Dispersion Estimates for One-Dimensional Schr\"odinger Equations with Singular Potentials

We derive a dispersion estimate for one-dimensional perturbed radial Schr\"odinger operators. We also derive several new estimates for solutions of the underlying differential equation and investigate the behavior of the Jost function near the edge of the continuous spectrum.

More specifically, our goal is to provide dispersive decay estimates for these equations. To this end we recall (e.g., [24,Sect. 9.7]) that for ∞ 0 x|q(x)|dx < ∞ the operator H has a purely absolutely continuous spectrum on (0, ∞) plus a finite number of eigenvalues in (−∞, 0]. At the edge of the continuous spectrum there could be a resonance (or an eigenvalue if l > 1 2 ) if there is a corresponding bounded solution. Then our main results read as follows:  Here P c (H) is the orthogonal projection in L 2 (R + ) onto the continuous spectrum of H.
This result will follow from the corresponding low energy result Theorem 3.2 (see also Theorem 3.1) with the high energy result Theorem 3.3. We also remark that the decay rate is optimal (see below).
On the whole line such results have a long tradition and we refer to Weder [26], Goldberg and Schlag [12], Egorova, Kopylova, Marchenko and Teschl [8] (for the discrete case see [7]) as well as the reviews [14,22]. On the half line the case l = 0 was treated by Weder [27]. The case of general l but without potential was recently considered in Kovařík and Truc [19] (see also [10,11] for related results). While our overall strategy looks quite similar to the classical case l = 0, the details are much more delicate at several points: the first problem stems from the fact that only one solution will be bounded near x = 0 while the other one will have a singularity if l > 0. In particular, in this case the Jost solutions will have a singularity near x = 0 and the expression of the regular solution (which is in the domain of our operator near x = 0) in terms of the Jost solutions (i.e., the scattering relations) can no longer be used to obtain useful estimates. The second problem is that the simple group structure of the exponential functions breaks down for Bessel functions which requires novel strategies to handle the Born series expansion of the resolvent. And of course one has to work much harder to get some estimates, which are trivial for trigonometric functions, for Bessel functions. In particular, our present paper should also be understood as a contribution to understanding the properties of solutions of the underlying spectral problem. In this respect we would like to emphasize that the behavior of the Jost function near the bottom of the essential spectrum is still not understood satisfactorily, and for this very reason the resonant case had to be excluded from our main theorem. This is definitely a gap which should be filled.
Finally, we mention that one of the motivation to study (1.1) is the fact that it arises naturally when discussing the n-dimensional Schrödinger equation However, it is important to emphasize that this is not the only motivation since operators of the type in (1.1) are the prototypical example of strongly singular Schrödinger operators which have attracted considerable interest recently (see e.g. [15,16,17,18] and the references therein) or as examples in other physical and mathematical models (see e.g. [2,5]). Nevertheless, and since a lot is known about dispersive estimates for (1.4) (see the reviews [14,22] already mentioned above), it seems worth while to discuss what these estimates imply for (1.1). To this end recall (see e.g. Example 1.5 in [28]) that if V (x) = q(x), x = |x|, is radially symmetric, then H n will be reduced by the spherical harmonics (cf. [20]) which are an orthonormal basis of eigenfunctions of the Laplace-Beltrami operator ∆ S n−1 , span L 2 (R n ) = l,m H l,m and give rise to the decomposition In particular, an estimate of the type Here L 2 (R + ; x α ), α ∈ R denotes the standard L 2 space with the weight x α . In the special case l = 0 we get and hence which generalizes Theorem 2.4 from [19] where the case without potential and with the weight (1 + x) s was established. For conditions on V for (1.5) to hold we refer again to the above mentioned survey articles [14,22]. At this point we only note that it of course holds in the case without potential where the time evolution is given by Moreover, the time evolution of H n,l can be obtained by projecting e i∆t to the corresponding spherical harmonics. For example, in three dimensions one obtains where we have chosen m = 0, x = (0, 0, x) and used as well as [21,Eq. 18.17.19] for the last integral. Here J ν (z) is the Bessel function of order ν and P l (z) are the Legendre polynomials. This should again be compared with [17,Eq. (3.23)]. In particular, for l = 0 we have [e −itH3,0 ](x, y) ∞ = √ π|t| 3/2 ), which shows that the decay in our main Theorem 1.1 is optimal.

Properties of solutions
In this section we will collect some properties of the solutions of the underlying differential equation required for or main results.
We start with two lemmas containing estimates for the Green's function of the unperturbed equation and the regular solution φ(z, x) (see, e.g., [ where J ν and Y ν are the usual Bessel and Neumann functions (see Appendix B).

3)
and 15]). Assume (2.1). Then φ(z, x) satisfies the integral equation Moreover, φ is entire in z for every x > 0 and satisfies the estimate We also need the following estimates. and Proof. The first inequality follows from the identity (see (B.8)) along with the bound (2.3). Before proving (2.7), let us mention that where H (2.8) Consider the function Step (i): |ξ| ≤ |η| ≤ 1. Let us estimate the function Similarly, if l > 1/2, then If |l| < 1/2, then using (B.1) and (B.2) we obtain Finally, for l = 1/2 we get Summarizing the above, we find that the function G l admits the following estimate Step (ii): |ξ| ≤ 1 ≤ |η|. First, we get as implied by (B.4) and (B.5). If l > 1/2, we get For |l| < 1/2 we obtain And finally, for l = 1/2, we get Summarizing the above, we find that the function G l admits the following estimate Step (iii): 1 ≤ |η| ≤ |ξ|. To deal with the remaining case we shall use the second equality in (2.8) and the asymptotic expansions of Hankel functions (B.6)-(B.7): as |η|, |ξ| → ∞. Therefore, we get Combining all these estimates for the function G l with the equality after straightforward calculations we arrive at (2.7).
and satisfies the estimate Proof. The proof is based on the successive iteration procedure. As in the proof of Lemma 2.2 in [15], set The series is absolutely convergent since Similarly, let us show that ∂ k φ(k 2 , x) given by (2.14) satisfies (2.10). Using (2.12) and (2.6), we can bound the first summand in (2.14) as follows Next, using induction, one can show that the second summand admits a similar bound and hence we finally get This immediately implies the convergence of (2.13) and, moreover, the estimate from which (2.11) follows under the assumption (2.1).
Furthermore, by [9,4], the regular solution φ admits a representation by means of transformation operators.
where the kernel B satisfies the estimate for all 0 < y < x and j ∈ {0, 1}.
In particular, this lemma immediately implies the following useful result.
Proof. If f ∈ L ∞ (R + ), then using the estimate (2.16) we get which proves the claim.
Remark 2.7. Note that B is a bounded operator on L 2 ((0, a)) for all a > 0. However, the estimate (2.16) allows to show that its norm behaves like O(a) as a → ∞ and hence B might not be bounded on L 2 (R + ).

2.2.
The singular Weyl function. The singular Weyl function m : 2). Note that the corresponding singular m-functionm is given bỹ in this case. Moreover, it was shown in [16], [17] that the singular m-function (2.17) admits the following integral representation Here κ l := ⌊ l 2 + 3 4 ⌋, the function E is real entire, and ρ : R → R is a nondecreasing function satisfying The operator H is unitarily equivalent to multiplication by the independent variable in L 2 (R, dρ) and thus ρ is called the spectral function and dρ is the spectral measure. Indeed, one has F : and its inverse mapping F −1 : L 2 (R, dρ) → L 2 (R + ) given bŷ Here "l.i.m." denotes the limit in the corresponding L 2 -norm. Then, for any Borel function f , one has F f (H)F −1 equal to multiplication by f (λ).
We also remark that the value of κ l in (2.18) is the best possible one as the following extension of Marchenko's asymptotic formula shows.
Theorem 2.8 ( [18]). Suppose that q satisfies (2.1) and m is the singular m-function (2.17). Then there is a real entire function E such that in any nonreal sector, Moreover, the spectral function satisfies

20)
where ρ l (λ) = 1 π(l + 3 2 ) Note that the formula (2.20) was first announced in [13]. For extensions of Theorem 2.8 to the case when q is a distribution in H −1 loc we refer to [6].
2.3. The Jost solution. In this subsection, we assume that the potential q belongs to the Marchenko class, i.e., in addition to (2.1), q also satisfies Recall that under these assumptions on q the spectrum of H is purely absolutely continuous on (0, ∞) with an at most finite number of eigenvalues λ n ∈ (−∞, 0]. Next we need some estimates for the Weyl solution ψ(z, x). We begin with some basic properties of the unperturbed Bessel equation in which case the Weyl solution is given by which is analytic in Im k > 0 and continuous in Im k ≥ 0. Here H The analog of Lemma 2.1 reads: |l| < 1 2 . (2.23) A solution f (k, ·) to τ y = k 2 y satisfying the following asymptotic normalization as x → ∞, is called the Jost solution. In the case q = 0 we have (cf. (B.6)) Lemma 2.10. Assume (2.21). Then f (k, x) satisfies the integral equation If l > −1/2, then for all x > 0, f (·, x) is analytic in the upper half plane and can be continuously extended to the real axis away from k = 0 and Moreover, the function h(k, x) := e −ikx f (k, x) satisfies the estimates

26)
and Proof. The proof is based on the standard successive iteration scheme (see, e.g., [3,Chap. I.5]). Set x, y)f n−1 (k, y)q(y)dy for all n ∈ N. The series is absolutely convergent since (2.28) The latter also proves (2.5).
The proof of (2.26) is given in Appendix B. It remains to prove (2.27). First, notice that h solves the following equation

Then setting
we need to show that it satisfies the integral equation Therefore, using (2.28), we can bound the first summand in (2.30) as follows Next, using induction, one can show that the second summand admits a similar bound and hence we finally get This immediately implies the convergence of (2.29) and, moreover, the estimate from which (2.27) follows under the assumption xq(x) ∈ L 1 (R + ).
Furthermore, by [9,4], the Jost solution f admits a representation by means of transformation operators.
In particular, this lemma immediately implies the following useful result.
Proof. If f ∈ L ∞ (R + ), then using the estimate (2.33) we get which proves the claim. and we also set (2.36) In particular, the Weyl m-function (2.17) is given by Note that both f (k) and g(k) are analytic in the upper half plane and f (k) has simple zeros at iκ n = √ λ n ∈ C + .
We also will need the behavior of F and F ′ near zero.
Proof. Since f (k, x) can only be a multiple of φ(k 2 , x) if k = 0, their Wronskian f (k) can only vanish at 0. Moreover, the singular Weyl function must satisfy |m(z)| ≤ C λ |z − λ| near every λ ∈ R, which follows from its integral representation (2.18). Hence we obtain from (2.41) and (2.38) as claimed. Proof. Using (2.42), we get The integral converges absolutely for all k = 0. Indeed, by (2.22) and (2.11), we obtain Using (2.5) and (2.23), we get the following estimates for the first summand: Now the claim follows.

Dispersive decay
In this section we prove the dispersive decay estimate (1.3) for the Schrödinger equation (1.1). In order to do this, we divide the analysis into a low and high energy regimes. In the analysis of both regimes we make use of variants of the van der Corput lemma (see Appendix A), combined with a Born series approach for the high energy regime.
3.1. The low energy part. For the low energy regime, it is convenient to use the following well known representation of the integral kernel of e −itH P c (H), where the integral is to be understood as an improper integral; in the last equality we have usedφ (k, Note that which follows from (2.3), (2.5) and ∂ kφ (k, x) = (l + 1)sgn(k)|k| l φ(k 2 , x) + |k| l+1 ∂ k φ(k 2 , x) together with (2.6), (2.11). We begin with the following estimate.
Theorem 3.1. Assume (2.1) and (2.45). Let also χ ∈ C ∞ c (R) with supp(χ) ⊂ (−k 0 , k 0 ) and suppose there is neither a resonance nor an eigenvalue at 0, that is Proof. By symmetry we can assume x ≤ y. We want to apply the van der Corput Lemma A.1 with c = 0 and Note that Our assumption F (0) = 0 together with Lemma 2.13 imply |F (k)| ≥ ε > 0 for all k ∈ R and hence 1/F ∞ < ∞ in view of Lemma 2.13. Using (3.3) we infer which holds for all x and y with some uniform constant C > 0. Moreover, and it suffices to bound the two terms from above on compact sets. In fact, it suffices to consider the first term since the second one follows from (3.6) and Lemma 2.16.
The estimate for the first term follows from (3.3) and (3.4) since It remains to apply the van der Corput lemma.
To get rid of the dependence of x and y in (3.5) we make use of the transformation operators (2.15) and (2.32). x max(2,l+1) |q(x)|dx < ∞. (3.7) Let also χ ∈ C ∞ c (R) with supp(χ) ⊂ (−k 0 , k 0 ) and suppose there is neither a resonance nor an eigenvalue at 0, that is F (0) = 0. Then Proof. Assume that 0 < x ≤ 1 ≤ y. We proceed as in the previous proof but use Lemma 2.5 and Lemma 2.11 to write Indeed, for all k ∈ R \ {0}, φ(k, ·) is bounded at infinity and admits the representation (2.37) by means of Jost solution f (k, ·) and f (−k, ·). Therefore, by Lemma 2.11, φ(k, y) = (I + K y )φ l (k, y) for all k ∈ R \ {0}. By symmetry A(k) = A(−k) and hence our integral reads Our aim is to use Lemma A.2 (plus the remarks after this lemma) and hence we need to show that the individual parts of A(k) coincide with a function which is the Fourier transform a finite measure. In particular, we can redefine A(k) for k < 0. To this end note thatφ l (k 2 , x) = J(|k|x), where , r ≥ 0.
Note that J(r) ∼ r l+1 as r → 0 and J(r) = 2 π cos(r − l+1 2 ) + O(r −1 ) as r → +∞ (see (B.4)). Moreover, J ′ (r) ∼ r l as r → 0 and J ′ (r) = 2 π sin(r − l+1 2 ) + O(r −1 ) as r → +∞ (see (B.9)). In particular,J(r) := J(r) − 2 π cos(r − l+1 2 ) is in H 1 (R + ). Moreover, we can define J(r) for r < 0 such that it is locally in H 1 and J(r) = 2 π cos(r − l+1 2 ) for r < −1. By construction we then haveJ ∈ H 1 (R) and thus J is the Fourier transform of an integrable function. Moreover, cos(r − l+1 2 ) is the Fourier transform of the sum of two Dirac delta measures and so J(r) is the Fourier transform of a finite measure. By scaling, the total variation of the measures corresponding to J(kx) is independent of x. Since the same is true for χ(k 2 )|F (k)| −2 by Lemma 2.16, an application of Lemma A.2 shows But by Fubini we have I(t, x, y) = (1 + B x )(1 + K y )Ĩ(t, x, y) and the claim follows since both B : L ∞ ((0, 1)) → L ∞ ((0, 1)) and K : L ∞ ((1, ∞)) → L ∞ ((1, ∞)) are bounded in view of Corollary 2.6 and Corollary 2.12, respectively. By symmetry, we immediately obtain the same estimate if 0 < y ≤ 1 ≤ x. The case min(x, y) ≥ 1 can be proved analogously, we only need to to write 3.2. The high energy part. For the analysis of the high energy regime we use the following -also well known-alternative representation: where R H (ω) = (H − ω) −1 is the resolvent of the Schrödinger operator H and the limit is understood in the strong sense [24]. We recall that the Green's function is given by Note also that Fix k 0 > 0 and let χ : R → [0, ∞) be a C ∞ function such that The purpose of this section is to prove the following estimate. Then Our starting point is the fact that the resolvent R H of H can be expanded into the Born series where R l stands for the resolvent of the unperturbed radial Schrödinger operator.
To this end we begin by collecting some facts about R l . Its kernel is given Lemma 3.4. The function r l (k, x, y), l > − 1 2 , can be written as with a measure whose total variation satisfies ρ l,x,y ≤ C(l).
Here ρ * is the complex conjugated measure.
Proof. Let x ≤ y and k ≥ 0. Write We continue J(r), H(r) to the region r < 0 such that they are continuously differentiable and satisfy are in H 1 (R). In fact, they are continuously differentiable and hence it suffices to look at their asymptotic behavior. For r < −1 they are zero and for r > 1 they are O(r −1 ) and their derivative is O(r −1 ) as can be seen from the asymptotic behavior of the Bessel and Hankel functions (see Appendix B). Hence both J(r) and H(r) are Fourier transforms of finite measures. By scaling the total variation of the measures corresponding to J(kx), H(ky) are independent of x, y, respectively.
Hence it remains to consider the Fourier transform First observe that F x,y (p) = 1 for all x ≤ y. Therefore, F x,y L 1 = F 1,y/x L 1 . Hence it suffices to consider the Fourier transform of as k → ∞. This implies that h η,l ∈ H 1 (R) and henceĥ η,l ∈ L 1 (R). According to (A.1), it remains to show that the family h η,l is uniformly bounded in H 1 (R) with respect to η ∈ (0, 1]. Clearly, for all k ∈ R and hence h η,l L 2 ≤ h 0,l L 2 . Noting that for all l > − 1 2 , we get . The latter implies that h ′ η,l L 2 are uniformly bounded. Remark 3.5. (i) For l ∈ N 0 the situation is somewhat simpler and we can write where P l,x,y (p) is a polynomial of degree 2l − 1 which is symmetric in x and y. Explicitly, P 0,x,y (p) = 0, P 1,x,y (p) = − p xy , P 2,x,y (p) = 3p p 2 − x 2 − y 2 2x 2 y 2 P 3,x,y (p) = − 3p 5 p 2 − x 2 2 + 2y 2 3x 2 − 5p 2 + 5y 4 8x 3 y 3 and one can verify the claim explicitly.
(ii) We have the following recursion Now we are in position to finish the proof of the main result.
The kernel of the operator R l (k 2 + i0)(−q R l (k 2 + i0)) n is given by r l (k; y i , y i+1 )r l (k; y n , y)dy 1 · · · dy n .
Applying Fubini's theorem, we can integrate in k first and hence we need to obtain a uniform estimate of the oscillatory integral Consider the function f n (k) = χ(k 2 ) k 2k0 −n . Clearly, f 0 is the Fourier transform of a measure ν 0 satisfying ν 0 ≤ C. For n ≥ 1, f n is H 1 with f n H 1 ≤ π −1/2 C(1+n). Hence by Lemma A.2 and Lemma 3.4 we obtain This proves Theorem 3.3.
Our proof will be based on the following variant of the van der Corput lemma which can be shown as in [ where α = |α| (R) denotes the total variation of α and C 2 is the Constant from the van der Corput lemma.
We also need the following simple fact. If f ∈ H 1 (R), then f is in the Wiener algebra A and