An optimal Lp-bound on the Krein spectral shift function

(Birmingham, November 10–12, 2000)

Barry Simon and D. H.

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

\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}

where F is any non-negative convex function on [0, ∞) with F (0) = 0 and μ_{j}(C) are the singular values of C.

The Krein spectral shift function

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 Theorem 1 (Combes,Hislop, and Nakamura) One has the L^{p}-bound

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

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

Proof: Write C := A − B = ∑_{j=1}^{∞} μ_{j}(C)⟨ϕ_{j}, .⟩ψ_{j} and B_{n} := B + ∑_{j=1}^{n} μ_{j}(C)⟨ϕ_{j}, .⟩ψ_{j}.

Then ζ_{B_{n+1} ,B_{n}} is the spectral shift function of a rank one pair. Hence

\begin{aligned}∫ |ζ_{B_{n+1} ,B_{n}}|^{p} & = ∫ |ζ_{B_{n+1} ,B_{n}}|^{p−1}|ζ_{B_{n+1} ,B_{n}}| \\ & ≤ ∫ |ζ_{B_{n+1} ,B_{n}}| ≤ μ_{n}. \end{aligned}

Use the triangle inequality

∥ζ_{A,B}∥_{p} = ∥∑ ζ_{B_{n+1} ,B_{n}}∥_{p} ≤ ∑ ∥ζ_{B_{n+1} ,B_{n}}∥_{p}

to sum this up.

A special spectral shift function

Let C be a positive trace class operator with eigenvalues μ_{j}. The spectral shift function for the pair C, 0 is simply given by

\begin{matrix}ξ_{C,0}(λ) = n if μ_{n+1} ≤ λ < μ_{n} \\ ξ_{C,0}(λ) = 0 if λ < 0 or λ ≥ μ_{1}. \end{matrix}

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} ).

• In addition, if F is monotone increasing, then

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

Main Result

The above example ζ_{C,0} is an extreme case:

Theorem 2 (Barry Simon, 100DM) 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),

\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}

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.

Corollary 3 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}.

Remark:

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

Proof: With μ(n) := μ_{n} − μ_{n+1} ≥ 0 rewrite

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

The right-hand side is the l^{p}-norm of the function n → n^{p} in the weighted l^{p}-space l^{p}(μ). Write n = 1+(n−1) and use Minkowski’s inequality 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 ; ∑ ; n=1 μ_{n}^{1/p} + ( ∞ ; ∑ ; n=N (n−N)^{p}μ(n))^{1/p}.

The Proof

Let m_{f}(t) := |{λ : |f(λ)| > t}|. We will write m_{A,B} for the distribution function of ξ_{A,B}. Lemma 4 (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.

Proof: Set (x − s)_{+} := sup {0, x − s}. Then

∫_{s}^{∞} m_{f}(t) dt = ∫ (|f(λ)| − s)_{+} dλ  (3)

for all s ≥ 0. Write |ξ_{A,B}| = |ξ_{A,B+C_{n}} + ξ_{B+C_{n},B} | ≤ |ξ_{A,B+C_{n}} | + n, with C_{n} := ∑_{j=1}^{n} μ_{j}(C)⟨ϕ_{j}, .⟩ψ_{j}. Thus

(|ξ_{A,B}(λ)| − n)_{+} ≤ |ξ_{A,B+C_{n}} (λ)|. Using (3), we get

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

Lemma 5 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: Let F′ be the left derivative of F, F′(0) := 0. Define ν_{F} by

ν_{F} ([a, b)) := F′(b) − F′(a).

Then F′(s) = ν_{F} ([0, s)). Calculate

∫_{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).

Lemma 5 gives

\begin{aligned}∫ F (|f(λ)|) dλ & = ∫_{0}^{∞} ∫ (|f(λ)| − u)_{+} dλν_{F} (du) \\ & = ∫_{0}^{∞} \underbrace{∫_{u}^{∞} m_{f}(u) du}_{=:Q_{f}(u)} ν_{F} (du) \end{aligned}

Hence we have

Lemma 6 Let F be any non-negative, convex function F on [0, ∞) which vanishes at zero. Given two functions f and g, 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.

Lemma 7 Suppose that g takes only values in ℕ_{0}. Then the inequality Q_{f}(n) ≤ Q_{g}(n) for n ∈ ℕ_{0} implies

Q_{f}(t) ≤ Q_{g}(t) for all t ≥ 0.

Proof: Q_{f} and Q_{g} are convex AND Q_{g} is linear on [n, n+1]. The claim follows from convexity.

Proof of the Theorem: Given A and B, let D = |A − B| and C be any non-negative trace class operator with

∞ ; ∑ ; j=n μ_{j}(D) ≤ ∞ ; ∑ ; j=n μ_{j}(C) for all n ∈ ℕ.

The Basic Lemma shows

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

(4)

Lemma 7 then implies that (4) extends from ℕ_{0} to all positive real n. Once one has that, Lemma 6 proves

∫ F (|ξ_{A,B}(λ)|) dλ ≤ ∫ F (|ξ_{C,0}(λ)|) dλ.

Fubini-Tonelli implies summation by parts^{∗}

∞ ; ∑ ; j=1 F (j)(μ_{j} − μ_{j+1} )

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

= _{1≤n≤j}^{∑} ( \underbrace{F (n) − F (n−1)}_{≥0} )( \underbrace{μ_{j} − μ_{j+1}}_{≥0} )

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

= ∞ ; ∑ ; n=1 (F (n) − F (n−1))μ_{n},

since ∑_{j=n}^{∞} (μ_{j} − μ_{j+1} ) telescopes to μ_{n}.

∗Or Riemann integral = Lebesgue integral!

Remark:

If

μ_{n} = \frac{1}{n^{p} ln(n + 1)^{α}}

then

∞ ; ∑ ; n=1 μ_{n}^{1/p} < ∞

if and only if α > p. Whereas

∞ ; ∑ ; n=1 (n^{p} − (n−1)^{p})μ_{n} < ∞

if and only if α > 1.