The inverse problem for a spectral asymmetry function of the Schrödinger operator on a finite interval
B. Malcolm Brown, Karl Michael Schmidt, Stephen P. Shipman, Ian Wood
aa r X i v : . [ m a t h . SP ] S e p The inverse problem for a spectral asymmetry functionof the Schr¨odinger operator on a finite interval
B. Malcolm Brown and Karl Michael Schmidt and Stephen P. Shipman and Ian Wood , Cardiff University and Louisiana State University and University of Kent
Abstract.
For the Schr¨odinger equation − d u/dx + q ( x ) u = λu on a finite x -interval, there isdefined an “asymmetry function” a ( λ ; q ), which is entire of order 1 / λ . Our mainresult identifies the classes of square-integrable potentials q ( x ) that possess a common asymmetryfunction. For any given a ( λ ), there is one potential for each Dirichlet spectral sequence. Key words: spectral theory; Schr¨odinger operator; inverse spectral problem; entire function; asymmetryfunction
MSC:
Consider the spectral Schr¨odinger equation − d u/dx + q ( x ) u = λu on the x -interval [0 ,
1] with q real valuedand square-integrable. Let c ( x, λ ; q ) and s ( x, λ ; q ) be a pair of fundamental solutions satisfying c (0 , λ ; q ) = 1 , s (0 , λ ; q ) = 0 ,c ′ (0 , λ ; q ) = 0 , s ′ (0 , λ ; q ) = 1 , in which the prime denotes differentiation with respect to x . In this article, we are interested in the followingentire spectral function associated with q : a ( λ ; q ) := (cid:0) c (1 , λ ; q ) − s ′ (1 , λ ; q ) (cid:1) . This is called the spectral asymmetry function for the potential q ( x ), or simply its asymmetry function.When the potential is understood, we may suppress writing the dependence of these functions on q .The connection of the asymmetry function to asymmetry of the potential is seen as follows. Define˜ q ( x ) = q (1 − x ); then, by considering the transfer function taking Cauchy data at 0 to Cauchy data at 1, wesee that c (1 , λ ; ˜ q ) = s ′ (1 , λ ; q ), so that a ( λ ; q ) := (cid:0) c (1 , λ ; q ) − c (1 , λ ; ˜ q ) (cid:1) . This shows that the asymmetry function vanishes identically as a function of λ whenever q is symmetricabout the midpoint of [0 , a ( λ ; q ) vanishes identically if and only if q is symmetric. An abbreviatedproof of this appears in [14, Lemma 4]; a more detailed proof is given in [11, Theorem 2], along with severalother properties of the asymmetry function. This motivates the idea of the asymmetry class of potentials associated to a given asymmetry function a ( λ )—it is the set of all potentials q that possess a ( λ ) as itsasymmetry function, that is, all q such that a ( λ ; q ) = a ( λ ). School of Computer Science and Informatics, Cardiff University, BrownBM@cardiff.ac.uk School of Mathematics, Cardiff University, SchmidtKM@cardiff.ac.uk Department of Mathematics, Louisiana State University @ Baton Rouge, [email protected], orcid School of Mathematics, Statistics and Actuarial Science, University of Kent, [email protected] N ( λ ; q ) = 1 s (1 , λ ; q ) " − c (1 , λ ; q ) 11 − s ′ (1 , λ ; q ) , which maps Dirichlet data of a solution of − d u/dx + q ( x ) u = λu on [0 ,
1] to the Neumann data of thesolution: N ( λ )[ u (0) , u (1)] t = [ u ′ (0) , − u ′ (1)] t . It is a meromorphic function with poles at the roots of s ( λ ; q ),which are the Dirichlet eigenvalues of the Schr¨odinger operator − d /dx + q ( x ) on [0 , b ( λ ) = (cid:0) c (1 , λ ; q ) + c (1 , λ ; ˜ q ) (cid:1) coincide with the functions u − ( λ ) and u + ( λ ) in [6, p. 494; Lemma 4.1], where the authors characterize the spectra of Hill’s operator( − d /dx + q ( x ) with periodic potential) as certain admissible sequences of intervals on the line. Thefunctions a and b are also identical to δ and ∆, respectively, in [15, p. 2].The purpose of the present work is to characterize the set of all asymmetry functions and, given a fixedasymmetry function a , to identify its asymmetry class of potentials, which, as defined above, is the set of allpotentials q that possess that a as its asymmetry function. The main result is Theorem 8 in section 4.1, whichstates that, for each asymmetry function a in a certain Hilbert space of entire functions, the asymmetry classof a contains one potential for each admissible sequence of Dirichlet eigenvalues.The analysis in this paper draws primarily upon deep work in spectral theory of the Schr¨odinger operatoron the interval by E. Trubowitz, H. P. McKean, and J. P¨oschel.In what follows, to avoid cumbersome notation, we will often suppress the dependence of functions onthe potential q . The following function spaces will be used in this article. L [0 ,
1] = n q : [0 , → R | Z q ( x ) dx < ∞ o ℓ ( Z ) = n α : Z → R | ∞ X n = −∞ α n < ∞ o S = n α ∈ ℓ | ( π n + α n ) n ∈ N is strictly increasing o ℓ ( N ) = n α : N → R | ∞ X n =1 n α n < ∞ o E = n φ : C → C | φ entire, order ≤
1, type ≤ φ ( R ) ⊂ R , Z R | φ ( λ ) | dλ < ∞ o E = n φ : C → C | φ entire, order ≤ /
2, type ≤ φ ( R ) ⊂ R , Z ∞ | φ ( λ ) | λ dλ < ∞ o The property of the asymmetry function most relevant to this paper is the fact that it is in the class E andthat its evaluation on a Dirichlet spectral sequence is in ℓ . Proposition 1.
Let q ∈ L and a sequence ( µ n ) n ∈ N , µ n = n π + c n , with ( c n ) n ∈ N a real-valued boundedsequence, be given. The asymmetry function a ( λ ; q ) lies in the class E , and the sequence ( a ( µ n ; q )) n ∈ N liesin the class ℓ .Proof. Since q is real-valued, both c ( x, λ ; q ) and s ( x, λ ; q ) are real-valued for real λ , and thus a is real-valuedon the real line. The function c (1 , λ ; q ) is an entire function of λ of order 1 / c (1 , λ ) = cos √ λ + λ – 12 c ( λ ) + R ( λ ) , (2.1)2here c ( λ ) = sin √ λ Z q ( x ) dx + Z sin (cid:0) √ λ (1 − x ) (cid:1) q ( x ) dx (2.2)and R is an entire function such that z e −| Im √ z | R ( z ) ( z ∈ C ) is bounded (cf. [2, (1.1.15)]; [9, pp. 14–15]).The function c (˜ q ; 1 , λ ) can be written identically to (2.1) except with q ( x ) replaced with q (1 − x ) in theexpression of c ( λ ). Thus we obtain a ( λ ) = 12 √ λ Z sin (cid:0) √ λ (1 − x ) (cid:1) q o ( x ) dx + ˜ R ( λ )= − √ λ Z − sin( √ λ y ) q o (cid:0) y +12 (cid:1) dy + ˜ R ( λ ) , (2.3)in which q o ( x ) = ( q ( x ) − q (1 − x )) / q , and ˜ R is an entire function such that z e −| Im √ z | ˜ R ( z )( z ∈ C ) is bounded. If we abbreviate f ( z ) = − Z − sin( zy ) q o (cid:0) y +12 (cid:1) dy ( z ∈ C ) , then by the Paley-Wiener theorem, f is an entire function of exponential type 1 that is square-integrableover parallels to the real axis. Hence, setting ˜ a ( λ ) = f ( √ λ ) / √ λ , we have Z ∞ | ˜ a ( λ ) | √ λ dλ = Z ∞−∞ | ˜ a ( z ) z | dz < ∞ . Thus ˜ a ∈ E . Moreover, ˜ R ∈ E , and as E is a vector space, it follows that a = ˜ a + ˜ R ∈ E .For the given sequence µ n , √ µ n = nπ + O ( n − ) and √ µ n − = ( nπ ) − + O ( n − ); and evaluation of (2.3)at λ = µ n yields a ( µ n ) = − nπ Z − sin( nπy ) q o (cid:0) y +12 (cid:1) dy + O ( n − ) . (2.4)The integrals in (2.4) form the sequence of (sine) Fourier coefficients of a square-integrable function, whichtherefore is a square-summable sequence; this places the sequence ( a ( µ n )) n ∈ N in ℓ .The asymmetry function has the following additional properties, which are proved in [11, Theorem 2].1. The potential q is symmetric if and only if a ( λ ; q ) ≡ N ( λ ; q ) and N ( λ ; q ) commute if and only if a ( λ ; q ) = a ( λ ; q ).3. The Dirichlet spectrum of − d /dx + q ( x ), together with a ( λ ; q ), determine q ∈ L [0 ,
1] uniquely.4. c ′ (1 , λ ; q ) a ( λ ; q ) = − Z q o ( x ) c ( x, λ ; q ) c ( x, λ ; ˜ q ) dx , where q o ( x ) = (cid:0) q ( x ) − q (1 − x ) (cid:1) .Property (3) will emerge as a result of the main Theorem 8. The question of whether a complex function from a certain class can be uniquely determined from its valueson a given discrete set of points is fundamental in interpolation and sampling theory, going all the way backto Shannon’s famous sampling theorem [10] based on work of Whittaker [12]. For functions sampled on thewhole complex plane, there are many results relating the growth of the function to the required density ofthe point set, see, e.g. [5, 7]. For functions sampled only at points on the real line, the classical result on therequired density of the discrete set is [3, Theorem 3.3]. In our case, we will be sampling functions in the class E on the Dirichlet spectrum of a Sturm-Liouville operator. This set does not satisfy the density assumptionsof any of the results mentioned above. However, the class E involves an additional decay assumption on3he entire functions being sampled which will enable us to prove the desired result. Theorem 6 is the keyresult in this section, as it tells us that the asymmetry function is completely determined by its values atthe Dirichlet spectrum of a potential q ∈ L [0 , § e ( ω ) := s ′ (1 , ω ) − iωs (1 , ω ) (3.5)is a de Branges function since it satisfies the inequality in the following lemma. See [1, Ch 2 §
19] for thetheory of de Branges spaces.
Lemma 2.
For all ω ∈ R , | e ( ω ) | = 1 + o (1) ( | ω | → ∞ ) , (3.6) and if e ( ω ) = 0 , then ω = 0 and s ′ (1 ,
0) = 0 .If the least Dirichlet eigenvalue µ of − d /dx + q ( x ) on [0 , is positive, then for all ω ∈ C with Im ω > , | e ( ω ) | > | e (¯ ω ) | . (3.7) Proof.
According to the estimates in [4, § § ω ∈ R , | e ( ω ) | = s ′ (1 , ω ) + ω s (1 , ω ) = 1 + o (1) ( | ω | → ∞ ) . (3.8)Since s (1 , λ ) and s ′ (1 , λ ) are real when λ is real, e ( ω ) = 0 for ω ∈ R implies that s ′ (1 , ω ) = ωs (1 , ω ) = 0.Since s and s ′ cannot simultaneously vanish, ω = 0, and thus s ′ (1 ,
0) = 0.The proof of the inequality is a modification of the proof of [8, § s ( x, λ ) is real for λ ∈ R , we have s ( x, ¯ λ ) = s ( x, λ ). For Im ω > | e ( ω ) | − | e (¯ ω ) | = 4 Im (cid:0) ω s (1 , ω ) s ′ (1 , ¯ ω ) (cid:1) (3.9)= 4 Im ω Z [ s ( x, ω ) s ′ ( x, ¯ ω ) ] ′ dx (3.10) > ω Z ( | s ′ ( x, ω ) | + q ( x ) | s ( x, ω ) | ) dx (3.11) ≥ (4 Im ω ) µ (cid:13)(cid:13) s ( · , ω ) (cid:13)(cid:13) L . (3.12)The last inequality comes from the Rayleigh quotient inequality for the quadratic form of the Dirichletoperator − d /dx + q ( x ).To the function e ( ω ) is associated a de Branges space B = B q , which is a reproducing-kernel Hilbertspace. It consists of all entire functions f such that Z R (cid:12)(cid:12)(cid:12)(cid:12) f ( λ ) e ( λ ) (cid:12)(cid:12)(cid:12)(cid:12) dλ < ∞ (3.13)and there exists a real number C f such that, for all ω with Im ω > (cid:12)(cid:12)(cid:12)(cid:12) f ( ω ) e ( ω ) (cid:12)(cid:12)(cid:12)(cid:12) , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f (¯ ω ) e ( ω ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < C f √ Im ω . (3.14)The inner product B [ · , · ] = B q [ · , · ] in B is that of L ( R , dλ/ | e ( λ ) | ); that is, through restriction of functionsin B to R , B is identified with a closed subspace of this weighted L space. The reproducing kernel is1 α ( β ) := e ( α ) e ( β ) − e ( α ) e ( β )2 i ( α − β ) = α s (1 , α ) s ′ (1 , β ) − β s ′ (1 , α ) s (1 , β ) α − β . (3.15)4he singularity at β = ¯ α is removable; indeed, 1 α is entire. Therefore, for all f ∈ B and all ω ∈ C , f ( ω ) = B [ f, ω ] = Z R f ( β )1 ω ( β ) dβ | e ( β ) | . (3.16)The function 1 α ( β ) is K ( w, z ) in [1, §
19 Theorem 19].We assume now, by adding a suitable constant to the potential, that s ′ (1 , = 0 , s (1 , = 0 . (3.17)Let ( µ k ) k ∈ N denote the sequence of Dirichlet eigenvalues associated to the operator − d /dx + q ( x ) on [0 , ω ( ±√ µ j ) = ω s (1 , ω ) s ′ (1 , µ j ) ω ∓ √ µ j , (3.18)so if β = ±√ µ j and ω ∈ {±√ µ k : k ∈ N } \ { β } , then 1 ω ( β ) = 0; and, as ω → ±√ µ j ,1 ω ( ±√ µ j ) → ±√ µ j s ′ (1 , µ j ) ∂∂ω s (1 , ω ) (cid:12)(cid:12)(cid:12) ω = ±√ µ j = 2 µ j s ′ (1 , µ j ) ∂∂λ s (1 , µ j ) = 2 µ j l j , (3.19)where we have used s ′ (1 , µ j ) ∂∂λ s (1 , µ j ) = l j [9, § l j = Z s ( x, µ j ) dx ( j ∈ N )being the Dirichlet-Dirichlet norming constants. For suitable functions f and g , define A [ f, g ] = A q [ f, g ] := f (0) g (0) s ′ (1 , s (1 ,
0) + ∞ X j =1 µ j l j (cid:0) f ( −√ µ j ) g ( −√ µ j ) + f ( √ µ j ) g ( √ µ j ) (cid:1) . Define the space A = A q to consist of all functions f defined on { } ∪ {±√ µ j } j ∈ N such that A [ f, f ] < ∞ . (3.20)The following two lemmas comprise an analog of [8, § Lemma 3.
Let the potential q ∈ L [0 , satisfy µ > , s ′ (1 , = 0 , and s (1 , = 0 . Then for all α and β in C , A q [1 α , β ] = B q [1 α , β ] . (3.21) Proof.
From (3.18) it follows for α, β = ±√ µ j that1 α ( ±√ µ j )1 β ( ±√ µ j ) = αs (1 , α ) s ′ (1 , µ j ) α ∓ √ µ j βs (1 , β ) s ′ (1 , µ j ) β ∓ √ µ j = αβ s (1 , α ) s (1 , β ) s ′ (1 , µ j ) ( ±√ µ j − α )( ±√ µ j − β ) , and further 1 α ( √ µ j )1 β ( √ µ j ) + 1 α ( −√ µ j )1 β ( −√ µ j )= αβs (1 , α ) s (1 , β ) s ′ (1 , µ j ) (cid:18) √ µ j − α )( √ µ j − β ) + 1( −√ µ j − α )( −√ µ j − β ) (cid:19) = 2 αβ s (1 , α ) s (1 , β ) s ′ (1 , µ j ) µ j + αβ ( µ j − α )( µ j − β ) . α (0) 1 β (0) = s (1 , α ) s (1 , β ) s ′ (1 , . This yields, by partial fraction decomposition, A [1 α , β ] = s ′ (1 , s (1 , s (1 , α ) s (1 , β ) + ∞ X j =1 αβl j s (1 , α ) s (1 , β ) s ′ (1 , µ j ) µ j + αβµ j ( µ j − α ) ( µ j − β )= s ′ (1 , s (1 , s (1 , α ) s (1 , β ) ++ αβ s (1 , α ) s (1 , β ) ∞ X j =1 s ′ (1 , µ j ) l j (cid:18) αβµ j + 1 α ( α − β )( µ j − α ) + 1 β ( β − α )( µ j − β ) (cid:19) . Now we consider the resolvent kernel for the Dirichlet boundary value problem, R ( x, y ; ω ) = 1 W (cid:26) u ( x, ω ) v ( y, ω ) if y ≤ xu ( y, ω ) v ( x, ω ) if x ≤ y, where v is a solution that satisfies the boundary condition at 0, u is a solution that satisfies the boundarycondition at 1 and W is the Wronskian of u, v. Here we can take v = s and u = s (1 , ω ) c − c (1 , ω ) s. Then W = (cid:12)(cid:12)(cid:12)(cid:12) s (1 , ω ) c − c (1 , ω ) s ss (1 , ω ) c ′ − c (1 , ω ) s ′ s ′ (cid:12)(cid:12)(cid:12)(cid:12) = s (1 , ω ) , so R ( x, y ; ω ) = 1 s (1 , ω ) (cid:26) ( s (1 , ω ) c ( x, ω ) − c (1 , ω ) s ( x, ω )) s ( y, ω ) if y ≤ x ( s (1 , ω ) c ( y, ω ) − c (1 , ω ) s ( y, ω )) s ( x, ω ) if x ≤ y. Now R may not be differentiable on the diagonal, but off the diagonal we find ∂ ∂x∂y R ( x, y ; ω ) = 1 s (1 , ω ) (cid:26) ( s (1 , ω ) c ′ ( x, ω ) − c (1 , ω ) s ′ ( x, ω )) s ′ ( y, ω ) if y < x ( s (1 , ω ) c ′ ( y, ω ) − c (1 , ω ) s ′ ( y, ω )) s ′ ( x, ω ) if x < y, which extends continuously to the diagonal, giving in particular ∂ ∂x∂y R (1 , ω ) = 1 s (1 , ω ) (cid:0) s (1 , ω ) c ′ (1 , ω ) − c (1 , ω ) s ′ (1 , ω ) (cid:1) s ′ (1 , ω ) = − s ′ (1 , ω ) s (1 , ω ) , using the Wronskian of c and s .On the other hand, the resolvent kernel can be expressed in terms of the normalized eigenfunctions as R ( x, y ; ω ) = ∞ X j =0 s ( x, µ j ) s ( y, µ j )( µ j − ω ) l j , and differentiating with respect to x and y here and equating the two formulae for the resolvent kernel, wededuce ∞ X j =0 s ′ (1 , µ j ) ( µ j − ω ) l j = − s ′ (1 , ω ) s (1 , ω ) . A [1 α , β ] = s (1 , α ) s (1 , β ) ∞ X j =1 s ′ (1 , µ j ) l j (cid:18) µ j + β ( α − β )( µ j − α ) + α ( β − α )( µ j − β ) (cid:19) ++ s ′ (1 , s (1 , s (1 , α ) s (1 , β )= s (1 , α ) s (1 , β ) (cid:18) − s ′ (1 , s (1 , − βα − β s ′ (1 , α ) s (1 , α ) − αβ − α s ′ (1 , β ) s (1 , β ) (cid:19) ++ s ′ (1 , s (1 , s (1 , α ) s (1 , β )= − βα − β s (1 , β ) s ′ (1 , α ) − αβ − α s (1 , α ) s ′ (1 , β ) (3.22)= 1 α ( β ) = B [1 α , β ] , Since A [1 α , β ] and B [1 α , β ] are analytic in α and β , the equality can be extended to α, β = ±√ µ j . Thiscompletes the proof. Lemma 4.
Let the potential q ∈ L [0 , satisfy µ > , s ′ (1 , = 0 , and s (1 , = 0 . The Hilbert spaces A q and B q are isomorphic through the maps B → A :: f f (cid:12)(cid:12) { }∪{±√ µ j } j ∈ N (3.23) and A → B :: g f, f ( ω ) = A [ g, ω ] . (3.24) Proof.
We follow the proof of [8, Lemma 2]. Considering Lemma 3, we only have to show that the functions1 ω span B and their restrictions span A in the sense of their linear algebraic spans being dense. To see thatthe functions 1 ω ( ω ∈ C ) span B observe that, for all f ∈ B , B [ f, ω ] = 0 for all ω implies that f = 0 in B because of the reproducing-kernel property (3.16) of 1 ω . For A , the computations after equation (3.18) showthat the restrictions of the functions 1 ω to the set { } ∪ {√ µ j } j ∈ N for ω ∈ { } ∪ {√ µ j } j ∈ N span A . Lemma 5.
Let the potential q ∈ L [0 , satisfy µ > , s ′ (1 , = 0 , and s (1 , = 0 . The space A q withinner product A [ f, g ] is a Hilbert space that contains the same functions as the classical space ℓ ( Z ) ; and thenorms of A and ℓ are equivalent. The Hilbert space B q contains the same functions as E ; and the norms of B and L ( R ) are equivalent.Proof. Using the estimates in [4, § § l j = Z s ( x, µ j ) = Z sin ( √ µ j x ) µ j dx + O ( µ − / j ) = 12 µ j + O ( µ − / j ) (3.25)as j → ∞ . Therefore 0 < µ j l j = 2 + O ( µ − / j ) , and the first statement follows.We now turn to the claim about B and consider1 ω ( z ) = e ( ω ) e ( z ) − e (¯ ω ) e (¯ z ) − i ( z − ¯ ω ) . The singularity at z = ¯ ω is removable, so 1 ω is entire. In view of the definition (3.5) of e ( z ), since zs (1 , z )and s ′ (1 , z ) are of order 1 and type 1, 1 ω is of order 1 and type 1. Moreover, for z ∈ R , the functions7 s (1 , z ) and s ′ (1 , z ) are bounded, so 1 ω ( z ) ≃ ( z − ¯ ω ) − ∈ L . Therefore, 1 ω ∈ E . By the proof of [8,Lemma 2], the functions 1 ω span B and since E is closed, we have B = span { ω } ⊆ E . We next show that
E ⊆ B . Lemma 2 shows that the L -norm is equivalent to B [ f, f ]. As f ∈ E ⊆ L ,we have that B [ f, f ] < ∞ . For f to lie in the de Branges space, it remains to show that for ω ∈ C + we have | f ( ω ) | | e ( ω ) | ≤ C Im ω for some constant C > f ∈ E , by the Paley-Wiener Theorem, ˆ f is supported in [ − , f ( ω ) = 1 √ π Z − e − iωx ˆ f ( x ) dx. Thus, setting Im ω = b , by Cauchy-Schwarz, | f ( ω ) | ≤ π Z − | ˆ f ( x ) | dx · Z − e bx dx = O (cid:18) sinh 2 bb (cid:19) . (3.26)By [9, Theorem 3], s (1 , ω ) = sin( ω ) ω + O (cid:18) e b | ω | (cid:19) ,ωs (1 , ω ) = sin( ω ) + O (cid:18) e b | ω | (cid:19) ,s ′ (1 , ω ) = cos( ω ) + O (cid:18) e b | ω | (cid:19) ,e ( ω ) = cos( ω ) − i sin( ω ) + O (cid:18) e b | ω | (cid:19) = e − iω (cid:18) O (cid:18) | ω | (cid:19)(cid:19) . Therefore, e ( ω ) − = e iω (cid:18) O (cid:18) | ω | (cid:19)(cid:19) , so | e ( ω ) | − = e − b (cid:18) O (cid:18) | ω | (cid:19)(cid:19) , and combined with (3.26), we have that (cid:12)(cid:12)(cid:12)(cid:12) f ( ω ) e ( ω ) (cid:12)(cid:12)(cid:12)(cid:12) = O (cid:18) sinh 2 bb e − b (cid:19) = O (cid:18) b (cid:19) . Thus, the required estimate is satisfied away from the real axis. By Lemma 2, 1 / | e ( ω ) | is bounded on thereal axis, and since f is square integrable, this implies that the estimate holds on C + . Theorem 6.
Let q ∈ L [0 , be given, and let ( µ j ) j ∈ N be the Dirichlet spectral sequence of − d /dx + q ( x ) on [0 , . The restriction of functions in E to ( µ j ) j ∈ N is a bounded linear bijection from E and ℓ ( N ) . Theinverse is effected through the following interpolation formula: For λ ∈ C , φ ( λ ) = ∞ X j =1 φ ( µ j ) Y i = j µ i − λµ i − µ j . (3.27)8 roof. Since the space E is invariant under shifts φ ( λ ) φ ( λ + c ), with c ∈ R , it suffices to prove thetheorem for potentials q such that µ > s ′ (1 , = 0, and s (1 , = 0, as this can be accomplished byshifting the potential by a constant. Let q satisfy these three spectral properties for the rest of the proof.The transformation φ f, f ( ω ) = − ωφ ( ω ) (3.28)is a bijection from the space of entire functions of order ≤ / ≤ ≤ ≤
1. It maps E onto the space of odd functions in E because Z R | f ( ω ) | dω = 2 Z ∞ | ωφ ( ω ) | dω = Z ∞ | φ ( λ ) | λ / dλ. (3.29)Now consider ( a j ) j ∈ Z ( − µ − / j a j ) j ∈ N (3.30)where a − j = − a j ; this is an isomorphism from the odd subspace of ℓ ( Z ) to ℓ ( N ) because √ µ j ∼ πj as j → ∞ . Therefore, we have the following succession of bounded linear bijections: From E to the oddfunctions in E given by (3.28), then to odd functions in B by Lemma 5, to odd sequences in A by Lemma4, to odd sequences in ℓ ( Z ) by Lemma 5, and finally to ℓ ( N ) by (3.30). This amounts to restriction offunctions in E to ( µ j ) j ∈ N , as φ
7→ − ωφ ( ω )
7→ − ωφ ( ω ) (cid:0) √ µ j φ ( µ j ) , −√ µ j φ ( µ j ) (cid:1) (cid:0) √ µ j φ ( µ j ) , −√ µ j φ ( µ j ) (cid:1) ( φ ( µ j )) , proving the first statement of the theorem.To prove the interpolation, use Lemma 3, noting that f (0) = 0, for ω ∈ R , − ωφ ( ω ) = f ( ω ) = B [ f, ω ] = A [ f, ω ] (3.31)= ∞ X j =1 µ j ℓ j (cid:16) f ( −√ µ j )1 ω ( −√ µ j ) + f ( √ µ j )1 ω ( √ µ j ) (cid:17) (3.32)= ∞ X j =1 µ j ℓ j (cid:18) f ( −√ µ j ) ω s (1 , ω ) s ′ (1 , µ j ) ω + √ µ j + f ( √ µ j ) ω s (1 , ω ) s ′ (1 , µ j ) ω − √ µ j (cid:19) (3.33)= ∞ X j =1 µ j ℓ j √ µ j φ ( µ j ) (cid:0) ωs (1 , ω ) s ′ (1 , µ j ) (cid:1) (cid:18) ω + √ µ j − ω − √ µ j (cid:19) (3.34)= ∞ X j =1 ℓ j φ ( µ j ) µ j − ω (cid:0) ω s (1 , ω ) s ′ (1 , µ j ) (cid:1) (3.35)= ω ∞ X j =1 φ ( µ j ) s (1 , ω )˙ s (1 , µ j )( µ j − ω ) (3.36)= − ω ∞ X j =1 φ ( µ j ) Y i = j µ i − ω µ i − µ j . (3.37)We have used the identity ℓ j = s ′ (1 , µ j ) ˙ s (1 , µ j ) , (3.38)where ˙ s = ds/dλ , and the representation of the entire function s (1 , λ ) of order 1 / φ ( λ ) in the theorem follows. The goals of this section are (1) to establish a bijection between square integrable potentials q and pairs(( µ n ) n ∈ N , a ) of spectral sequences and asymmetry functions; and (2) to establish the analyticity of thiscorrespondence. The first part rides on an inverse spectral theory of P¨oschel and Trubowitz [9] for theDirichlet Schr¨odinger operator on an interval. 9 .1 Bijective correspondence Let µ ( q ) = ( µ n ( q )) n ∈ N be the Dirichlet eigenvalue sequence for a potential q ∈ L [0 , µ on L consists of all real strictly increasing sequences ( µ n ) n ∈ N of the form µ n = π n + c + σ n , (4.39)in which c ∈ R and ( σ n ) n ∈ N is in ℓ . The set of sequences ( σ n ) n ∈ N such that (4.39) is strictly increasing isan open set in ℓ , and it is denoted by S . The spectral data introduced in [9] are the constants κ n ( q ) := log | s ′ ( µ n ( q ) , q ) | = log(( − n s ′ ( µ n ( q ) , q )) . (4.40)The sequence κ ( q ) = ( κ n ( q )) n ∈ N lies in the space ℓ . It is proved in [9] that the correspondence L → R × S × ℓ :: q ( x ) (cid:0) c, ( σ n ) n ∈ N , ( κ n ) n ∈ N (cid:1) (4.41)is bijective and analytic.By considering the Wronskian, we see that c ( µ n ( q ); q ) s ′ ( µ n ( q ); q ) = 1. Therefore, the sequence ( κ n ( q )) n ∈ N is related to a ( λ ; q ) by evaluation at the Dirichlet spectrum of q , α n ( q ) := a ( µ n ( q ); q ) = (cid:0) c ( µ n ( q ); q ) − s ′ ( µ n ( q ); q ) (cid:1) = (cid:0) s ′ ( µ n ( q ); q ) − − s ′ ( µ n ( q ); q ) (cid:1) = ( − n (cid:0) e − κ n ( q ) − e κ n ( q ) (cid:1) = ( − n +1 sinh κ n ( q ) . (4.42)The correspondence κ n ( q ) ↔ α n ( q ) is a bijection of ℓ , and therefore the data ( α n ( q )) n ∈ N can be used insteadof ( κ n ( q )) n ∈ N , that is, the correspondence L → R × S × ℓ :: q ( x ) (cid:0) c, ( σ n ) n ∈ N , ( α n ) n ∈ N (cid:1) (4.43)is bijective.Let p ∈ L [0 ,
1] be given, and denote by M ( p ) the set of potentials isospectral to p , M ( p ) := (cid:8) q ∈ L [0 ,
1] : µ ( q ) = µ ( p ) (cid:9) . (4.44)Proposition 1, the interpolation theorem, and the inverse spectral theory provide the following diagram. M ( p ) −→ E −→ ℓ −→ M ( p ) q a ( a ( µ n ( p ))) n ∈ N q (4.45)The restriction map E → ℓ is a bijection by Theorem 6. The composite map from M ( p ) → M ( p ) is theidentity map [9, Ch.3,Theorems 4,5]; and the composite map M ( p ) → ℓ is surjective [9, Ch.4,Theorems 1*,3].This makes the map M ( p ) → E a bijection, resulting in the following theorem. Theorem 7.
The set of asymmetry functions a ( λ ; q ) , as q runs over L [0 , , is equal to E ; and for eachisospectral set M ( p ) , the correspondence q a ( · ; q ) is a bijection between M ( p ) and E . As a corollary, we obtain the main result of the paper, as announced in the abstract.
Theorem 8.
The set of asymmetry functions a ( λ ; q ) , as q runs over L [0 , , is equal to E . Let a ∈ E begiven. The set of potentials possessing a ( λ ) as its asymmetry function consists of one q ∈ L [0 , for eachDirichlet spectral sequence. .2 Analyticity We now establish the bi-analyticity of the map between potentials q ∈ L [0 ,
1] and spectral data ( c, ( σ n ) n ∈ N , a ) ∈ R × S × E . Theorem 9.
The map L [0 , −→ R × S × E :: q −→ (cid:0) c, ( σ n ) n ∈ N , a (cid:1) , from potentials q to the spectrum ( µ n = c + σ n ) n ∈ N and asymmetry function a of the Dirichlet Schr¨odingeroperator − d /dx + q ( x ) on [0 , is bi-analytic. ( S is defined before (4.40).) This will be proved through the following maps: L [0 , ←→ R × S × ℓ ←→ R × S × ℓ ←→ R × S × E q ←→ (cid:0) c, ( σ n ) n ∈ N , ( κ n ) n ∈ N (cid:1) ←→ (cid:0) c, ( σ n ) n ∈ N , ( α n ) n ∈ N (cid:1) ←→ (cid:0) c, ( σ n ) n ∈ N , a (cid:1) (4.46)Analyticity of the map given by the first arrow is due to P¨oschel and Trubowitz [9]. The analyticity impliedby the second arrow is very easy because α n = ( − n +1 sinh κ n . The rest of this section establishes analyticityof the map given by the last arrow from right to left. Because of the inverse function theorem for analyticfunctions in Banach spaces (see [13, p. 1081] or [9, p. 142]), analyticity in one direction implies analyticity inthe other.Theorem 6 provides an identification of E with ℓ through evaluation on the sequence ( π n ) n ∈ N . Thisis a bounded linear bijection of Hilbert spaces and is therefore analytic. Denote the values of a ( λ ) on thissequence by ( a n ) n ∈ N , a n := a ( π n ) . (4.47)Thus the analyticity of the map given by the third arrow pointing leftward is established by proving thatthe map (cid:0) c, ( σ n ) n ∈ N , ( a n ) n ∈ N (cid:1) (cid:0) c, ( σ n ) n ∈ N , ( a ( n π + c + σ n )) n ∈ N (cid:1) (4.48)from R × S × ℓ to R × S × ℓ is analytic.By [9, Theorem 3, p.138], analyticity of a map between an open subset of a complex Banach space toanother Hilbert space is equivalent to the map satisfying two properties simultaneously, (1) weak analyticityof each projection to the elements of an orthonormal basis of the target space, and (2) local boundedness. Wetherefore work with the complexification of R × ℓ × ℓ , namely ( R × ℓ × ℓ ) ⊗ C = ( R × ℓ × ℓ )+ i ( R × ℓ × ℓ ).We care only about analyticity on the open subset R × S × ℓ of the real subspace of this complex Banachspace. This open subset is contained in an open subset C × S C × ( ℓ ⊗ C ) of ( R × ℓ × ℓ ) ⊗ C , in which S C = [ σ ∈S U σ , (4.49)with U σ being an open neighborhood of σ in ℓ ⊗ C with U σ ∩ ℓ ⊂ S so that S C ∩ ℓ = S . (4.50)Weak analyticity of each coordinate is due to the calculation ddh (cid:0) c + h ˆ c, ( σ n + h ˆ σ n ) n ∈ N , (( a + h ˆ a )( n π + c + h ˆ c + σ n + h ˆ σ n )) n ∈ N (cid:1) = (cid:0) ˆ c, (ˆ σ n ) n ∈ N , ((ˆ c + ˆ σ n ) a ′ ( n π + c + σ n ) + ˆ a ( n π + c + σ n )) n ∈ N (cid:1) . (4.51)Now we need to prove local boundedness, which is the content of Proposition 10. By applying Theorem 6to E and i E , an asymmetry function a ∈ E C is recovered from ( a n ) n ∈ N through interpolation, a ( λ ) = ∞ X j =1 a j Y k = j π k − λπ k − π j (cid:0) a j = a ( π j ) (cid:1) . (4.52)11herefore, we expect the formula α n = a ( π n + c + σ n ) = ∞ X j =1 a j Y k = j π k − π n − c n π k − π j , (4.53)in which c n = c + σ n , to extend the map (4.48) to one from C × S C × ( ℓ ⊗ C ) to itself. Proposition 10.
The map ( c, ( σ n ) n ∈ N , ( a n ) n ∈ N ) ( c, ( σ n ) n ∈ N , ( α n ) n ∈ N ) , (4.54) with α n defined in (4.53) is a locally bounded map from C × S C × ( ℓ ⊗ C ) to itself. The proof of the proposition rests on two lemmas. The definition of α n can be expressed as nα n = ∞ X j =1 ja j K ( n, j ) , (4.55)in which K ( n, j ) = nj Y k = j π k − π n − c n π k − π j = Y k = n (cid:16) − c n π k − π n (cid:17) , n = jnj · − c n π n − π j Y k j,n } π k − π n − c k π k − π j , n = j. (4.56) Lemma 11. If j = n + c n /π , then K ( n, j ) = ( − j +1 njj − n − c n /π · sinc π p n + c n /π (4.57)= ( − n + j +1 j j − n − c n /π · e n π j p n + c n /π , (4.58) in which e n = d n sinc d n n , d n = c n /π p c n π n , (4.59) and the argument of the square root is taken to lie in ( − π, π ] .If n + c n /π = j with j > , then K ( n, j ) = nj , (4.60) and if n + c n /π = 0 , K ( n, j ) = ( − j +1 nj . (4.61) Remark 12. K ( n, j ) is an entire function of c n as a complex variable. In the first expression of theproposition, c n = π ( j − n ) is a removable singularity because the sinc function vanishes there.2. Note that we have e n ∼ c n /π as n → ∞ , while when n + c n /π = 0 , we have d n = − n π , and thus e n = 0 .Proof. To prove Lemma 11, write K ( n, j ) as K ( n, j ) = nj Y k = j (cid:18) − n + γ n k (cid:19) Y k = j (cid:18) − j k (cid:19) − (4.62)12ith γ n = c n /π and use the formula ∞ Y k =1 (cid:18) − z k (cid:19) = sin πzπz = sinc πz. (4.63)Excluding k = j from the product yields Y k = j (cid:18) − z k (cid:19) = (cid:18) − z j (cid:19) − sin πzπz = j sin πzπz ( j − z ) , (4.64)and thus Y k = j (cid:18) − j k (cid:19) = lim z → j j sin πzπz ( j − z ) = j lim z → j cos πzj − z = ( − j +1 , (4.65)and we obtain, for the second infinite product in K ( n, j ), Y k = j (cid:18) − j k (cid:19) − = 2( − j +1 . (4.66)If j = n + c n /π , then for the first product, the foregoing calculation gives Y k = j (cid:18) − n + γ n k (cid:19) = ( − j +1 , (4.67)and putting this together with (4.66) yields the result. If j = n + c n /π , Y k = j (cid:18) − n + γ n k (cid:19) = (cid:18) − n + γ n j (cid:19) − ∞ Y k =1 (cid:18) − n + γ n k (cid:19) = j j − n − γ n sin π p n + γ n π p n + γ n . (4.68)Using p n + γ n = n + d n / ( nπ ), we obtainsin π p n + γ n = ( − n sin d n n = ( − n d n n sinc d n n = ( − n e n n (4.69)and therefore Y k = j (cid:18) − n + γ n k (cid:19) = j j − n − γ n ( − n e n πn p n + γ n . (4.70)Putting this together with (4.66) yields the result. Lemma 13.
Let C and { c n } n ∈ N be such that | c n | < C for all n ∈ N . There is a constant C such that | K ( n, n ) | < C for all n ∈ N (4.71) and such that X n,j ∈ N , n = j | K ( n, j ) | < C. (4.72) Proof.
Write K ( n, j ) as K ( n, j ) = ( − n + j +1 π κ nj , κ nj := jj − n − γ n e n p n + γ n . (4.73)13hen j = n + r ( n ∈ N , r ∈ N ), κ nj = n + rr (2 n + r ) − γ n e n p n + γ n := α rn , (4.74)and when n = j + r ( j ∈ N , r ∈ N ), κ nj = − jr (2 j + r ) + γ j + r e j + r p ( j + r ) + γ j + r := β rj , (4.75)and on the diagonal, κ nn = α n = β n = − e n γ n p γ n /n . (4.76)Now let | c n | < C for all n ∈ N , or | γ n | < γ = C /π . Let E be such that | e n | < E for all n ∈ N .First, let’s examine K ( n, n ). We have κ nn = − d n γ n p γ n /n sinc d n n (4.77)= − π p γ n /n p γ n /n sinc d n n . (4.78)If n > C /π , then | γ n /n | < / d n lie in a strip R + i ( − a, a ) about the real line, in which thesinc function is bounded. So, for some A > | κ nn | < A. (4.79)From Lemma 11, we have K ( n, n ) = ( − n n γ n sinc π p n + γ n . (4.80)For each n ∈ N , K ( n, n ) is an entire function of γ n , for sinc π p n + γ n vanishes whenever γ n does. Therefore,there is a constant C n such that | K ( n, n ) | < C n whenever | γ n | < γ . Taking C to be the maximum of 2 A/π and the constants C n over n ≤ γ , we obtain | K ( n, n ) | < C ∀ n ∈ N , | c n | < c. (4.81)Next, consider j = n + r with r ≥
1. We have rn α rn = n + r n + r − γ n /r e n p γ n /n . (4.82)For all r ∈ N and n > γ + 1, | rn α rn | < √ E, (4.83)and therefore | K ( n, n + r ) | < √ Eπ rn ( r ∈ N , n > γ + 1) . (4.84)Next, write rK ( n, n + r ) = ( − n + r +1 n ( n + r )2 n + r − γ n /r sinc π p n + γ n . (4.85)By analyticity of sinc π p n + γ n in γ n and the restriction | γ n | < γ , there exists C > ≤ n ≤ γ , then (cid:12)(cid:12)(cid:12) sinc π p n + γ n (cid:12)(cid:12)(cid:12) < C. (4.86)14ince lim r →∞ n ( n + r )2 n + r − γ n /r = 2 n, (4.87)there exists r ∈ N such that, if r > r and n is such that 1 ≤ n ≤ γ + 1, (cid:12)(cid:12)(cid:12)(cid:12) n ( n + r )2 n + r − γ n /r (cid:12)(cid:12)(cid:12)(cid:12) < γ ) . (4.88)Thus we obtain | K ( n, n + r ) | < C (1 + γ ) r ( r > r , n ≤ γ + 1) . (4.89)Because of the analyticity of K ( n, n + r ) in γ n and the constraint | γ n | < γ , a bound like (4.89) (with adifferent constant) holds also for the finite set of pairs ( n, r ) with 1 ≤ r ≤ r and 1 ≤ n ≤ γ , and weobtain | K ( n, n + r ) | < C r ( r ∈ N , n ≤ γ + 1) . (4.90)For each r ∈ N , X n ∈ N | K ( n, n + r ) | < r ( γ + 1) C + C X n>γ +1 n ! . (4.91)Thus, there is a constant C such that, if | c n | < c for all n ∈ N , then X r,n ∈ N | K ( n, n + r ) | < C. (4.92)Now consider n = j + r with r ≥
1. We have rj β rj = − j j + r + γ j + r /r e j + r p (1 + r /j ) + γ j + r /j (4.93)Similarly to the previous case, for all r ∈ N and j > γ + 1, we obtain | rj β rj | < √ E, (4.94)and therefore | K ( j + r, j ) | < √ Eπ rj ( r ∈ N , j > γ + 1) . (4.95)Now consider rK ( j + r, j ) = ( − j j + r ) j j + r + γ j + r /r sinc π q ( j + r ) + γ j + r . (4.96)If r is large enough, then ( j + r ) + γ j + r > j , so that the sinc factor is bounded by 1 in absolutevalue. Then we argue as before and obtain C such that, for all r ∈ N and j such that 1 ≤ j ≤ γ , (cid:12)(cid:12)(cid:12)(cid:12) sinc π q ( j + r ) + γ j + r (cid:12)(cid:12)(cid:12)(cid:12) < C. (4.97)By an identical argument as above, with n replaced by j , we obtain r ∈ N such that, if r > r and n is suchthat 1 ≤ n ≤ γ + 1, (cid:12)(cid:12)(cid:12)(cid:12) j + r ) j j + r + γ j + r /r (cid:12)(cid:12)(cid:12)(cid:12) < γ ) , (4.98)and hence | K ( j + r, j ) | < C (1 + γ ) r ( r > r , j ≤ γ + 1) . (4.99)15imilarly to above, this bound is extended to all r , | K ( j + r, j ) | < C r ( r ∈ N , j ≤ γ + 1) . (4.100)In the end, we obtain a constant C such that, if | c n | < c for all n ∈ N , then X r,j ∈ N | K ( j + r, j ) | < C. (4.101)This, together with (4.92) yields the statement of the proposition. Proof of Proposition 10.
The local boundedness of the map( c, ( σ n ) n ∈ N , ( a n ) n ∈ N ) ( α n ) n ∈ N (4.102)from C × S C × ( ℓ ⊗ C ) to ℓ ⊗ C is implied by the following: For each C , there exists C such that, for all c, ( σ n ) , ( a n ) with | c | < C , k ( σ n ) k ℓ < C and k ( a n ) k ℓ < C , one has k ( α n ) k ℓ < C .Given the conditions | c | < C and k ( σ n ) k ℓ < C , Lemma 13, together with Young’s generalized inequal-ity, implies that the linear map ( a n ) ( α n ) is bounded from ℓ ⊗ C to itself with bound 2 C . The additionalcondition k ( a n ) k ℓ < C then implies k ( α n ) k ℓ < CC .According to the discussion after the statement of the theorem, Proposition 10 completes the proof ofTheorem 9. Acknowledgement.
This material is based upon work supported by the National Science Foundationunder Grant No. DMS-1814902 and by UK EPSRC under grant EP/P005985/1.
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