Annals of Mathematics, 152 (2000), 593–643
A new approach to inverse spectral theory, II. General real potentials and the connection to the spectral measure
By Fritz Gesztesy and Barry Simon
Abstract
We continue the study of the A-amplitude associated to a half-line d2 2 ∞ Schr odinger operator, dx2 + q in L ((0,b)),R b . A is related to the Weyl- 2 a 2 (2a ε) Titchmarsh m-function via m( )= 0 A( )e d +O(e ) for all ε>0. We discuss ve issues here. First, we extend the theory to general q in L1((0,a)) for all a, including q’s which are limit circle at in nity. Second, we prove the followingR relation between√ the A-amplitude and the spectral measure ∞ 1 : A( )= 2 ∞ 2 sin(2 ) d ( ) (since the integral is divergent, this formula has to be properly interpreted). Third, we provide a Laplace trans- form representation for m without error term in the case b<∞. Fourth, we discuss m-functions associated to other boundary conditions than the Dirichlet boundary conditions associated to the principal Weyl-Titchmarsh m-function. Finally, we discuss some examples where one can compute A exactly.
1. Introduction
In this paper we will consider Schr odinger operators d2 (1.1) + q dx2 in L2((0,b)) for 0 This material is based upon work supported by the National Science Foundation under Grant No. DMS-9707661. The government has certain rights in this material. 1991 Mathematics Subject Classi cation. Primary: 34A55, 34B20; Secondary: 34L05, 47A10. Key words and phrases. Inverse spectral theory, Weyl-Titchmarsh m-function, spectral measure. 594 FRITZ GESZTESY AND BARRY SIMON Case 1. b<∞. We suppose q ∈ L1((0,b)). We then pick h ∈ R ∪ {∞} and add the boundary condition at b 0 (1.2) u (b )+hu(b )=0, where h = ∞ is shorthand for the Dirichlet boundary condition u(b )=0. For Cases 2–4, b = ∞ and Z a (1.3) |q(x)| dx < ∞ for all a<∞. 0 Case 2. q is “essentially” bounded from below in the sense that Z a+1 (1.4) sup max( q(x), 0) dx < ∞. a>0 a Examples include q(x)=c(x+1) for c>0 and all ∈ R or q(x)= c(x+1) for all c>0 and 0. Case 3. (1.4) fails but (1.1) is limit point at ∞ (see [6, Ch. 9]; [33, Sect. X.1] for a discussion of limit point/limit circle), that is, for each z ∈ C+ = {z ∈ C | Im(z) > 0}, 00 (1.5) u + qu = zu has a unique solution, up to a multiplicative constant, which is L2 at ∞.An example is q(x)= c(x +1) for c>0 and 0 < 2. Case 4. (1.1) is limit circle at in nity; that is, every solution of (1.5) 2 is L ((0, ∞)) at in nity if z ∈ C+. We then pick a boundary condition by picking a nonzero solution u0 of (1.5) for z = i. Other functions u satisfying the associated boundary condition at in nity then are supposed to satisfy 0 0 (1.6) lim [u0(x)u (x) u (x)u(x)]=0. x→∞ 0 Examples include q(x)= c(x +1) for c>0 and >2. The Weyl-Titchmarsh m-function, m(z), is de ned for z ∈ C+ as follows. Fix z ∈ C+. Let u(x, z) be a nonzero solution of (1.5) which satis es the boundary condition at b. In Case 1, that meansR u satis es (1.2); in Case 4, it ∞ | |2 ∞ satis es (1.6); and in Cases 2–3, it satis es R u(x, z) dx < for some (and hence for all) R 0. Then, u0(0 ,z) (1.7) m(z)= + u(0+,z) and, more generally, u0(x, z) (1.8) m(z,x)= . u(x, z) INVERSE SPECTRAL THEORY, II 595 0 ∂m m(z,x) satis es the Riccati equation (with m = ∂x ), 0 (1.9) m (z,x)=q(x) z m(z,x)2. m is an analytic function of z for z ∈ C+, and moreover: Case 1. m is meromorphic in C with a discrete set 1 < 2 < of poles on R (and none on ( ∞, 1)). Case 2. For some ∈ R, m has an analytic continuation to C\[ ,∞) with m real on ( ∞, ). Case 3. In general, m cannot be continued beyond C+ (there exist q’s where m has a dense set of polar singularities on R). Case 4. m is meromorphic in C with a discrete set of poles (and zeros) on R with limit points at both +∞ and ∞. Moreover, if z ∈ C+ then m(z,x) ∈ C+; so m satis es a Herglotz representation theorem, Z 1 (1.10) m(z)=c + d ( ), R z 1+ 2 where is a positive measure called the spectral measure, which satis es Z d ( ) (1.11) < ∞, R 1+| |2 1 (1.12) d ( ) = w-lim ↓ Im(m( + iε)) d , ε 0 where w-lim is meant in the distributional sense. All these properties of m are well known (see, e.g. [23, Ch. 2]). In (1.10), c (which is equal to Re(m(i))) is determined by the result of Everitt [10] that for each ε>0, (1.13) m( 2)= + o(1) as | |→∞with + ε 596 FRITZ GESZTESY AND BARRY SIMON Theorem 1.1. There exists a function A( ) for ∈ [0,b) so that A ∈ L1((0,a)) for all a This result was proven in Cases 1 and 2 in [35]. Thus, one of our purposes here is to prove this result if one only assumes (1.3) (i.e., in Cases 3 and 4). Actually, in [35], (1.15) was proven in Cases 1 and 2 for real with | |→∞. Our proof under only (1.3) includes Case 2 in the general -region ∈ arg( ) ( 2 + ε, ε) and, as we will remark, the proof also holds in this region for Case 1. Remark. At rst sight, it may appear that Theorem 1.1 as we stated it does not imply the real result of [35], but if the spectral measure of (1.10) ∈ ∞ ∈ R | | has supp( ) [a, ) for some a , (1.15) extends to all in arg( ) < 2 ε, | | a + 1. To see this, one notes by (1.10) that m0(z) is bounded away from [a, ∞) so one has the a priori bound |m(z)| C|z| in the region Re(z) Here is a result from [35] which we will need: Theorem 1.2 ([35, Theorem 2.1]). Let q ∈ L1((0, ∞)). Then there exists a function A( ) on (0, ∞) so that A q is continuous and satis es (1.16) such 1 k k that for Re( ) > 2 q 1, Z ∞ (1.17) m( 2)= A( )e 2 d . 0 1 k k Remark. In [35], this is only stated for real with > 2 q 1, but (1.16) | | k k2 k k implies that A( ) q( ) q 1 exp( q 1) so the right-hand side of (1.17) 1 k k converges to an analytic function in Re( ) > 2 q 1. Since m(z) is analytic C\ ∞ { ∈ C | 1 k k } in [ , ) for suitable , we have equality in Re( ) > 2 q 1 by analyticity. Theorem 1.1 in all cases follows from Theorem 1.2 and the following result which we will prove in Section 3. INVERSE SPECTRAL THEORY, II 597 Theorem 1.3. Let q1,q2 be potentials de ned on (0,bj) with bj >a for j =1, 2. Suppose that q1 = q2 on [0,a]. Then in the region arg( ) ∈ | | ( 2 + ε, ε), K0, we have that 2 2 (1.18) |m1( ) m2( )| Cε, exp( 2aRe( )), R x+ | | where Cε, depends only on ε, , and sup0 x a( x qj(y) dy), where >0 is any number so that a + bj, j =1, 2. Remarks. 1. An important consequence of Theorem 1.3 is that if q1(x) = q2(x) for x ∈ [0,a], then A1( )=A2( ) for ∈ [0,a]. Thus, A( ) is only a function of q on [0, ]. At the end of the introduction, we will note that q(x) is only a function of A on [0,x]. 2. This implies Theorem 1.1 by taking q1 = q and q2 = q [0,a] and using Theorem 1.2 on q2. ∈ 3. Our proof implies (1.18) on a larger region than arg( ) ( 2 + ε, ε). Basically, we will need Im( ) C1 exp( C2| |)ifRe( ) →∞. We will obtain Theorem 1.3 from the following pair of results. Theorem 1.4. Let q be de ned on (0,a+ ) and q ∈ L1((0,a+ )). Then ∈ | | ∈ in any region arg( ) ( 2 + ε, ε), K0, we have for all x [0,a] that 2 (1.19) |m( ,x)+ | Cε, , R x+ | | where Cε, depends only on ε, and sup0 x a( x q(y) dy). Theorem 1.5. Let q1 = q2 on [0,a] and suppose m1 and m2 obey (1.19) for x ∈ [0,a]. Then in the same -region, 2 2 (1.20) |m1( ) m2( )| 2Cε, exp( (Re( ))(2a 2Cε, )). We will prove Theorem 1.5 in Section 2 using the Riccati equation and Theorem 1.4 in Section 3 by following ideas of Atkinson [3]. In Sections 5–9, we turn to the connection between the spectral measure d and the A-amplitude. Our basic formula says that Z ∞ √ 1 (1.21) A( )= 2 2 sin(2 ) d ( ). ∞ In this formula, if gives nonzero weight to ( ∞, 0], we interpret ( √ 1 2 √ if =0, (1.22) 2 sin(2 )= 1 ( ) 2 sinh(2 )if <0, √ 1 consistent with the fact that 2 sin(2 ) de ned on (0, ∞) extends to an entire function of . 598 FRITZ GESZTESY AND BARRY SIMON The integralR in (1.21) is not convergent. Indeed, the asymptotics (1.13) R 3 2 2 imply that 0 d ( ) 3 R so (1.21) is never absolutely convergent. As we will see in Section 9, it is never even conditionally convergent in case b<∞ (and also in many cases with b = ∞). So (1.21) has to be suitably interpreted. In Sections 5–7, we prove (1.21) as a distributional relation, smeared in ∈ ∞ ∞ on both sides by a function f C0 ((0, )). This holds for all q’s in Cases 1–4. In Section 8, we prove an Abelianized version of (1.21), viz., Z ∞ √ ε 1 (1.23) A( )= 2 lim e 2 sin(2 ) d ( ) ε↓0 ∞ at any point, , of Lebesgue continuity for q. (1.23) is only proven for a restricted class of q’s including Case 1, 2 and those q’s satisfying q(x) Cx2,x R for some R>0, C>0, which are always in the limit point case at in nity. We will use (1.23) as our point of departure for relating A( ) to scattering data at the end of Section 8. In order to prove (1.21) for nite b, we need to analyze the nite b case extending (1.15) to all a including a = ∞ (by allowing A to have and 0 singularities at multiples of b). This was done in [35] for real and positive and a<∞. We now need results in the entire region Re( ) K0, and this is what we do in Section 4. Explicitly, we will prove Theorem 1.6. In Case 1, there are An,Bn for n =1, 2,..., and a function A( ) on (0, ∞) with (i) |An| C. (ii) |Bn| Cn. R a | | | | 1 (iii) 0 A( ) d C exp(K0 a ) so that for Re( ) > 2 K0: X∞ X∞ Z ∞ 2 2 bn 2 bn 2 (1.24) m( )= An e Bne A( )e d . n=1 n=1 0 R 0 In Section 6, we will use (1.21) to obtain a priori bounds on R d ( )as R →∞. Section 9 includes further discussion of the signi cance of (1.21) and the connection between A and the Gel0fand-Levitan transformation kernel. Sections 10 and 11 present a few simple examples where one can compute A explicitly. One of the examples, when combined with a general comparison INVERSE SPECTRAL THEORY, II 599 theorem, allows us to prove the general bound |A( )| 1 ( )e2 ( ), | | 1 where ( ) = sup0 x q(x) 2 and this lets us extend (1.17) to bounded q. In the appendix we discuss analogs of (1.15) for the other m-functions that arise in the Weyl-Titchmarsh theory. While we will not discuss the theory in detail in this paper, we end this introduction by recalling the major thrust of [35] — the connection between A and inverse theory (which holds for the principal m-function but not for the m-functions discussed in the appendix). Namely, there is an A( , x) function associated to m(z,x)by Z a (1.25) m( 2,x)= A( , x)e 2 d + O˜(e 2 ) 0 for a F.G. would like to thank C. Peck and T. Tombrello for the hospitality of Caltech where this work was done. 600 FRITZ GESZTESY AND BARRY SIMON 2. Using the Riccati equation As explained in the introduction, the Riccati equation and a priori control on mj allow one to obtain exponentially small estimates on m1 m2 (Theo- rem 1.5). Proposition 2.1. Let m1(x),m2(x) be two absolutely continuous func- tions on [a, b] so that for some Q ∈ L1((a, b)), 0 2 ∈ (2.1) mj(x)=Q(x) mj(x) ,j=1, 2,x (a, b). Then Z ! b [m1(a) m2(a)]=[m1(b) m2(b)] exp [m1(y)+m2(y)] dy . a Proof. Let f(x)=m1(x) m2(x) and g(x)=m1(x)+m2(x). Then 0 f (x)= f(x) g(x), from which it follows that Z b f(x)=f(b) exp g(y) dy . x As an immediate corollary, we have the following (this implies Theo- rem 1.3) 2 Theorem 2.2. Let mj(x, ) be functions de ned for x ∈ [a, b] and ∈ K some region of C. Suppose that for each in K, mj is absolutely continuous in x and satis es (N.B.: q is the same for m1 and m2), 0 2 2 2 2 mj(x, )=q(x)+ mj(x, ) ,j=1, 2. Suppose C is such that for each x ∈ [a, b] and ∈ K, 2 (2.2) |mj(x, )+ | C, j =1, 2, then 2 2 (2.3) |m1(a, ) m2(a, )| 2C exp[ 2(b a)[Re( ) C]]. 3. Atkinson’s method Theorem 2.2 places importance on a priori bounds of the form (2.2). Fortunately, by modifying ideas of Atkinson [3], we can obtain estimates of this form as long as Im( ) is bounded away from zero. INVERSE SPECTRAL THEORY, II 601 Throughout this section, b ∞and q ∈ L1((0,a)) for all a 0 2 2 u (b , )+hu(b , )=0 (|h| < ∞) or 2 u(b , )=0 (h = ∞) and secondly, b = ∞, and either q limit point at in nity or q limit circle with some boundary condition picked at b. Then take u to be an L2 solution of (3.1). In either case, u can be chosen analytic in although the bounds in Propositions 3.1 and 3.2 below do not require that. Atkinson’s method allows us to estimate |m( 2)+ | in two steps. We 2 will x some a Proposition 3.1. There is a C>0 depending only on q and a universal constant E>0 so that if Re( ) C and Im( ) =0,6 then 2 | | (3.4) |m( 2) m ( 2)| E e 2aRe( ). 0 |Im( )| In fact, one can take Z a 2 3 2 12 C = max a 1 ln(6), 4 |q(x)| dx ,E= . 0 5 Proposition 3.2. There exist constants D1 and D2 (depending only on a and q), so that for Re( ) >D1, 2 |m0( )+ | D2. 602 FRITZ GESZTESY AND BARRY SIMON Indeed, one can take Z a D1 = D2 =2 |q(x)| dx. 0 These propositions together with Theorem 1.2 yield the following explicit form of Theorem 1.3. Theorem 3.3. Let q1,q2 be de ned on (0,bj) with bj >afor j =1, 2. Suppose that q1 = Rq2 on [0,a]. Pick so that a + min(b1,b2) and let x+ | | 1 = sup0 x a;j=1,2( x qj(y) dy). Then if Re( ) max(4 , ln(6)) and Im( ) =0,6 we have that 2 2 |m1( ) m2( )| 2F ( ) exp( 2a[Re( ) F ( )]), where 2 864 | | F ( )=2 + e 2 Re( ) . 5 |Im( )| Remarks. 1. To obtain Theorem 1.3, we need only note that in the region ∈ | | arg( ) ( 2 + ε, ε), K0, F ( ) is bounded. 2. We need not require that arg( ) < ε to obtain F bounded. It su ces, for example, that Re( ) |Im( )| e Re( ) for some <2 . 3. For F to be bounded, we need not require that arg( ) > 2 + ε.It su ces that |Im( )| Re( ) ln[|Im( )|] for some >(2 ) 1. Unfortu- nately, this does not include the region Im( 2)=c, Re( 2) →∞, where Re( ) goes to zero as | | 1. However, as Re( 2) →∞, we only need that |Im( 2)| 2 | | ln(| |). As a preliminary to the proof of Proposition 3.1, we have Lemma 3.4. Let A, B, C, D ∈ C so that AD BC =1and so that 6 6 C D =0= Im( D ). Let f be the fractional linear transformation A + B f( )= . C + D Then f[R ∪ {∞}] is a circle of diameter 1 C (3.5) |D| 2 Im = |Im(CD )| 1. D | | C R ∪ {∞} Remark. If D = 0 or Im( D ) = 0, then f[ ] is a straight line. 1 0 Proof. Consider rst g( )= a +1 = a+ 1 . Then g(0) = 0 and g (0)=1, so g[R ∪ {∞}] is a circle tangent to the real axis. The other point on the 1 1 i R ∪ {∞} imaginary axis has = Re(a) with g( Re(a) )= Im(a) so diam(g[ ]) 1 = |Im(a)| . INVERSE SPECTRAL THEORY, II 603 Now write (using AD BC =1) B B f( )= + = + . CD + D2 D 2 C D D [ D +1] | | i Thus letting a = C/D, g( )= a +1 and writing D = D e , we have that B f( )=e 2i |D| 2g( )+ . D B/D is a translation and e 2i a rotation, and neither changes the diameter of a circle. So diam(f[R ∪ {∞}]) = |D| 2diam(g[R ∪ {∞}]). Now let ϕ(x, 2), (x, 2) solve (3.1) with 2 0 2 (3.6a) ϕ(0+, )=0,ϕ(0+, )=1, 2 0 2 (3.6b) (0+, )=1, (0+, )=0. De ne (a, 2) 0(a, 2) (3.7) f( )= . ϕ(a, 2) ϕ0(a, 2) 0 0 Lemma u (a, 2) u (0, 2) 3.5. If u solves (3.1) and u(a, 2) = , then u(0, 2) = f( ) with f given by (3.7). 0 0 0 ϕ (a, 2) (a, 2) u (0, 2) Proof. Let T = . Then T = ϕ(a, 2) (a, 2) u(0, 2) u0(a, 2) by linearity of (3.1). By constancy of the Wronskian, T has u(a, 2) determinant 1 and thus ! 2 0 2 (a, ) (a, ) T 1 = ϕ(a, 2) ϕ0(a, 2) and so u0(0, 2) (a, 2)u0(a, 2) 0(a, 2)u(a, 2) = u(0, 2) ϕ(a, 2)u0(a, 2)+ϕ0(a, 2)u(a, 2) ! u0(a, 2) = f . u(a, 2) Corollary 3.6. 2 2 1 (3.8) |m( ) m0( )| . |Im( ϕ(a, 2) ϕ0(a, 2))| 604 FRITZ GESZTESY AND BARRY SIMON 2 Proof. We consider the case Im( ) < 0. Let m1( ,x) be any solution of 0 2 2 2 2 0 2 m1( ,x)=q(x)+ m1( ,x). Then Im(m1( ,x)) 2 2 = 2Re( )Im( ) 2 Im(m1( ,x))Re(m1( ,x)). It follows that at a point 0 2 ∈ where Im(m1) = 0, that Im(m1) < 0. Thus if Im(m1( ,y)) = 0 for y [0,a], 2 2 then Im(m1( ,x)) < 0 for x ∈ (y, a]. Thus Im(m1( ,a)) 0 implies 2 2 Im(m1( , 0)) > 0, so f maps C+ onto a circle in C+. Since m0( ,a)= 2 and m( ,a) are in C+, both points are in C+ and so at x = 0, both lie in- side the disc bounded by f[R ∪ {∞}]. By det(T ) = 1 and Lemma 3.4, (3.8) holds. Proof of Proposition 3.1. By (3.8), we need to estimate ϕ(a, 2). De ne 2 2 0 2 2 w(x, ) = Im( ϕ(x, ) ϕ (x, )). Then, w(0+, )=0and by a standard Wronskian calculation, w0(x, 2)= Im( 2) |ϕ(x, 2)|2 = 2Re( )Im( )|ϕ(x, 2)|2. Thus, Z a 0 (3.9) |Im( ϕ(a, 2) ϕ (a, 2))| = 2Re( )|Im( )| |ϕ(x, 2)|2 dx. 0 ϕ(x, 2) satis es the following integral equation ([5, §I.2]), Z sinh( x) x sinh( (x y)) (3.10) ϕ(x, 2)= + q(y)ϕ(y, 2) dy. 0 De ne (x, 2)= e xRe( )ϕ(x, 2). Then, by (3.10) and | sinh( )| e| |Re( ), ∈ R, " # Z xRe( ) x sinh( x) e 1 (3.11) ϕ(x, 2) sup | (y, 2)| |q(y)| dy. | | 0 y x | | 0 Moreover, (3.10) becomes (x, 2)=e xRe( ) sinh( x) Z x sinh( (x y)) + e (x y)Re( )q(y) (y, 2) dy, 0 which implies that Z x 1 (3.12) | (x, 2)| 1+ |q(y)|| (y, 2)| dy. 0 | | Pick so that Z a (3.13) | | 4 |q(y)| dy. 0 Then (3.12) implies 1 sup | (x, 2)| 1+ sup | (x, 2)| 0 x a 4 0 x a INVERSE SPECTRAL THEORY, II 605 so that 4 (3.14) sup | (x, 2)| . 0 x a 3 Using (3.13) and (3.14) in (3.11), we get xRe( ) sinh( x) 1 e (3.15) ϕ(x, 2) . 3 | | | | | | 1 |Re(z)| |Re(z)| Now sinh(z) sinh( Re(z) )= 2 [e e ], so (3.15) implies 1 (3.16) |ϕ(x, 2)| [exRe( ) 3e xRe( )]. 6| | Now suppose (3.17) aRe( ) ln(6). a | 2 | 1 xRe( ) Thus for x 2 , (3.16) implies ϕ(x, ) 12| | e and we obtain Z a 1 1 (3.18) |ϕ(y, 2)|2 dy e2aRe( )[1 e aRe( )] a 288| |2 Re( ) 2 5 1 1 e2aRe( ). 6 288| |2 Re( ) Putting together (3.8), (3.9), and (3.18), we see that if (3.13) and (3.17) hold, then 3 1 |m( 2) m ( 2)| 288| |2 e 2aRe( ). 0 5 |Im( )| Proof of Proposition 3.2. Let Z a (3.19) = |q(y)| dy. 0 Let z(x, 2) solve (3.1) with boundary conditions z(a, 2)=1,z0(a, 2) = , and let z0(x, 2) (x, 2)= + . z(x, 2) 2 Then the Riccati equation for m0( ,a) becomes 0 (3.20) (x, 2)=q(x) (x, 2)2 +2 (x, 2) and we have (3.21) (a, 2)=0. Thus (x, 2) satis es Z a (3.22) (x, 2)= e 2 (y x)[q(y) 2(y, 2)] dy. x 606 FRITZ GESZTESY AND BARRY SIMON 2 | 2 | De ne (x, ) = supx y a (y, ) . Since Re( ) > 0, (3.22) implies that 2 1 1 2 2 (3.23) (x, ) + 2 (Re( )) (x, ) . Suppose that (3.24) Re( ) > 2 . Then (3.23) implies 2 1 1 2 2 (3.25) (x, ) < + 4 (x, ) . (3.25) implies (x, 2) =26 . Since (a, 2) = 0 and is continuous, we 2 2 2 conclude that (0+, ) < 2 ,so| (0+, )| < 2 ; hence |(m0( )+ | 2 . Remark. There is an interesting alternate proof of Proposition 3.2 that 2 has better constants. It begins by noting that m0( )isthem-function for the potential which is q(x) for x a and 0 for x>a. Thus Theorem 1.2 applies. 1 k k So using the bounds (1.16) for A, we see immediately that for Re( ) > 2 q 1, k k2 2 q 1 |m0( )+ | kqk1 + , [2Re( ) kqk1] R k k a | | where q 1 = 0 q(y) dy. 4. Finite b representations with no errors Theorem 1.2 says that if b = ∞ and q ∈ L1((0, ∞)), then (1.17) holds, a Laplace transform representation for m without errors. It is, of course, of direct interest that such a formula holds, but we are especially interested in a particular consequence of it — namely, that it implies that the formula (1.15) with error holds in the region Re( ) >K0 with error uniformly bounded in Im( ); that is, we are interested in Theorem ∈ 1 ∞ 1 k k 4.1. If q L ((0, )) and Re( ) > 2 q 1, then for all a: Z a (4.1) m( 2)+ + A( )e 2 d "0 # k k k k2 a q 1 q 1e 2aRe( ) kqk1 + e . 2Re( ) kqk1 Proof. An immediate consequence of (1.17) and the estimate | | k k2 k k A( ) q( ) q 1 exp( q 1). INVERSE SPECTRAL THEORY, II 607 Our principal goal in this section is to prove an analog of this result in the case b<∞. To do so, we will need to rst prove an analog of (1.17) in the case b<∞ — something of interest in its own right. The idea will be to mimic the proof of Theorem 2 from [35] but use the nite b, q(0)(x)=0, x 0 Green’s function where [35] used the in nite b Green’s function. The basic idea is simple, but the arithmetic is a bit involved. We will start with the h = ∞ case. Three functions for q(0)(x)=0, d2 2 1 x 0 are signi cant. First, the kernel of the resolvent ( dx2 + ) with u(0+)=u(b ) = 0 boundary conditions. By an elementary calculation (see, e.g., [35, §5]), it has the form " # sinh( x ) e x> e (2b x>) (4.2) G(0) (x, y, 2)= < , h=∞ 1 e 2 b with x< = min(x, y), x> = max(x, y). The second function is (0) x (2b x) (0) 2 ∂Gh=∞ 2 e e (4.3) ∞(x, ) lim (x, y, )= h= y↓0 ∂y 1 e 2 b (0) 2 (0) and nally (notice that h=∞(0+, )=1and h=∞ satis es the equations 00 = 2 and (b , 2) = 0): 0 + e 2 b (4.4) m(0) ( 2)= (0) (0 , 2)= . h=∞ h=∞ + 1 e 2 b In (4.4), prime means d/dx. Fix now q ∈ C∞((0,b)). The pair of formulas 0 ! ! ! n 2 1 X∞ 2 1 2 1 d 2 n d 2 d 2 2 + q + = ( 1) 2 + q 2 + dx n=0 dx dx and ∂2G(x, y, 2) m( 2) = lim x 608 FRITZ GESZTESY AND BARRY SIMON and for n 2, Z Z (4.8) b b 2 n Mn( ; q)=( 1) dx1 ... dxn q(x1) ...q(xn) 0 0 nY 1 (0) 2 (0) 2 (0) 2 h=∞(x1, ) h=∞(xn, ) Gh=∞(xj,xj+1, ). j=1 The precise region of convergence is unimportant since we will eventu- ally expand regions by analytic continuation. For now, we note it certainly converges in the region real with 2 > kqk∞. We want to write each term in (4.5) as a Laplace transform. We begin with (4.6), using (4.4) 2 b X∞ 2 e (4.9) M ( 2; q)= = 2 e 2j b. 0 1 e 2 b j=1 Next, note by (4.3) that X∞ X∞ (0) 2 (x+2bj) (2bj+(2b x)) (4.10) h=∞(x, )= e e , j=0 j=0 so X∞ (0) 2 2 (2x+2bj) (4.11) h=∞(x, ) = e (j +1) j=0 X∞ X∞ + e (2bj+(4b 2x))(j +1) 2 je 2b j; j=0 j=1 hence, "Z # Z b X∞ ∞ 2 2b j 2 (4.12) M1( ; q)=2 q(x) dx je A1( )e d , 0 j=1 0 where (4.13) q( ), 0 To manipulate Mn for n 2, we rst rewrite (4.10) as X∞ (0) 2 (0),(j) 2 (4.14) h=∞(x, )= (x, ) j=0 where (0),(j) 2 j (4.15) (x, )=( 1) exp( Xj(x)), INVERSE SPECTRAL THEORY, II 609 with ( x + bj, j =0, 2,..., (4.16) X (x)= j b x + bj, j =1, 3,..., and then for n 2 X∞ 2 2 (4.17) Mn( )= Mn,j,p( ), j,p=0 where (4.18) Z Z b b 2 n Mn,j,p( )=( 1) dx1 ... dxn q(x1) ...q(xn) 0 0 nY 1 (0),(j) 2 (0),(p) 2 (0) 2 (x1, ) (xn, ) Gh=∞(xj,xj+1, ). j=1 Next use the representation from [35], Z x+y sinh( x ) 1 < e x> = e ` d`, 2 |x y| to rewrite (4.2) as Z " # x+y ` (2b `) (0) 2 1 e e Gh=∞(x, y, )= 2 b d` 2 |x y| 1 e Z Z 1 1 = e ` d` e ` d`, 2 S+(x,y) 2 S (x,y) where [∞ S+(x, y)= [|x y| +2nb, x + y +2nb] n=0 and [∞ S (x, y)= [2b(n +1) x y, 2b(n +1) |x y|]. n=0 Each union consists of disjoint intervals although the two unions can overlap. The net result is that Z ∞ (0) 2 1 ` (4.19) Gh=∞(x, y, )= U(x, y, `)e d`, 2 0 where U is +1, 1, or 0. The exact values of U are complicated — that |U| 1 is all we will need. 610 FRITZ GESZTESY AND BARRY SIMON Plugging (4.19) in (4.18), we obtain Z Z Z Z n+j+p b b ∞ ∞ 2 ( 1) Mn,j,p( )= n 1 dx1 ... dxn d`1 ... d`n 1 2 0 0 0 0 nY 1 q(x1) ...q(xn) U(xj,xj+1,`j) j=1 exp( [`1 + + `n 1 + Xj(x1)+Xp(xn)]). 1 Letting = 2 [`1 + + `n 1 + Xj(x1)+Xp(xn)] and changing from d`n 1 to d (since n 2, there is an `n 2), we see that Z ∞ 2 2 (4.20) Mn,j,p( )= An,j,p( )e d , 1 2 b(j+p) where (4.21) Z Z Z ( 1)n+j+p b b An,j,p( )= n 2 dx1 ... dxn d`1 ...d`n 2 2 0 0 R(x1,...xn,`1,...,`n 2) nY 2 U(xj,xj+1,`j)U(xn 1,xn, 2 `1 ...`n 2 Xj(x1) Xp(xn)), j=1 where R(x1,...,xn,`1,...,`n 2) is the region (4.22) R(x1,...,xn,`1,...,`n 2) ( ) nX 2 = (`1,...,`n 2) `i 0 and Xj(x1)+Xp(xn)+ `k 2 . k=1 1 1 In (4.20), the integral starts at 2 b(j + p) since 2 [Xj(x1)+Xp(xn)] and (4.16) implies that Xj(x) bj. For each value of x, R is contained in the P n 2 (2 )n 2 { | } simplex (`1,...,`n 2) `i 0 and k=1 `k 2 which has volume (n 2)! . This fact and |U| 1 employed in (4.21) imply ! Z n b n 2 (4.23) |An,j,p( )| |q(x)| dx . 0 (n 2)! Moreover, by (4.20), 1 (4.24) An,j,p( )=0 if < 2 b(j + p). For any xed , the number of pairs (j, p) with j, p =0, 1, 2 ... so that 1 1 2 2 > 2 b(j + p)is 2 ([ b ] + 1)([ b ] + 2); thus, Z ∞ 2 2 (4.25) Mn( )= An( )e d , 0 with (2 + b)(2 +2b) n 2 (4.26) |A ( )| kqkn. n 2b2 (n 2)! 1 INVERSE SPECTRAL THEORY, II 611 As in [35], we can sum on n from 2 to in nity and justify extending the result to all q ∈ L1((0,b)). We therefore obtain Theorem 4.3 (Theorem 1.6 for h = ∞). Let b<∞, h = ∞, and ∈ 1 1 k k q L ((0,b)). Then for Re( ) > 2 q 1, we have that X∞ X∞ Z ∞ 2 2 bj 2 bj 2 (4.27) m( )= Aj e Bje A( )e d , 0 where j=1 j=1 (i) Aj =2. R b (ii) Bj = 2j 0 q(x) dx. | | (2 +b)(2 +2b) k k2 k k (iii) A( ) A1( ) 2b2 q 1 exp( q 1) with A1 given by (4.13). In particular, Z a 2 |A( )| d C(b, kqk1)(1 + a ) exp(akqk1). 0 As in the proof of Theorem 4.1, this implies Corollary ∈ 1 ∞ 1 k k 4.4. If q L ((0, )) and Re( ) 2 q 1 + ε, then for all a ∈ (0,b), b<∞, we have that Z a m( 2)+ + A( )e 2 d C(a, ε)e 2aRe( ), 0 where C(a, ε) depends only on a and ε (and kqk1) but not on Im( ). Remark. One can also prove results for a>bif b<∞ but this is the result we need in the next section. The case h = 0 (Neumann boundary conditions at b) is almost the same. (4.2)–(4.4) are replaced by " # sinh( x ) e x> + e (2b x>) (4.28) G(0) (x, y, 2)= < , h=0 1+e 2 b e x + e (2b x) (4.29) (0) (x, 2)= , h=0 1+e 2 b e 2 b (4.30) m(0) ( 2)= . h=0 1+e 2 b The only change in the further arguments is that U can now take the values 0, 1, and 2so|U| 2. That means that (4.26) becomes (2 + b)(2 +2b) (2 )n 2 |A ( )| 2 kqkn. n,h=0 2b2 (n 2)! 1 The net result is 612 FRITZ GESZTESY AND BARRY SIMON Theorem 4.5 (Theorem 1.6 for h = 0). Let b<∞, h =0,and q ∈ 1 L ((0,b)). Then for Re( ) > kqk1, (4.27) holds, where j (i) Aj =2( 1) . R j+1 b (ii) Bj =2( 1) j 0 q(x) dx. | | (2 +b)(2 +2b) k k2 k k (iii) A( ) A1( ) b2 q 1 exp(2 q 1) with A1 given by A1,h=0( ) q( ), 0 An analog of Corollary 4.4 holds, but we will wait for the general h ∈ R case to state it. Finally, we turn to general |h| < ∞. In this case, (4.2)–(4.4) become (0) 2 sinh( x<) (0) 2 (4.31) Gh (x, y, )= h (x>, ), " # e x + (h, )e (2b x) (4.32) (0)(x, 2)= , h 1+ (h, )e 2b (h, )e 2 b (4.33) m(0)( 2)= +2 , h 1+ (h, )e 2 b where h (4.34) (h, )= . + h To analyze this further, we need Laplace transform formulas for . Proposition 4.6. The following formulas hold in the -region h + Re( ) > 0. R ∞ (2 +2h) (i) (h, )=1 4h 0 e d . P R m m j m (4h)j ∞ j 1 (2 +2h) (ii) (h, ) =1+ j=1( 1) j (j 1)! 0 e d . R 2 ∞ (2 +2h) (iii) (h, )= 2h +4h 0 e d . P j+1 R ∞ (iv) (h, )m = 2mh 1 m ( 1)j[ m +2 m ] (4h) j 1 4 j=1 j j+1 (j 1)! 0 (2 +2h) m e d , where m+1 is interpreted as 0. INVERSE SPECTRAL THEORY, II 613 Proof. Straightforward algebra. Rewriting (4.33) as X∞ (0) 2 m+1 m 2m b mh ( )= +2 ( 1) e m=1 and then using Proposition 4.6(iv), we nd that X∞ (0) 2 m 2m b (4.35) mh ( )= 2 ( 1) e m=1 X∞ Z ∞ m+1 2m b 2 4 ( 1) mhe A0,h( )e d , m=1 2b where X∞ Xm 1 m 2( 2mb)h j (4.36) A ( )= ( 1) ∞ ( )e ( 1) 0,h 2 [2mb, ) m" =1 ! !# j=1 m m ( 2mb)j 1 +2 (4h)j+1 . j j +1 (j 1)! j 1 j 1 | | P Using the crudeP estimates (4h) ( 2mb) [2m,b)( )/(j 1)! exp(4 h ), m m m m m m j=1 j 2 , j=1 j+1 2 , and m /2b, we see that 3 (4.37) |A ( )| exp ln(2) exp(6|h| ). 0,h 2 2b 2b R b 2 2 A similar analysis of 0 q(x) 0,n(x, ) dx shows that (4.38) Z Z ! b b X∞ (0) 2 2 m+1 2b m q(x) h (x, ) dx = q(x) dx 2 ( 1) me 0 0 Z m=1 Z b ∞ 2 2 q( )e d A1,h( )e d , 0 b where A1,h satis es for suitable constants C1 and C2 (4.39) 1 |A1,h( )| C1 exp(C2(|h| +1+b ) ) [|q( nb)| + |q(n +1)b )|] for nb <(n +1)b. Finally, using (4.31) and Proposition 4.6, we write Z ∞ (0) 2 1 ` (4.40) Gh (x, y, )= U(x, y, h, `)e d`, 2 0 where 1 (4.41) |U(x, y, h, `)| C3 exp(C4(|h| +1+b )`) 614 FRITZ GESZTESY AND BARRY SIMON for suitable constants C3 and C4. From it, it follows that Z Z b b 2 n Mn( ; q)=( 1) dx1 ... dxn q(x1) ...q(xn) 0 0 nY 1 (0) 2 (0) 2 (0) 2 h (x1, ) h (xn, ) Gh (xj,xj+1, ) j=1 Z ∞ 2 = An,h( )e d , 0 where n 2 |A ( )| C 2 exp(C (|h| +1+b 1) ) kqkn,n 2. n,h 5 6 (n 2)! 1 We conclude Theorem 4.7 (Theorem 1.6 for general |h| < ∞). Let b<∞, |h| < ∞, ∈ 1 1 k k | | 1 and q L ((0,b)). Then for Re( ) > 2 D1[ q 1 + h + b +1] for a suitable universal constant D1, (4.27) holds, where j (i) Aj =2( 1) . R j+1 b (ii) Bj =2( 1) j[2h + 0 q(x) dx]. 2 (iii) |A( ) q( )| kqk exp( kqk1) if | | Hence we immediately get Corollary 4.8. Fix b<∞, q ∈ L1((0,b)), and |h| < ∞. Fix a 5. The relation between A and : Distributional form, I. Our primary goal in the next ve sections is to discuss a formula which formally says that Z ∞ √ 1 (5.1) A( )= 2 2 sin(2 ) d ( ), ∞ where for 0, we de ne ( √ 1 2 √ if =0, 2 sin(2 )= 1 ( ) 2 sinh(2 )if <0. INVERSE SPECTRAL THEORY, II 615 In a certain sense which will become clear, the left-hand side of (5.1) should be A( ) A( )+ 0( ). To understand (5.1) at a formal level, note the basic formulas, Z ∞ (5.2) m( 2)= A( )e 2 d , 0 Z ∞ 1 (5.3) m( 2) = Re(m(i)) + d ( ), ∞ + 2 1+ 2 and Z ∞ √ 2 1 1 2 (5.4) ( + ) =2 2 sin(2 )e d , 0 which is an elementary integral if >0 and >0. Plug (5.4) into (5.3), formally interchange order of integrations, and (5.2) should only hold if (5.1) does. However, a closer examination of this procedure reveals that the inter- change of order of integrations is not justi ed and indeedR (5.1) is not true as R 2 3 a simple integral since, as we will see in the next section, d ( ) R 2 , 0 R→∞ 3 which implies that (5.1) is not absolutely convergent. We will even see (in §9) that the integral sometimes fails to be conditionally convergent. Our primary method for understanding (5.1) is as a distributional state- ment, that is, it will hold when smeared in for in (0,b). We prove this in this section if q ∈ L1((0, ∞)) or if b<∞. In Section 7, we will extend this to all q (i.e., all Cases 1–4) by a limiting argument using estimates we prove in Section 6. The estimates themselves will come from (5.1)! In Section 8, we will prove (5.1) as a pointwise statement where the integral is de ned as an Abelian limit. Again, estimates from Section 6 will play a role. ∞ ∞ ∈ 1 ∈ ∞ Suppose b< or b = and q L ((0,b)). Fix a 616 FRITZ GESZTESY AND BARRY SIMON 0 Thus, since f(0+)=f (0+) = 0, in fact, f has support away from 0 and a, Z Z a a 2 b 2 0 (5.8) A( )f( ) d = √ g (2 , 0,a)e f( ) d 0 2 0 Z 2a 1 b 0 = √ g ( , 0,a)e f d 2 0 2 Z 1 = √ g(y, 0,a)F (y, 0) dy, 2 R where we have used the unitarity of b and Z 2a 1 ( 0+iy) (5.9) F (y, 0)=√ e f d 2 0 2 Z 2 a = √ e2 ( 0+iy)f( ) d . 2 0 Notice that 2(a ε) 0 2 1 (5.10) |F (y, 0)| Ce (1 + |y| ) since f is smooth and supported in (0,a ε) for some ε>0. By Theorem 4.1 and Corollary 4.8, 2 2 2a 0 (5.11) |ma( ( 0 + iy) ) m( ( 0 + iy) )| Ce for large 0, uniformly in y. From (5.8), (5.10), and (5.11), one concludes that Lemma ∈ ∞ ∈ 1 5.1. Let f C0 ((0,a)) with 0 2 N As a function of y, for 0 xed, the alpha integral is O((1 + y ) ) for all N because f is C∞. Now de ne Z 2 d ( ) (5.13)m ˜ R( )= cR + 2 , R + R → d ( ) ∞ where cR is chosen so thatm ˜ R m. Because R 2 < , the convergence R→∞ 1+ is uniform in y for 0 xed and su ciently large. Thus in (5.12) we can replace →∞ ↑∞ m byR mR and take a limit ( rst R and then 0 ). Since f(0+)=0, the dy cR d -integrand is zero. Moreover, we can now interchange the dy d INVERSE SPECTRAL THEORY, II 617 and d ( ) integrals. The result is that (5.14) Z Z a A( )f( ) d = lim lim d ( ) ↑∞ →∞ 0 0 R R "Z " Z ## a 1 e2 iy dy 2 0 d e f( ) 2 . 0 R ( 0 + iy) + In the case at hand, d is bounded below, say K0. As long as we take >K , the poles of ( + iy)2 + occur in the upper half-plane 0 0 0 √ y = i 0 . Closing the contour in the upper plane, we nd that if K0, Z √ 2 iy 1 e dy sin(2 ) 2 0 √ 2 = 2e . R ( 0 + iy) + Thus (5.14) becomes Z Z "Z √ # a a sin(2 ) A( )f( ) d = 2 lim lim f( ) √ d d ( ). ↑∞ →∞ 0 0 R R 0 2 1 0 has dropped out and theR integral is bounded by C(1 + ) , so we take →∞ d ( ) ∞ the limit as R since R 1+ 2 < . We have therefore proven the following result. Theorem ∈ ∞ ∞ 5.2. Let f C0 ((0,a)) with a We will need to strengthen this in two ways. First, we want to allow a>b if b<∞. As long as A is interpreted as a distribution with and 0 functions at = nb, this is easy. We also want to allow f to have a nonzero derivative at = 0. The net result is Theorem ∈ ∞ R ∈ R 5.3. Let f C0 ( ) with f( )= f( ), and either b<∞ or q ∈ L1((0, ∞)) with b = ∞. Then Z "Z √ # Z ∞ sin(2 ) ∞ (5.16) 2 f( ) √ d d ( )= A˜( )f( ) d , R ∞ ∞ where A˜ is the distribution ˜ 0 (5.17a) A( )= (0,∞)( )A( ) ( ∞,0)( )A( )+ ( ) if b = ∞ and 618 FRITZ GESZTESY AND BARRY SIMON ˜ 0 (5.17b) A( )= (0,∞)( )A( ) ( ∞,0)( )A( )+ ( ) X∞ + Bj[ ( 2bj) ( +2bj)] j=1 X∞ 1 0 0 + 2 Aj[ ( 2bj)+ ( +2bj)] j=1 if b<∞, where Aj,Bj are h dependent and given in Theorems 4.3, 4.5, and 4.7. The proof is identical to the argument above. f(0) is still 0 but since f 0(0) =6 0, we carry it along. ∞ (0) (0) 1 √ Example. Let b = , q (x)=0,x 0. Then d ( )= [0,∞)( ) d . Thus, Z "Z √ # ∞ ∞ sin(2 ) (5.18) 2 f( ) √ d d (0)( ) ∞ ∞ Z Z 2 ∞ ∞ √ = f( ) sin(2 ) d d . 0 ∞ √ k2 R Next, changeR variables by k =2 , that is, = 4 , and then change from ∞ 1 ∞ 0 dk to 2 ∞ dk to obtain (recall f( )= f( )) Z Z 1 ∞ ∞ (5.18) = f( ) sin( k) d kdk 2 ∞ ∞ Z 1 ∞ = √ ik f (k) dk 2 ∞ 0 = f (0) Z ∞ 0 = f( ) ( ) d , ∞ as claimed in (5.16) and (5.17a) since A(0)( )=0, 0. R R 6. Bounds on 0 d ( ) R As we will see, (1.13) implies asymptotic results on R d ( ) and (5.1) R √ R 0 b will show that ∞ e d ( ) < ∞ for all b>0 and more (for remarks on the history of the subject, see the end of this section). It follows from (5.3) that Z d ( ) Im(m(ia)) = a ,a>0. R 2 + a2 INVERSE SPECTRAL THEORY, II 619 Thus, Everitt’s result (1.13) (which also follows from our results in §§2 and 3) implies that Z 1 d ( ) 1 lim a 2 =2 2 . a→∞ R 2 + a2 Standard Tauberian arguments (see, e.g., [34, §III.10], which in this case shows 3 1 1 1 that on even functions R 2 d ( ) → | | 2 d ) then imply R R→∞ 2 Theorem 6.1. Z R 3 2 (6.1) lim R 2 d ( )= . R→∞ R 3 Remarks. 1. This holds in all cases (1–4) we consider here, including some with supp(d ) unbounded below. R R R 0 R R 2. Since we will see ∞ d is bounded, we can replace R by 0 in (6.1). We will need the following a priori bound that follows from Proposi- tions 3.1 and 3.2 Proposition 6.2. Let d be the spectral measure for a Schrodinger op- erator inR Cases 1–4. Fix a ∈ C Proof. ByR Propositions 3.1 and 3.2, we can nd C1 and z1 + depending a | | only on a and 0 q(y) dy so that |m(z1)| C1. Thus, Z d ( ) Im(m(z1)) C1 2 2 = , R ( Re(z1)) + (Im(z1)) Im(z1) Im(z1) and Z " # 2 2 d C1 ( Re(z1)) + (Im(z1)) 2 sup 2 Ca. R 1+ Im(z1) ∈R 1+ R √ 0 2 Our main goal in the rest of this section will be to bound ∞ e d ( ) for any 620 FRITZ GESZTESY AND BARRY SIMON Proposition 6.3. Let b = ∞ and let d be the spectral measure for a problem of types 2–4. Let d R,h be the spectral measure for the problem with b = R<∞,h, and potential equal to q(x) for x R. Then there exists h(R) so that → d R,h(R) d , R→∞ when smeared with any function f of compact support. In particular, if f 0, then Z Z f( ) d ( ) lim f( ) d R,h(R)( ). R R→∞ R This result implies that we need only obtain bounds for b<∞ (where we have already proven (5.15)). Lemma 6.4. If 1 has support in [ E0, ∞), E0 > 0, then Z Z 0 √ √ 0 d ( ) E0 2 1 (6.3) e d 1( ) e (1 + E0 ) . ∞ ∞ 1+ 2 Proof. Obvious. R ∞ 1 Now let f be xed in C0 ((0, 1)) with f 0 and 0 f(y) dy = 1. Let f 0 ( )=f( 0). Let d 2 be the spectral measure for some problem with b 0 +1 and let d 1 be the spectral measure for the problem with b = 0 +1, ∞ h = , and the same potentialR on [0, 0 + 1]. Then, by Theorem 1.3, A1( )= A ( ) for ∈ [0, +1]so 0+1 f ( )[A ( ) A ( )] d = 0, and thus by 2 0 0 0 1 2 Theorem 5.2, Z (6.4) G ( )[d 1( ) d 2( )]=0, R 0 where Z √ +1 0 sin(2√ ) (6.5) G 0 ( )= f 0 ( ) d . 0 Lemma | | 6.5. (i) For 0, G 0 ( ) 2(1 + 0). R | | 2 1 1 | 000 | 2 (ii) G 0 ( ) 8 0 f (u) du := C0 for >0. √ | | 2( 0+1) (iii) For 0, G 0 ( ) 2( 0 +1)e . √ (iv) For 0, G ( ) √1 [e2 0 1]. 0 2 √ √ | | || | | Proof. (i) Since sin(x) R x , sin(2 )/ 2 . Thus, since supp(f ) [ , + 1] and 0+1 f ( ) d =1,|G ( )| 2(1 + ). 0 0 0 0 0 0 0 √ √ 1 d3 (ii) 1 d 3 cos(2 ) = sin(2 ), so this follows upon integrating (2 2 )3 by parts repeatedly. INVERSE SPECTRAL THEORY, II 621 (iii), (iv) For y 0, Z sinh(y) 1 y = cosh(u) du y y 0 1 u u y so 2 e cosh u e e ,0 u y implies ey 1 sinh(y) ey. 2y y RThis implies (iii) and (iv) given supp(f 0 ) [ 0, 0 + 1], f 0 ( ) 0, and 0+1 f ( ) d =1. 0 0 We can plug in these estimates into (6.4) to obtain Z 0 h √ i 1 2 0 √ e 1 d 2( ) T1 + T2 + T3, ∞ 2 where, Z d j( ) Tj = max(4(1 + 0), 2C0) 2 ,j=1, 2, Z R 1+ 0 √ 2( 0+1) T3 = 2( 0 +1)e d 1( ), ∞ and we have used 2 1 , 0 1, 1+ 2 1 2 , 1. 2 1+ 2 Thus, Propositions 6.2, 6.3 and Lemma 6.4 together with Z u 1 e 1 = eyu dy e(1 )u u 0 for any u>0 and any ∈ R imply Theorem 6.6. Let be the spectral measure for some problem of the types 2–4. Let (6.6) Z 0+1 | 0 |2 | |2 ∈ ∞ E( 0):= inf ( ϕn(x) + q(x) ϕ(x) dx) ϕ C0 ((0, 0 + 1)), Z0 0+1 |ϕ(x)|2 dx 1 . 0 Then for all >0 and 0 > 0, (6.7)Z 0 √ √ 2(1 ) 0 2 2( 0+1) E( 0) 0 e d ( ) C1(1 + 0)+C2(1 + E( 0) )e , ∞ 622 FRITZ GESZTESY AND BARRY SIMON R 1 | | where C1,C2 only depend on 0 q(x) dx. In particular, Z 0 √ (6.8) eB d ( ) < ∞ ∞ for all B<∞. 2 2 As a special case, suppose q(x) C(x+1) . Then E( 0) C( 0 +2) and we see that Z √ 0 2 B D2B (6.9) e d ( ) D1e . ∞ This implies Theorem 6.7. If d is the spectral measure for a potential which satis es (6.10) q(x) Cx2,x R for some R>0, C>0, then for ε>0 su ciently small, Z 0 (6.11) e ε d ( ) < ∞. ∞ Remarks. 1. Our proof shows in terms of the D2 of (6.9), one only needs that ε<1/4D2. 2. Our proof implies that if 1 lim max(0, q(x))=0, x→∞ x2 then (6.11) holds for all ε>0. Proof. (6.9) implies that Z 2 n 2 D2B Bn d ( ) D1e e . (n+1)2 Taking B = n/2D2, we see that Z 2 n n2 4D d ( ) D1e 2 . (n+1)2 Thus, Z Z 0 X∞ n2 2 e ε d ( ) eε(n+1) d ( ) ∞ 2 n=0 (n+1) X∞ 2 2 n ε(n+1) 4D D1e e 2 < ∞ n=0 if ε<1/4D2. INVERSE SPECTRAL THEORY, II 623 Remark. If in addition q ∈ L1((0, ∞)), then the corresponding Schr o- dinger operator is bounded from below and hence d has compact support on ( ∞, 0]. This fact will be useful in the scattering theoretic context at the end of Section 8. The estimate (6.8), in the case of non-Dirichlet boundary conditions at x =0+, appears to be due to Marchenko [26]. Since it is a fundamental ingredient in the inverse spectral problem, it generated considerable attention; see, for instance, [12], [18], [19], [20], [22], [27], [28, §2.4]. The case of a Dirichlet boundary at x =0+ was studied in detail by Levitan [20]. These authors, in addition to studying the spectral asymptotics of ( )as ↓ ∞, were also particularly interested in the asymptotics of ( )as ↑∞and established Theorem 6.1 (and (A.9)). In the latter context, we also refer to Bennewitz [4], Harris [16], and the literature cited therein. In contrast to these activities, we were not able to nd estimates of the type (6.7) (which implies (6.8)) and (6.11) in the literature. 7. The relation between A and : Distributional form, II We can now extend Theorem 5.2 to all cases. Theorem ∈ ∞ ∞ ∞ 7.1. Let f C0 ((0, )) and suppose b = . Assume q satis es (1.3) and let d be the associated spectral measure and A the associated A-function. Then (5.16) and (5.17) hold. ∈ ∞ Proof. Suppose f C0 ((0,a)). For R>a, we can nd h(R)sod R,h(R) → d (by Proposition 6.3) weakly. By Proposition 6.2 we have uniform R→∞ R R √ ∞ 2 1 0 2a( bounds on 0 (1+ ) d R,h(R) and by Theorem 6.6 on ∞ e d R,h(R). 2 1 Since√ the integral in (5.15) is bounded by C(1 + ) for >0 and by Ce2a for 0, the right-hand side of (5.15) converges as R →∞to the d integral. By Theorem 1.3, A is independent of R for ∈ (0,a) and R>a, so the left-hand side of (5.15) is constant. Thus, (5.15) holds for d . 8. The relation between A and , III: Abelian limits ∈ ∞ R ∈ R For f C0 ( ), de ne for Z √ ∞ sin(2 ) (8.1) Q(f)( )= f( ) √ d ∞ 624 FRITZ GESZTESY AND BARRY SIMON and then Z (8.2) T (f)= 2 Q(f)( ) d ( ) R Z ∞ (8.3) = A˜( )f( ) d . ∞ ∈ ∞ R We have proven in (5.16), (5.17) that for f C0 ( ), the two expressions (8.2), (8.3) de ne the same T (f). We only proved this for odd f’s but both integrals vanish for even f’s. We will use (8.2) to extend to a large class of f, ∈ ∞ R but need to exercise some care not to use (8.3) except for f C0 ( ). Q(f) can be de ned as long as f satis es k| | (8.4) |f( )| Cke , ∈ R for all k>0. In particular, a simple calculation shows that √ h i 1 ( )2/ε sin(2 0 ) ε (8.5) f( )=( ε) 2 e 0 ⇒ Q(f)( )= √ e . We use f( , 0,ε) for the function f in (8.5). For 0, repeated integrations by parts show that " # 3 d f (8.6) |Q(f)( )| C(1 + 2) 1 kfk + , 1 d 3 1 1 where k k1 represents the L (R)-norm. Moreover, essentially by repeating the calculation that led to (8.5), we see that for 0, | | 2 (8.7) |Q(f)( )| Ceε e+ /εf . ∞ We conclude R Proposition 2 1 ∞ R 8.1. If R d ( )(1 + ) < (always true!) and 0 ε ∞ e 0 d ( ) < ∞ (see Theorem 6.7 and the remark following its proof ), 2 then using (8.2), T ( ) can be extended to C3(R)f’s that satisfy e /ε0 f ∈ ∞ R d3f ∈ 1 R L ( ) for some ε0 > 0 and d 3 L ( ), and moreover, " # 3 d f 2 (8.8) |T (f)| C + e /ε0 f 3 ∞ d 1 ||| ||| := C f ε0 . R 0 ε Next, x 0 and ε0 > 0 so that ∞ e 0 d ( ) < ∞.If0<ε<ε0, ||| ||| ∞ ∈ ∞ R f( , 0,ε) satis es f ε0 < so we can de ne T (f). Fix g C0 ( ) with ||| ||| → ↓ g :=1on( 2 0, 2 0). Then f( , 0,ε)(1 g) ε0 0asε 0. So lim T (f( , 0,ε)) = lim T (gf( , 0,ε)). ε↓0 ε↓0 INVERSE SPECTRAL THEORY, II 625 For gf, we can use the expression (8.3). f is approximately ( 0)so standard estimates show if 0 is a point of Lebesgue continuity of A˜( ), then Z ∞ f( , 0,ε)g( )A˜( ) d → A˜( 0). ∞ ε↓0 Since A q is continuous, points of Lebesgue continuity of A exactly are points of Lebesgue continuity of q. We have therefore proven Theorem 8.2. Suppose either b<∞ and q ∈ L1((0,b)) or b = ∞, and then either q ∈ L1((0, ∞)) or q ∈ L1((0,a)) for all a<∞ and q(x) Cx2,x R for some R>0, C>0. Let 0 ∈ (0,b) and be a point of Lebesgue continuity of q. Then Z √ ε sin(2 0 ) (8.9) A( 0)= 2 lim e √ d ( ). ε↓0 R We brie y illustrate the rate of convergence as ε ↓ 0 in (8.9) in√ the special (0) (0) 1 case where q (x)=0,x 0. Then d ( )=√ [0,∞)( ) d and formula 3.9521 of [15] (changing variables to k = 0) yield Z ∞ √ (8.10) A(0)( )= 2 1 lim e ε sin(2 ) d ε↓0 0 ! 2 1 3 = 2 2 lim ε 2 exp =0, 0. ε↓0 ε Finally, we specialize (8.9) to the scattering theoretic setting. Assuming q ∈ L1((0, ∞); (1 + x) dx), the corresponding Jost solution f(x, z) is de ned by √ √ Z ∞ 0 sin( z (x x )) 0 0 0 √ (8.11) f(x, z)=ei zx √ q(x )f(x ,z) dx , Im( z ) 0 x z √ and the corresponding Jost function, F ( z ), and scattering matrix, S( ), 0, then read √ (8.12) F ( z )=f(0+,z), √ √ (8.13) S( )=F ( )/F ( ), 0. The spectrum of the Schr odinger operator in L2((0, ∞)) associated with the d2 di erential expression dx2 + q(x) and a Dirichlet boundary condition at x =0+ is simple and of the type { 2 } ∪ ∞ j < 0 j∈J [0, ). 626 FRITZ GESZTESY AND BARRY SIMON Here J is a nite (possibly empty) index set, j > 0, j ∈ J, and the essential spectrum is purely absolutely continuous. The corresponding spectral measure explicitly reads ( √ √ 1|F ( )| 2 d , 0, (8.14) d ( )= P 2 j∈J cj ( + j ) d , < 0, where k 2 k 2 ∈ (8.15) cj = ϕ( , j ) 2 ,jJ 2 are the norming constants associated with the eigenvalues j = j < 0. Here ϕ(x, z) (which has been introduced in (3.6a) and (3.10)) and f(x, z) in (8.11) 2 ∈ are linearly dependent precisely for z = j , j J. Since ! Z √ Y 2 1 ∞ ( 0) d 0 |F ( )| = 1+ j exp P , 0, 0 j∈J 0 R ∞ where P 0 denotes the principal value symbol and ( ) the corresponding scattering phase shift, that is, S( ) = exp(2i ( )), ( ) → 0, the scattering ↑∞ data { 2 } ∪{ } j ,cj j∈J S( ) 0 uniquely determine the spectral measure (8.14) and hence A( ). Inserting (8.14) into (8.9) then yields the following expression for A( ) in terms of scat- tering data. Theorem 8.3. Suppose that q ∈ L1((0, ∞); (1 + x) dx). Then X 1 (8.16) A( )= 2 cj j sinh(2 j) j∈J Z ∞ √ √ 2 1 lim e ε |F ( )| 2 sin(2 ) d ε↓0 0 at points 0 of Lebesgue continuity of q. Remark. In great generality |F (k)|→1ask →∞, so one cannot take the limit√ in ε inside the√ integral in (8.16). In general, though, one can can replace |F ( )| 2 by (|F ( )| 2 1) X( ) and ask if one can take a limit there. As long as q is C2((0, ∞)) with q00 ∈ L1((0, ∞)), it is not hard to see that as →∞ q(0) X( )= + O( 2). 2 INVERSE SPECTRAL THEORY, II 627 Thus, if q(0) = 0, then X 1 (8.17) A( )= 2 cj j sinh(2 j) j∈J Z ∞ √ √ 2 1 (|F ( )| 2 1) sin(2 ) d . 0 The integral in (8.17) is only conditionally convergent if q(0) =0.6 We note that in the present case where q ∈ L1((0, ∞); (1 + x) dx), the representation (1.17) of the m-function in terms of the A( )-amplitude was considered in a paper by Ramm [31] (see also [32, pp. 288–291]). 9. The relation between A and , IV: Remarks Here is a totally formal way of understanding why (5.1) is true. We start with the basic representation without errors, Z ∞ 2 2 2 (9.1) m( )=m0( ) A( )e d . 0 Pretend we can analytically continue from real to = ik (at which point 2 is k2 + i0). Then Z ∞ 2 2 2i k (9.2) m(k + i0) = m0(k + i0) A( )e d . 0 This normally cannot be literally true. In many cases, A( ) →∞at in nity (although for the case q(x) = constant > 0, which we discuss later, it is true). But this is only a formal argument. Taking imaginary parts and using for , 0 > 0 that Z ∞ 2 (9.3) sin(2 k) sin(2 0k) dk = ( 0) 0 8 R ∞ i k (which follows from ∞ e dk =2 ( )), we conclude that for 0 > 0, Z ∞ 4 2 (9.4) A( 0)= sin(2 0k)Im(m0(k + i0)) dk 0 Z ∞ √ Im(m0( + i0)) d = 2 sin(2 0 ) √ , 0 which, given (1.12), is just (5.1). As explained in [35], a motivation for A is the analogy to the m-function for a tridiagonal Jacobi matrix. For this point of view, the relation (5.1) is an important missing link. The analog of (1.7) in the discrete case is X∞ n (9.5) m(z)= n+1 . n=0 z 628 FRITZ GESZTESY AND BARRY SIMON The coe cients of n of the Taylor series at in nity are the analog of A( ). In this case, the spectral measure is nite and of nite support (if the Jacobi matrix is bounded) and Z d ( ) (9.6) m(z)= R z so that (9.5) implies that Z n (9.7) n = d ( ). R (5.1) should be then thought of as the analog of (9.7) for the continuum case. Perhaps the most important consequence of (9.7) is the implied positivity condition of the ’s — explicitly, that XN n+mam an 0 m,n=0 N+1 for all (a0,...,aN ) ∈ C . Recall (see, e.g., Gel0fand-Vilenkin [13, §II.5]) that Krein proved the fol- lowing fact: Theorem 9.1. A continuous even function f on R has the property that Z (9.8) f(x y)ϕ(y) ϕ(x) dxdy 0 R2 ∈ ∞ R for all even functions ϕ C0 ( ) if andR only if there are nite positive mea- ∞ ∞ a ∞ sures d 1 and d 2 on [0, ) so that 0 e d 2( ) < for all a>0 and so that Z ∞ Z ∞ (9.9) f(x)= cos( x) d 1( )+ cosh( x) d 2( ). 0 0 Using the extension in Gel0fand-Vilenkin to distributional f (cf. [13, §II.6.3, Theorem 5]), one obtains Theorem 9.2. Let A˜( ) be the distribution of Theorem 5.3. Let B( )= A˜0( ) be the distributional derivative of A˜. Then Z Z (9.10) B( ) ϕ( )ϕ( ) d d 0 R R ∈ ∞ R ˜ for all even ϕ C0 ( ). Moreover, if A is a distribution related to a signed measure, d , by (5.16), then (9.10) is equivalent to the positivity of the measure d . INVERSE SPECTRAL THEORY, II 629 As discussed in [13], the measures d j in (9.9) may not be unique. Our theory illuminates this fact. If q is in the limit circle case at in nity, then distinct boundary conditions lead to distinct spectral measures but the same A-function, so the same A˜ and the same B = A˜0. Thus, we have additional examples of nonuniqueness.R The growth restrictions on f which guarantee cx2 uniqueness in (9.9) (e.g., R e f(x) dx < ∞ for all c>0) are not unrelated to the standard q(x) Cx2 that leads to the limit point case at ∞ for the d2 Schr odinger di erential expression dx2 + q(x). Next we turn to the relation between A and the Gel0fand-Levitan trans- formation kernel L in [12]. For the function L(x, y) associated to Dirichlet boundary conditions at x = 0, satisfying (cf. (3.6a), (3.10)) √ Z x sin( zx) 0 0 0 √ = ϕ(x, z)+ L(x, x )ϕ(x ,z) dx , z 0 we claim that ∂ (9.11) A( )= 2 L(2 , y) . ∂y y=0+ We will rst proceed formally without worrying about regularity conditions. Detailed discussions of transformation operators can be found, for instance, in [11], [21, Ch. 1], [22], [24], [25], [26], [28, Ch. 1], [30, Ch. VIII], [36], [37], and, in the particular case of scattering theory, in [2, Chs. I and V], [8], and [29]. 2 Let d ( ) be the spectral measure for d + q(x) and dx2 √ (0) 1 (9.12) d ( )= [0,∞)( ) d d2 the spectral measure for dx2 (both corresponding to the Dirichlet boundary condition parameter h = ∞ at x = 0), and de ne d = d d (0). Then L is de ned as follows [12]. Let Z √ √ ∞ [1 cos( x)][1 cos( y)] (9.13) F (x, y)= d ( ) ∞ 2 and Z √ √ ∂2F ∞ sin( x) sin( y) (9.14) k(x, y)= (x, y) “=” d ( ), ∂x∂y ∞ where the nal “=” is formal since the integral may not converge absolutely. L satis es the following nonlinear Gel0fand-Levitan equation, Z y 0 0 0 (9.15) k(x, y)=L(x, y)+ L(x, x )L(y, x ) dx , 0 Z x 1 0 0 (9.16) L(x, 0+)=0,L(x, x)= q(x ) dx . 2 0 Thus, formally by (9.15) and (9.16), ∂k ∂L (9.17) (x, y) = (x, y) , ∂y y=0+ ∂y y=0+ 630 FRITZ GESZTESY AND BARRY SIMON and then by (9.14) Z √ ∞ ∂k sin( x) (9.18) (x, y) “=” √ d ( ), ∞ ∂y y=0+ which, by (5.1), says that (9.11) holds. Alternatively, one can derive (9.11) as follows. Suppose Q ∈ L1((0, ∞)) coincides with q on the interval [0, ], is real-valued, and of compact support. √ 0 Denote by fQ(x, z), FQ( z ), and LQ(x, x ) the Jost solution, Jost function, and transformation kernel (satisfying (9.15), (9.16)) associated with Q. Then (cf. [5, §V.2]), √ Z ∞ √ fQ(x, z) i zx 0 i zx0 0 (9.19) √ = e + LQ(x ,x)e dx , FQ( z ) x and fQ(x, z) (9.20a) uQ(x, z)= √ ,uQ(0+,z)=1, FQ( z ) 2 (9.20b) uQ( ,z) ∈ L ((0, ∞)),z∈ C\R is the unique Weyl solution association with Q. Thus, the normalization of uQ in (9.20a), (9.19), LQ(0+, 0+) = 0, and (1.7) then yield Z ! √ ∞ √ 0 ∂ 0 i zx0 0 (9.21) mQ(z)=uQ(0+,z)=i z + LQ(x ,x) e dx . 0 ∂x x=0+ Identifying z = 2, x0 =2 , a comparison with (1.17) then implies AQ( )= 2LQ,y(2 , 0+). Since by Theorem 1.3 and the following remark, A( ) only depends on q(x)= Q(x) for x ∈ [0, ], and L(x, y) depends on q(x0)=Q(x0) for x0 ∈ [0, (x + y)/2] with 0 y x 2 (cf. [5, eq. (III.1.11)], [28, pp. 19, 20]), one concludes (9.11). Next, we want to note that (5.1) sometimes does not represent a condi- tionally convergent integral; that is, Z R √ 1 (9.22) A( )= 2 lim 2 sin(2 ) d ( ) R→∞ ∞ can fail. Indeed, it even fails in the case b<∞, h = ∞, and q(0)(x)=0, 0 x b. For in that case (see (4.4)), + e 2 b m(0)( 2)= . 1 e 2 b INVERSE SPECTRAL THEORY, II 631 Straightforward residue calculus then implies that X∞ (0) (9.23a) d ( )= wn (E En), n=1 with 2n2 (9.23b) E = n b2 and 2 2n2 (9.23c) w = n b3 R (the reader might want to check that this is consistent with R d ( ) 0 R→∞ 2 3/2 3 R ). Thus, Z R √ X 1 (0) 2 n 2 sin(2 ) d ( )= sin(2 n/b) ∞ b2 n bR1/2/ is not conditionally convergent as R →∞. Given the known asymptotics for the eigenvalues and weights when b<∞ (cf., e.g., [23, §1.2]), one can see that (9.22) never holds if b<∞. There are also cases with b = ∞, where it is easy to see the integral cannot be conditionally convergent. If q(x)=x , >0 then WKB analysis (see, e.g., [38, §7.1]) shows that En = [Cn + O(1)] , n↑∞ 1 1 1 1 1/ where = 2 + and wn = CEn (1 + o(1)). As long as >2, 1 2 wnEn → ∞, and so the integral is not conditionally convergent. n→∞ Another canonical scenario displaying this phenomenon is provided by the scattering theoretic setting discussed at the end of Section 8. In fact, assuming q ∈ L1((0, ∞); (1 + x) dx), one sees that (9.24) |F (k)| = 1+o(k 1) k↑∞ (cf. [5, eq. II.4.13] and apply the Riemann-Lebesgue lemma. Actually, one only needs q ∈ L1((0, ∞)) for the asymptotic results on F (k)ask ↑∞but we will ignore this re nement in the following.) A comparison of (9.24) and (8.16) then clearly demonstrates the necessity of an Abelian limit in (8.16). Even replacing√ d in (8.9) by√d = d d (0) (cf. (8.10)); that is, e ectively replacing |F ( )| 2 by [|F ( )| 2 1] in (8.16) still does not necessarily produce an absolutely convergent integral in (8.16). 632 FRITZ GESZTESY AND BARRY SIMON The latter situation changes upon increasing the smoothness properties of q since, for example, assuming q ∈ L1((0, ∞); (1 + x) dx), q0 ∈ L1((0, ∞)), yields |F (k)| 2 1=O(k 2) k↑∞ as detailed high-energy considerations (cf. [14]) reveal. Indeed as we saw at the end of Section 8, if q00 ∈ L1((0, ∞)), then the integral one gets is absolutely convergent if and only if q(0)=0. Unlike the oscillator-like cases, though, the integrals in the scattering the- ory case are conditionally convergent. These examples allow us to say something about the following question raised by R. del Rio [9]. Does |m( 2)+ | stay bounded as z = 2 moves along the curve Im(z)=a0 > 0 with Re(z) →∞? In general, the answer is no. The m( 2) of (4.4) has G( 2):=|m( 2)+ | =2| ||e 2 b|/|1 e 2 b| so q n 2 G(( ) + ia0)/ |En|→∞ b n→∞ 2 (En =( n/b) , n ∈ N denoting the corresponding eigenvalues) showing |m( 2)| is not even bounded by C| | on the curve. Similarly, in the case 2 q(x)=x , one infers that |m( )+ |/| | is unbounded at En + ia0 as long as >2. As a nal issue related to the representation (5.1), we discuss the issue of bounds on A when |q(x)| Cx2. We have two general bounds on A: the estimate of [35] (see ((1.16)), Z Z 2 (9.25) |A( ) q( )| |q(y)| dy exp |q(y)| dy , 0 0 and the estimate we will prove in the next section (Theorem 10.2), ( ) (9.26) |A( )| I (2 ( )), 1 | | | |1/2 where ( ) = sup0 x q(x) and I1( ) is the modi ed Bessel function of order one (cf., e.g., [1, Ch. 9]). Since ([1, p. 375]) x (9.27) 0 I1(x) e ,x 0, we conclude that √ √ 2 (9.28) |A( )| Ce2 C if |q(x)| Cx2. This is a pointwise bound related to the integral bounds on A( ) implicit in Lemma 6.5. INVERSE SPECTRAL THEORY, II 633 10. Examples, I: Constant q We begin with the case b = ∞, q(x)=q0, x 0, with q0 a real constant. We claim Theorem 10.1. If b = ∞ and q(x)=q0, x 0, then if q0 > 0, 1 q 2 1 (10.1) A( )= 0 J (2 q 2 ), 1 0 where J1( ) is the Bessel function of order one (cf., e.g., [1, Ch. 9]); and if q0 < 0, 1 ( q0) 2 1 (10.2) A( )= I (2 ( q ) 2 ), 1 0 with I1( ) the corresponding modi ed Bessel function. Proof. We use the following formula ([15, 6.6233]), Z √ ∞ 2 2 dx a + b a (10.3) e axJ (bx) = ,a>0,b∈ R, 1 x b Z0 √ ∞ 2 2 ax dx a b a (10.4) e I1(bx) = ,a>0, |b| This example is especially important because of a monotonicity property: Theorem 10.2. Let |q1(x)| q2(x) on [0,a] with a min(b1,b2). | | Then A1( ) A2( ) on [0,a]. In particular, for any q satisfying sup0 x |q(x)| < ∞, we have that ( ) (10.5) |A( )| I (2 ( )), 1 634 FRITZ GESZTESY AND BARRY SIMON where 1 (10.6) ( ) = sup (|q(x)| 2 ). 0 x In particular, (9.27) implies (10.7) |A( )| 1 ( )e2 ( ), and if q is bounded, 1 1 (10.8) |A( )| 1kqk∞2 exp(2 kqk∞2 ). Proof. Since A( ) is only a function of q on [0, ), we can suppose that b1 = b2 = ∞ and q1(x)=q2(x)=0forx>a. By a limiting argument, we can ∞ suppose that qj are C ([0,a]). We can then use the expansion of [35, §2], Z X∞ n ( 1) (10.9) A( )= q( )+ n 2 q(x1) ...q(xn) n=2 2 Rn( ) dx1 ...dxn d`1 ...d`n 2, where Rn( ) is a complicated region on {x, `} space that is q independent (given by (2.19) from [35]). The monotonicity result follows immediately from this expression. (10.5) then follows from (10.2), and (10.7), (10.8) from (9.27). Remarks. 1. The expansion X∞ 1 2n+1 ( 2 z) I1(x)= n=0 n!(n + 1)! allows one to compute exactly the volume of the region Rn( ) of [35], viz., 2n 2 |R ( )| = . n (n 1)!n! | | 2n 2 The bounds in [35] only imply Rn( ) (n 2)! and are much worse than the actual answer for large n! 2. For small, (10.7) is a poor estimate and one should use (9.25) which 2 2 2kqk∞ implies that |A( ) kqk∞ + kqk∞e . This lets us prove ∞ Theorem 10.3. Let h = ∞ and q ∈ L ((0, ∞)). Suppose 2 > kqk∞. Then Z ∞ (10.10) m( 2)= A( )e 2 d 0 (with a convergent integral and no error term). Proof. Let qn = q [0,n](x). Let mn, An be the m-function and A-amplitude, respectively, for qn. Then INVERSE SPECTRAL THEORY, II 635 (1) mn(z) → m(z) for z ∈ C\[ kqk∞, ∞). (2) An( ) → A( ) pointwise (since An( )=A( )ifn> ). 1 (3) (10.10) holds for qn since qn ∈ L ((0, ∞)) (see Theorem 1.2). 1 1 1 2 2 (4) |An( )| kqk∞ exp(2 kqk∞). This is (10.8). 2 2 (5) |An( )| kqk∞[1 + kqk∞ exp( kqk∞)]. This is (9.25). The dominated convergence theorem thus implies that (10.10) holds for q ∈ L∞((0, ∞)). Remarks. 1. If inf supp(d )= E0 with E0 > 0, then m(z) has a singu- 2(E ε) larity at z = E0 so we cannot expect that |A( )| Ce 0 for any ε>0. Thus, A must grow exponentially as →∞. One might naively guess that if inf supp(d )=E0 with E0 > 0, then A( ) decays exponentially, but this is false in general. For example, if q(x)=q0 > 0, then by (10.1) for large, 1 1 3 1 2 4 2 2 2 A( ) q0 cos(2q0 + 4 )+O( ) by the known asymptotics of J1 ([1, p. 364]). 2. For q(x)=q0 > 0, A( ) → 0as →∞. This leads one to ask if perhaps A( ) → 0 for all cases where supp( ) [0, ∞) or at least if q(x) > 0. It would be interesting to know the answer even for the harmonic oscillator. 3. We have proven exponential bounds on A( )as →∞for the cases ∈ 1 ∞ ∈ ∞ ∞ 1 ∞ ∞ ∞ q L ((0, )) and q L ((0,R )), but not even for L ((0, )) + L ((0, )). x+1 | | One might guess that supx>0( x q(y) dy) su ces for such a bound. 11. Examples, II: Bargmann potentials Our second set of examples involves Bargmann potentials (cf., e.g., [5, §§IV.3 and VI.1]), that is, potentials q ∈ L1((0, ∞); (1 + x) dx) such that the associated Jost function F (k) (cf. (8.12)) is a rational function of k.We explicitly discuss two simple examples and then hint at the general case. (0) Case 1. F (k)=(k i 1)/(k + i 1). Thus, d ( )=d ( )on[0, ∞) 2 and there is a single eigenvalue at energy = 1. There is a single norming constant, c1, and it is known (cf. [5, §VI.1]) that Z 2 x d c1 2 (11.1) q(x)= 2 2 ln 1+ 2 sinh ( 1y) dy . dx 1 0 In (5.1), the >0 contribution to A( ) is the same as in the free case, and so it yields zero contribution to A (cf. (8.10)). Thus, Z 0 q | | 1 | | 2 A( )= 2c1 2 sinh(2 ) ( + 1) d ∞ 636 FRITZ GESZTESY AND BARRY SIMON and hence 2c1 (11.2) A( )= sinh(2 1). 1 Note that q(0+) = A(0+) = 0 (verifying q(0+) = A(0+)). Case 2. F (k)=(k + i )/(k + i ), >0, 0. It is known (cf. [5, p. 87]) that e 2 x (11.3) q(x)= 8 2 . + 2 x 2 (1+( + )e ) The case = 0 corresponds to q(x)= 2 2/ cosh2( x) (the one-soliton po- tential on its odd subspace). We claim that 2 2 (11.4) m( 2)= , + for clearly, m( 2) is analytic in C\[0, ∞) and satis es m( 2)= +O( 1) and at = ik (i.e., E = 2 = k2 + i0), " # " # 2 2 k2 + 2 k Im(m(k2 + i0)) = k k = k = 2 + k2 k2 + 2 |F (k)|2 consistent with (8.14). Thus, uniqueness of m given d and the asymptotics proves (11.4). Since Z ∞ 1 =2 e 2 e 2 d , + 0 (11.4) and uniqueness of the inverse Laplace transform implies that (11.5) A( )=2( 2 2)e 2 . Notice that q(0+) = A(0+) = 2( 2 2) and the odd soliton ( =0) corresponds to A( )= 2 2, a constant. Remark. Thus, we see that A( ) equal to a negative constant is a valid A-function. However, A( ) a positive constant, say, A0 > 0, is not since then Im(m(k + i0)) = k A0/k is negative for k>0 small. In the case of a general Bargmann-type potential q(x), one considers a Jost function of the form ! Y k i k Y k + i (11.6) F (k)= j ` , k + i j k + i 0 k + i ` j∈Je `∈Jr j > 0,j∈ Je, 0 0, ` > 0, ` > 0,`∈ Jr, 0 ` =6 `, for all `, ` ∈ Jr, ` =6 j for all ` ∈ Jr,j∈ Je, INVERSE SPECTRAL THEORY, II 637 with Je (resp. Jr) a nite (possibly empty) index set associated with the eigen- 2 values j = j < 0 (resp. resonances) of q, and 0 0 associated with a possible zero-energy resonance of q. Attaching norming constants cj > 0to 2 ∈ the eigenvalues j = j , j Je of q, one then obtains √ √ 1[|F ( )| 2 1] d , 0 (11.7) d ( ) d (0)( )= P 2 j∈Je cj ( + j ) d , < 0 P √ 1 2 1 `∈J ∪{0} A`( + ` ) d , 0 = P r 2 j∈Je cj ( + j ) d , < 0. Here d (0) denotes the spectral measure (9.12) for the free case q(0)(x)=0, x 0, and 0 =0, Q Q 2 2 2 2 1 ∈ ∪{ }( ) ( ) ,`∈ J ∪{0}, > 0, m Jr 0 m ` n ` r 0 n∈Jr∪{0} n=6 ` A = Q Q ` 2 2 2 2 1 ∈ m∈Jr ( m ` ) ( n ` ) ,`Jr, 0 =0, n∈Jr n=6 ` and A0 =0if 0 =0. Next, observing the spectral representation for the free Green’s function associated with q(0)(x)=0,x 0 and a Dirichlet boundary condition at x =0 , one computes + √ √ Z ∞ √ 1 sin(√ x) sin(√ y) 1 x sinh( y) (11.8) 2 d = e ,x y. 0 + Taking into account A(0)( )=0, 0 according to (8.10) (hence subtracting d (0) in (11.7) will have no e ect on computing A( ) using (8.16)), the y- ↓ derivative of the integral (11.8) at y =0+ combined withR an Abelian limit√ ε 0 ∞ ε yields precisely the prototype of integral (viz., limε↓0+ 0 e sin(2 )( + 2) 1 d ) needed to compute A( ) upon inserting (11.7) into (8.16). The net result then becomes X X 1 2 ` (11.9) A( )= 2 cj j sinh(2 j) 2 A`e . j∈Je `∈Jr∪{0} The corresponding potential q(x) can be computed along the lines indicated in [5, Ch. IV] and is known to be continuous on [0, ∞). Hence (11.9) holds for all 0. More precisely, the condition q ∈ L1((0, ∞); (1 + x) dx) imposes certain restrictions on the possible choice of ` > 0, ` > 0 in (11.6) in order 2 to avoid isolated singularities of the type 2(x x0) in q(x). Away from such isolated singularities, (11.7) inserted into the Gel0fand-Levitan equation yields a C∞ potential q (in fact, a rational function of certain exponential functions and their x-derivatives) upon solving the resulting linear algebraicP system of equations. In particular, one obtains q(0+)=A(0+)= 2 `∈Jr∪{0} A`. 638 FRITZ GESZTESY AND BARRY SIMON Appendix A. The Bh function Throughout this paper, we have discussed the principal m-function, m(z) given by (1.7). This is naturally associated to Dirichlet boundary conditions because the spectral measure d of (1.10) is a spectral measure for an operator H, with u(0+) = 0 boundary conditions. For h ∈ R, there are subsidiary m-functions, mh(z), associated to 0 (A.1) u (0+)+hu(0+)=0 boundary conditions. Our goal in this section is to present Laplace transform asymptotics for mh(z). One de nes mh(z)by (A.2) mh(z)=[hm(z) 1]/[m(z)+h]. That this is associated to the boundary (A.1) is hinted at by the fact that 0 m(z)+h = 0 if and only if u (0+,z)+hu(0+,z) = 0. The function h 1 F ( )= h + h satis es (i) Fh : C+ → C+, where C+ = {z ∈ C | Im(z) > 0}, (1+h2) (ii) Fh( )=h +h , 1 1 (iii) Fh( ) Fh( 0)=( 0)( + h) ( 0 + h) . This implies 0 (A.3) (i ) Im(mh(z)) > 0ifIm(z) > 0, 2 0 2 (1 + h ) 2 (A.4) (ii ) mh( )=h + + O( ), | |→∞ 0 2 (0) 2 2 (A.5) (iii ) mh( ) mh ( )=o( ), | |→∞ and 2 (0) 2 3 (A.6) mh( ) mh ( )=O( )ifq is bounded near x =0, | |→∞ where h +1 (A.7) m(0)( 2)= h h is the free (i.e., q(x)=0,x 0) mh function (= Fh( )). In (A.4)–(A.6), the | |→∞ asymptotics hold as with 2 + ε INVERSE SPECTRAL THEORY, II 639 where, by a Tauberian argument, Z R 2 2(1 + h ) 1 (A.9) d h( ) R 2 R R→∞ and d h is the spectral measure for the Schr odinger operator with (A.1) bound- ary conditions. The appendix of [35] discusses the calculus for functions of the form Z a 1+ 1 Q( )e 2 d + O˜(e 2a ), 0 with Q ∈ L1((0,a)). This calculus and Theorem 1.1 of this paper immediately imply Theorem A.1. For any Schrodinger problem of types (1)–(4), we have 1 a function Bh( ) in L ((0,a)) so that for any a Remarks. 1. If m( 2) has a representation of type (1.17) with no error 1 ∞ 2 term (e.g., if b = ∞ and q ∈ L (R)orq ∈ L (R)), mh( ) has a represen- tation with no error term, although the new representation will converge in Re( ) >Kh with Kh dependent on h. Similarly, there is a formula without error term if b<∞ with 0 and singularities at = nb. 2. (A.10) implies that if q is continuous at 0+, the following asymptotics hold: 2 3 4 2 1 2 h +1 h + h h + h 2 q(0) 3 (A.11) mh( )=h + + + + o( ). | |→∞ 2 3 Of course, one can derive this from the de nition (A.2) of mh(z) and the known 2 asymptotics of m(z). For systematic expansions of mh( )as| |→∞,we refer, for instance, to [7], [17] and the literature cited therein. 3. Bh( ) is analogous to A( ) but we are missing the local rst-order, q- independent, di erential equation that A satis es. We have found an equation 0 for Bh( , x) but it is higher than order one and contains q(x) and q (x). By following our idea in Sections 5–8, one obtains Theorem A.2. Z √ 1 (A.12) Bh( )=2 2 sin(2 ) d h( ), R 640 FRITZ GESZTESY AND BARRY SIMON (0) (0) (0) where d h( )=d h( ) d h ( ), with d h ( ) the spectral measure of mh (z); explicitly, 2 1 1 1+h 2 [0,∞)( ) +h2 d , h 0, d (0)( )= h i h 2 1 2 2 1 1+h 2 2(1 + h )h ( + h )+ [0,∞)( ) +h2 d , h > 0. As in Section 5, (A.12) is interpreted in the distributional sense. In analogy to (9.11), one derives ∂ B ( )= 2 L (2 , 0 ), h ∂ h + where Z x √ 0 0 0 cos( zx)=ϕh(x, z)+ Lh(x, x )ϕh(x ,z) dx , 0 00 with ϕh(x, z) satisfying ϕh(x, z)=(q(x) z)ϕh(x, z) and 0 ϕh(0+,z)=1,ϕh(0+,z)=h. Finally, we compute Bh( ) when q(x)=q0 > 0, x 0 and h =0 (a similar result holds if q0 < 0 with modi ed Bessel functions instead). Theorem A.3. If b = ∞, q(x)=q0 > 0, x 0, and h =0,then 1/2 q 1 1 (A.13) B ( )= 0 J (2q 2 ) 2q J (2q 2 ). h=0 1 0 0 2 0 In particular ([1, p. 364]), 3 4 1 2q0 2 1 Bh=0( ) cos 2q + O . →∞ 1 1 0 2 2 4 Proof. Let us make the q0 dependence explicit by writing Bh( ; q0). We start by noting that " # Z ∞ 2 2 1 1 (A.14) e Bh=0( ; q0) d = 2 1 0 ( + q0) 2 on account of (A.10) (or the version with no error term). Thus, Z ∞ 2 2 ∂Bh=0 1 (A.15) e ( ; q0) d = . 2 3 0 ∂q0 2 ( + q0) 2 Now ([15, 6.6232]) Z ∞ 3 x +1 2 (2 ) ( + 2 ) (A.16) e J ( x)x dx = . 1 + 3 0 2 ( 2 + 2) 2 INVERSE SPECTRAL THEORY, II 641 Taking the derivative with respect of in (A.16), setting = 1, and using J 1(x)= J1(x), we obtain Z ∞ 2 x 1 (A.17) e J1( x)xdx= 2 2 1/2 2 2 3/2 . 0 ( + ) ( + ) On the other hand ([15, 6.6231]), Z ∞ x 1 (A.18) e J0( x) dx = 2 2 1/2 . 0 ( + ) (A.15)–(A.18) show that ∂ 1 1 1 2 2 2 (A.19) Bh=0( ; q0)=J0(2q0 ) 2q0 J1(2q0 ). ∂q0 Now ([15, 8.4723]) d x J (x)=x J (x), dx 1 so (A.19) implies that the derivatives of the two sides of (A.13) are equal. Since both sides vanish at q0 = 0, (A.13) holds. University of Missouri, Columbia, MO E-mail address: [email protected] California Institute of Technology, Pasadena, CA E-mail address: [email protected] References [1] M. Abramowitz and I. A. Stegun (eds.), Handbook of Mathematical Functions, 9th print- ing, Dover, New York, 1972. [2] Z. S. Agranovich and V. A. 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