AN OPTIMAL L^{p}-BOUND ON THE KREIN SPECTRAL SHIFT FUNCTION

DIRK HUNDERTMARK^{1} AND BARRY SIMON^{1,2}

Dedicated to the memory of Tom Wolff, analyst extraordinaire and friend

Abstract. Let ξ_{A,B} be the Krein spectral shift function for a pair of operators A, B, with C = A − B trace class. We establish the bound

∫ F (|ξ_{A,B}(λ)|) dλ ≤ ∫ F (|ξ_{|C|,0}(λ)|) dλ = ∑_{j=1}^{∞} [F (j) − F (j − 1)]μ_{j}(C),

where F is any non-negative convex function on [0, ∞) with F (0) = 0 and μ_{j}(C) are the singular values of C. Specializing to F (t) = t^{p}, p ≥ 1 this improves a recent bound of Combes, Hislop, and Nakamura.

1. Introduction

Let A, B be bounded self-adjoint operators such that their difference A − B is trace class. The Krein spectral shift function ξ_{A,B} for the pair A, B is determined by

tr(f(A) − f(B)) = ∫ f′(λ)ξ_{A,B}(λ) dλ

for all functions f ∈ C_{0}^{∞}(ℝ) and ξ(λ) = 0 if |λ| is large enough. The two bounds

∫ |ξ_{A,B}(λ)| dλ ≤ tr(|A − B|)  (1)

and

|ξ_{A,B}(λ)| ≤ n  if A − B is rank n  (2)

are well known; see, for example, [5] or [6]. The Krein spectral shift function can also be defined for unbounded self-adjoint operators A, B and enjoys the same properties as long as their difference is trace class. The results of this paper extend to general unbounded operators A and B (as long as their difference is trace class) but for simplicity, we will suppose that A and B are bounded. For applications

Date: September 12, 2000. 1991 Mathematics Subject Classification. 47A30, 81Q10. Key words and phrases. Krein spectral shift function, L^{p}-bounds. 1 Department of Mathematics 253–37, California Institute of Technology, Pasadena, CA 91125,

U.S.A.; E-mail: dirkh@caltech.edu, bsimon@caltech.edu.

2 Supported in part by NSF grant DMS-9707661. \begin{matrix}c \\ ◯ \end{matrix}2000 by the authors. Reproduction of this article, in its entirety, by any means is permitted

for non-commercial purposes.

To appear in J. d’Analyse Math.

of the spectral shift function in scattering theory, see, for example, the survey article [1].

The spectral shift function also found applications in the theory of random Schrödinger operators. Kostrykin and Schrader [3, 4] constructed a spectral shift density for random Schrödinger operators with Anderson-type potentials. More recently, Combes, Hislop, and Nakamura [2] realized that L^{p}-bounds on the Krein spectral shift function can serve as a basic tool for a proof of Hölder continuity of the integrated density of states for a large class of continuous random Schrödinger operators. In terms of the singular values of the difference C = A − B, their bound reads

∥ξ_{A,B}∥_{p} := ( ∫ |ξ_{A,B}(λ)|^{p} dλ)^{1/p} ≤ ∑_{j=1}^{∞} μ_{j}(C)^{1/p}  (3)

for 1 ≤ p < ∞. Note that (3) includes the endpoint cases (1) and (2) for p = 1 and and in the limit p → ∞, respectively.

To see what type of bound is the correct one for an L^{p}-bound on the Krein spectral shift function, we consider a special case: Let C be a positive trace class operator with eigenvalues μ_{j}. Calculating

tr[f(C) − f(0)] = ∑_{j=1}^{∞} ∫_{0}^{μ_{j}} f′(λ) dλ,

we see that the spectral shift function for the pair C, 0 is simply given by

ξ_{C,0}(λ) = n if μ_{n+1} ≤ λ < μ_{n}  and  ξ_{C,0}(λ) = 0 if λ < 0 or λ ≥ μ_{1}. (4)

For cases like this where A and B are finite rank, it is known that the spectral shift function is just the difference of the dimensions of the spectral subspaces — which leads immediately to another way of seeing why (4) is true. In particular, ξ_{C,0} enjoys the following important properties:

• ξ_{C,0} takes only values in ℕ_{0} (or ℤ if C is not non-negative).

• For any non-negative function F on [0, ∞) with F (0) = 0, we have

∫ F (|ξ_{C,0}(λ)|) dλ = ∑_{j=1}^{∞} F (j)(μ_{j} − μ_{j+1}).  (5)

• In addition, if F is monotone increasing, then

∫ F (|ξ_{C,0}(λ)|) dλ = ∑_{j=1}^{∞} [F (j) − F (j−1)]μ_{j}.  (6)

The first two claims follow immediately from (4). Formally, the last claim follows from the second by summation by parts. However, since

∑_{j=1}^{N} F (j)(μ_{j} − μ_{j+1}) = ∑_{j=1}^{N} (F (j) − F (j−1))μ_{j} − F (N)μ_{N+1},

this usually poses some growth restriction on F (j) to do the limit N → ∞. We prefer not to do this but use positivity instead. Notice

∑_{j=1}^{∞} F (j)(μ_{j} − μ_{j+1}) = ∑_{j=1}^{∞} ∑_{n=1}^{j} (F (n) − F (n−1))(μ_{j} − μ_{j+1})

= ∑_{1≤n≤j} (F (n) − F (n−1))(μ_{j} − μ_{j+1}).

By the assumptions on F and μ_{j}, all terms in this double sum are non-negative. Hence we can use the Fubini-Tonelli theorem to freely interchange the summation and conclude

∑_{j=1}^{∞} F (j)(μ_{j} − μ_{j+1}) = ∑_{n=1}^{∞} (F (n) − F (n−1)) ∑_{j=n}^{∞} (μ_{j} − μ_{j+1})

= ∑_{n+1}^{∞} (F (n) − F (n−1))μ_{n},

where we used the fact that the last sum telescopes and μ_{n} → 0 as n → ∞. In particular, the right-hand side of (5) and (6) is finite if and only if the other is and then they are equal. Below we will use this type of argument to freely do summation by parts without a priori bounds on the boundary terms. Alternatively, one could consider the case of finite rank operators C = A − B first and then use some approximation arguments.

Our main result is that the above example (6) is an extreme case:

Theorem 1. Let F be a non-negative convex function on [0, ∞) vanishing at zero. Given a non-negative compact operator C with singular values μ_{j}(C), we have

\begin{aligned}∫ F (|ξ_{A,B}(λ)|) dλ & ≤ ∫ F (|ξ_{C,0}(λ)|) dλ \\ & = ∑_{j=1}^{∞} [F (j) − F (j−1)]μ_{j}(C) \end{aligned}  (7)

for all pairs of bounded operators A, B with ∑_{j=n}^{∞} μ_{j}(|A − B|) ≤ ∑_{j=n}^{∞} μ_{j}(C) for all n ∈ ℕ. In particular, this is the case if |A − B| ≤ C.

Remark. Moreover, if F is strictly convex, the above inequality is strict if either the modulus of ξ_{A,B} takes non-integer values on a set of positive Lebesgue measure or one does not have equality in Lemma 3 below. However, it seems to be difficult to find necessary and sufficient conditions on A and B alone for the case of equality in (7).

Specializing to F (t) = t^{p} for some p ≥ 1 we get as a corollary,

Corollary 2. Let ξ_{A,B} be the Krein spectral shift function for the pair A, B. In terms of the singular values μ_{j} of the difference A − B, we have the L^{p}-bound

∥ξ_{A,B}∥_{p} ≤ ∥ξ_{|A−B|,0}∥_{p} = ( ∑_{n=1}^{∞} [n^{p} − (n−1)^{p}]μ_{n})^{1/p}.  (8)

Remarks. i) There are two different ways to see that (8) is, indeed, stronger than the bound (3) by Combes et al. First, the direct argument: Rewrite

( ∑_{n=1}^{∞} [n^{p} − (n−1)^{p}]μ_{n})^{1/p} = ( ∑_{n=1}^{∞} n^{p}(μ_{n} − μ_{n+1}))^{1/p}

and consider the right-hand side as the l^{p}-norm of the function n → n^{p} in the weighted l^{p}-space l^{p}(μ) with measure μ(n) := μ_{n} − μ_{n+1} for n ∈ ℕ. Write n = 1+(n−1) and use Minkowski’s inequality for the l^{p}(μ)-norm to get

( ∑_{n=1}^{∞} n^{p}μ(n))^{1/p} ≤ ( ∑_{n=1}^{∞} μ(n))^{1/p} + ( ∑_{n=2}^{∞}(n−1)^{p}μ(j))^{1/p}

= μ_{1}^{1/p} + ( ∑_{n=2}^{∞}(n−1)^{p}μ(n))^{1/p}

≤ ∑_{n=1}^{N} μ_{n}^{1/p} + ( ∑_{n=N}^{∞}(n−N)^{p}μ(n))^{1/p},

where, for the last inequality, we repeated the first step N times. Using monotone convergence for the limit N → ∞, we conclude

( ∑_{n=1}^{∞} [n^{p} − (n−1)^{p}]μ_{n})^{1/p} ≤ ∑_{n=1}^{∞} μ_{n}^{1/p}.  (9)

For the soft argument, note that, according to (6),

( ∑_{n=1}^{∞} [n^{p} − (n−1)^{p}]μ_{n})^{1/p} = ∥ξ_{C,0}∥_{p},

where C is any non-negative compact operator with singular values μ_{n}. The bound (3) for ξ_{C,0} immediately implies the inequality (9). Of course, the direct argument is, in some sense, a reformulation of the inductive proof of Combes et al.

ii) Our result shows that if ∑_{n=1}^{∞} n^{p−1}μ_{n} < ∞, then ξ_{A,B} ∈ L^{p}(ℝ). Note that this cannot be improved as far as only conditions on μ_{n} are used. It is also strictly better than the result by Combes et al. For example, if μ_{n} = n^{−p} log(n + 2)^{−α}, then Combes et al. require α > p to conclude ξ_{A,B} ∈ L^{p}(ℝ), while our result only needs α>1.

2. Two Proofs

We want to give two proofs of our main result, Theorem 1, both depending on different aspects of the problem. First, some notation. For a complex-valued function f, let m_{f} be its distribution function, that is, m_{f}(t) := |{λ : |f(λ)| > t}|, with |A| the Lebesgue measure of a Borel set A ⊂ ℝ. We will write m_{A,B} for the distribution function of ξ_{A,B}. The following lemma is the core of both proofs:

Lemma 3 (Basic Lemma). With C = A − B, we have for all n ∈ ℕ_{0}

∫_{n}^{∞} m_{A,B}(t) dt ≤ ∑_{j=n+1}^{∞} μ_{j}(C) = ∫_{n}^{∞} m_{|C|,0}(t) dt.

Remark. Setting (x − s)_{+} := sup{0, x − s}, we have

∫_{s}^{∞} m_{f}(t) dt = ∫ (|f(λ)| − s)_{+} dλ  (10)

for all s ≥ 0. Hence Lemma 3 is equivalent to

∫ (|ξ_{A,B}(λ)| − n)_{+} dλ ≤ ∑_{j=n+1}^{∞} μ_{j}(C) = ∫ (|ξ_{|C|,0}(λ)| − n)_{+} dλ.

Proof of Lemma 3: By the formula (4) for ξ_{|C|,0},

∫ (|ξ_{|C|,0}|(λ) − n)_{+} dλ = ∑_{j=n+1}^{∞}(j − n)(μ_{j} − μ_{j+1})

= ∑_{l=n+1}^{∞} ∑_{j=l}^{∞}(μ_{j} − μ_{j+1}) = ∑_{l=n+1}^{∞} μ_{l},

proving the second part of the assertion in the lemma.

For the inequality, let ψ_{j}, ϕ_{j} be two sets of orthonormal vectors such that A−B = ∑_{j=1}^{∞} μ_{j}⟨ϕ_{j}, ·⟩ψ_{j}. Note that ϕ_{j} = ±ψ_{j} since A and B are self-adjoint. Set C_{0} := 0 and C_{n} := ∑_{j=1}^{n} μ_{j}⟨ϕ_{j}, ·⟩ψ_{j} for n ∈ ℕ. The spectral shift function is transitive. In particular,

ξ_{A,B} = ξ_{A,A+C_{n}} + ξ_{A+C_{n},B}.

By (2) we know |ξ_{A,A+C_{n}}| ≤ n. Thus

(|ξ_{A,B}(λ)| − n)_{+} ≤ |ξ_{A+C_{n},B}(λ)|

and hence, by (10) and (1):

∫_{n}^{∞} m_{A,B}(t) dt = ∫ (|ξ_{A,B}(λ)| − n)_{+} dλ ≤ tr(C − C_{n}) = ∑_{j=n+1}^{∞} μ_{j}.

In the following we will write m and ξ for m_{A,B}, ξ_{A,B}, respectively.

First proof of Theorem 1: For n ∈ ℕ_{0}, put a_{n} := F (n + 1) − F (n), so we have F (n) = ∑_{0≤l<n} a_{l}. The a_{n} are monotone increasing due to the convexity of F, that is, a_{n} − a_{n−1} ≥ 0 for all n ∈ ℕ. Furthermore, set

x_{n} := ∫_{n}^{∞} m(t) dt

and observe that, due to (10),

∫_{n<|ξ|≤n+1} (|ξ_{A,B}(λ)| − n) dλ = x_{n} − x_{n+1} − m(n+1).  (11)

Look at

∫ F (|ξ(λ)|) dλ = ∑_{n=0}^{∞} ∫_{n<|ξ|≤n+1} F (|ξ(λ)|) dλ.  (12)

By convexity of F, we have F (s) ≤ F (n) + (s−n)(F (n+1) − F (n)) for s ∈ [n, n+1]. Plugging this into (12), we infer

∫ F (|ξ(λ)|) dλ

≤ ∑_{n=0}^{∞} [F (n) ∫_{n<|ξ|≤n+1} dλ + a_{n} ∫_{n<|ξ|≤n+1}(|ξ(λ)| − n) dλ]

= ∑_{n=0}^{∞} [( ∑_{l=0}^{n−1} a_{l})(m(n) − m(n+1)) + a_{n}(x_{n} − x_{n+1} − m(n + 1))] (by (11))

= ∑_{l=0}^{∞} a_{l} ∑_{n=l+1}^{∞} (m(n) − m(n+1)) + ∑_{n=0}^{∞} a_{n}(x_{n} − x_{n+1} − m(n+1))

= ∑_{l=0}^{∞} a_{l}m(l+1) + ∑_{n=0}^{∞} a_{n}(x_{n} − x_{n+1} − m(n+1))

= ∑_{n=0}^{∞} a_{n}(x_{n} − x_{n+1}).  (13)

Note that if F is strictly convex, we have equality in this inequality if and only if ξ takes only integer values! Using that the terms in this sum are non-negative, we can again reorder the summations freely to conclude, setting a_{−1} := 0,

\begin{aligned}∫ F (|ξ(λ)|) dλ & ≤ ∑_{n=0}^{∞} a_{n}(x_{n} − x_{n+1}) = ∑_{0≤l≤n}(a_{l} − a_{l−1})(x_{n} − x_{n+1}) \\ & = ∑_{l=0}^{∞} (a_{l} − a_{l−1})x_{l} ≤ ∑_{l=0}^{∞} (a_{l} − a_{l−1}) ∑_{j=l+1} μ_{j} \\ & = ∑_{j=1}^{∞} (F (j) − F (j−1))μ_{j}, \end{aligned}  (14)

where in the last inequality we used positivity of the increments a_{l} − a_{l−1} (aka convexity of F) and Lemma 3. This proves Theorem 1, including the case of equality since, if F is strictly convex and ξ takes non-integer values on a set of positive measure, the inequality in (13) is strict. We will come back to this question in the second proof.

For the second proof we need some preparatory lemmas:

Lemma 4. For any non-negative, convex function F on [0, ∞) which vanishes at zero, there exists a non-negative, locally finite measure ν_{F} on [0, ∞) such that

F (t) = ∫_{0}^{∞}(t − u)_{+}ν_{F} (du) for all t ≥ 0.

F is strictly convex if and only if ν_{F} is strictly positive, that is, ν_{F} ([a, b]) > 0 for all 0 ≤ a < b.

Proof. Of course, this proof is well known. We want to give a short proof for completeness. Define F (t) to be zero for negative t — it then becomes convex on all of ℝ. The assumptions on F show that the left derivative F′ exists everywhere, is non-negative, monotone increasing, and continuous from the left. Hence it defines a measure ν_{F} by setting ν_{F} ([a, b)) := F′(b) − F′(a). That this formula involves intervals that include a is a consequence of the left continuity of F′ as we have defined it. Note that in this formula F′(0) = 0 since it is the left derivative. The measure ν_{F} has a point mass at 0 which is precisely the right derivative of F at zero. This measure is strictly positive on [0, ∞) if and only if F′ is strictly increasing on [0, ∞), that is, if and only if F is strictly convex. Moreover, by construction, F′(s) = ν_{F} ([0, s)) for all s ≥ 0. Calculating

∫_{0}^{∞}(t − u)_{+}ν_{F} (du) = ∫_{[0,t)} ∫_{u}^{t} ds ν_{F} (du) = ∫_{0}^{t} ν_{F} ([0, s)) ds

= ∫_{0}^{t} F′(s) ds = F (t)

shows the assertion.

Some more notation: Recall that for a function f, m_{f}(t) := |{λ | |f(λ)| > t}| and define

Q_{f}(s) := ∫_{s}^{∞} m_{f}(t) dt = ∫ (|f(λ)| − s)_{+} dλ.

Lemma 5. Let F be any non-negative, convex function F on [0, ∞) which vanishes at zero. Given two functions f and g, we have that Q_{f} ≤ Q_{g} implies

∫ F (|f(λ)|) dλ ≤ ∫ F (|g(λ)|) dλ.

Moreover, if F is strictly convex and Q_{f} < Q_{g} on a set of positive Lebesgue measure, then the inequality above is strict.

Proof. From Lemma 4 we infer

∫ F (|f(λ)|) dλ = ∫_{0}^{∞} Q_{f}(u) ν_{F} (du),

which concludes the proof.

Lemma 6. Suppose that g takes only values in an unbounded, discrete set S ⊂ [0, ∞) with 0 ∈ S. Then the inequality Q_{f}(s) ≤ Q_{g}(s) for s ∈ S implies Q_{f}(t) ≤ Q_{g}(t) for all t ∈ [0, ∞).

Proof. Let S = {0 = s_{0} < s_{1} < s_{2} < ⋯ }. Notice that Q_{f} is always convex and, furthermore, Q_{g} is linear on [s_{j}, s_{j+1}]. The claim follows from convexity since, by assumption, Q_{f}(s) ≤ Q_{g}(s) for s ∈ S.

Now we come to the

Second proof of Theorem 1: Given bounded operators A and B, let D = |A−B| and C any non-negative, compact, trace class operator with ∑_{j=n}^{∞} μ_{j}(D) ≤ ∑_{j=n}^{∞} μ_{j}(C) for all n ∈ ℕ. The basic Lemma 3 shows

Q_{ξ_{A,B}}(n) ≤ Q_{ξ_{|D|,0}}(n) ≤ Q_{ξ_{C,0}}(n) for all n ∈ ℕ_{0}.  (15)

Lemma 6 then implies that (15) extends from ℕ_{0} to all positive real n. Once one has that, Lemma 5 proves (7).

If |ξ_{A,B}| takes non-integer values on a set of positive Lebesgue measure, then there exists n ∈ ℕ_{0} such that Q_{ξ_{A,B}} is strictly smaller than Q_{ξ_{|D|,0}} on the interval (n, n+1) . This shows that if F is strictly convex, equality can only hold as long as |ξ_{A,B}| is integer-valued.

Acknowledgment: We thank Rowan Killip for a refreshing discussion.

References

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