ) − Ψ( t, x )) + Z t −∞ K ( x + y )(Φ( t, y ) + Ψ( t, y )) dy = 0 . Using this equality, we haveΩ( t, z ) = − Z ∞ t (cid:18)Z t −∞ K ( x + y )(Φ( t, y ) + Ψ( t, y )) dy (cid:19) e izx dx = − Z ∞−∞ K ( x ) e izx dx Z t −∞ (Φ( t, y ) + Ψ( t, y )) e − izy dy + Z t −∞ Z t −∞ K ( x + y )(Φ( t, y ) + Ψ( t, y )) dy e izx dx = − Z ∞−∞ K ( x ) e izx dx Z t − t (Φ( t, y ) + Ψ( t, y )) e − izy dy − Z ∞−∞ K ( x ) e izx dx Z − t −∞ (Φ( t, y ) + Ψ( t, y )) e − izy dy − Z t −∞ (Φ( t, x ) − Ψ( t, x )) e izx dx for ℑ ( z ) > c . Because Φ( t, x ) = Ψ( t, x ) = 1 for x < − t , we haveΩ( t, z ) = Θ( z ) (cid:18) e izt iz − Z t − t (Φ( t, y ) + Ψ( t, y )) e − izy dy (cid:19) − Z t − t (Φ( t, x ) − Ψ( t, x )) e izx dx for ℑ ( z ) > c . The right-hand side extends Ω( t, z ) to C \ Z . From this equality,Θ( z )Ω( t, − z ) = (cid:18) − e − izt iz − Z t − t (Φ( t, y ) + Ψ( t, y )) e izy dy (cid:19) − Θ( z ) Z t − t (Φ( t, x ) − Ψ( t, x )) e − izx dx (3.1)for z ∈ C \ Z by (1.3). Adding (1.8) and (1.9),(Φ( t, x ) + Ψ( t, x )) + Z t −∞ K ( x + y )(Φ( t, y ) − Ψ( t, y )) dy = 2 . Integrating this with respect to x from − t to ∞ after multiplying by e izx both side, Z ∞− t (Φ( t, x ) + Ψ( t, x )) e izx dx + Z ∞− t Z t −∞ K ( x + y )(Φ( t, y ) − Ψ( t, y )) dy e izx dx = − e − izt iz . The integration R ∞− t R t −∞ can be replaced by R ∞−∞ R t − t , since R t −∞ K ( x + y )(Φ( t, y ) − Ψ( t, y )) dy = 0 and Φ( t, x ) − Ψ( t, x ) = 0 for x < − t . Thus, Z ∞− t (Φ( t, x ) + Ψ( t, x )) e izx dx + Θ( z ) Z t − t (Φ( t, y ) − Ψ( t, y )) e − izy dy = − e − izt iz Combining this with (3.1),Θ( z )Ω( t, − z ) = Z ∞ t (Φ( t, y ) + Ψ( t, y )) e izy dy. (3.2)The left-hand side extends the right-hand side to C \ Z . On the other hand, A ( t, z ) + iB ( t, z ) = iz E ( z ) Z ∞ t (Φ( t, x ) − Ψ( t, x )) e izx dx = iz E ( z )Ω( t, z ) ,A ( t, z ) − iB ( t, z ) = − iz E ( z ) Z ∞ t (Φ( t, x ) + Ψ( t, x )) e izx dx = i ( − z )2 E ( − z )Ω( t, − z ) (3.3)by (1.11) and (3.2). These two formula extend A ( t, z ) and B ( t, z ) to C \ Z , and showthat A ( t, z ) is even and that B ( t, z ) is odd.If F ( z ) = F ( f ( x ))( z ) for ℑ ( z ) ≫ F (¯ z ) = F ( f ( − x ))( z ) and F ( − z ) = F ( f ( − x ))( z )for ℑ ( z ) ≪
0. Therefore, if f ( x ) (resp. if ( x )) is real-valued, F ( z ) is continued to ananalytic function in C \Z , and F ( − z ) = F ( z ) (resp. F ( − z ) = − F ( z )), then F (¯ z ) = F ( z )holds for C \ Z . From this argument and Proposition 2.1 (3), A ( t, z ) = A ( t, ¯ z ) and B ( t, z ) = B ( t, ¯ z ) hold. (cid:3) Proof of (2) and (3).
By Proposition 2.3, formulas (2.10), (2.11), and Proposition2.2 (3), A ( t, z ) and B ( t, z ) are differentiable with respect to t . Therefore, it remains toshow that ( ∂/∂t ) A ( t, z ) = zm ( t ) B ( t, z ) and ( ∂/∂t ) B ( t, z ) = − zm ( t ) − A ( t, z ). Using(1.12) and (2.11), we have ∂∂t A ( t, z ) = − iz E ( z ) (cid:18) − Ψ( t, t ) e izt + Z ∞ t ∂∂t Ψ( t, x ) e izx dx (cid:19) = − iz E ( z ) (cid:18) − Ψ( t, t ) e izt − Ψ( t, t )Φ( t, t ) Z ∞ t ∂∂x Φ( t, x ) e izx dx (cid:19) = − iz E ( z ) (cid:18) Ψ( t, t )Φ( t, t ) iz Z ∞ t Φ( t, x ) e izx dx (cid:19) = z Ψ( t, t )Φ( t, t ) B ( t, z ) = zm ( t ) B ( t, z ) . The second equality is proved in a similar way. (cid:3)
Proof of (4).
For t ≤
0, we have A ( t, z ) = A ( z ) cos( tz ) + B ( z ) sin( tz ) ,B ( t, z ) = − A ( z ) sin( tz ) + B ( z ) cos( tz ) (3.4)by (2.1), (1.10) and (1.11). In particular, Theorem 1.1 (4) holds. (cid:3) Proof of Theorems 1.2 and 1.3
Lemma 4.1.
Let E be a function satisfying (K1) ∼ (K3). Define K f = lim t →∞ K [ t ] f in pointwise convergence for f in the space C ∞ c ( R ) of all compactly supported smoothfunction on R . Then the Fourier integral formula ( FK f )( z ) = Θ( z ) ( F f )( − z ) (4.1) holds for ℑ ( z ) > c . Suppose that Θ = E ♯ /E is an inner function in C + in addition to(K1) ∼ (K3). Then K f belongs to L ( R ) for f ∈ C ∞ c ( R ) , and the linear map f K f N INVERSE PROBLEM FOR A CLASS OF DIAGONAL HAMILTONIANS 11 extends to the isometry K : L ( R ) → L ( R ) satisfying K = id and (4.1) holds for z ∈ R \ Z . Moreover, (4.1) holds for z ∈ ( C + ∪ R ) \ Z , if f ∈ L ( R ) has a support in ( −∞ , t ] for some t ∈ R .Proof. This is proved by almost the same argument as [15, Lemma 3.2], because thedifference of condition (K1) between this paper and [15] is not essential in the proof. (cid:3)
Here we recall basic properties of the Hardy spaces. The Hardy space H = H ( C + )in C + is defined to be the space of all analytic functions f in C + endowed with norm k f k H := sup v> R R | f ( u + iv ) | du < ∞ . The Hardy space ¯ H = H ( C − ) in thelower half-plane C − is defined in the similar way. As usual, we identify H and ¯ H with subspaces of L ( R ) via nontangential boundary values on the real line such that L ( R ) = H ⊕ ¯ H , where L ( R ) has the inner product h f, g i = R R f ( u ) g ( u ) du . TheHardy space H is a reproducing kernel Hilbert space, in particular the point evaluationfunctional f f ( z ) is continuous for each z ∈ C + and it is represented by k z ( w ) =(2 πi ) − (¯ z − w ) − ∈ H as h f, k z i = f ( z ). Lemma 4.2.
Let θ be an analytic function defined in C + . Suppose that θf ∈ H forevery f ∈ H . Then the pointwise multiplication operator M θ : f θf is bounded on H and θ ∈ H ∞ .Proof. We find that M θ is bounded from the closed graph theorem and continuity ofpoint evaluations. Let k z ∈ H be the vector representing the point evaluation at z ∈ C + . Then, we have h M θ f, k z i = h θf, k z i = θ ( z ) f ( z ) = h f, θ ( z ) k z i . for f ∈ H . Therefore, the adjoint M ∗ θ acts on k z as M ∗ θ k z = θ ( z ) k z . This implies k z isan eigenvector of M ∗ θ with eigenvalue θ ( z ). Hence, | θ ( z ) | ≤ k M θ k op for every z ∈ C + .This shows that θ ( z ) is uniformly bounded on C + . (cid:3) Proposition 4.1.
Suppose that E satisfies (K1) ∼ (K3). Then the following are equiva-lent: (1) Θ = E ♯ /E is an inner function in C + . (2) K [ t ] converges as t → ∞ with respect to the operator norm k · k op . (3) K [ t ] f converges as t → ∞ with respect to k · k L ( R ) for all f ∈ L ( R ) . (4) there exists M > such that k K [ t ] k op ≤ M for every t > .Proof. We prove the implication (2) ⇒ (1). Suppose that K = lim t →∞ K [ t ] exists withrespect to the operator norm. Then K f ∈ L ( R ) for any f ∈ L ( R ), where we understandas K f ( x ) = Z ∞−∞ K ( x + y ) f ( y ) dy := l . i . m t →∞ 0) andtake { f n } n ⊂ C ∞ c ( R ) such that lim n →∞ f n = f in L ( R ). Then,( FK f )( z ) = lim n →∞ ( FK f n )( z ) = lim n →∞ lim t →∞ ( FK [ t ] f n )( z ) , since K and F are bounded. On the right-hand side,lim t →∞ ( FK [ t ] f n )( z ) = Θ( z )( F f n )( − z )for ℑ ( z ) > c by Lemma 4.1. Therefore, ( FK f )( z ) = Θ( z )( F f )( − z ) for ℑ ( z ) > c . Onthe other hand, K f ∈ L (0 , ∞ ) by (K3). Thus ( FK f )( z ) defines an analytic function in C + . Hence ( FK f )( z ) = Θ( z )( F f )( − z ) which conclude Θ H ⊂ H . Hence Θ ∈ H ∞ byLemma 4.2. Therefore, by (1.4), Θ is an inner function in C + .We prove the implication (1) ⇒ (3). If Θ is an inner function in C + , for any f ∈ L ( R ), L f := F − M Θ FJ f is defined and belongs to L ( R ), L f ( x ) = R K ( x + y ) f ( y ) dy := l . i . m T →∞ R T − T K ( x + y ) f ( y ) dy , and k L f k = k f k by the argument similar to the proof of[15, Lemma 3.2], where J f ( x ) = f ( − x ). We have K [ t ] = P t LP t . Therefore, L f − K [ t ] f = L (1 − P t ) f + (1 − P t ) LP t f → t → ∞ ) . Hence K [ t ] f converges to L f , and L = K .The implication (3) ⇒ (2) is a consequence of the Banach–Steinhaus theorem. Theimplication (2) ⇒ (4) is trivial. Finally, we show that (4) implies (3). Let t > s > f ∈ L , and let P t be the projection to L ( −∞ , t ). Then, K [ t ] f − K [ s ] f = P t K ( P t − P s ) f − ( P t − P s ) K ( P t − P s ) f + ( P t − P s ) KP t f , and thus k K [ t ] f − K [ s ] f k ≤ k K [ t ]( P t − P s ) f k + k ( P t − P s ) K [ t ]( P t − P s ) f k + k ( P t − P s ) KP t f k≤ M k ( P t − P s ) f k + k ( P t − P s ) KP t f k . The first term of the right-hand side is smaller as t > s > t > s > a > a ≤ k ( P t − P s ) KP t f k = Z ts (cid:12)(cid:12)(cid:12)(cid:12)Z t −∞ K ( x + y ) f ( y ) dy (cid:12)(cid:12)(cid:12)(cid:12) dx holds for some t > s > s for arbitrary s . Then we can take a strictly increasingnumbers s < s < . . . such that a ≤ R s n +1 s n (cid:12)(cid:12)(cid:12)R s n +1 −∞ K ( x + y ) f ( y ) dy (cid:12)(cid:12)(cid:12) dx holds. For suchnumbers, a ≤ Z s n +1 s n (cid:12)(cid:12)(cid:12)(cid:12)Z s n +1 −∞ K ( x + y ) f ( y ) dy (cid:12)(cid:12)(cid:12)(cid:12) dx ≤ Z s n +1 s n (cid:12)(cid:12)(cid:12)(cid:12)Z s −∞ K ( x + y ) f ( y ) dy (cid:12)(cid:12)(cid:12)(cid:12) dx + Z s n +1 s n (cid:12)(cid:12)(cid:12)(cid:12)Z s n +1 s K ( x + y ) f ( y ) dy (cid:12)(cid:12)(cid:12)(cid:12) dx ≤ Z s n +1 s n (cid:12)(cid:12)(cid:12)(cid:12)Z s −∞ K ( x + y ) f ( y ) dy (cid:12)(cid:12)(cid:12)(cid:12) dx + k ( P s n +1 − P s n ) K [ s n +1 ]( P s n +1 − P s ) f k ≤ Z s n +1 s n (cid:12)(cid:12)(cid:12)(cid:12)Z s −∞ K ( x + y ) f ( y ) dy (cid:12)(cid:12)(cid:12)(cid:12) dx + M k ( P s n +1 − P s ) f k . The second term of the right-hand side is small for large s . Therefore, by replacing a if necessary, a ≤ Z s n +1 s n (cid:12)(cid:12)(cid:12)(cid:12)Z s −∞ K ( x + y ) f ( y ) dy (cid:12)(cid:12)(cid:12)(cid:12) dx ( n = 0 , , , . . . ) . This implies k K [ s n +1 ]( P s f ) k ≥ k K [ s ]( P s f ) k + na , which contradicts the uniformboundedness (4). (cid:3) Proposition 4.2. We can arrange the eigenvalues of the family { K [ t ] } t such that λ +1 ( t ) ≥ λ +2 ( t ) ≥ . . . are positive eigenvalues, λ − ( t ) ≤ λ − ( t ) ≤ . . . are negative eigen-values, where eigenvalues are repeated as many times as multiplicities, and each | λ ± i ( t ) | is a continuous nondecreasing function of t .Proof. We find that λ + j ( t ) is a nondecreasing function of t by the min–max principlefor positive eigenvalues λ + j ( t ) = min V j max f ∈ V ⊥ j {h K [ t ] f, f i / h f, f i} , where V j runs all ( j − L ( −∞ , t ) and V ⊥ j stands for the orthogonal complement of V j , because the maximum part of the formula is a nondecreasing function of t by theinclusion L ( −∞ , s ) ⊂ L ( −∞ , t ) for s < t obtained by the extension by zero. Also λ − j ( t ) is a nonincreasing function of t by applying the above argument to − K [ t ]. Since N INVERSE PROBLEM FOR A CLASS OF DIAGONAL HAMILTONIANS 13 K [ t ] is a Hilbert–Schmidt operator, P j λ + j ( t ) + P j λ − j ( t ) = R t − t R t − t | K ( x + y ) | dxdy .Therefore, for sufficiently small h > X j ( λ + j ( t + h ) − λ + j ( t ) ) + X j ( λ − j ( t + h ) − λ − j ( t ) )= (cid:18)Z t + ht Z t + h − t − h + Z − t − t − h Z t + h − t − h + Z t + ht Z t − t + Z − t − t − h Z t − t (cid:19) | K ( x + y ) | dxdy ≤ h Z t − t | K ( x ) | dx, and hence λ ± j ( t + h ) − λ ± j ( t ) ≤ h R t − t | K ( x ) | dx . This implies that each λ ± j ( t ) is rightcontinuous. The left continuity of each λ ± j ( t ) is proved by the same argument. (cid:3) Proof of Theorems 1.2 and 1.3. Recall that the operator norm of a compact self-adjoint operator is equal to the maximum of absolute values of eigenvalues. Then,condition (K5) with τ = ∞ implies that k K [ t ] k op < C + by Proposition 4.1, and Theorem 1.2 is proved. If we supposethat E satisfies (K1) ∼ (K3) and (K6) and that Θ is an inner function in C + , both ± K [ t ] for every t > k K [ t ] k op < C + by Proposition 4.1,and Theorem 1.3 is proved. (cid:3) Theory of model subspaces In this section, we review basic notions and properties of model subspaces and deBranges spaces in preparation for the next section according to Havin–Mashreghi [8,9], Baranov [1] for model subspaces and de Branges [2], Romanov [12], Winkler [19],Woracek [20] for de Branges spaces.5.1. Model subspaces. For an inner function θ , a model subspace (or coinvariant sub-space) K ( θ ) is defined by the orthogonal complement K ( θ ) = H ⊖ θH , (5.1)where θH = { θ ( z ) F ( z ) | F ∈ H } . It has the alternative representation K ( θ ) = H ∩ θ ¯ H . (5.2)A model subspace K ( θ ) is a reproducing kernel Hilbert space with respect to the norminduced from H . The reproducing kernel of K ( θ ) is j ( z, w ) = 1 − θ ( z ) θ ( w )2 πi (¯ z − w ) ( z, w ∈ C + ) , that is, h f, j ( z, · ) i H = f ( z ) for f ∈ K ( θ ) and z ∈ C + .5.2. De Branges spaces. For E ∈ HB , the set H ( E ) := { f | f is entire, f /E and f ♯ /E ∈ H } forms a Hilbert space under the norm k f k H ( E ) := k f /E k H . The Hilbert space H ( E ) iscalled the de Branges space generated by E . The de Branges space H ( E ) is a reproducingkernel Hilbert space consisting of entire functions endowed with the reproducing kernel J ( z, w ) = E ( z ) E ( w ) − E ♯ ( z ) E ♯ ( w )2 πi (¯ z − w ) ( z, w ∈ C + ) . The reproducing formula h f, J ( z, · ) i H ( E ) = f ( z ) for f ∈ H ( E ) and z ∈ C + remains truefor z ∈ R if θ = E ♯ /E is analytic in a neighborhood of z . If an inner function θ in C + is meromorphic in C , it is called a meromorphic innerfunction . It is known that every meromorphic inner function is expressed as θ = E ♯ /E by using some E ∈ HB . If θ is a meromorphic inner function such that θ = E ♯ /E , themodel subspace K ( θ ) is isomorphic and isometric to the de Branges space H ( E ) as aHilbert space by K ( θ ) → H ( E ) : f ( z ) E ( z ) f ( z ).As developed in [3, 4, 5, 6], the Hankel type operator K with the kernel K ( x + y ) isuseful to study model subspaces K ( θ ) or de Branges spaces H ( E ) via Fourier analysis.6. Proof of Theorem 1.5 We suppose that E satisfies (K1) ∼ (K5) with τ = ∞ throughout this section. ThenΘ = E ♯ /E is an inner function in C + by Theorem 1.2 (thus Z ⊂ R ) and f K f definesan isometry of L ( R ) satisfying K = id by Lemma 4.1. These properties are essentialin the argument of the section. For convenience, we put A ( t, z ) = m ( t ) − A ( t, z ) , B ( t, z ) = m ( t ) B ( t, z ) . (6.1) Lemma 6.1. For ℑ ( z ) > c , we have A ( t, z ) = 12 E ( z ) (cid:18) e izt + Z ∞ t φ + ( t, x ) e izx dx (cid:19) , − i B ( t, z ) = 12 E ( z ) (cid:18) e izt − Z ∞ t φ − ( t, x ) e izx dx (cid:19) . (6.2) Proof. Using (2.10) and (2.11), e izt + Z ∞ t φ + ( t, x ) e izx dx = − iz Z ∞ t Ψ( t, x )Ψ( t, t ) e izx dx,e izt − Z ∞ t φ − ( t, x ) e izx dx = − iz Z ∞ t Φ( t, x )Φ( t, t ) e izx dx. Hence we obtain (6.2) by definition (1.11). The convergence of integrals are justified byProposition 2.2 (3). (cid:3) Formulas for reproducing kernels of model subspaces. Let V t be the Hilbertspace of all functions f such that both f and K f are square integrable functions havingsupports in [ t, ∞ ): V t = L ( t, ∞ ) ∩ K ( L ( t, ∞ )) . If t ≥ F ( V t ) is a closed subspace of H , because the Fourier transform provides anisometry of L ( R ) up to a constant such that H = F L (0 , ∞ ) and ¯ H = F L ( −∞ , V t is a closed subspace of L (0 , ∞ ) by definition.Therefore, F ( V t ) is a reproducing kernel Hilbert space, since the point evaluation map F F ( z ) is continuous in F ( V t ) for each z ∈ C + as well as H . Lemma 6.2. We have F ( V ) = K (Θ) .Proof. If f ∈ V , F f and FK f belong to the Hardy space H . On the other hand, wehave ( FK f )( z ) = Θ( z )( F f )( − z ) by Lemma 4.1. This implies ( F f )( z ) = Θ( z )( FK f )( − z )by (1.3). Therefore, F f belongs to K (Θ) by (5.2). Conversely, if F ∈ K (Θ), thereexists f ∈ L (0 , ∞ ) and g ∈ L ( −∞ , 0) such that F ( z ) = ( F f )( z ) = Θ( z )( F g )( z ).We have ( F g )( − z ) = Θ( z )( F f )( − z ) by (1.3) again. Here ( F g )( − z ) = ( F g − )( z ) for g − ( x ) = g ( − x ) ∈ L (0 , ∞ ), and Θ( z )( F f )( − z ) = ( FK f )( z ) as above. Hence K f belongsto L (0 , ∞ ), and thus f ∈ V . (cid:3) Lemma 6.3. We have V t = { } for every t ∈ R . N INVERSE PROBLEM FOR A CLASS OF DIAGONAL HAMILTONIANS 15 Proof. The case of t = 0 is Lemma 6.2. The orthogonal complement V ⊥ t of V t in L ( R )is equal to the closure of L ( −∞ , t ) + K ( L ( −∞ , t )). Therefore, if t < V t contains L ( t, − t ) and thus = { } , since K ( L ( −∞ , t )) ⊂ L ( − t, ∞ ) by (K3). We suppose that t > L ( −∞ , t ) + K ( L ( −∞ , t )) is closed. If w = u + K v for some u, v ∈ L ( −∞ , t ), we have P t w = u + K [ t ] v and P t K w = K [ t ] u + v by K = id. Itis solved as u = (1 − K [ t ] ) − ( P t − K [ t ] P t K ) w and v = (1 − K [ t ] ) − ( P t K − K [ t ] P t ) w ,since both ± K [ t ]. Therefore, if w n = u n + K v n converges in L ( R ), it implies that both u n and v n also converge in L ( R ), and hence the space L ( −∞ , t ) + K ( L ( −∞ , t )) is closed. Hence we obtain V ⊥ t = L ( −∞ , t ) + K ( L ( −∞ , t )) . (6.3)To prove V t = { } , it is sufficient to show that L ( −∞ , t ) + K ( L ( −∞ , t )) is a properclosed subspace of L ( R ). Suppose that f ∈ K ( L ( −∞ , t )). Then the restriction of f to ( t, ∞ ) is continuous on ( t, ∞ ). In fact, if f ( x ) := ( K g )( x ) = R t − x K ( x + y ) g ( y ) dy forsome g ∈ L ( −∞ , t ), the continuity of K implies the continuity of f as | f ( x + δ ) − f ( x ) | = (cid:12)(cid:12)(cid:12)(cid:12)Z t − x − δ K ( x + δ + y ) g ( y ) dy − Z t − x K ( x + y ) g ( y ) dy (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z x − x − δ | K ( x + δ + y ) || g ( y ) | dy + Z t − x | K ( x + δ + y ) − K ( x + y ) || g ( y ) | dy ≤ (cid:18)Z x − x − δ | K ( x + δ + y ) | dy + C x · δ · | t + x | (cid:19) k g k ≤ δ (cid:18) max ≤ y ≤ x | K ( y ) | + C x · δ · | t + x | (cid:19) k g k , where 0 < δ < x and we used the mean value theorem and the Schwartz inequality.Hence, if f ∈ L ( −∞ , t ) + K ( L ( −∞ , t )), f is continuous on ( t, ∞ ). Thus L ( −∞ , t ) + K ( L ( −∞ , t )) is a proper closed subspace of L ( R ). (cid:3) Lemma 6.4. Let z ∈ C + . Then the integral equations a tz + KP t a tz = e z + K e z , (6.4) b tz − KP t b tz = e z − K e z (6.5) have unique solutions, where e z ( x ) = exp( izx ) and P t is the projection to L ( −∞ , t ) .Moreover, for ℑ ( z ) > c , a tz ( t ) = 2 A ( t, z ) E ( z ) , b tz ( t ) = − i B ( t, z ) E ( z ) . (6.6) Proof. Note that K e z makes sense as a function, because ( K e z )( x ) = e − izx Θ( z ) holds if ℑ ( z ) > c by (K1), ( K e z )( x ) = ( K >t e z )( x ) + R t − x K ( x + y ) e izy dy holds if 0 < ℑ ( z ) ≤ c by (K3), and >t e z ∈ L ( R ). First, we show the uniqueness of the solution. Aftersubtracting P t e z + KP t e z (resp. P t e z − KP t e z ) from both sides of (6.4) (resp. (6.5)),multiplying by P t on both sides, we find that equations (6.4) and (6.5) have uniquesolutions a tz = e z + K (1 − P t ) e z − K (1 + K [ t ]) − P t K (1 − P t ) e z ,b tz = e z − K (1 − P t ) e z − K (1 − K [ t ]) − P t K (1 − P t ) e z , respectively. By multiplying both sides of a tz = e z + K e z − KP t a tz and K ( x + t ) − R t −∞ K ( x + y ) φ + ( t, y ) dy = φ + ( t, x ), integrating the obtained equation with respect to x from −∞ to t , and using the symmetry of the kernel K ( x + y ), we obtain Z t −∞ a tz ( x ) K ( t + x ) dx = Z t −∞ φ + ( t, x )( e izx + ( K e z )( x )) dx. Combining this with (6.4), a tz ( t ) = e izt + ( K e z )( t ) − Z t −∞ φ + ( t, x )( e izx + ( K e z )( x )) dx. (6.7)If we suppose that ℑ ( z ) > c , ( K e z )( x ) = e − izx Θ( z ) holds, F ( z ) := F ( φ + ( t, · ))( z ) isdefined by Proposition 2.2 (3), and F ( z ) = 2 A ( t, z ) E ( z ) − e izt + Z t −∞ φ + ( t, x ) e izx dx (6.8)by (6.2). On the other hand, by taking the Fourier transform of both sides of (2.6), F ( z ) + Θ( z ) Z t −∞ φ + ( t, x ) e − izx dx = e − izt Θ( z ) . (6.9)Substituting (6.8) into (6.9), and arranging, we find that the right-hand side of (6.7)equals 2 A ( t, z ) /E ( z ). Hence the formula on the left of (6.6) holds. The the formula onthe right of (6.6) is proved by a similar argument. (cid:3) Integral equations (6.4) and (6.5) are solved as a tz ( x ) = a xz ( x ) + Z xt a sz ( s ) φ + ( s, x ) ds, b tz ( x ) = b xz ( x ) − Z xt b sz ( s ) φ − ( s, x ) ds. In fact, differentiating both sides of (6.4) with respect to t , ∂∂t a tz ( x ) + Z t −∞ K ( x + y ) ∂∂t a tz ( y ) dy = ( − a tz ( t )) K ( x + t ) . This shows that ( − a tz ( t )) − ( ∂/∂t ) a tz ( x ) is a solution of (2.6), hence it equals to φ + ( t, x ).Then, we obtain a tz by integrating ( ∂/∂t ) a tz ( x ) = − a tz ( t ) φ + ( t, x ) with respect to t . Weobtain b tz by a way similar to a tz .6.2. Proof of Theorem 1.5 (1). Let z ∈ C + . Then f F f ( z ) is a continuousfunctional on V t , since the point evaluation F F ( z ) is continuous on H and theFourier transform F is an isometry between L (0 , ∞ ) and H up to a constant. Therefore,there exists an unique vector Y tz ∈ V t such that R ∞ f ( x ) Y tz ( x ) dx = ( F f )( z ) holds for all f ∈ V t by the Riesz representation theorem. Let j ( t ; z, w ) be the reproducing kernel of F ( V t ). Then, j ( t ; z, w ) = 12 π h Y tw , Y tz i , (6.10)where h f, g i = R R f ( u ) g ( u ) du as above. In fact, for F = F f ∈ F ( V t ), h F ( w ) , h Y tw , Y tz ii H = Z R F ( w ) (cid:18)Z R Y t − ¯ w ( x ) Y tz ( x ) dx (cid:19) dw = Z R F ( w ) (cid:18)Z R e − i ¯ wx Y tz ( x ) dx (cid:19) dw = 2 π Z R f ( x ) e − i ¯ wx Y tz ( x ) dx = 2 πF ( z ) . On the other hand, Y tz is the orthogonal projection of ≥ t ( x ) e izx ∈ L ( R ) to V t . By(6.3), there exists unique vectors u tz and v tz in L ( −∞ , t ) such that [ t, ∞ ) e z = Y tz + u tz + K v tz . (6.11)Using the solutions a tz and b tz of (6.4) and (6.5), equation (6.11) is solved as Y tz = (1 − P t ) 12 ( a tz + b tz ) . (6.12)This equality is proved as follows. Put U tz = ( a tz + b tz ) / − e z and V tz = ( a tz − b tz ) / U tz + KP t V tz = 0 and V tz + KP t U tz = K (1 − P t ) e z by (6.4). By multiplying K on N INVERSE PROBLEM FOR A CLASS OF DIAGONAL HAMILTONIANS 17 both sides of the second equation, (1 − P t ) e z = K (1 − P t ) V tz + P t U tz + KP t V tz . Therefore, Y tz = K (1 − P t ) V tz , u tz = P t U tz and v tz = P t V tz . Moreover, Y tz = K (1 − P t ) V tz = K V tz − KP t V tz = ((1 − P t ) e z − P t U tz ) + U tz = (1 − P t )( e z + U tz ) . Hence (6.12) holds. By differentiating both sides of (6.4) with respect to x , we obtain ∂∂x a tz ( x ) − Z t −∞ K ( x + y ) ∂∂y a tz ( y ) dy = − K ( x + t ) a tz ( t ) + iz ( e izx − K e izx )Multiplying (6.5) by iz and then subtracting from this equation, we find that the function( − a tz ( t )) − (( ∂/∂x ) a tz ( x ) − izb tz ( x )) solves (2.7). Therefore, by the uniqueness of thesolution of (2.7), ∂∂x a tz ( x ) − izb tz ( x ) = − a tz ( t ) φ − ( t, x ) . We also obtain ∂∂x b tz ( x ) − izb tz ( x ) = b tz ( t ) φ + ( t, x ) . by a similar argument. Adding these, ∂∂x ( a tz ( x ) + b tz ( x )) − iz ( a tz ( x ) + b tz ( x )) = b tz ( t ) φ + ( t, x ) − a tz ( t ) φ − ( t, x ) . Hence, if ℑ ( z ) > c and ℑ ( w ) > c , − i ( z + w ) Z ∞ Y tz ( x ) e iwx dx = − i ( z + w ) Z ∞ t 12 ( a tz ( x ) + b tz ( x )) e iwx dx = 12 b tz ( t ) (cid:18) e iwt + Z ∞ t φ + ( t, x ) e iwx dx (cid:19) + 12 a tz ( t ) (cid:18) e iwt − Z ∞ t φ − ( t, x ) e iwx dx (cid:19) = − i B ( t, z ) E ( z ) A ( t, w ) E ( w ) − i A ( t, z ) E ( z ) B ( t, w ) E ( w )by (6.2) and (6.6). Therefore,12 π h Y tw , Y tz i = 12 π Z Y tw ( x ) Y tz ( x ) dx = 12 π Z Y tw ( x ) e − i ¯ zx dx = 1 E ( − ¯ z ) E ( w ) B ( t, w ) A ( t, − ¯ z ) + A ( t, w ) B ( t, − ¯ z ) π ( w − ¯ z )= 1 E ( z ) E ( w ) A ( t, z ) B ( t, w ) − A ( t, w ) B ( t, z ) π ( w − ¯ z )by (6.10), (6.1), and Theorem 1.1 (1). Hence we obtain (1.13) by (6.10) under therestrictions ℑ ( z ) > c and ℑ ( w ) > c , but (1.13) holds for all z, w ∈ C + by analyticcontinuation. Hence we complete the proof. (cid:3) Proof of Theorem 1.5 (2). Suppose that j ( t ; z, w ) ≡ t > 0. Then V t = { } , since j ( t ; z, w ) is the reproducing kernel of F ( V t ). This contradicts Lemma6.3, and thus j ( t ; z, w ) t > 0. We have | j ( t ; z, w ) | = 2 π |h Y tw , Y tz i| ≤ k Y tz k · k Y tw k for fixed z, w ∈ C + by (1.13), where k · k = k · k L ( R ) . Therefore, for Theorem 1.5(2), it is sufficient to show that k Y tz k → t → ∞ for a fixed z ∈ C + . We have k Y tz k = h Y tz , Y tz i = ( F Y tz )( z ) = h Y tz , Y z i , since V t ⊂ V . Therefore, h Y tz , Y z i = (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ t Y tz ( x ) Y z ( x ) dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ k Y tz k (cid:18)Z ∞ t | Y z ( x ) | dx (cid:19) / by the Cauchy–Schwarz inequality. Hence, k Y tz k ≤ R ∞ t | Y z ( x ) | dx . This shows that k Y tz k → t → ∞ , since Y z ∈ L (0 , ∞ ). (cid:3) Model subspaces for general t .Theorem 6.1. Let A ( t, z ) and B ( t, z ) be as in Theorem 1.1 and define E ( t, z ) = A ( t, z ) − iB ( t, z ) , Θ( t, z ) = E ( t, ¯ z ) E ( t, z ) (6.13) for t ∈ R . Then Θ( t, z ) is an inner function in C + .Proof. Note that Z ⊂ R , since Θ is an inner function in C + by the assumption. FromTheorem 1.1 (1), we have E ( t, ¯ z ) = A ( t, z ) + iB ( t, z ). Therefore, | Θ( t, z ) | = 1 for almostall z ∈ R . Suppose that t ≥ J ( t ; z, w ) = E ( z ) E ( w ) j ( t ; z, w ). Then, (1.13) istransformed as J ( t ; z, w ) = E ( t, z ) E ( t, w ) − E ( t, ¯ z ) E ( t, ¯ w )2 πi (¯ z − w ) (6.14)by using (6.13) and E ( t, ¯ z ) = A ( t, z ) + iB ( t, z ). On the other hand, we have J ( t ; z, w ) − J ( s ; z, w ) = 1 π Z st A ( t, z ) A ( t, w ) 1 γ ( t ) dt + 1 π Z st B ( t, z ) B ( t, w ) γ ( t ) dt (6.15)for any t < s < ∞ and z, w ∈ C + in a way similar to the proof of Lemma 2.1 of [7].Taking w = z ∈ C + in (6.15) and then tending s to ∞ , we have J ( t ; z, z ) = 1 π Z ∞ t | A ( t, z ) | γ ( t ) dt + 1 π Z ∞ t | B ( t, z ) | γ ( t ) dt (6.16)by Theorem 1.5 (2), where γ ( t ) = m ( t ) > 0. This shows that | Θ( t, z ) | < z ∈ C + , since the left-hand side equals to ( | E ( t, z ) | − | E ( t, ¯ z ) | ) / (2 π ℑ ( z )). HenceΘ( t, z ) is an inner function in C + . For t ≤ 0, we have E ( t, z ) = E ( z ) e izt , Θ( t, z ) = Θ( z ) e − izt (6.17)by (3.4). Hence Θ( t, z ) is an inner function in C + , since the factors Θ( z ) and e − izt areboth inner functions in C + . (cid:3) Equality (6.16) also shows the following. Corollary 6.1. A ( t, z ) and B ( t, z ) are square integrable at t = ∞ for each z ∈ C + . Recall that R ∞ α γ ( t ) − dt < ∞ is a part of the sufficient condition in [2, Theorem 41]for the existence of the solution of the canonical system for H ( t ) = diag(1 /γ ( t ) , γ ( t ))on [ α, ∞ ). We find that it is necessary if E (0) = A (0) = 0. In fact, we have J ( t ; 0 , 0) = π − A (0) R ∞ t γ ( t ) − dt by taking z = 0 in (6.16). Therefore, γ ( t ) − is L at t = ∞ .By comparing the kernels of F ( V t ) and K (Θ( t, z )), we obtain the following relation. Corollary 6.2. For t ≥ , F ( V t ) = E ( t, z ) E ( z ) K (Θ( t, z )) . (6.18)If t < 0, the space F ( V t ) is no longer a subspace of H = F ( L (0 , ∞ )), since V t contains L ( t, − t ) as found in the proof of Lemma 6.3, in particular, F ( V t ) can not be a modelsubspace, but the right-hand side of (6.18) can be extended to negative t ’s. We foundabove that Θ( t, z ) is an inner function in C + . The kernel of K (Θ( t, z )) is J ( t ; z, w ) = A ( z ) B ( w ) − A ( w ) B ( z ) π ( w − ¯ z ) cos( t ( w − ¯ z )) − ( A ( z ) A ( w ) + B ( z ) B ( w )) sin( t ( w − ¯ z )) π ( w − ¯ z ) N INVERSE PROBLEM FOR A CLASS OF DIAGONAL HAMILTONIANS 19 by (6.17). If θ and θ are inner functions in C + , K ( θ θ ) = K ( θ ) ⊕ θ K ( θ ) by [10,Lemma 2.5]. Hence, E ( t, z ) E ( z ) K (Θ( t, z )) = e itz ( K (Θ) ⊕ ΘPW − t ) , where PW a = K ( e iaz ) ( a > 0) is the Paley–Wiener space, which consists of the entirefunctions of exponential type at most a the restrictions of which to the real line R arein L ( R ). Therefore, F ( V t ) is a shift of a model subspace.7. Related differential equations From Theorem 1.4 and the second equation of (2.10) and (2.11), we find that Φ( t, x )and Ψ( t, x ) are characterized as the unique solution of the Cauchy problem: Φ t ( t, x ) + γ ( t )Ψ x ( t, x ) = 0 , Ψ t ( t, x ) + γ ( t ) − Φ x ( t, x ) = 0 ,γ ( t ) = Ψ( t, t ) / Φ( t, t ) (= Φ( t, t ) − = Ψ( t, t ) ) , Φ(0 , x ) = 1 − Z x K ( y ) dy, Ψ(0 , x ) = 1 + Z x K ( y ) dy (7.1)for ( t, x ) ∈ [0 , τ ) × R . In this formulation, (K5) is understood as a statement about theexistence of a global solution. We should remark that γ ( t ) is not a given function in(7.1) different from usual Cauchy problem for hyperbolic first-order systems. It wouldbe interesting to study the inverse problem for Hamiltonian systems in terms of this(unusual) Cauchy problem.On the other hand, if we note that Φ( t, x ) and Ψ( t, x ) have the second derivativefor both variables by Propositions 2.2 and 2.3, we find that they satisfy the followingdamped wave equations (or wave equations with time-dependent dissipation) ( Φ tt ( t, x ) − Φ xx ( t, x ) − µ ( t )Φ t ( t, x ) = 0 , Ψ tt ( t, x ) − Ψ xx ( t, x ) + 2 µ ( t )Ψ t ( t, x ) = 0by the first line of (7.1), where 2 µ ( t ) = γ ( t ) ′ /γ ( t ). Moreover, definition (1.11) derivesthe Schr¨odinger equations ( A tt ( t, z ) + z A ( t, z ) − µ ( t ) A t ( t, z ) = 0 ,B tt ( t, z ) + z B ( t, z ) + 2 µ ( t ) B t ( t, z ) = 0 . These are of course directly proved by Theorem 1.1 (3). Taking z = 0 in Theorem1.1 (3), we have A ( t, 0) = A (0) = E (0) and B ( t, 0) = 0 for each t < τ . Therefore, if E (0) = A (0) = 0 and (K5) holds for τ = ∞ , j ( t ; 0 , z ) = ( πzE ( z )) − B ( t, z ) → t → ∞ by Theorem 1.5 (2). Hence B ( t, z ) → t → ∞ , and Φ( t, x ) → t → ∞ . This factwould be interesting if we recall the following J. Wirth’s results. He studied the Cauchyproblem for a damped wave equation u tt − u xx + bu t = 0, u (0 , · ) = u , u t (0 , · ) = u withpositive time-depending dissipation b = b ( t ). He showed that if tb ( t ) → ∞ as t → ∞ ,1 /b ∈ L (0 , ∞ ), u ∈ W s, , and u ∈ W s − , , the solution u ( t, x ) tends in W s, to a realanalytic function u ( ∞ , x ) = lim t →∞ u ( t, x ), which is nonzero except at most one u foreach u , where W s, is the Sobolev space on (0 , ∞ ) ([21, p. 76, Result 3]).From the result of Wirth, it is naturally asked whether A ( t, z ) and Ψ( t, z ) tend tofunctions as t → ∞ . This problem is also interesting from the view point of the so-calledconnection formula for solutions of canonical systems: (cid:20) A ( t, z ) B ( t, z ) (cid:21) = M ( t, s ; z ) (cid:20) A ( s, z ) B ( s, z ) (cid:21) , but nothing is known about the behavior of A ( t, z ) or Φ( t, z ) when t → ∞ . References [1] A. Baranov, Isometric embeddings of the spaces K Θ in the upper half-plane, . Math. Sci. (NewYork) (2001), no. 5, 2319–2329, Function theory and partial differential equations.[2] L. de Branges, Hilbert spaces of entire functions, Prentice-Hall, Inc., Englewood Cliffs, N.J. C. R. Math. Acad. Sci. Paris (2002), no. 8, 689–692.[4] , On Fourier and zeta(s), Forum Math. (2004), no. 6, 789–840.[5] , Entrelacement de co-Poisson, Ann. Inst. Fourier (Grenoble) (2007), no. 2, 525–602.[6] , Scattering, determinants, hyperfunctions in relation to Γ(1 − s ) / Γ( s ), http://arxiv.org/abs/math/0602425 ,[7] H. 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