Upper bounds for the resonance counting function

of Schr � odinger operators in odd dimensions

Richard Froese

Department of Mathematics University of British Columbia Vancouver, British Columbia, Canada

Abstract

The purpose of this note is to provide a simple proof of the sharp polynomial upper bound for the resonance counting function of a Schr� odinger operator in odd dimensions. At the same time we generalize the result to the class of super-exponentially decreasing potentials.

Mathematics subject classi�cation numbers: primary: 47A10 47A40 secondary: 81U05

Introduction

Let H = −Δ + V be a Schr� odinger operator acting in L^{2}(ℝ^{n}), n odd, whose potential V is super-exponentially decreasing. By de�nition, this means that for every N there is a constant C such that

|V (x)| ≤ Ce^{−N|x|}.

Let R(k) = (H − k^{2})^{−1} be the resolvent of H, initially de�ned for k in the upper half plane. Let R_{V} (k) denote the weighted resolvent

R_{V} (k) = V ^{\frac{1}{2}} R(k)|V |^{\frac{1}{2}} .

When V is super-exponentially decreasing and n is odd, R_{V} (k) has a compact operator valued meromorphic continuation to the entire complex plane. Resonances are poles in this meromorphic continuation.

We are interested in upper bounds for the counting function

n(r) = #{resonances k : |k| < r}.

Here is what is known about the large r behaviour of n(r).

When n = 1 and V has compact support

n(r) = \frac{2}{π} diam(supp(V))r + o(r).

This result is due to Zworski [Z1]. For some super-exponentially decaying potentials in one dimension there is a comparable result [F]

n(r) = Cr^{ρ} + o(r^{ρ}),

where ρ is the order of growth of the Fourier transform of V. For a class of radially symmetric potentials in dimensions greater than one Zworski [Z2] proves

n(r) = C_{n}radius(supp(V))r^{n} + o(r^{n}).

This exhausts the examples of Schr� odinger operators for which the �rst term in an asymptotic expansion is known.

For a general compactly supported potential in dimensions greater than one, only upper bounds (of polynomial type) are known:

n(r) ≤ Cr^{n}.

The �rst polynomial bound (r^{n+1}) was obtained by Melrose [Me1]. The sharp bound (r^{n}) was �rst obtained by Zworski [Z3]. For potentials decreasing like exp(−|x|^{1+ϵ}), S� a Barreto and Zworski [SZ1] prove

n(r) ≤ Cr^{(1+1/ϵ)n}.

In we generalize this to super-exponentially decaying potentials. The Fourier transform of a potential in this class is an entire function in ℂ^{n}. Our bound is given in terms of the growth of the Fourier transform. We will show that if V̂(k) grows like exp(Φ(|k|)), then

n(r) ≤ CΦ^{n}(cr).

Although polynomial lower bounds have not been established in general, it is known that in�nitely many resonances exist. This was shown in three dimensions by Melrose [Me2], and in any odd dimension by S� a Barreto and Zworski [SZ2]. The existence of in�nitely many resonances was previously only known under positivity conditions on the potential (see the references in [SZ2]).

There is a substantial literature on closely related problems, involving, for example, metric perturbations of the Laplacian or even dimensions. We mention here only the paper of Vodev [V] which was in�uential in our thinking. Further references can be found in the review article [Z4] and the book [Me2].

It is worth pointing out that there are many other de�nitions of resonances in the literature. The theory of dilation and translation analyticity give rise to de�nitions that do not require the potential to decrease so rapidly.

Meromorphic continuation and the scattering operator

We begin by using standard Birman-Schwinger identities to show that R_{V} (k) has a meromorphic continuation. Let R_{0}(k) = (−Δ − k^{2})^{−1} be the free resolvent and de�ne R_{0V} (k) = V ^{\frac{1}{2}} R_{0}(k)|V |^{\frac{1}{2}} . For Im k > 0, the resolvent equation

R_{0}(k) − R(k) − R(k)V R_{0}(k) = 0

implies

(1 − R_{V} (k))(1 + R_{0V} (k)) = 1.

For Im k large, the norm of R_{0V} (k) is small, so 1 + R_{0V} (k) is invertible and

1 − R_{V} (k) = (1 + R_{0V} (k))^{−1}  ()

The operator R_{0V} (k) has an explicit integral kernel given by V ^{\frac{1}{2}} (x)G_{0}(x, y, k)|V |^{\frac{1}{2}} (y), where G_{0} is the free Greens function. Using this representation, it is easy to see that in odd dimensions R_{0V} (k) has a compact operator valued analytic continuation to the entire complex plane (except for a pole at zero in dimension 1). Thus, by the meromorphic Fredholm theorem [S], the left side of () de�nes a meromorphic continuation for R_{V} (k).

From this formula we can see that the resonances are precisely those values of k for which R_{0V} (k) has an eigenvalue −1. Equivalently, resonances are precisely the zeros of the analytic function det(1 + R_{0V} (k))�provided R_{0V} (k) is trace class. Unfortunately, this only happens when n = 1. In higher dimensions, it turns out that R_{0V}^{p} (k) for p > n/2 is trace class. If −1 is an eigenvalue for R_{0V} (k), then ±1 is an eigenvalue for R_{0V}^{p} (k). Thus the set of resonances is contained in the set of zeros of the function det(1 ± R_{0V}^{p} (k)) for p > n/2 and ±(−1)^{p} = 1. This function is entire, except for poles arising from eigenvalues. One can therefore estimate the number of resonances by estimating the growth of this function. This approach is used in previous work.

Our approach is to �rst multiply 1 + R_{0V} (k) by a suitable invertible operator and then take the determinant. For k in the lower half plane, −k is in the upper half plane, and so by () the operator 1 + R_{0V} (−k) is invertible (with inverse 1 − R_{V} (−k)) except at the �nitely places where k^{2} is an eigenvalue of H. De�ne

Q(k) = (1 + R_{0V} (−k))^{−1}(1 + R_{0V} (k))

We will show that the determinant of Q(k) is well de�ned. Then resonances are precisely the zeros of det Q(k) in the lower half plane. The disadvantage of our regularization is that det Q(k) is not entire, but has poles in the upper half plane. (Clearly −k is pole whenever k is a zero.) This makes it necessary to include an estimate of the scattering phase in our proof.

In fact, we will show presently that det Q(k) = det S(−k) where S(k) is the scattering operator, con�rming that resonances are precisely equal to scattering poles. This equivalence is well known, but we could not resist giving this simple proof.

Begin with the classical Green’s function identity

G_{0}(x, y, k) − G_{0}(x, y, −k) = c(k) ∫_{S^{n−1}} e^{ik⟨ω,x−y⟩}dω,  ()

where

c(k) = \frac{iπk^{n−2}}{(2π)^{n}} .

This can be proven using the representation of G_{0} as a Fourier transform [see Me2], or by applying Green’s theorem to the functions f(z) = G_{0}(x, z, k), g(z) = G_{0}(z, y, −k) in a sphere of radius r and letting r tend to in�nity.

The identity () for real k can be written as an operator equation. Let π_{k} denote the operator that takes a function on ℝ^{n} to its Fourier transform restricted to a sphere of radius k. Then () can be written

R_{0}(k) = R_{0}(−k) + c(k)π_{k}^{∗}π_{k}.

This has to be interpreted as an equation involving operators between Besov spaces, since none of the operators are bounded on L^{2}(ℝ^{n}) for real k. However, if we multiply on the left by V ^{\frac{1}{2}} and on the right by |V |^{\frac{1}{2}} then we do obtain an equation involving operators on L^{2}(ℝ^{n}), namely

R_{0V} (k) = R_{0V} (−k) + c(k)F_{V}^{T} (k)F_{|V |}(k)  ()

where the operator F_{|V |}(k) : L^{2}(ℝ^{n}) → L^{2}(S^{n−1}) is the formal product π_{k}|V |^{\frac{1}{2}} is given by (F_{|V |}(k)ψ)(ω) = ∫_{ℝ^{n}} e^{ik⟨x,ω⟩}|V |^{\frac{1}{2}} (x)ψ(x)d^{n}x,

and the operator F_{V}^{T} (k) : L^{2}(S^{n−1}) → L^{2}(ℝ^{n}) is the formal product V ^{\frac{1}{2}} π_{k}^{∗} given by

(F_{V}^{T} (k)ϕ)(x) = V ^{\frac{1}{2}} (x) ∫_{S^{n−1}} e^{ik⟨x,ω⟩}ϕ(ω)dω

Since all the operators in the equation () have analytic continuations to complex k, the equation remains valid for all k ∈ ℂ.

We now return to Q(k). Using () this operator can be rewritten

Q(k) = (1 + R_{0V} (−k))^{−1}(1 + R_{0V} (k))
= (1 + R_{0V} (−k))^{−1}(1 + R_{0V} (−k) + c(k)F_{V}^{T} (k)F_{|V |}(k))
= 1 + c(k)(1 + R_{0V} (−k))^{−1}F_{V}^{T} (k)F_{|V |}(k))
= 1 + c(k)(1 − R_{V} (−k))F_{V}^{T} (k)F_{|V |}(k)

It follows from the estimates on singular values below that the second term on the left side is trace class, so we may take the determinant. Using the identity det(1 + AB) = det(1 + BA) gives

det Q(k) = det(1 + c(k)F_{|V |}(k)(1 − R_{V} (−k))F_{V}^{T} (k)).

The operators on the left are now operators on L^{2}(S^{n−1}). Now

c(k)F_{|V |}(k)(1 − R_{V} (−k))F_{V}^{T} (k) = c(k)π_{k}(V − V R(−k)V)π_{k}^{∗} = T (−k),  ()

is exactly the expression from stationary scattering theory for the T matrix. Thus

det Q(k) = det(1 + T (−k)) = det S(−k),

as claimed.

Upper bounds

We now state our main theorem.

Theorem Suppose that V is a super-exponentially decaying potential with

V̂(z) ≤ Ce^{Φ(|z|)}

Then

n(r) ≤ CΦ^{n}(cr) + O(Φ^{n−1}(cr))

for some constants c and C.

The proof will be broken up into a series of lemmas. We begin with a simple estimate.

Lemma Let ϕ(k) = det(1 + T (k)) where T (k) is a trace-class operator valued analytic function in the closed upper half plane, where 1 + T (s) is unitary for s ∈ ℝ, and where T (0) = 0. Let n(t) denote the number of zeros of ϕ(k) in a half disk in the upper half plane of radius t. De�ne

N(r) = ∫_{0}^{r} \frac{n(t)}{t}dt.

Then

N(r) ≤ \frac{1}{2π} ∫_{0}^{r} t^{−1} ∫_{−t}^{t} ∥T′(s)∥_{1}dsdt + \frac{1}{2π} ∫_{0}^{π} ln |ϕ(re^{iθ})|dθ

Here ∥ · ∥_{1} denotes the trace norm.

Proof: Integrating along a contour enclosing the half disk, we have

n(t) = \frac{1}{2πi} ∮ \frac{ϕ′(k)}{ϕ(k)}dk

= \frac{1}{2π} Im ∫_{−t}^{t} \frac{ϕ′(s)}{ϕ(s)}ds + \frac{1}{2π} ∫_{0}^{π} t\frac{d}{dt} ln |ϕ(te^{iθ})|dθ

≤ \frac{1}{2π} ∫_{−t}^{t} |ϕ′(s)|ds + \frac{1}{2π} ∫_{0}^{π} t\frac{d}{dt} ln |ϕ(te^{iθ})|dθ

We used the fact that |ϕ(s)| = 1 for real s. Dividing by t and integrating, we �nd

N(r) ≤ \frac{1}{2π} ∫_{0}^{r} t^{−1} ∫_{−t}^{t} |ϕ′(s)|dsdt + \frac{1}{2π} ∫_{0}^{π} ln |ϕ(re^{iθ})|dθ

We used ϕ(0) = 1 to evaluate the second term. Since ϕ′(s) = ϕ(s)tr((1 + T (s))^{−1}T′(s)), and |ϕ(s)| = ∥(1 + T (s))^{−1}∥ = 1 for real s, it follows that for real s

|ϕ′(s)| ≤ ∥((1 + T (s))^{−1}T′(s)∥_{1}
≤ ∥(1 + T (s))^{−1}∥∥T′(s)∥_{1}
= ∥T′(s)∥_{1}.

This completes the proof.

To apply this lemma, we must estimate the trace norm of T′ along the real axis, and the growth of ϕ = det(1 + T) in the upper half plane when T is given by (). Lemma Let V be a super-exponentially decaying potential and let T (k) be given by (). Then for s ∈ ℝ

∥T′(s)∥_{1} ≤ C|s|^{n−2}

Proof: The operator T (s) is a product c(s)F_{|V |}(s)(1 − R_{V} (−s))F_{V}^{T} (s). We estimate each term and its derivative. To begin, we have

\begin{aligned}|c(s)| & ≤ C|s|^{n−2} \\ |c′(s)| & ≤ C|s|^{n−3} \end{aligned}

It follows from the representation () that

\begin{aligned}∥1 − R_{V} (−s)∥ & ≤ C \\ ∥R′_{V} (−s)∥ & ≤ C \end{aligned}

Using the explicit integral kernels for F_{V}^{T} (s) and F_{|V |}(s) it easy to estimate the Hilbert-Schmidt norms

\begin{aligned}∥F_{V}^{T} (s)∥_{2}^{2} & = ∫_{S^{n−1}} ∫_{ℝ^{n}} |e^{is⟨ω,x⟩}V ^{\frac{1}{2}} (x)|^{2}dxdω ≤ C \\ ∥F_{V}^{T ′}(s)∥_{2}^{2} & = ∫_{S^{n−1}} ∫_{ℝ^{n}} |i⟨ω, x⟩e^{is⟨ω,x⟩}V ^{\frac{1}{2}} (x)|^{2}dxdω ≤ C. \end{aligned}

The same estimates hold for ∥F_{|V |}(s)∥_{2} and ∥F′_{|V |}(s)∥_{2}. The proof is completed by using the Leibnitz rule to write T′(s), and the estimate

∥AB∥_{1} ≤ ∥A∥2^{\frac{1}{2}} ∥B∥2^{\frac{1}{2}}

It remains to estimate the growth of ϕ(k) for complex k.

Lemma Suppose V is a super-exponentially decreasing potential, and let T (k) be given by (). Then for k in the upper half plane, T (k) is trace class. Let ϕ(k) = det(1 + T (k)). Suppose that the Fourier transform of V satis�es the growth estimate

V̂(z) ≤ Ce^{Φ(|z|)}

for some positive, increasing function Φ(x). Then

|ϕ(k)| ≤ exp(δ^{−(n−1)}Φ^{n}((2 + ϵ)|k|) + O(Φ^{n−1}((2 + ϵ)|k|)))

for any ϵ > 0, and some constant C.

Proof: We will use Weyl’s estimate

|ϕ(k)| = | det(1 + T (k))| ≤ ∏_{j} (1 + μ_{j}(T (k)))

and therefore must estimate the singular values of T (k). Using the estimate

μ_{j}(AB) ≤ ∥A∥μ_{j}(B)

and

μ_{j}(AB) = μ_{j}(BA)

(twice) we �nd

\begin{aligned}μ_{j}(T (k)) & ≤ ∥c(−k)(1 − R_{V} (k)∥μ_{j}(F_{V}^{T} (−k)F_{|V |}(−k)) \\ & ≤ C|k|^{n−2}μ_{j}(V_{k}). \end{aligned}  ()

Where the operator V_{k} = F_{|V |}(−k)F_{V}^{T} (−k) is an integral operator on L^{2}(S^{n−1}) with integral kernel V̂(−k(ω − ω′)). To estimate the singular values we will use the following bound without proof. (It follows from the analyticity of V̂.) Let L_{ω} denote the positive Laplacian on S^{n−1} in the variables ω. Then for any ϵ > 0 there is a constant C such that

|L_{ω}^{p/2}V̂(k(ω − ω′))| ≤ C^{p}p!e^{Φ((2+ϵ)|k|)}

Summing the Taylor expansion for the exponential, this gives, for δ < C^{−1},

|e^{δL \frac{1}{2} ; ω} V̂(k(ω − ω′))| ≤ (1 − δC)^{−1}e^{Φ((2+ϵ)|k|)}

Since the left side is the integral kernel for the operator e^{δL \frac{1}{2} ; ω} V_{k}, this implies that

∥e^{δL \frac{1}{2} ; ω} V_{k}∥ ≤ Ce^{Φ((2+ϵ)|k|)}

(The constants C may change from line to line.) Thus

μ_{j}(V_{k}) = μ_{j}(e^{−δL \frac{1}{2} ; ω} e^{δL \frac{1}{2} ; ω} V_{k})

≤ ∥e^{δL \frac{1}{2} ; ω} V_{k}∥μ_{j}(e^{−δL \frac{1}{2} ; ω})

≤ Ce^{Φ−δj^{1/(n−1)}}

Here Φ = Φ((2 + ϵ)|k|). Using (), we get the same bound for μ_{j}(T (k)), if we increase ϵ slightly. Thus T (k) is trace class, and

| det(1 + T (k))| ≤ ∏(1 + Ce^{Φ−δj^{1/(n−1)}}).

This product is easily estimated by breaking it into two pieces. For j ≤ (Φ/δ)^{n−1} we obtain

∏_{j≤(Φ/δ)^{n−1}}(1 + Ce^{Φ−δj^{1/(n−1)}}) ≤ ∏_{j≤(Φ/δ)^{n−1}}(C + 1)e^{Φ−δj^{1/(n−1)}}

= (C + 1)^{(Φ/δ)^{n−1}}e^{Φ^{n}/δ^{n−1}} exp( ∑_{j=1}^{(Φ/δ)^{n−1}} −δj^{1/(n−1)})

= e^{Φ^{n}/δ^{n−1}+O(Φ^{n−1})}

For j > (Φ/δ)^{n−1} we have

∏_{j>(Φ/δ)^{n−1}}(1 + Ce^{Φ−δj^{1/(n−1)}}) ≤ exp( ∑_{j>(Φ/δ)^{n−1}} Ce^{Φ−δj^{1/(n−1)}})

≤ exp(Ce^{Φ} ∑_{j>(Φ/δ)^{n−1}} e^{−δj^{1/(n−1)}})

The sum appearing in this formula can be estimated by an integral.

\begin{aligned}∑_{j>(Φ/δ)^{n−1}} e^{−δj^{1/(n−1)}} & ≤ e^{−δ(Φ/δ)} + ∫_{(Φ/δ)^{n−1}}^{∞} e^{−δx^{1/(n−1)}}dx \\ & ≤ e^{−Φ} + Ce^{−Φ}Φ^{n−2} \end{aligned}

Thus the product for large j satis�es the bound

∏_{j>(Φ/δ)^{n−1}}(1 + Ce^{Φ−δj^{1/(n−1)}}) ≤ e^{CΦ^{n−2}}.

Combining the estimates for small and large j completes the proof.

We can now prove the main theorem.

Proof of : Let T (k) be given by (). We must count the zeros of ϕ(k) = det(1 + T (k)) in a half disk of radius r in the upper half plane. The operator T (k) is analytic in the upper half plane, except for possibly �nitely many poles, which won’t affect the counting function. Since 1 + T (k) = S(k) is the scattering operator, we know that 1 + T (k) is unitary for k real, and that T (0) = 0. Therefore applies. Using to estimate ∥T′(k)∥_{1} and to estimate ln |ϕ(k)| yields

N(r)

\begin{matrix}≤ C ∫_{0}^{r} t^{−1} ∫_{−t}^{t} |s|^{n−2}dsdt + (2π)^{−1} ∫_{0}^{π} (δ^{−(n−1)}Φ^{n}((2 + ϵ)r) + O(Φ^{n−1}((2 + ϵ)r))) dθ \\ ≤ Cr^{n−1} + 2^{−1}δ^{−(n−1)}Φ^{n}((2 + ϵ)r) + O(Φ^{n−1}((2 + ϵ)r)) \end{matrix}

By looking at V̂ along the imaginary axis, we see that Φ must grow at least as fast as r. (For compactly supported V ’s we have Φ(r) = Cr.) Thus we can ignore the �rst term. This gives

N(r) ≤ 2^{−1}δ^{−(n−1)}Φ^{n}((2 + ϵ)r) + O(Φ^{n−1}((2 + ϵ)r))

To get an estimate on n(r) from this, note that for any s > 1, since n(r) is monotone,

n(r) = n(r)(ln s)^{−1} ∫_{r}^{sr} \frac{1}{t}dt

≤ (ln s)^{−1} ∫_{r}^{sr} \frac{n(t)}{t}dt

= (ln s)^{−1}(N(sr) − N(r)) ≤ (ln s)^{−1}N(sr).

This completes the proof.

References

[F] Richard Froese, Asymptotic distribution of resonances in one dimension, J. Diff. Eqns. 137 no. 2 (1997) 251�272.

[Me1] Richard Melrose, Polynomial bound on the number of scattering poles, J. Functional Analysis, 53, 1983, 287�303

[Me2] Richard Melrose, Geometric scattering theory, Cambridge University Press, 1995

[S] Barry Simon, Trace Ideals and their Applications, London Mathematical Society Lecture Note Series 35, Cambridge Univ. Press, UK, 1979

[SZ1] Ant� onio S� a Barreto and Maciej Zworski, Existence of resonances in three dimensions, Commun. Math. Phys. 173 (1995) no. 2 401�415

[SZ2] Ant� onio S� a Barreto and Maciej Zworski, Existence of resonances in potential scattering, Commun. Pure Appl. Math. 49 (1996) no. 12 1271�1280

[Z1] Maciej Zworski, Distribution of poles for scattering on the real line, J. of Funct. Anal., 73 (2) , (1987), 277�296

[Z2] Maciej Zworski, Sharp polynomial bounds on the number of scattering poles of radial potentials, J. of Funct. Anal., 82 (2) , (1989), 370�403 [Z3] Maciej Zworski, Sharp polynomial bounds on the number of scattering poles, Duke Math. Jour, 59 (2) , (1989), 311�323

[Z4] Maciej Zworski, Counting scattering poles, Spectral and scattering theory (M. Ikawa, ed.), Marcel Dekker.

[V] Georgi Vodev, Sharp polynomial bounds on the number of scattering poles of perturbations of the Laplacian, Comm. Math. Phys. 146, (1992), 205�216.