On Absence of Threshold Resonances for Schrodinger¨ and Dirac Operators

DISCRETE AND CONTINUOUS doi:10.3934/dcdss.2020243 DYNAMICAL SYSTEMS SERIES S Volume 13, Number 12, December 2020 pp. 3427–3460

ON ABSENCE OF THRESHOLD RESONANCES FOR SCHRODINGER¨ AND DIRAC OPERATORS

Fritz Gesztesy∗ Department of Mathematics Baylor University One Bear Place #97328 Waco, TX 76798-7328, USA Roger Nichols Department of Mathematics The University of Tennessee at Chattanooga 415 EMCS Building, Dept. 6956 615 McCallie Avenue Chattanooga, TN 37403-2504, USA

Abstract. Using a unified approach employing a homogeneous Lippmann- Schwinger-type equation satisfied by resonance functions and basic facts on Riesz potentials, we discuss the absence of threshold resonances for Dirac and Schrödingeroperators with sufficiently short-range interactions in general space dimensions. More specifically, assuming a sufficient power law decay of potentials, we derive the absence of zero-energy resonances for massless Dirac operators in space dimensions n > 3, the absence of resonances at ±m for massive Dirac operators (with mass m > 0) in dimensions n > 5, and recall the well-known case of absence of zero-energy resonances for Schrödingeroperators in dimension n > 5.

1. Introduction. Happy Birthday, Gisèle,we sincerely hope that our modest con- tribution to threshold resonances of Schrödingerand Dirac operators will create some joy. The principal purpose of this paper is a systematic investigation of threshold resonances, more precisely, the absence of the latter, for massless and massive Dirac operators, and for Schrödingeroperators in general space dimensions n ∈ N, n > 2, given sufficiently fast decreasing interaction potentials at infinity (i.e., short-range interactions). In the case that (−∞,E0) and/or (E1,E2), Ej ∈ R, j = 0, 1, 2, are essential (resp., absolutely continuous) spectral gaps of a self-adjoint unbounded operator A in some complex, separable Hilbert space H, then typically, the numbers E0 and/or the numbers E1,E2 are called threshold energies of A. The intuition behind the term “threshold” being that if the coupling constant of an appropriate interaction potential is varied, then eigenvalues will eventually “emerge” out of the continuous

2020 Mathematics Subject Classiﬁcation. Primary: 35J10, 35Q41, 45P05, 47A11, 47G10; Sec- ondary: 35Q40, 47A10, 81Q10. Key words and phrases. Threshold resonances, threshold eigenvalues, Schr¨odingeroperators, Dirac operators. ∗ Corresponding author: Fritz Gesztesy.

3427 3428 FRITZ GESZTESY AND ROGER NICHOLS spectrum of the operator A and enter the spectral gap (−∞,E0) and/or (E0,E1). The precise phenomenon behind this intuition is somewhat involved, depending, in particular, on the (possibly singular) behavior of the resolvent of A at the points Ej, j = 0, 1, 2. In the concrete case of Dirac and Schrödinger operators with sufficient short-range interactions at hand, this behavior is well understood (see, e.g., [56], [57], [59] and the literature cited therein) and E0 = 0 is then a threshold for Schrödingeroperators h, with σess(h) = [0, ∞) (cf. (2.16)–(2.18)), whereas ±m are thresholds for massive Dirac operators H(m) corresponding to mass m > 0, with σess(H(m)) = (−∞, −m] ∪ [m, ∞) (cf. (2.9)–(2.11)). The case of massless Dirac operators H is more intricate as σess(H) = R (cf. (2.4), (2.6)), and hence H exhibits no spectral gap. However, since the potential coefficients tend to zero at infinity, 0 is still a threshold point of H that is known to possibly support an eigenvalue and/or a resonance. Threshold resonances for Dirac and Schrödingeroperators then are associated with distributional solutions ψ of hψ = 0, or Ψ±m of H(m)Ψ±m = ±mΨ±m, respectively, distributional solutions Ψ of HΨ = 0, that do not belong to the domains of h, H(m), respectively H, but these functions are “close” to belonging to the re- spective operator domains in the sense that they (actually, their components) belong p n to L (R ) for p ∈ (p0, ∞) ∪ {∞} for appropriate p0 > 2. More precisely, we then recall in our principal Section3 that Schrödingeroperators have no zero-energy resonances in dimensions n > 5 (a well-known result, cf. Remark 3.5( iii)), that massless Dirac operators have no zero-energy resonances in dimension n > 3 (the case n = 3 was well-known, cf. Remark 3.9( iii)), and that massive Dirac operators have no resonances at ±m in dimensions n > 5. We prove these results in a unified manner employing a homogeneous Lippmann–Schwinger-type equation satisfied by threshold resonance functions and the use of basic properties of Riesz potentials. While our power law decay assumptions of the potentials at infinity are not optimal, optimality was not the main motivation for writing this paper. Instead, we wanted to present a simple, yet unified, approach to the question of absence of threshold resonances with the added bonus that our results in the massless case for n > 4 and in the massive case appear to be new. Since we explicitly permit matrix-valued potentials in the (massless and massive) Dirac case, we note that electric and magnetic potentials are included in our treatment (cf. Remark 2.2). We also emphasize that threshold resonances and/or threshold eigenvalues of course profoundly influence the (singularity) behavior of resolvents at threshold energies, the threshold behavior of (fixed energy, or on-shell) scattering matrices, as well as dispersive (i.e., large time) estimates for wave functions. These issues have received an abundance of attention since the late 1970’s in connection with Schrödingeroperators, and very recently especially in the context of (massless and massive) Dirac operators (the case of massless Dirac operators having applications to graphene). While a complete bibliography in this connection is clearly beyond the scope of our paper, we refer to [2]–[11], [12, Ch. 4], [13], [15]–[21], [28]–[43], [45], [46], [50]–[54], [58]–[62], [65]–[73], [76]–[78], and the literature therein to demonstrate some of the interest generated by this circle of ideas. We also mention the relevance of threshold states in connection with the Witten index for non-Fredholm operators, see, for instance, [22]–[26]. In Section2 we present the necessary background for Dirac and Schrödinger operators, including a description of Green’s functions (resp., matrices) and their ABSENCE OF THRESHOLD RESONANCES 3429 threshold asymptotics in the free case, that is, in the absence of interaction potentials in terms of appropriate Hankel functions. AppendixA discusses the failure of our technique to produce essential boundedness (of components) of threshold resonances and eigenfunctions for space dimensions sufficiently large. We conclude this introduction with some comments on the notation employed in this paper: If H denotes a separable, complex Hilbert space, IH represents the N identity operator in H, but we simplify this to IN in the finite-dimensional case C , N ∈ N. If T is a linear operator mapping (a subspace of) a Hilbert space into another, then dom(T ) and ker(T ) denote the domain and kernel (i.e., null space) of T . The spectrum, point spectrum (the set of eigenvalues), and the essential spectrum of a closed linear operator in H will be denoted by σ( · ), σp( · ), and σess( · ), respectively. Similarly, the absolutely continuous and singularly continuous spectrum of a self- adjoint operator in H are denoted by σac( · ) and σsc( · ). The Banach space of bounded linear operators on a separable complex Hilbert space H is denoted by B(H). If p ∈ [1, ∞) ∪ {∞}, then p0 ∈ [1, ∞) ∪ {∞} denotes its conjugate index, that is, p0 := (1 − 1/p)−1. If Lebesgue measure is understood, we simply write Lp(M), M ⊆ Rn measurable, n ∈ N, instead of the more elaborate notation Lp(M; dnx). b · c denotes the floor function on R, that is, bxc characterizes the largest integer n less than or equal to x ∈ R. Finally, if x = (x1, . . . , xn) ∈ R , n ∈ N, then we abbreviate hxi := (1 + |x|2)1/2.

2. Some background material. This preparatory section is primarily devoted to various results on Green’s functions for the free Schr¨odingeroperator (i.e., minus the Laplacian) and free (massive and massless) Dirac operators (i.e., in the absence of interaction potentials) in dimensions n ∈ N, n > 2. To rigorously deﬁne the case of free massless n-dimensional Dirac operators, n > 2, we now introduce the following set of basic hypotheses assumed for the remainder of this manuscript.

Hypothesis 2.1. Let n ∈ N, n > 2. b(n+1)/2c (i) Set N = 2 and let αj, 1 6 j 6 n, αn+1 := β, denote n + 1 anti- commuting Hermitian N × N matrices with squares equal to IN , that is, ∗ αj = αj, αjαk + αkαj = 2δj,kIN , 1 6 j, k 6 n + 1. (2.1) (ii) Introduce in [L2(Rn)]N the free massless Dirac operator n X 1,2 n N H0 = α · (−i∇) = αj(−i∂j), dom(H0) = [W (R )] , (2.2) j=1 where ∂j = ∂/∂xj, 1 6 j 6 n.

(iii) Next, consider the self-adjoint matrix-valued potential V = {V`,m}16`,m6N satisfying for some ﬁxed ρ ∈ (1, ∞), C ∈ (0, ∞), ∞ n N×N −ρ n V ∈ [L (R )] , |V`,m(x)| 6 Chxi for a.e. x ∈ R , 1 6 `, m 6 N. (2.3) Under these assumptions on V , the massless Dirac operator H in [L2(Rn)]N is deﬁned via 1,2 n N H = H0 + V, dom(H) = dom(H0) = [W (R )] . (2.4) 3430 FRITZ GESZTESY AND ROGER NICHOLS

We recall that 2 n N 2 n N 1,2 n N 1,2 n N [L (R )] = L (R ; C ), [W (R )] = W (R ; C ), etc. (2.5) 2 n N Then H0 and H are self-adjoint in [L (R )] , with essential spectrum covering the entire real line, σess(H) = σess(H0) = σ(H0) = R, (2.6) a consequence of relative compactness of V with respect to H0. In addition,

σac(H0) = R, σp(H0) = σsc(H0) = ∅. (2.7) We also recall that the massive free Dirac operator in [L2(Rn)]N associated with the mass parameter m > 0 is of the form (with β = αn+1) 1,2 n N H0(m) = H0 + mβ, dom(H0(m)) = [W (R )] , m > 0, (2.8) and the corresponding interacting massive Dirac operator in [L2(Rn)]N is given by 1,2 n N H(m) = H0(m) + V = H0 + mβ + V, dom(H(m)) = [W (R )] , m > 0. (2.9) In this case,

σess(H(m)) = σess(H0(m)) = σ(H0(m)) = (−∞, −m] ∪ [m, ∞), m > 0, (2.10) and

σac(H0(m)) = (−∞, −m] ∪ [m, ∞), σp(H0(m)) = σsc(H0(m)) = ∅, m > 0. (2.11) In the special one-dimensional case n = 1, one can choose for α1 either a real constant or one of the three Pauli matrices. Similarly, in the massive case, β would typically be a second Pauli matrix (different from α1). For simplicity we confine ourselves to n ∈ N, n > 2, in the following. Employing the relations (2.1), one observes that 2 2 2,2 n N H0 = −IN ∆, dom H0 = [W (R )] . (2.12) Remark 2.2. Since we permit a (sufficiently decaying) matrix-valued potential V in H, this includes, in particular, the case of electromagnetic interactions introduced via minimal coupling, that is, V describes also special cases of the form, 1,2 n N H(v, A) := α·(−i∇−A)+vIN = H0 +[vIN −α·A], dom(H(v, A)) = [W (R )] , (2.13) where (v, A) represent the electromagnetic potentials on Rn, with v : Rn → R, ∞ n n ∞ n v ∈ L (R ), A = (A1,...,An), Aj : R → R, Aj ∈ L (R ), 1 6 j 6 n, and for some fixed ρ > 1, C ∈ (0, ∞), −ρ n |v(x)| + |Aj(x)| 6 Chxi , x ∈ R , 1 6 j 6 n. (2.14)

The analogous remark applies of course to the massive case with H0 replaced by H0(m), m > 0. We continue with the self-adjoint Laplacian in L2(Rn), 2,2 n h0 = −∆, dom(h0) = W (R ), (2.15) and the interacting Schr¨odingeroperator in L2(Rn) given by 2,2 n h = h0 + v, dom(h) = W (R ), (2.16) where v ∈ L∞(Rn), v real-valued, satisﬁes the estimate (2.14) (for simplicity). In this case, σess(h) = σess(h0) = σ(h0) = [0, ∞), (2.17) ABSENCE OF THRESHOLD RESONANCES 3431 and

σac(h0) = [0, ∞), σp(h0) = σsc(h0) = ∅. (2.18)

The Green’s function of h0, denoted by g0(z; · , · ), is of the form,

−1 g0(z; x, y) := (h0 − zI) (x, y)  −1/2 iz1/2|x−y| (i/2)z e , n = 1, z ∈ C\{0}, = −1/2 (2−n)/2 (1) 1/2 (i/4) 2πz |x − y| H(n−2)/2 z |x − y| , n > 2, z ∈ C\{0}, 1/2 n Im z > 0, x, y ∈ R , x 6= y, (2.19) and for z = 0, n > 3,

1 2−n g0(0; x, y) = |x − y| (n − 2)ωn−1 (2.20) −1 −n/2 2−n n = 4 π Γ((n − 2)/2)|x − y| , n > 3, x, y ∈ R , x 6= y.

(1) Here Hν ( · ) denotes the Hankel function of the ﬁrst kind with index ν > 0 (cf. [1, n/2 Sect. 9.1]) and ωn−1 = 2π /Γ(n/2) (Γ( · ) the Gamma function, cf. [1, Sect. 6.1]) represents the area of the unit sphere Sn−1 in Rn. As z → 0, g0(z; · , · ) is continuous on the oﬀ-diagonal for n > 3,

1 2−n lim g0(z; x, y) = g0(0; x, y) = |x − y| , z→0 (n − 2)ωn−1 z∈C\{0} (2.21) n x, y ∈ R , x 6= y, n ∈ N, n > 3, but blows up for n = 1 as

−1/2 −1 1/2 2 g0(z; x, y) = (i/2)z − 2 |x − y| + O z |x − y| , x, y ∈ , (2.22) z→0 R z∈C\{0} and for n = 2 as

1 1/2 2 1 g0(z; x, y) = − ln z |x − y|/2 1 + O z|x − y| + ψ(1) z→0 2π 2π z∈C\{0} (2.23) 2 2 + O |z||x − y| , x, y ∈ R , x 6= y. Here ψ(w) = Γ0(w)/Γ(w) denotes the digamma function (cf. [1, Sect. 6.3]). In connection with the free massive Dirac operator H0(m) = H0 + m β, m > 0, one computes,

−1 2 2 −1 (H0(m) − zI) = (H0(m) + zI) H0(m) − z I (2.24) 2 2 −1 = (−iα · ∇ + m β + zI) h0 − (z − m )I IN , employing 2 2 H0(m) = (h0 + m I)IN . (2.25) Assuming

2 21/2 n m > 0, z ∈ C\(R\[−m, m]), Im z − m > 0, x, y ∈ R , x 6= y, n ∈ N, n > 2, (2.26) 3432 FRITZ GESZTESY AND ROGER NICHOLS and exploiting (2.24), one thus obtains for the Green’s function G0(m, z; · , · ) of H0(m), −1 G0(m, z; x, y) := (H0(m) − zI) (x, y) −1 (2−n)/2 2−n = i4 (2π) |x − y| (m β + zIN ) 2 21/2 (n−2)/2 (1) 2 21/2 × z − m |x − y| H(n−2)/2 z − m |x − y| (x − y) − 4−1(2π)(2−n)/2|x − y|1−n α · |x − y| 2 21/2 n/2 (1) 2 21/2 × z − m |x − y| Hn/2 z − m |x − y| . (2.27) Here we employed the identity ([1, p. 361]), (1) 0 (1) −1 (1) Hν (ζ) = −Hν+1(ζ) + ν ζ Hν (ζ), ν, ζ ∈ C. (2.28) We also recall the asymptotic behavior (cf. [1, p. 360], [53, p. 723–724]) (1) 2 H0 (ζ) = (2i/π)ln(ζ) + O |ln(ζ)||ζ| , (2.29) ζ→0 ζ∈C\{0} ( min(ν,−ν+2) (1) ν −ν O |ζ| , ν∈ / N, Hν (ζ) = −(i/π)2 Γ(ν)ζ + ν −ν+2 (2.30) ζ→0 O |ln(ζ)||ζ| + O ζ , ν ∈ N, ζ∈C\{0} Re(ν) > 0, (1) 1/2 −1/2 i[ζ−(νπ/2)−(π/4)] Hν (ζ) = (2/π) ζ e , ν 0, Im(ζ) 0. (2.31) ζ→∞ > > Equations (2.29), (2.30) reveal the facts (still assuming (2.26)), −1 −n/2 2−n lim G0(m, z; x, y) = 4 π Γ((n − 2)/2)|x − y| (m β ± mIN ) (2.32) z→±m z∈C\{±m} (x − y) + i2−1π−n/2Γ(n/2) α · , m > 0, x, y ∈ n, x 6= y, n ∈ , n 3, |x − y|n R N > −1 2 2 G0(m, z; x, y) = −(4π) ln z − m (m β ± mI2) z→±m z∈C\{±m} (x − y) − (2π)−1ln(|x − y|)(m β ± mI ) + i(2π)−1 α · (2.33) 2 |x − y|2 2 2 2 2 2 + O z − m ln z − m , m > 0, x, y ∈ R , x 6= y. 2 2 2 2 2 (Here the remainder term O z − m ln z − m depends on x, y ∈ R , but this is of no concern at this point.) In particular, G0(m, z; · , · ) blows up logarithmically as z → ±m in dimensions n = 2, just as g0(z, · , · ) does as z → 0. By contrast, the massless case is quite diﬀerent and assuming n z ∈ C+, x, y ∈ R , x 6= y, n ∈ N, n > 2, (2.34) one computes in the case m = 0 for the Green’s function G0(z; · , · ) of H0, −1 G0(z; x, y) := (H0 − zI) (x, y) −1 (2−n)/2 2−n (n−2)/2 (1) = i4 (2π) |x − y| z [z|x − y|] H(n−2)/2(z|x − y|)IN (2.35) (x − y) − 4−1(2π)(2−n)/2|x − y|1−n[z|x − y|]n/2H(1) (z|x − y|) α · . n/2 |x − y| ABSENCE OF THRESHOLD RESONANCES 3433

The Green’s function G0(z; · , · ) of H0 continuously extends to z ∈ C+. In addition, in the massless case m = 0, the limit z → 0 exists1,

lim G0(z; x, y) := G0(0 + i 0; x, y) z→0, z∈ \{0} C+ (2.36) (x − y) = i2−1π−n/2Γ(n/2) α · , x, y ∈ n, x 6= y, n ∈ , n 2, |x − y|n R N > and no blow up occurs for all n ∈ N, n > 2. This observation is consistent with the suﬃcient condition for the Dirac operator H = H0 + V (in dimensions n ∈ N, n > 2), with V an appropriate self-adjoint N × N matrix-valued potential, having no eigenvalues, as derived in [54, Theorems 2.1, 2.3]. The following remark is primarily of a heuristic nature; at this point it just serves as a motivation for our detailed discussion of absence of threshold resonances for Schr¨odingerand Dirac operators in Section3.

Remark 2.3. The asymptotic behavior, for some dn ∈ (0, ∞), 1−n n kG0(0 + i 0; x, y)kB( N ) = dn|x − y| , x, y ∈ , x 6= y, n ∈ , n 2, C z→0, R N > z∈C+\{0} (2.37) combined with the homogeneous Lippmann–Schwinger-type integral equation in (3.79), formally “implies” the absence of zero-energy resonances (cf. Section3 for a detailed discussion) of H for n ∈ N, n > 3, for sufficiently fast decaying short-range potentials V at infinity, as | · |1−n lies in L2(Rn; dnx) near infinity if and only if n > 3. This is consistent with observations in [7], [12, Sect. 4.4], [13], [16], [67], [68], [78] for n = 3 (see also Remark 3.9( iii)). This should be contrasted with the behavior of Schrödingeroperators where

1 2−n lim g0(z; x, y) = g0(0; x, y) = |x − y| , z→0 (n − 2)ωn−1 z∈C\{0} (2.38) n x, y ∈ R , x 6= y, n ∈ N, n > 3, “implies” the absence of zero-energy resonances of h = h0 + v for n ∈ N, n > 5, again for sufficiently fast decaying short-range potentials v at infinity, as | · |2−n 2 n n lies in L (R ; d x) near infinity if and only if n > 5, as observed in [50]. Since H0 has no spectral gap, σ(H0) = R, but h0 has the half-line (−∞, 0) in its resolvent set, a comparison of h0 with the massive free Dirac operator H0(m) = H0 +m β, m > 0, with spectral gap (−m, m), replacing the energy z = 0 by z = ±m, is quite natural and then exhibits a similar logarithmic blowup behavior as z → 0 in dimensions n = 2. Finally, the leading asymptotic behavior | · |2−n of the free massive Green’s matrix G0(m, ±m; · , 0) in (2.32) near infinity then “implies” the absence of threshold resonances of H(m) = H0(m) + V for n ∈ N, n > 5, as in the case of Schrödingeroperators h. Section3 is devoted to a rigorous treatment of the formal statements in this remark. Next, we also recall some basic facts on Lp-properties of Riesz potentials (see, e.g., [74, Sect. V.1]):

1 Our choice of notation 0 + i 0 in G0(0 + i 0; x, y) indicates that the limit limz→0 is performed in the closed upper half-plane C+. 3434 FRITZ GESZTESY AND ROGER NICHOLS

Theorem 2.4. Let n ∈ N, α ∈ (0, n), and introduce the Riesz potential operator Rα,n as follows:

−α/2 −1 n α−n (Rα,nf)(x) = (−∆) f (x) = γ(α, n) d y |x − y| f(y), ˆ n R (2.39) γ(α, n) = πn/22αΓ(α/2)/Γ((n − α)/2), for appropriate functions f (see below ). p n (i) Let p ∈ [1, ∞) and f ∈ L (R ). Then the integral (Rα,nf)(x) converges for (Lebesgue ) a.e. x ∈ Rn. (ii) Let 1 < p < q < ∞, q−1 = p−1 − αn−1, and f ∈ Lp(Rn). Then there exists Cp,q,α,n ∈ (0, ∞) such that

q n p n kRα,nfkL (R ) 6 Cp,q,α,nkfkL (R ). (2.40) We also note the β function-type integral (cf. [74, p. 118]),

n α−n β−n d y |ek − y| |y| = γ(α, n)γ(β, n)/γ(α + β, n), ˆ n R 0 < α < n, 0 < β < n, α + β < n, (2.41)

ek = (0,..., 1 ,..., 0), 1 6 k 6 n, |{z} k and more generally, the Riesz composition formula (see [27, Sects. 3.1, 3.2]),

dny |x − y|α−n|y − w|β−n = [γ(α, n)γ(β, n)/γ(α + β, n)]|x − w|α+β−n, ˆ n R (2.42) n 0 < α < n, 0 < β < n, 0 < α + β < n, x, w ∈ R . For later use in Section3, we recall the following estimate taken from [36, Lemma 6.3].

n Lemma 2.5. Let n ∈ N and x1, x2 ∈ R . If k, ` ∈ [0, n), ε, β ∈ (0, ∞), with k + ` + β > n, and k + ` 6= n, then n −k −β−ε −` d y |x1 − y| hyi |y − x2| ˆ n R ( − max{0,k+`−n} (2.43) |x1 − x2| , |x1 − x2| 6 1, 6 Cn,k,`,β,ε · − min{k,`,k+`+β−n} |x1 − x2| , |x1 − x2| > 1, where Cn,k,`,β,ε ∈ (0, ∞) is an x1, x2-independent constant. We conclude this section by recalling the following interesting results of McOwen [61] and Nirenberg–Walker [64], which provide necessary and suﬃcient conditions for the boundedness of certain classes of integral operators in Lp(Rn): Theorem 2.6. Let n ∈ N, c, d ∈ R, c + d > 0, p ∈ (1, ∞), and p0 = p/(p − 1). The following items (i) and (ii) hold. (i) If −c (c+d)−n −d n 0 Kc,d(x, y) = |x| |x − y| |y| , x, y ∈ R , x 6= x , (2.44) p n then the integral operator Kc,d in L (R ) with integral kernel Kc,d( · , · ) in (2.44) is bounded if and only if c < n/p and d < n/p0. (ii) If −c (c+d)−n −d n 0 Kec,d(x, y) = (1 + |x|) |x − y| (1 + |y|) , x, y ∈ R , x 6= x , (2.45) ABSENCE OF THRESHOLD RESONANCES 3435

p n then the integral operator Kec,d in L (R ) with integral kernel Kec,d( · , · ) in (2.45) is bounded if and only if c < n/p and d < n/p0.

3. Nonexistence of threshold resonances. In this section we prove our principal results on nonexistence of threshold resonances in three cases: First, in the case of Schrödingeroperators in dimension n > 3; second, in the case of massless Dirac operators in dimensions n > 2; and third, in the case of massive Dirac operators in dimensions n > 3. n SchrödingerOperators in R , n > 3. We begin with the case of Schrödingerop- erators h in dimension n > 3 as defined in (2.15), and start by making the following assumptions on the potential v.

Hypothesis 3.1. Let n ∈ N, n > 3. Assume the a.e. real-valued potential v satisfies for some fixed C ∈ (0, ∞), ∞ n −4 n v ∈ L (R ), |v(x)| 6 Chxi for a.e. x ∈ R . (3.1) In addition, we suppose that 1/2 1/2 v = v1v2 = |v| uv|v| , (3.2) 1/2 1/2 where v1 = |v| , v2 = uv|v| , uv = sgn(v). Here we abbreviated ( 1, v(x) > 0, n sgn(v(x)) = for a.e. x ∈ R . (3.3) −1, v(x) < 0, We continue with the threshold behavior, that is, the z = 0 behavior, of h: Definition 3.2. Assume Hypothesis 3.1. (i) The point 0 is called a zero-energy eigenvalue of h if hψ = 0 has a distributional solution ψ satisfying 2,2 n ψ ∈ dom(h) = W (R ) (3.4) (equivalently, ker(h) % {0}). (ii) The point 0 is called a zero-energy (or threshold ) resonance of h if 2 n 2 n −1 ker IL (R ) + v2(h0 + 0IL (R )) v1 % {0} (3.5) −1 and if there exists 0 6= φ ∈ ker IL2( n) + v2(h0 + 0I 2 n v1 such that ψ defined R L (R ) by −1 2 n ψ(x) = − (h0 + 0IL (R )) v1φ (x) −1 n 2−n (3.6) = [(n − 2)ωn−1] d y |x − y| v1(y)φ(y) ˆ n R n (for a.e. x ∈ R , n > 3) is a distributional solution of hψ = 0 satisfying 2 n ψ∈ / L (R ). (3.7) (iii) 0 is called a regular point for h if it is neither a zero-energy eigenvalue nor a zero-energy resonance of h. Additional properties of ψ are isolated in Theorem 3.3. While the point 0 being regular for h is the generic situation, zero-energy eigenvalues and/or resonances are exceptional cases. 3436 FRITZ GESZTESY AND ROGER NICHOLS

Next, we introduce the following convenient abbreviation (for x, y ∈ Rn, x 6= y): −1 2−n r0,0(x − y) = lim g(z; x, y) = g0(0; x, y) = [(n − 2)ωn−1] |x − y| z→0 z∈C\{0} (3.8) −1 −n/2 2−n = 4 π Γ((n − 2)/2)|x − y| , n > 3, and note that n −∆xr0,0(x − y) = −∆xg0(0; x, y) = δ(x − y), x, y ∈ R , x 6= y, n > 3, (3.9) in the sense of distributions. In the remainder of this section we will frequently apply [44, Prop. 6.10], that is, p r q the fact that 0 < p < q < r 6 ∞ implies L ∩ L ⊂ L . Theorem 3.3. Assume Hypothesis 3.1. (i) If n = 3, 4, there are precisely four possible cases: Case (I): 0 is regular for h. Case (II): 0 is a (necessarily simple2 ) resonance of h. In this case, the resonance function ψ satisﬁes q 3 ψ ∈ L (R ), q ∈ (3, ∞) ∪ {∞}, (3.10) q 4 ψ ∈ L (R ), q ∈ (2, ∞) ∪ {∞}, (3.11) 2 n n q n ∇ψ ∈ L (R ) , ∆ψ ∈ L (R ), q ∈ [2, ∞) ∪ {∞}, n = 3, 4, (3.12) 2 n ψ∈ / L (R ), n = 3, 4. (3.13) Case (III): 0 is a (possibly degenerate ) eigenvalue of h. In this case, the corre- 2,2 n sponding eigenfunctions ψ ∈ dom(h) = W (R ), n ∈ N, n > 3, of hψ = 0 also satisfy q n ψ, ∆ψ ∈ L (R ), q ∈ [2, ∞) ∪ {∞}, n = 3, 4, (3.14) 2 n n ∇ψ ∈ L (R ) , n = 3, 4. (3.15) Case (IV ): A possible mixture of Cases (II) and (III). (ii) If n ∈ N, n > 5, there are precisely two possible cases: Case (I): 0 is regular for h. Case (II): 0 is a (possibly degenerate ) eigenvalue of h. In this case, the corresponding eigenfunctions ψ ∈ dom(h) = W 2,2(Rn) of hψ = 0 also satisfy ( q n (n/(n − 2), ∞) ∪ {∞}, 5 6 n 6 7, ψ, ∆ψ ∈ L (R ), q ∈ (3.16) (2n/(n + 4), 2n/(n − 4)), n > 8,  (5/3, ∞), n = 5,  q n n (3/2, ∞), n = 6, ∇ψ ∈ L (R ) , q ∈ (3.17) (7/5, 14), n = 7,   (2n/(n + 4), 2n/(n − 6)), n > 8. In particular, there are no zero-energy resonances of h in dimension n > 5. (iii) The point 0 is regular for h if and only if 2 n 2 n −1 ker IL (R ) + v2(h0 + 0IL (R )) v1 = {0}. (3.18)

2One can show that if n = 3, 4, there is at most one resonance function in Case (II), see [8], [51], [52]. ABSENCE OF THRESHOLD RESONANCES 3437

Proof. Since g0(0; x, y), x 6= y, exists for all n > 3 (cf. (2.20)), the Birman– Schwinger eigenvalue equation

2 n 2 n 2 n −1 IL (R ) + v2(h0 − 0IL (R )) v1 φ0 = 0, 0 6= φ0 ∈ L (R ), (3.19) 1 n gives rise to a distributional zero-energy solution ψ0 ∈ Lloc(R ) of hψ0 = 0 in terms n of φ0 of the form (for a.e. x ∈ R , n > 3), −1 2 n ψ0(x) = − (h0 − 0IL (R )) v1φ0 (x) = −[r0,0 ∗ (v1φ0)](x) (3.20)

−1 −n/2 n 2−n = −4 π Γ((n − 2)/2) d y |x − y| v1(y)φ0(y), (3.21) ˆ n R φ0(x) = (v2ψ0)(x). (3.22)

In particular, one concludes that ψ0 6= 0. We note from the outset that ∆ψ = vψ with v ∈ L∞(Rn) yields ∆ψ ∈ Lr(Rn) r n whenever ψ ∈ L (R ) for some r > 1. Moreover, since also ∆ψ = v1φ, with ∞ n 2 n 2 n v1 ∈ L (R ) and φ ∈ L (R ), this yields ∆ψ ∈ L (R ), and, if in addition ψ is an eigenfunction of H, then of course ψ ∈ W 1,2(Rn). −2 Thus, one estimates, with |v1(y)| 6 chyi and some constant dn ∈ (0, ∞),

n 2−n −2 |ψ0(x)| 6 dn d y |x − y| hyi |φ0(y)| ˆ n R −2 = dnR2,n h · i |φ0( · )| (x) (3.23) −2 n = dnR2,n h · i |v2( · )||ψ0( · )| (x), x ∈ R . (3.24) An application of Theorem 2.6( ii) with c = 0, d = 2, p = p0 = 2, and the inequality 2 n 2 < n/2, combined with φ0 ∈ L (R ), then yield 2 n ψ0 ∈ L (R ), n > 5. (3.25) 2,2 n To prove that actually ψ0 ∈ dom(h) = W (R ), it suﬃces to argue as follows: 2 n ∆ψ0 = vψ0 ∈ L (R ), n > 5, (3.26)

∞ n 2 n in the sense of distributions since v ∈ L (R ) and ψ0 ∈ L (R ). Thus, also 2 n n ∇ψ0 ∈ [L (R )] , and hence, 2,2 n ψ0 ∈ W (R ), n > 5. (3.27)

Moreover, employing (3.20), one obtains for n > 3 that

− (∆ψ0)(x) = ∆x(r0,0 ∗ v1φ0)(x) = −v1(x)φ0(x), (3.28) and hence 2 n (∆ψ0) ∈ L (R ), n > 3, (3.29) ∞ n 2 n n since v1 ∈ L (R ) and φ0 ∈ L (R ). In addition, (3.22) yields for a.e. x ∈ R , n > 3,

−1 −n/2 n 1−n (x − y) (∇ψ0)(x) = 4 π (2 − n)Γ((n − 2)/2) d y |x − y| v1(y)φ0(y), ˆ n |x − y| R (3.30) 3438 FRITZ GESZTESY AND ROGER NICHOLS implying

n 1−n |(∇ψ0)(x)| 6 cn d y |x − y| |v1(y)||φ0(y)| e ˆ n R n 1−n −2 6 cn d y |x − y| hyi |φ0(y)| (3.31) b ˆ n R for some ecn, bcn ∈ (0, ∞). An application of Theorem 2.6( ii) with c = 0, d = 1, p = p0 = 2, and the condition 1 < n/2 then implies 2 n n (∇ψ0) ∈ L (R ) , n > 3. (3.32)

Returning to the estimate (3.23), and invoking the Riesz potential R2,n (cf. Theorem 2.4), one obtains (for some constant den ∈ (0, ∞))

n 2−n −2 |ψ0(x)| 6 dn d y |x − y| hyi |φ0(y)| ˆ n R −2 n 6 denR2,n hyi |φ0( · )| (x), x ∈ R , (3.33) and hence (2.40) implies (for some constant Cep,q,n ∈ (0, ∞)) −2 kψ0kLq ( n) den R2,n hyi |φ0( · )| q n R 6 L (R ) −2 Cep,q,n hyi |φ0( · )| p n 6 L (R ) −2 Cep,q,n hyi s n kφ0kkL2( n), (3.34) 6 L (R ) R −1 −1 −1 −1 1 1. In particular, p = qn/(n + 2q), 2n + 4q − qn > 0. (3.35) (a) The case n = 3: Then p = 3q/(3 + 2q) and p > 1 requires q ∈ (3, ∞), hence −2 p ∈ (1, 3/2), and s = 6q/(q + 6) > 2. Thus, (3.34) and h · i s 3 < ∞ imply L (R ) q 3 ψ0 ∈ L (R ), q ∈ (3, ∞). (3.36) To prove that a resonance function (resp., eigenfunction) satisﬁes ψ ∈ L∞(R3), −2 one applies (3.22) and the fact that |v1( · )| 6 ch · i for some constant c ∈ (0, ∞) in (3.21) to obtain

3 −1 −4 3 |ψ(x)| 6 de3 d y |x − y| hyi |ψ(y)|, x ∈ R , (3.37) ˆ 3 R for an appropriate x-independent constant de3 ∈ (0, ∞). By Hölder’sinequality (with conjugate exponents p = 5/4 and p0 = 5), 4/5 1/5 3 −5/4 −5 3 5 3 |ψ(x)| 6 de3 d y |x − y| hyi d y |ψ(y)| , x ∈ R . (3.38) ˆ 3 ˆ 3 R R The second integral on the right-hand side in (3.38) is finite since a resonance function (resp., eigenfunction) satisfies ψ ∈ L5(R3). An application of Lemma 2.5 with x1 = x, k = 5/4, ` = 0, β = 4, and ε = 1 then implies

4/5 3 5 3 |ψ(x)| 6 de3C3,5/4,0,4,1kψkL (R ), x ∈ R , (3.39) and hence it follows that ψ ∈ L∞(R3). ABSENCE OF THRESHOLD RESONANCES 3439

(b) The case n = 4: Then p = 4q/(4 + 2q) and p > 1 requires q ∈ (2, ∞), hence −2 p ∈ (1, 2), and s = q > 2. Thus, (3.34) and (1 + | · |) s 4 < ∞ implies L (R ) q 4 ψ0 ∈ L (R ), q ∈ (2, ∞). (3.40) To prove that a resonance function (resp., eigenfunction) satisﬁes ψ ∈ L∞(R4), −2 one applies (3.22) and the fact that |v1( · )| 6 ch · i for some constant c ∈ (0, ∞) in (3.21) to obtain

4 −2 −4 4 |ψ(x)| 6 de4 d y |x − y| hyi |ψ(y)|, x ∈ R , (3.41) ˆ 4 R for an appropriate x-independent constant de4 ∈ (0, ∞). By Hölder’sinequality (with conjugate exponents p = 3/2 and p0 = 3), 2/3 1/3 4 −3 −6 4 3 4 |ψ(x)| 6 de4 d y |x − y| hyi d y |ψ(y)| , x ∈ R . (3.42) ˆ 4 ˆ 4 R R The second integral on the right-hand side in (3.42) is finite since a resonance function (resp., eigenfunction) satisfies ψ ∈ L3(R4). An application of Lemma 2.5 with x1 = x, k = 3, ` = 0, β = 5, and ε = 1 then implies 2/3 4 3 4 |ψ(x)| 6 de4C4,3,0,5,1kψkL (R ), x ∈ R , (3.43) and it follows that ψ ∈ L∞(R4). (c) The case n > 5: Returning to (3.21), we employ the fact φ0 = v2ψ0 and hence obtain for some constants Den,Dn ∈ (0, ∞),

n 2−n −4 |ψ0(x)| 6 Den d y |x − y| hyi |ψ0(y)| ˆ n R −4 n = DnR2,n h · i |ψ0( · )| (x), x ∈ R , (3.44) and hence by Theorem 2.4( ii) −4 kψ0kLq ( n) Dn R2,n h · i |ψ0( · )| q n R 6 L (R ) −4 Cbp,q,n h · i |ψ0( · )| p n 6 L (R ) −4 Cbp,q,n h · i s n kψ0kL2( n) (3.45) 6 L (R ) R for some Cbp,q,n ∈ (0, ∞). Then p = qn/(n + 2q) > 1 yields q > n/(n − 2). Since s = 2qn/(2n+4q −qn) and one needs s > 1 and hence 2n+4q −qn > 0, this implies q < 2n/(n − 4). Thus, q ∈ (n/(n − 2), 2n/(n − 4)) and hence s > 2. The condition −4 h · i s n < ∞ then implies q > 2n/(n + 4). Altogether one obtains L (R ) ( (n/(n − 2), 2n/(n − 4)), 5 n 7, q ∈ 6 6 (3.46) (2n/(n + 4), 2n/(n − 4)), n > 8, and thus, ( q n (n/(n − 2), 2n/(n − 4)), 5 6 n 6 7, ψ0 ∈ L (R ), q ∈ (3.47) (2n/(n + 4), 2n/(n − 4)), n > 8. ∞ n Since ∆ψ0 = vψ0 and v ∈ L (R ), (3.47) also implies ( q n (n/(n − 2), 2n/(n − 4)), 5 6 n 6 7, ∆ψ0 ∈ L (R ), q ∈ (3.48) (2n/(n + 4), 2n/(n − 4)), n > 8, 3440 FRITZ GESZTESY AND ROGER NICHOLS and hence (3.16). Employing φ0 = v2ψ0 in (3.31) yields

n 1−n −4 |(∇ψ0)(x)| 6 Cen d y |x − y| hyi |ψ0(y)| ˆ n R −4 n = CnR1,n h · i |ψ0( · )| (x), x ∈ R , (3.49) for some Cen,Cn ∈ (0, ∞). Thus, Theorem 2.4( ii) yields r n |∇ψ0| ∈ L (R ), r = qn/(n − q), q < n, (3.50) with q given as in (3.47). Working out the details yields  (5/2, ∞), n = 5,  r n (2, ∞), n = 6, |∇ψ0| ∈ L (R ), r ∈ (3.51) (7/4, 14), n = 7,   (2n/(n − 2), 2n/(n − 6)), n > 8. On the other hand, applying Theorem 2.6( ii) to the ﬁrst line in (3.49) with c = 0, d = 1, 1 < n/q0 = n(q − 1)/q, and hence q > n/(n − 1) yields ( q n (n/(n − 2), 2n/(n − 4)), 5 6 n 6 7, |∇ψ0| ∈ L (R ), q ∈ (3.52) (2n/(n + 4), 2n/(n − 4)), n > 8. A comparison of (3.51) and (3.52) implies (3.17). To prove essential boundedness of eigenfunctions ψ in dimensions n ∈ {5, 6, 7}, −2 one notes that a combination of (3.21), (3.22), and the fact that |v1( · )| 6 ch · i for some constant c ∈ (0, ∞), yields

n 2−n −4 n |ψ(x)| 6 den d y |x − y| hyi |ψ(y)|, x ∈ R , n > 5, (3.53) ˆ n R for an appropriate x-independent constant den ∈ (0, ∞). To prove that the right- hand side in (3.53) is essentially bounded, we make use of (3.47) and consider the cases n ∈ {5, 6, 7} next. By (3.47), ψ ∈ Lq(Rn) for all q ∈ (n/(n − 2), 2n/(n − 4)). (3.54) Note that q satisfies (3.54) if and only if its conjugate exponent q0 = q/(q − 1) satisfies q0 ∈ (2n/(n + 4), n/2). (3.55) 0 0 0 Choose q0 ∈ (2n/(n + 4), n/(n − 2)) ⊂ (2n/(n + 4), n/2). Then q0 := q0/(q0 − 1) ∈ (n/(n − 2), 2n/(n − 4)) and ψ ∈ Lq0 (Rn). Applying Hölder’sinequality with 0 conjugate exponents q0 and q0 on the right-hand side of (3.53), one obtains 0 1/q0 1/q0 0 0 n (2−n)q −4q n q0 n |ψ(x)| 6 den d y |x − y| 0 hyi 0 d y |ψ(y)| , x ∈ R . ˆ n ˆ n R R (3.56)

The second integral on the right-hand side in (3.56) if ﬁnite since ψ ∈ Lq0 (Rn). To estimate the ﬁrst integral on the right-hand side in (3.56), one applies Lemma 2.5 0 0 with the choices x1 = x, k = k0 := (n − 2)q0, ` = 0, β = β0 := 4q0 − [4n/(n + 4)], and ε = ε0 := 4n/(n + 4), to obtain 0 1/q0 n |ψ(x)| d C kψk q0 n , x ∈ . (3.57) 6 en n,k0,0,β0,ε0 L (R ) R ABSENCE OF THRESHOLD RESONANCES 3441

0 One notes that the hypotheses of Lemma 2.5 are satisﬁes since k = (n − 2)q0 < (n − 2) · n/(n − 2) = n, while 0 0 −4q0 −β0−4n/(n+4) β = 4q0 − 4n/(n + 4) > 8n/(n + 4) − 4n/(n + 4) > 0, h · i = h · i , (3.58) and 0 0 0 0 k+`+β = nq0 +2q0 −4n/(n+4) > nq0 +4n/(n+4)−4n/(n+4) = nq0 > n. (3.59) ∞ n The inequality in (3.57) implies ψ ∈ L (R ) for 5 6 n 6 7. In Remark A.1 we will illustrate why the same line of reasoning fails for n > 8. Finally, we will prove that if ker(h) % {0} then one necessarily also obtains that 2 n 2 n −1 ker IL (R ) + v2(h0 + 0IL (R )) v1 % {0}. Indeed, if 0 6= ψ0 ∈ ker(h), then 2 n 2 n φ0 := v2ψ0 = uvv1ψ0 ∈ L (R ) and hence v1φ0 ∈ L (R ). Then, hψ0 = 0 yields ∆ψ0 = vψ0 = v1v2ψ0 = v1φ0. Thus, an application of (3.9) yields for all n > 3, −1 2 n − ∆ ψ0 + (h0 + 0IL (R )) v1φ0 (x) = −[∆ψ0](x) − ∆x[r0,0 ∗ (v1φ0)](x)

= −v(x)ψ0(x) + v1(x)φ0(x)

= −v(x)ψ0(x) + v(x)ψ0(x) = 0. Consequently, −1 2 n − ∆ ψ0 + (h0 + 0IL (R )) v1φ0 = 0. (3.60) 2 n N Since ψ0 ∈ [L (R )] , and by exactly the same arguments employed in (3.23)– −1 2 n 2 n (3.25) also (h0 + 0IL (R )) (v1φ0) ∈ L (R ), one concludes that −∆ψe0 = 0 in the sense of distributions, where −1 2 n 2 n ψe0 = ψ0 + (h0 + 0IL (R )) v1φ0 ∈ L (R ). (3.61) 2 n 2,2 n Thus, also ∇ψe0 ∈ L (R ) implying ψe0 ∈ W (R ). But then ψe0 = 0 since ker(h0) = {0}. Hence, −1 2 n ψ0 = −(h0 + 0IL (R )) v1φ0, (3.62) implying, φ0 6= 0, and −1 2 n 0 = v2ψe0 = v2ψ0 + v2(h0 + 0IL (R )) v1φ0 2 n 2 n −1 = IL (R ) + v2(h0 + 0IL (R )) v1 φ0, (3.63) that is, 2 n 2 n −1 0 6= φ0 ∈ ker IL (R ) + v2(h0 + 0IL (R )) v1 . (3.64) This concludes the proof.

Remark 3.4. Employing ∆ψ = vψ with v ∈ Lr(Rn) for r > n/4 and ψ ∈ Lq(Rn) s n for q ∈ (3, ∞) ∪ {∞}, respectively, ∆ψ = v1φ, with v1 ∈ L (R ) for s > n/2 and φ ∈ L2(Rn), an application of H¨older’sinequality yields additional Lp(Rn)- properties of ∆ψ, but we omit further details at this point. Remark 3.5. (i) For basics on the Birman–Schwinger principle in an abstract context, especially, if the energy parameter in the Birman–Schwinger operator belongs to the resolvent set of the unperturbed operator, we refer to [48] (cf. also [14], [47]) and the extensive literature cited therein. In the concrete case of Schr¨odingeroperators, relations (3.78), (3.80) are discussed at length in [8], [19], [20], [31], [37], [41], [42], [46], [50]–[53], [62], [76] (see also the list of references quoted therein). 3442 FRITZ GESZTESY AND ROGER NICHOLS

(ii) In physical notation (see, e.g., [63, footnote 3 on p. 300] for details), the zero- energy resonances in Cases (II) and (IV ) for n = 3, 4, are s-wave resonances (i.e., corresponding to angular momentum zero) in the case where V is spherically symmetric (see also the discussion in [59]). (iii) As mentioned in Remark 2.3, the absence of zero-energy resonances is well- known in dimensions n > 5, see [50]. (iv) For discussions of the threshold behavior of resolvents of Schr¨odingeroperators in dimensions n = 1, 2 (the cases not studied in this paper) we refer to [17]–[21], [37], [53], [59], [62], [76]. The cases n > 3 are treated in [8], [31], [35], [41], [42], [46], [50]–[53], [59], [62], [77]. n Massless Dirac Operators in R , n > 2. Next, we turn to the case of massless Dirac operators H in dimension n > 2 as deﬁned in (2.4), and start by making the following assumptions on the matrix-valued potential V .

Hypothesis 3.6. Let n ∈ N, n > 2. Assume the a.e. self-adjoint matrix-valued potential V = {V`,m}16`,m6N satisﬁes for some C ∈ (0, ∞), ∞ n N×N V ∈ [L (R )] , (3.65) −2 n |V`,m(x)| 6 Chxi for a.e. x ∈ R , 1 6 `, m 6 N.

In addition, alluding to the polar decomposition of V ( · )(i.e., V ( · ) = UV ( · )|V ( · )|) in the following symmetrized form (cf. [49]), we suppose that ∗ 1/2 1/2 ∗ 1/2 1/2 V = V1 V2 = |V | UV |V | , where V1 = V1 = |V | ,V2 = UV |V | . (3.66) We continue with the threshold behavior, that is, the z = 0 behavior, of H: Deﬁnition 3.7. Assume Hypothesis 3.6. (i) The point 0 is called a zero-energy eigenvalue of H if HΨ = 0 has a distributional solution Ψ satisfying 1,2 n N Ψ ∈ dom(H) = [W (R )] (3.67) (equivalently, ker(H) % {0}). (ii) The point 0 is called a zero-energy (or threshold ) resonance of H if

−1 ∗ ker I 2 n N + V (H − (0 + i0)I 2 n N ) V {0}, (3.68) [L (R )] 2 0 [L (R )] 1 % −1 ∗ and if there exists 0 6= Φ ∈ ker I 2 n N + V (H − (0 + i0)I 2 n N ) V [L (R )] 2 0 [L (R )] 1 such that Ψ deﬁned by −1 ∗ Ψ(x) = − (H − (0 + i0)I 2 n N ) V Φ (x) 0 [L (R )] 1 −1 −n/2 n −n ∗ (3.69) = −i2 π Γ(n/2) d y |x − y| [α(x − y)]V1(y) Φ(y) ˆ n R n (for a.e. x ∈ R , n > 2) is a distributional solution of HΨ = 0 satisfying 2 n N Ψ ∈/ [L (R )] . (3.70) (iii) 0 is called a regular point for H if it is neither a zero-energy eigenvalue nor a zero-energy resonance of H. Additional properties of Ψ are isolated in Theorem 3.8. While the point 0 being regular for H is the generic situation, zero-energy eigenvalues and/or resonances are exceptional cases. ABSENCE OF THRESHOLD RESONANCES 3443

Next, we introduce the following convenient abbreviation (for x, y ∈ Rn, x 6= y):

R0,0(x − y) = lim G0(z; x, y) = G0(0 + i0; x, y) z→0 z∈C+\{0} (x − y) = i2−1π−n/2Γ(n/2) α · |x − y|n ( (2π)−1iα[∇ ln(|x − y|)], n = 2, = x (3.71) −iα[∇xg0(0; x, y)], n > 3.

Theorem 3.8. Assume Hypothesis 3.6. (i) If n = 2, there are precisely four possible cases: Case (I): 0 is regular for H. Case (II): 0 is a (possibly degenerate3 ) resonance of H. In this case, the resonance functions Ψ satisfy

q 2 2 2 2 2×2 Ψ ∈ [L (R )] , q ∈ (2, ∞) ∪ {∞}, ∇Ψ ∈ [L (R )] , 2 2 2 (3.72) Ψ ∈/ [L (R )] .

Case (III): 0 is a (possibly degenerate ) eigenvalue of H. In this case, the corre- 1,2 2 2 sponding eigenfunctions Ψ ∈ dom(H) = W (R ) of HΨ = 0 also satisfy

q 2 2 Ψ ∈ [L (R )] , q ∈ [2, ∞) ∪ {∞}. (3.73)

Case (IV ): A possible mixture of Cases (II) and (III). (ii) If n ∈ N, n > 3, there are precisely two possible cases: Case (I): 0 is regular for H. Case (II): 0 is a (possibly degenerate ) eigenvalue of H. In this case, the corre- 1,2 n N sponding eigenfunctions Ψ ∈ dom(H) = W (R ) of HΨ = 0 also satisfy  (3/2, ∞) ∪ {∞}, n = 3, q n N  Ψ ∈ L (R ) , q ∈ (4/3, 4), n = 4, (3.74)  (2n/(n + 2), 2n/(n − 2)), n > 5.

In particular, there are no zero-energy resonances of H in dimension n > 3. (iii) The point 0 is regular for H if and only if

−1 ∗ ker I 2 n N + V (H − (0 + i0)I 2 n N ) V = {0}. (3.75) [L (R )] 2 0 [L (R )] 1

Proof. Since G0(0 + i0; x, y), x 6= y, exists for all n > 2 (cf. (2.36)), the Birman– Schwinger eigenvalue equation

−1 ∗ 2 n N I 2 n N + V (H − (0 + i0)I 2 n N ) V Φ = 0, 0 6= Φ ∈ [L ( )] , [L (R )] 2 0 [L (R )] 1 0 0 R (3.76)

3One can show that if n = 2, the degeneracy in Case (II) is at most two, see [33]. 3444 FRITZ GESZTESY AND ROGER NICHOLS

1 n N gives rise to a distributional zero-energy solution Ψ0 ∈ [Lloc(R )] of HΨ0 = 0 in n terms of Φ0 of the form (for a.e. x ∈ R , n > 2), −1 ∗ Ψ (x) = − (H − (0 + i0)I 2 n N ) V Φ (x) 0 0 [L (R )] 1 0 ∗ = −[R0,0 ∗ (V1 Φ0)](x) (3.77)

−1 −n/2 n −n ∗ = −i2 π Γ(n/2) d y |x − y| [α(x − y)]V1(y) Φ0(y), (3.78) ˆ n R −1 −n/2 n −n ∗ = −i2 π Γ(n/2) d y |x − y| [α(x − y)]V1(y) V2(y)Ψ0(y), (3.79) ˆ n R Φ0(x) = (V2Ψ0)(x). (3.80)

In particular, one concludes that Ψ 6= 0. Thus, one estimates, with kV ( · )k N×N 0 1 C 6 −1 ch · i and some constant dn ∈ (0, ∞),

n 1−n −1 kΨ (x)k N d d y |x − y| hyi kΦ (y)k N 0 C 6 n 0 C ˆ n R (3.81) −1 n = d R h · i kΦ ( · )k N (x), x ∈ . n 1,n 0 C R An application of Theorem 2.6( ii) with c = 0, d = 1, p = p0 = 2, and the inequality 2 n 1 < n/2, combined with kΦ ( · )k N ∈ L ( ), then yield 0 C R 2 n 2 n N kΨ ( · )k N ∈ L ( ) and hence, Ψ ∈ [L ( )] , n 3. (3.82) 0 C R 0 R > 1,2 n N To prove that actually Ψ0 ∈ dom(H) = [W (R )] , it suﬃces to argue as follows: 2 n N iα · ∇Ψ0 = −V Ψ0 ∈ [L (R )] (3.83) ∞ n N×N 2 n N in the sense of distributions since V ∈ [L (R )] and Ψ0 ∈ [L (R )] . Given 1,2 n N the fact dom(H0) = W (R ) (cf. (2.2)), one concludes 1,2 n N Ψ0 ∈ W (R ) , n > 3. (3.84)

Returning to the estimate (3.81), and invoking the Riesz potential R1,n (cf. Theorem 2.4), one obtains (for some constant den ∈ (0, ∞))

n 1−n −1 kΨ (x)k N d d y |x − y| hyi kΦ (y)k N 0 C 6 n 0 C ˆ n R −1 n d R h · i kΦ ( · )k N (x), x ∈ , (3.85) 6 en 1,n 0 C R and hence (2.40) implies (for some constant Cep,q,n ∈ (0, ∞)) −1 kΨ0k[Lq ( n)]N den R1,n h · i kΦ0( · )k N q n R 6 C L (R ) −1 Cep,q,n h · i kΦ0( · )k N p n 6 C L (R ) −1 Cep,q,n h · i s n kkΦ0( · )k N kL2( n) 6 L (R ) C R −1 = Cep,q,n h · i s n kΦ0k[L2( n)]N , (3.86) L (R ) R −1 −1 −1 −1 1 1. In particular, p = qn/(n + q), 2n + 2q − qn > 0. (3.87) (a) The case n = 2: Then one can choose q ∈ (2, ∞), hence p = 2q/(q + 2) ∈ (1, 2), −1 and s = q > 2. Thus, (3.86) and h · i s 2 < ∞ imply L (R ) q 2 N Ψ0 ∈ [L (R )] , q ∈ (2, ∞). (3.88) ABSENCE OF THRESHOLD RESONANCES 3445

Recalling R0,0( · ) in (3.71), this implies

−iα · ∇xR0,0(x − y) = −∆xg0(0; x, y)IN = δ(x − y)IN , n (3.89) x, y ∈ R , x 6= y, n > 2, in the sense of distributions. Here we abused notation a bit and denoted also in the case n = 2, −1 2 g0(0; x, y) = −(2π) ln(|x − y|), x, y ∈ R , x 6= y, n = 2. (3.90) Thus, one obtains ∗ ∗ iα · (∇Ψ)(x) = −iα · ∇x[R0,0 ∗ (V1 Φ0)](x) = −iα · ∇x[−i(α · ∇xg0 ∗ (V1 Φ0)](x) ∗ ∗ 2 2 2 = (−∆xg0IN ) ∗ (V1 Φ0)](x) = (V1 Φ0)(x) ∈ [L (R )] , (3.91) 2 2 2×2 2 2 n proving ∇Ψ0 ∈ [L (R )] , upon employing the fact that [α · p] = IN |p| , p ∈ R . To prove that Ψ ∈ [L∞(R2)]2 in (3.72) and (3.73), one applies (3.80) to the −1 2×2 inequality in (3.81), and then employs the condition kV2( · )kC 6 Ch · i for some constant C ∈ (0, ∞) to obtain

2 −1 −2 2 2 2 kΨ0(x)kC 6 de2 d y |x − y| hyi kΨ0(y)kC , x ∈ R , (3.92) ˆ 2 R where de2 ∈ (0, ∞) is an appropriate x-independent constant. By H¨older’sinequality, 2/3 1/3 2 −3/2 −3 2 3 2 2 kΨ0(x)kC 6 de2 d y |x − y| hyi d y kΨ0(y)k 2 , x ∈ R . ˆ 2 ˆ 2 C R R (3.93) 3 2 2 The second integral on the right-hand side in (3.93) is ﬁnite since Ψ0 ∈ [L (R )] . Choosing x1 = x, k = 3/2, ` = 0, β = 2, and ε = 1 in Lemma 2.5, one infers that

2 −3/2 −3 2 d y |x − y| hyi 6 C2,3/2,0,2,1, x ∈ R . (3.94) ˆ 2 R ∞ 2 2 Hence, the containment Ψ0 ∈ [L (R )] follows from (3.93) and (3.94). 1,2 n N (b) The case n > 3: By (3.84) we know that Ψ0 ∈ W (R ) . Employing the fact that Φ0 = V2Ψ0 in the ﬁrst line of (3.85), one obtains

n 1−n −2 kΨ (x)k N D d y |x − y| hyi kΨ (y)k N 0 C 6 en 0 C ˆ n R −2 n = D R h · i kΨ ( · )k N (x), x ∈ , (3.95) n 1,n 0 C R for some constants Den,Dn ∈ (0, ∞). Thus, as in (3.86), (2.40) implies for n > 3, −2 kΨ0k[Lq ( n)]N Dn R1,n h · i kΨ0( · )k N q n R 6 C L (R ) −2 Dep,q,n h · i kΨ0( · )k N p n 6 C L (R ) −2 Dep,q,n h · i s n kkΨ0( · )k N kL2( n) 6 L (R ) C R −2 = Dep,q,n h · i s n kΨ0k[L2( n)]N , (3.96) L (R ) R −1 −1 −1 −1 1 1, for some constant Dep,q,n ∈ (0, ∞). In particular, one again has p = qn/(n + q) and 2n + 2q − qn > 0. The latter condition implies q < 2n/(n − 2). The requirement p > 1 results in q > n/(n−1), and the condition s > 1 yields q > 2n/(3n−2) which, 3446 FRITZ GESZTESY AND ROGER NICHOLS

−2 however, is superseded by q > p > 1. Moreover, the requirement h · i s n < L (R ) ∞ yields q > 2n/(n + 2). Putting it all together implies (3.74). To prove the containment Ψ ∈ [L∞(R3)]4 in (3.74), one applies (3.80) to the inequality in (3.81) to obtain

3 −2 −2 3 4 4 kΨ0(x)kC 6 d3 d y |x − y| hyi kΨ0(y)kC , x ∈ R . (3.97) ˆ 3 R By Hölder’sinequality (with conjugate exponents q0 = 27/20 and q = 27/7), one infers that 20/27 7/27 3 −27/10 −27/10 3 2/77 4 kΨ0(x)kC 6 d3 d y |x − y| hyi d y kΨ0(y)k 4 , ˆ 3 ˆ 3 C R R 3 x ∈ R . (3.98) 27/7 3 4 The second integral on the right-hand side in (3.98) is finite since Ψ0 ∈ [L (R )] , and the first integral on the right-hand side in (3.98) may be estimated by taking x1 = x, k = 27/10, ` = 0, β = 2, and ε = 7/10 in Lemma 2.5,

27 27 3 − 10 − 10 3 d y |x − y| hyi 6 C3, 27 ,0,2,7/10, x ∈ R \{0}. (3.99) ˆ 3 10 R ∞ 3 4 Hence, the containment Ψ0 ∈ [L (R )] follows from (3.98) and (3.99). We will illustrate in Remark A.2 that the same line of reasoning fails for n > 4. Finally (following the proof of [33, Lemma 7.4]), we show that if ker(H) % {0} −1 ∗ then also ker I 2 n N + V (H − (0 + i0)I 2 n N ) V {0}. Indeed, if [L (R )] 2 0 [L (R )] 1 % 2 n N ∗ 0 6= Ψ0 ∈ ker(H), then Φ0 := V2Ψ0 = UV V1Ψ0 ∈ [L (R )] and hence V1 Φ0 ∈ 2 n N ∗ ∗ [L (R )] . Then, HΨ0 = 0 yields iα · ∇Ψ0 = V Ψ0 = V1 V2Ψ0 = V1 Φ0. Thus, applying (3.71), (3.89)–(3.90) once again, one obtains for all n > 2, −1 ∗ − iα · ∇ Ψ + (H − (0 + i0)I 2 n N ) V Φ (x) 0 0 [L (R )] 1 0 ∗ = −i[α · ∇Ψ0](x) − iα · ∇x[R0,0 ∗ (V1 Φ0)](x) ∗ = −i[α · ∇Ψ0](x) − iα · ∇x[−i(α · ∇xg0 ∗ (V1 Φ0)](x) ∗ = −i[α · ∇Ψ0](x) + (−∆xg0IN ) ∗ (V1 Φ0)](x) ∗ = −i[α · ∇Ψ0](x) + (V1 Φ0)(x)

= −V (x)Ψ0(x) + V (x)Ψ0(x) = 0. (3.100) Consequently, −1 ∗ − iα · ∇[Ψ + (H − (0 + i0)I 2 n N ) V Φ ] = 0, (3.101) 0 0 [L (R )] 1 0 implying −1 ∗ > Ψ + (H − (0 + i0)I 2 n N ) V Φ = c , (3.102) 0 0 [L (R )] 1 0 N 2 n N for some c ∈ C . Since Ψ0 ∈ [L (R )] , and by exactly the same arguments ∗ 2 n N employed in (3.81)–(3.82), also R0,0 ∗ (V1 Φ0) ∈ [L (R )] , one concludes that c = 0 and hence −1 ∗ Ψ = −(H − (0 + i0)I 2 n N ) V Φ . (3.103) 0 0 [L (R )] 1 0 Thus, Φ0 6= 0, and −1 ∗ 0 = V Ψ + V (H − (0 + i0)I 2 n N ) V Φ 2 0 2 0 [L (R )] 1 0 −1 ∗ = I 2 n N + V (H − (0 + i0)I 2 n N ) V Φ , (3.104) [L (R )] 2 0 [L (R )] 1 0 ABSENCE OF THRESHOLD RESONANCES 3447 that is, −1 ∗ 0 6= Φ ∈ ker I 2 n N + V (H − (0 + i0)I 2 n N ) V . (3.105) 0 [L (R )] 2 0 [L (R )] 1 This concludes the proof. Remark 3.9. (i) For basics on the Birman–Schwinger principle in the concrete case of massless Dirac operators, see [32], [33]. (ii) In physical notation, the zero-energy resonances in Cases (II) and (IV ) for n = 2 correspond to eigenvalues ±1/2 of the spin-orbit operator (cf. the operator S in [54], [55]) when V is spherically symmetric, see the discussion in [33]. (iii) As mentioned in Remark 2.3, the absence of zero-energy resonances is well- known in the three-dimensional case n = 3, see [7], [12, Sect. 4.4], [13], [16], [67], [68], [78]. In fact, for n = 3 the absence of zero-energy resonances has been shown −1−ε 3 under the weaker decay |Vj,k| 6 Chxi , x ∈ R , in [7]. The absence of zero- energy resonances for massless Dirac operators in dimensions n > 4 as contained in Theorem 3.8( ii) appears to have gone unnoticed in the literature. n Massive Dirac Operators in R , n > 3. Finally, we turn to the case of massive Dirac operators H(m), m > 0, in dimension n > 3 as deﬁned in (2.8), and start by making the following assumptions on the matrix-valued potential V .

Hypothesis 3.10. Let n ∈ N, n > 3. Assume the a.e. self-adjoint matrix-valued potential V = {Vj,k}16j,k6N satisﬁes for some C ∈ (0, ∞), ∞ n N×N V ∈ [L (R )] , (3.106) −4 n |Vj,k(x)| 6 Chxi for a.e. x ∈ R , 1 6 j, k 6 N.

In addition, alluding to the polar decomposition of V ( · )(i.e., V ( · ) = UV ( · )|V ( · )|) in the following symmetrized form (cf. [49]), we suppose that ∗ 1/2 1/2 ∗ 1/2 1/2 V = V1 V2 = |V | UV |V | , where V1 = V1 = |V | ,V2 = UV |V | . (3.107) We continue with the threshold behavior, that is, the z = ±m behavior, of H(m): Deﬁnition 3.11. Assume Hypothesis 3.6. (i) The point ±m is called a threshold eigenvalue of H(m) if H(m)Ψ = ±mΨ has a distributional solution Ψ satisfying 1,2 n N Ψ ∈ dom(H(m)) = [W (R )] (3.108)

(equivalently, ker(H ∓ mI 2 n N ) {0}). [L (R )] % (ii) The point ±m is called a threshold resonance of H(m) if −1 ∗ ker I 2 n N + V (H (m) − (±m + i0)I 2 n N ) V {0} (3.109) [L (R )] 2 0 [L (R )] 1 % and if there exists −1 ∗ 0 6= Φ ∈ ker I 2 n N + V (H (m) − (±m + i0)I 2 n N ) V (3.110) ±m [L (R )] 2 0 [L (R )] 1 such that Ψ±m deﬁned by −1 ∗ Ψ (x) = − (H (m) − (±m + i0)I 2 n N ) V Φ (x) ±m 0 [L (R )] 1 n −1 −n/2 2−n = − d y 4 π Γ((n − 2)/2)|x − y| (mβ ± mIN ) (3.111) ˆ n R (x − y) + i2−1π−n/2Γ(n/2)α · V (y)∗Φ (y) |x − y|n 1 ±m 3448 FRITZ GESZTESY AND ROGER NICHOLS

n (for a.e. x ∈ R , n > 2) is a distributional solution of H(m)Ψ±m = ±mΨ±m satisfying 2 n N Ψ±m ∈/ [L (R )] . (3.112) (iii) ±m is called a regular point for H(m) if it is neither a threshold eigenvalue nor a threshold resonance of H(m).

Additional properties of Ψ±m are isolated in Theorem 3.8. While the points ±m being regular for H(m) is the generic situation, threshold eigenvalues and/or resonances at energies ±m are exceptional cases. Next, we introduce the following convenient abbreviation (for x, y ∈ Rn, x 6= y):

R0,±m(x − y) = lim G0(m, z; x, y) = G0(m, ±m + i0; x, y) z→±m z∈C+\{±m} −1 −n/2 2−n = 4 π Γ((n − 2)/2)|x − y| (mβ ± mIN ) (x − y) + i2−1π−n/2Γ(n/2)α · |x − y|n

= g0(0; x, y)(mβ ± mIN ) + G0(0 + i0; x, y) = r0,0(x − y)(mβ ± mIN ) + R0,0(x − y), n ∈ N, n > 3. (3.113) Theorem 3.12. Assume Hypothesis 3.10. (i) If n = 3, 4, there are precisely four possible cases for each of ±m: Case (I): ±m is regular for H(m). Case (II): ±m is a (possibly degenerate) resonance of H(m). In this case the resonance functions Ψ±m satisfy ( q n N (3, ∞) ∪ {∞}, n = 3, Ψ±m ∈ [L (R )] , q ∈ (2, 4), n = 4, (3.114) 2 n N×n 2 n N ∇Ψ±m ∈ [L (R )] , Ψ±m ∈/ [L (R )] . Case (III): ±m is a (possibly degenerate ) eigenvalue of H(m). In this case the 1,2 n N corresponding eigenfunctions Ψ±m ∈ dom(H(m)) = W (R ) of H(m)Ψ±m = ±mΨ±m also satisfy ( q n N [2, ∞) ∪ {∞}, n = 3, Ψ±m ∈ [L (R )] , q ∈ (3.115) [2, 4), n = 4. Case (IV ): A possible mixture of Cases (II) and (III). (ii) If n ∈ N, n > 5, there are precisely two possible cases: Case (I): ±m is regular for H(m). Case (II): ±m is a (possibly degenerate ) eigenvalue of H(m). In this case, the 1,2 n N corresponding eigenfunctions Ψ±m ∈ dom(H(m)) = W (R ) of HΨ±m = ±mΨ±m also satisfy ( q n N (n/(n − 1), 2n/(n − 2)), 5 6 n 6 8, Ψ±m ∈ L (R ) , q ∈ (3.116) (2n/(n + 6), 2n/(n − 4)), n > 9. In particular, there are no resonances at energies ±m of H(m) in dimension n > 5. (iii) The point ±m is regular for H(m) if and only if −1 ∗ ker I 2 n N + V (H (m) − (±m + i0)I 2 n N ) V = {0}. (3.117) [L (R )] 2 0 [L (R )] 1 ABSENCE OF THRESHOLD RESONANCES 3449

Proof. Since G0(m, ±m + i0; x, y), x 6= y, exists for all n > 3 (cf. (3.113)), the Birman–Schwinger eigenvalue equation −1 ∗ 0 = I 2 n N + V (H (m) − (±m + i0)I 2 n N ) V Φ , (3.118) [L (R )] 2 0 [L (R )] 1 ±m 2 n N 0 6= Φ±m ∈ [L (R )] , 1 n N gives rise to a distributional solution Ψ±m ∈ [Lloc(R )] of the equation

H(m)Ψ±m = ±mΨ±m (3.119) n in terms of Φ±m of the form (for a.e. x ∈ R , n > 3), −1 ∗ Ψ (x) = − (H (m) − (±m + i0)I 2 n N ) V Φ (x) ±m 0 [L (R )] 1 0 ∗ = −[R0,±m ∗ (V1 Φ±m)](x) (3.120) n −1 −n/2 2−n = − d y 4 π Γ((n − 2)/2)|x − y| (mβ ± mIN ) (3.121) ˆ n R (x − y) + i2−1π−n/2Γ(n/2)α · V (y)∗Φ(y), (3.122) |x − y|n 1

Φ±m(x) = (V2Ψ±m)(x). (3.123)

In particular, one concludes that Ψ 6= 0. Using the inequality kV ( · )k N×N ±m 1 C 6 −2 n ch · i , one obtains the following estimate for some dn, den ∈ (0, ∞) (for a.e. x ∈ R , n > 3): n 2−n −2 kΨ (x)k N d d y |x − y| hyi kΦ (y)k N (3.124) ±m C 6 n ±m C ˆ n R n 1−n −2 + d y |x − y| hyi kΦ (y)k N ±m C ˆ n R −2 −2 d R h · i kΦ ( · )k N (x) + R h · i kΦ ( · )k N (x) . (3.125) 6 en 1,n ±m C 2,n ±m C The ﬁrst term on the right-hand side in (3.125) may be estimated as follows:

−2 −1 n 1−n −2 R h · i kΦ ( · )k N (x) = γ(1, n) d y |x − y| hyi kΦ (y)k N 1,n ±m C ±m C ˆ n R −1 n 1−n −1 n γ(1, n) d y |x − y| hyi kΦ (y)k N , x ∈ . (3.126) 6 ±m C R ˆ n R An application of Theorem 2.6( ii) with c = 0, d = 1, p = p0 = 2, and the inequality 2 n 1 < n/2, combined with kΦ ( · )k N ∈ L ( ), then yield ±m C R −2 2 n R h · i kΦ ( · )k N ∈ L ( ), n 3. (3.127) 1,n ±m C R > Similarly, the second term on the right-hand side in (3.125) may be estimated as follows: −2 −1 n 2−n −2 R h · i kΦ ( · )k N (x) = γ(2, n) d y |x − y| hyi kΦ (y)k N , 2,n ±m C ±m C ˆ n R n x ∈ R . (3.128) An application of Theorem 2.6( ii) with c = 0, d = 2, p = p0 = 2, and the inequality 2 n 2 < n/2, combined with kΦ ( · )k N ∈ L ( ), then yield ±m C R −2 2 n R h · i kΦ ( · )k N ∈ L ( ), n 5. (3.129) 2,n ±m C R > Thus, (3.125), (3.129), and (3.127) imply 2 n 2 n N kΨ ( · )k N ∈ L ( ) and hence, Ψ ∈ [L ( )] , n 5. (3.130) ±m C R ±m R > 3450 FRITZ GESZTESY AND ROGER NICHOLS

1,2 n N To prove that actually Ψ±m ∈ dom(H(m)) = [W (R )] for n > 5, it suffices to argue as follows: 2 n N iα · ∇Ψ±m = −V Ψ±m ± mΨ±m ∈ [L (R )] (3.131) ∞ n N×N 2 n N in the sense of distributions since V ∈ [L (R )] and Ψ±m ∈ [L (R )] . Given 1,2 n N the fact dom(H0(m)) = W (R ) (cf. (2.8)), one concludes that 1,2 n N Ψ±m ∈ W (R ) , n > 5. (3.132) (1) Returning to (3.125), one applies (2.40) to estimate (for some constant Cp,q,n ∈ (0, ∞)) −2 (1) −2 R1,n h · i kΦ±m( · )k N q n Cp,q,n h · i kΦ±m( · )k N p n C L (R ) 6 C L (R ) (1) −2 Cp,q,n h · i s n kΦ±m( · )k N 2 n 6 L (R ) C L (R ) (1) −2 = Cp,q,n h · i s n kΦ±mk[L2( n)]N , (3.133) L (R ) R −1 −1 −1 −1 1 1, so that, in particular, p = qn/(n + q), 2n + 2q − qn > 0. (3.134) (2) Similarly, one applies (2.40) to estimate (for some constant Cp,q,n ∈ (0, ∞)) −2 (2) −2 R2,n h · i kΦ±m( · )k N q n Cp,q,n h · i kΦ±m( · )k N p n C L (R ) 6 C L (R ) (2) −2 Cp,q,n h · i s n kΦ±m( · )k N 2 n 6 L (R ) C L (R ) (2) −2 = Cp,q,n h · i s n kΦ±mk[L2( n)]N , (3.135) L (R ) R −1 −1 −1 −1 1 1, so that, in particular, p = qn/(n + 2q), 2n + 4q − qn > 0. (3.136) 2 n N We note that if Ψ±m ∈ [L (R )] , then an application of (3.123) after the second (1) inequality in (3.133) yields (for some constant Cep,q,n ∈ (0, ∞)) −2 R1,n h · i kΦ±m( · )k N q n C L (R ) (1) −2 Cp,q,n h · i kΦ±m( · )k N p n 6 C L (R ) (1) −2 Cp,q,n h · i kV2( · )k N×N kΨ±m( · )k N p n 6 C C L (R ) (1) −4 Cep,q,n h · i kΨ±m( · ) N p n 6 C L (R ) (1) −4 Cep,q,n h · i s n kΨ±mk[L2( n)]N , (3.137) 6 L (R ) R −2 under the same conditions in (3.133) and (3.133). Note that kV ( · )k N×N ch · i 2 C 6 (2) has been used in (3.137). Similarly, one obtains (for some constant Cep,q,n ∈ (0, ∞)) −2 (2) −4 R2,n h · i kΦ±m( · )k N q n Cep,q,n h · i s n kΨ±mk[L2( n)]N (3.138) C L (R ) 6 L (R ) R under the same conditions in (3.135) and (3.136). (a) The case n = 3: The first condition in (3.134) implies p = 3q/(3 + q). The condition p > 1 requires q ∈ (3/2, ∞), and the condition s = 6q/(6 − q) > 1 further ABSENCE OF THRESHOLD RESONANCES 3451 requires q ∈ (3/2, 6). For q ∈ (3/2, 6), one infers p ∈ (1, 2) and s ∈ (2, ∞); the latter −2 s 3 ensures kh · i kL (R ) < ∞. Hence, one may take q ∈ (3/2, 6), p = 3q/(3 + q), s = 6q/(6 − q) (3.139) throughout (3.133) to obtain −2 q 3 4 R1,3 h · i kΦ±m( · )kC ∈ L (R ), q ∈ (3/2, 6). (3.140) The first condition in (3.136) implies p = 3q/(3 + 2q). The condition p > 1 requires −2 s 3 q ∈ (3, ∞), which implies s = 6q/(6 + q) ∈ (2, 6), ensuring kh · i kL (R ) < ∞. Hence, one may take q ∈ (3, ∞), p = 3q/(3 + 2q), s = 6q/(6 + q) (3.141) throughout (3.135) to obtain −2 q 3 4 R2,3 h · i kΦ±m( · )kC ∈ L (R ), q ∈ (3, ∞). (3.142) The containments in (3.140) and (3.142), combined with the estimate in (3.125) imply q 3 q 3 4 4 kΨ±m( · )kC ∈ L (R ) and hence, Ψ±m ∈ [L (R )] , q ∈ (3, 6). (3.143) 2 3 4×3 To prove that ∇Ψ±m ∈ [L (R )] , it suffices to argue as follows. By (3.120), (3.30), and (3.91), one computes

−1 −n/2 3 1−n (x − y) −iα · [∇Ψ±m](x) = iα · 4 π Γ((n − 2)/2) d y |x − y| ˆ 3 |x − y| R (3.144) ∗ ∗ 3 × V1 (y)Φ±m(y)[mβ ± mI4] + (V1 Φ±m)(x), x ∈ R . In addition,

3 1−n (x − y) ∗ d y |x − y| V1 (y)Φ±m(y) ˆ 3 |x − y| 4 R C 3 1−n ∗ 4×4 4 6 d y |x − y| kV1 (y)kC kΦ±m(y)kC ˆ 3 R 3 1−n −2 4 6 c d y |x − y| hyi kΦ±m(y)kC , (3.145) ˆ 3 R and an application of Theorem 2.6( ii) with c = 0, d = 1, p = p0 = 2 then implies

3 1−n −2 2 3 4 4 d y | · −y| hyi kΦ±m(y)kC ∈ [L (R )] . (3.146) ˆ 3 R ∗ 3 3 4 Since V1 Φ±m ∈ [L (R )] , the containment in (3.146) and the identity in (3.144) 2 3 4×3 imply ∇Ψ±m ∈ [L (R )] . 2 3 4 Finally, if Ψ±m ∈ [L (R )] , then together with (3.143), one obtains (3.115) for n = 3. If Ψ±m is a resonance function (resp., eigenfunction), then it has been shown q 3 4 that Ψ±m ∈ [L (R )] for all q ∈ (3, 6) (resp., q ∈ [2, 6)). To prove that Ψ±m ∈ ∞ 3 4 −2 4×4 [L (R )] , one applies (3.123) and the condition kV2( · )kC 6 Ch · i for some C ∈ (0, ∞) in (3.124) to obtain 3 −1 −4 4 4 kΨ±m(x)kC 6 de3 d y |x − y| hyi kΨ±m(y)kC ˆ 3 R 3 −2 −4 3 4 + d y |x − y| hyi kΨ±m(x)kC , x ∈ R . (3.147) ˆ 3 R 3452 FRITZ GESZTESY AND ROGER NICHOLS

Here de3 ∈ (0, ∞) is an appropriate x-independent constant. The ﬁrst integral on the right-hand side in (3.147) may be estimated with H¨older’sinequality (with conjugate exponents p = 5/4 and p0 = 5) as follows:

3 −1 −4 4 d y |x − y| hyi kΨ±m(y)kC (3.148) ˆ 3 R 4/5 1/5 3 −5/4 −5 3 5 3 6 d y |x − y| hyi d y kΨ±m(y)k 4 , x ∈ R . ˆ 3 ˆ 3 C R R The second integral on the right-hand side in (3.148) is finite since a resonance 5 3 4 function (resp., eigenfunction) satisfies Ψ±m ∈ [L (R )] . To estimate the first integral on the right-hand side in (3.148), ones applies Lemma 2.5 with the choices x1 = x, k = 5/4, ` = 0, β = 4, and ε = 1, so that

3 −5/4 −5 3 d y |x − y| hyi 6 C3,5/4,0,4,1, x ∈ R . (3.149) ˆ 3 R Along similar lines, the second integral on the right-hand side in (3.147) may be estimated with H¨older’s inequality (with conjugate exponents p = 5/4 and p0 = 5) as follows:

3 −2 −4 4 d y |x − y| hyi kΨ±m(y)kC (3.150) ˆ 3 R 4/5 1/5 3 −5/2 −5 3 5 3 6 d y |x − y| hyi d y kΨ±m(y)k 4 , x ∈ R . ˆ 3 ˆ 3 C R R An application of Lemma 2.5 with x1 = x, k = 5/2, ` = 0, β = 4, and ε = 1, yields

3 −5/2 −5 3 d y |x − y| hyi 6 C3,5/2,0,4,1, x ∈ R . (3.151) ˆ 3 R Therefore, combining (3.147), (3.148), (3.149), (3.150), and (3.151), one obtains

n 4/5 4/5 o 3 4 5 3 4 kΨ±m(x)kC 6 de3 C3,5/4,0,4,1 + C3,5/2,0,4,1 kΨ±mk[L (R )] , x ∈ R , (3.152) ∞ 3 4 and it follows that Ψ±m ∈ [L (R )] , settling the case n = 3. We will illustrate in Remark A.3 that the same line of reasoning fails for n > 4. (b) The case n = 4: The first condition in (3.134) implies p = 4q/(4 + q). The condition p > 1 requires q ∈ (4/3, ∞), and the condition s = 4q/(4 − q) > 1 further requires q ∈ (4/3, 4). For q ∈ (4/3, 4), one infers p ∈ (1, 2) and s ∈ (2, ∞); the latter −2 s 4 ensures kh · i kL (R ) < ∞. Hence, one may take q ∈ (4/3, 4), p = 4q/(4 + q), s = 4q/(4 − q) (3.153) throughout (3.133) to obtain −2 q 4 4 R1,4 h · i kΦ±m( · )kC ∈ L (R ), q ∈ (4/3, 4). (3.154) The first condition in (3.136) implies p = 2q/(2+q). The condition p > 1 requires q ∈ (2, ∞), and the condition on s in (3.135) reduces to s = q, so the requirement s > 1 is satisfied for all q ∈ (2, ∞). For q ∈ (2, ∞), p ∈ (1, 2), and s ∈ (2, ∞); the −2 s 4 latter ensures kh · i kL (R ) < ∞. Hence, one may take q ∈ (2, ∞), p = 2q/(2 + q), s = q (3.155) throughout (3.135) to obtain −2 q 4 4 R2,4 h · i kΦ±m( · )kC ∈ L (R ), q ∈ (2, ∞). (3.156) ABSENCE OF THRESHOLD RESONANCES 3453

The containments in (3.154) and (3.156), combined with the estimate in (3.125) imply

q 4 q 4 4 4 kΨ±m( · )kC ∈ L (R ) and hence, Ψ±m ∈ [L (R )] , q ∈ (2, 4). (3.157) 2 4 4×4 One proves ∇Ψ±m ∈ [L (R )] in a manner entirely analogous to (3.144)– 2 4 4 (3.146). Finally, if Ψ±m ∈ [L (R )] , then together with (3.157), one obtains (3.115) for n = 4. Therefore, the n = 4 case is settled. (c) The case n > 5: The ﬁrst condition in (3.134) implies p = qn/(n + q). The condition p > 1 requires q ∈ (n/(n − 1), ∞). The condition 2n + 2q − qn > 0 further requires q < 2n/(n − 2), and the condition s = 2qn/(2n + 2q − qn) > 1 −4 s n further requires q > 2n/(3n − 2). Moreover, kh · i kL (R ) < ∞ if and only if n/4 < s = 2qn/(2n + 2q − qn), which requires q > 2n/(n + 6). Thus, one may take ( (n/(n − 1), 2n/(n − 2)), 5 n 8, q ∈ 6 6 (2n/(n + 6), 2n/(n − 2)), n > 9, (3.158) p = qn/(n + q), s = 2qn/(2n + 2q − qn) throughout (3.137) to obtain ( −2 q n (n/(n − 1), 2n/(n − 2)), 5 6 n 6 8, R h · i kΦ ( · )k N ∈ L ( ), q ∈ 1,n ±m C R (2n/(n + 6), 2n/(n − 2)), n > 9. (3.159) The ﬁrst condition in (3.136) implies p = qn/(n + 2q). The condition p > 1 requires q ∈ (2/(n − 2), ∞). The condition 2n + 4q − qn > 0 further requires q < 2n/(n − 4), and the condition s = 2qn/(2n + 4q − qn) > 1 further requires −4 2 n q > 2n/(3n − 4). Moreover, kh · i kL (R ) < ∞ if and only if n/4 < s = 2qn/(2n + 4q − qn), which requires q > 2n/(n + 4). Hence, one may take q ∈ (2n/(n + 4), 2n/(n − 4)), p = qn/(n + 2q), s = 2qn/(2n + 4q − qn) (3.160) throughout (3.135) to obtain

−2 q n R h · i kΦ ( · )k N ∈ L ( ), q ∈ (2n/(n + 8), 2n/(n − 4)). (3.161) 2,n ±m C R The containments in (3.159) and (3.161), combined with the estimate in (3.125), imply

q n q n N kΨ ( · )k N ∈ L ( ) and hence, Ψ ∈ [L ( )] , ±m C R ±m R ( (n/(n − 1), 2n/(n − 2)), 5 n 8, (3.162) q ∈ 6 6 (2n/(n + 6), 2n/(n − 4)), n > 9. Finally (following the proof of [33, Lemma 7.4]), we show that if

ker(H(m) ∓ mI 2 n N ) {0}, (3.163) [L (R )] % then also

−1 ∗ ker I 2 n N + V (H (m) − (±m + i0)I 2 n N ) V {0}. (3.164) [L (R )] 2 0 [L (R )] 1 %

Indeed, if 0 6= Ψ ∈ ker(H(m) ∓ mI 2 n N ), then ±m [L (R )] 2 n N Φ±m := V2Ψ±m = UV V1Ψ±m ∈ [L (R )] , (3.165) 3454 FRITZ GESZTESY AND ROGER NICHOLS

∗ 2 n N and hence V1 Φ±m ∈ [L (R )] . Therefore, H(m)Ψ±m = ±mΨ±m yields

iα · ∇Ψ±m = (mβ ∓ mIN )Ψ±m + V Ψ±m ∗ = (mβ ∓ mIN )Ψ±m + V1 V2Ψ±m ∗ = (mβ ∓ mIN )Ψ±m + V1 Φ±m. (3.166) Hence,

−1 ∗ (−iα · ∇ + mβ ∓ mI ) Ψ + (H (m) − (∓m + i0)I 2 n N ) V Φ (x) N ±m 0 [L (R )] 1 ±m = [(−iα · ∇ + mβ ∓ mIN )Ψ±m](x) ∗ + (−iα · ∇ + mβ ∓ mIN )[R0,±m ∗ (V1 Φ±m)](x) ∗ ∗ = −(V1 Φ±m)(x) + [((−iα · ∇ + mβ ∓ mIN )R0,±m) ∗ (V1 Φ±m)](x) ∗ ∗ n = −(V1 Φ±m)(x) + (V1 Φ±m)(x) = 0, x ∈ R . (3.167)

2 n N In addition, Ψ±m ∈ [L (R )] , and the same arguments as those in (3.126)–(3.127) and (3.138)–(3.158) (which now extend to include n ∈ {3, 4}, due to the assumption 2 n N ∗ 2 n N that Ψ±m ∈ [L (R )] ) imply [R0,±m ∗ (V1 Φ±m)] ∈ [L (R )] , so that −1 ∗ 2 n N Ψ + (H (m) − (∓m + i0)I 2 n N ) V Φ ∈ [L ( )] . (3.168) ±m 0 [L (R )] 1 ±m R It follows that

−1 ∗ Ψ := Ψ + (H (m) − (∓m + i0)I 2 n N ) V Φ = 0; (3.169) e ±m ±m 0 [L (R )] 1 ±m otherwise, ±m is an eigenvalue of H0(m) and Ψe ±m is a corresponding eigenfunction. However, this contradicts the fact that the spectrum of H0(m) is purely absolutely continuous. Hence,

−1 ∗ Ψ = −(H (m) − (∓m + i0)I 2 n N ) V Φ . (3.170) ±m 0 [L (R )] 1 ±m

Thus, Φ±m 6= 0, and

−1 ∗ 0 = V Ψ + V (H (m) − (±m + i0)I 2 n N ) V Φ 2 ±m 2 0 [L (R )] 1 ±m −1 ∗ = I 2 n N + V (H (m) − (±m + i0)I 2 n N ) V Φ , (3.171) [L (R )] 2 0 [L (R )] 1 ±m that is,

−1 ∗ 0 6= Φ ∈ ker I 2 n N + V (H (m) − (±m + i0)I 2 n N ) V . (3.172) ±m [L (R )] 2 0 [L (R )] 1 This concludes the proof.

Remark 3.13. (i) For basics on the Birman–Schwinger principle in the concrete case of massive Dirac operators, see [32], [38]–[40], [57]. (ii) In physical notation (see, e.g., [54], [55], [58], [75, Sect. 4.6] for details), the threshold resonances at energies ±m in Cases (II) and (IV ) for n = 3 correspond to eigenvalues ±1 of the spin-orbit operator (cf, [54], [55], [75, eq. (4.105), p. 125]) in the case where V is spherically symmetric, see the discussion in [40]. (iii) The absence of threshold resonances for massive Dirac operators in dimensions n > 5 as contained in Theorem 3.12( ii) appears to have gone unnoticed in the literature. ABSENCE OF THRESHOLD RESONANCES 3455

Appendix A. Some remarks on L∞(Rn)-properties of threshold eigenfunctions. In this appendix we collect some negative results on L∞(Rn)-properties of threshold eigenfunctions. We hope to return to this issue at a later date. We start with the case of Schrödingeroperators: Remark A.1. We briefly indicate why the line of reasoning used in the proof of ∞ n ψ ∈ L (R ) for 3 6 n 6 7 in Theorem 3.3 is bound to fail for n > 8. q n Suppose n > 8. Then (3.53) continues to hold and by (3.47), ψ ∈ L (R ) for all q ∈ (2n/(n + 4), 2n/(n − 4)). (A.1) One notes that q satisfies (A.1) if and only if its conjugate exponent q0 := q/(q − 1) satisfies q0 ∈ (2n/(n + 4), 2n/(n − 4)). (A.2) Applying Hölder’sinequality with conjugate exponents q0 and q on the right-hand side in (3.53) yields 1/q0 1/q n (2−n)q0 −4q0 n q n |ψ(x)| 6 den d y |x − y| hyi d y |ψ(y)| , x ∈ R . ˆ n ˆ n R R (A.3)

The second integral on the right-hand side in (A.3) is finite for any choice of q0 which satisfies (A.2). When applying Lemma 2.5 to estimate the first integral on the right-hand side in (A.3), one must choose k = (n − 2)q0. The hypotheses in Lemma 2.5 require k ∈ [0, n). Hence, one must choose q0 so that (n − 2)q0 < n, that is, one must choose q0 < n/(n − 2). However, there is no q0 which simultaneously satisfies (A.2) and q0 < n/(n − 2) since 2n n n(n − 8) n 2n − = > 0 implies < . (A.4) n + 4 n − 2 (n − 2)(n + 4) n − 2 n + 4 Thus, q0 < n/(n − 2) yields q0 < 2n/(n + 4), so that q0 does not satisfy (A.2). Next, we illustrate the case of massless Dirac operators: Remark A.2. We briefly indicate why the line of reasoning used in the proof of ∞ n N Ψ ∈ L (R ) for n = 2, 3 in Theorem 3.8 is bound to fail for n > 4. We start with the case n = 4. Then the inequality in (3.81) implies, when combined with (3.80),

4 −3 −2 4 4 4 kΨ0(x)kC 6 de4 d y |x − y| hyi kΨ0(y)kC , x ∈ R . (A.5) ˆ 4 R q 4 4 In order to invoke the property Ψ0 ∈ [L (R )] , q ∈ (4/3, 4), one applies H¨older’s inequality with conjugate indices q0 and q on the right-hand side in (A.5) to obtain 1/q0 1/q 4 −3q0 −2q0 4 q 4 4 kΨ0(x)kC 6 de4 d y |x−y| hyi d y kΨ0(y)k 4 , x ∈ R . ˆ 4 ˆ 4 C R R (A.6) The second integral on the right-hand side in (A.6) is ﬁnite for any q ∈ (4/3, 4). The requirement q ∈ (4/3, 4) implies q0 ∈ (4/3, 4). (A.7) 3456 FRITZ GESZTESY AND ROGER NICHOLS

In order to apply Lemma 2.5 to the first integral on the right-hand side in (A.6), one would choose k = 3q0 and ` = 0. The hypotheses (viz., k ∈ [0, n)) of Lemma 2.5 require k = 3q0 < 4, (A.8) which is satisfied if and only if q0 < 4/3. However, the condition q0 < 4/3 is incompatible with (A.7). An analogous problem is also encountered for all other dimensions n > 5: Indeed, for n > 5, (3.80) and (3.81) combine to yield 1/q0 1/q n (1−n)q0 −2q0 n q kΨ (x)k N d d y |x − y| hyi d y kΨ (y)k , 0 C 6 en 0 N ˆ n ˆ n C R R n x ∈ R , (A.9) for conjugate exponents q0 and q. The second integral on the right-hand side in (A.9) is finite for all q ∈ (2n/(n + 2), 2n/(n − 2)). For q ∈ (2n/(n + 2), 2n/(n − 2)), one infers q0 ∈ (2n/(n + 2), 2n/(n − 2)). (A.10) To apply Lemma 2.5 to the first integral on the right-hand side in (A.9), one must choose k = (n − 1)q0. The condition k ∈ [0, n) in Lemma 2.5 requires (n − 1)q0 < n, which is equivalent to n q0 < . (A.11) n − 1 However, the condition in (A.11) is incompatible with (A.10) in the sense that there is no q0 ∈ [1, ∞) which satisfies both (A.10) and (A.11) simultaneously. Finally, we also illustrate the case of massive Dirac operators: Remark A.3. We briefly indicate why the line of reasoning used in the proof of ∞ 3 4 Ψ ∈ L (R ) in Theorem 3.12 is bound to fail for n > 4. For simplicity we focus again on the case n = 4. Applying (3.123) and the −2 4×4 condition kV2( · )kC 6 Ch · i for some C ∈ (0, ∞) in (3.124), one obtains 4 −2 −4 4 4 kΨ±m(x)kC 6 de4 d y |x − y| hyi kΨ±m(y)kC ˆ 4 R 4 −3 −4 4 4 + d y |x − y| hyi kΨ±m(x)kC , x ∈ R , (A.12) ˆ 4 R where de4 ∈ (0, ∞) is an appropriate x-independent constant. Applying Hölder’sinequality (with conjugate exponents p = 3/2 and p0 = 3) to the first integral on the right-hand side in (A.12), one obtains

4 −2 −4 4 d y |x − y| hyi kΨ±m(y)kC ˆ 4 R 2/3 1/3 4 −3 −6 4 3 4 6 d y |x − y| hyi d y kΨ±m(y)k 4 , x ∈ R . (A.13) ˆ 4 ˆ 4 C R R The second integral on the right-hand side in (A.13) is ﬁnite since a resonance 3 4 4 function (resp., eigenfunction) satisﬁes Ψ±m ∈ [L (R )] . By Lemma 2.5 with x1 = x, k = 3, ` = 0, and β = 5,

4 −3 −6 4 d y |x − y| hyi 6 C4,3,0,5,1, x ∈ R . (A.14) ˆ 4 R ABSENCE OF THRESHOLD RESONANCES 3457

Applying H¨older’sinequality (with conjugate indices q0 and q) to the second integral on the right-hand side in (A.12), one obtains

4 −3 −4 4 d y |x − y| hyi kΨ±m(y)kC (A.15) ˆ 4 R 1/q0 1/q 4 −3q0 −4q0 4 q 4 6 d y |x − y| hyi d y kΨ±m(y)k 4 , x ∈ R . ˆ 4 ˆ 4 C R R The second integral on the right-hand side in (A.15) is finite for all q ∈ (2, 4). The requirement q ∈ (2, 4) implies q0 ∈ (4/3, 2). (A.16) In order to apply Lemma 2.5 to the first integral on the right-hand side in (A.15), one would choose k = 3q0 and ` = 0. The hypotheses (viz., k ∈ [0, n)) of Lemma 2.5 require k = 3q0 < 4, (A.17) which is satisfied if and only if q0 < 4/3. However, the condition q0 < 4/3 is incompatible with (A.16).

Acknowledgments. We are indebted to Alan Carey, Will Green, Jens Kaad, Galina Levitina, Denis Potapov, and Fedor Sukochev for many helpful discussions on this subject. We also thank the referee for a critical reading of our manuscript.

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