AN OPTIMAL Lp-BOUND ON THE KREIN SPECTRAL SHIFT FUNCTION

AN OPTIMAL Lp-BOUND ON THE KREIN SPECTRAL SHIFT FUNCTION DIRK HUNDERTMARK1 AND BARRY SIMON1;2 Dedicated to the memory of Tom Wol, analyst extraordinaire and friend

Abstract. [Note: This paper supplants B. Simon's preprint, \Lp bounds on the Krein spectral shift," which has been withdrawn.] Let 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 Z

Z

F (jA;B ()j) d  F (jjCj;0 ()j) d =

1

X

j =1

F (j ) ? F (j ? 1)]j (C );

where F is any non-negative convex function on [0; 1) with F (0) = 0 and j (C ) are the singular values of C . Specializing to F (t) = tp , 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 dierence A ? B is trace class. The Krein spectral shift function A;B for the pair A, B is determined by Z

f 0 ()A;B () d for all functions f 2 C01 (R) and  () = 0 if jj is large enough. The two bounds Z jA;B ()j d  tr(jA ? B j) (1) tr(f (A) ? f (B )) =

and

jA;B ()j  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 de ned for unbounded self-adjoint operators A; B and enjoys the same properties as long as their dierence is trace class. The results of this paper extend to general unbounded operators A and B (as long as their dierence is trace class)

Date : September 12, 2000. 1991 Mathematics Subject Classi cation. 47A30, 81Q10. Key words and phrases. Krein spectral shift function, Lp -bounds. 1 Department of Mathematics 253{37, California Institute of Technology, Pasadena, CA 91125, U.S.A.; E-mail: [email protected], [email protected]. 2 Supported in part by NSF grant DMS-9707661.

c 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. 1

2

D. HUNDERTMARK AND B. SIMON

but for simplicity, we will suppose that A and B are bounded. For applications 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 Schrodinger operators. Kostrykin and Schrader [3, 4] constructed a spectral shift density for random Schrodinger operators with Anderson-type potentials. More recently, Combes, Hislop, and Nakamura [2] realized that Lp -bounds on the Krein spectral shift function can serve as a basic tool for a proof of Holder continuity of the integrated density of states for a large class of continuous random Schrodinger operators. In terms of the singular values of the dierence C = A ? B , their bound reads

kA;B kp :=

Z

1=p

jA;B ()jp d



1 X j=1

j (C )1=p

(3)

for 1  p < 1. Note that (3) includes the endpoint cases (1) and (2) for p = 1 and and in the limit p ! 1, respectively. To see what type of bound is the correct one for an Lp -bound on the Krein spectral shift function, we consider a special case: Let C be a positive trace class operator with eigenvalues j . Calculating tr[f (C ) ? f (0)] =

1Z X j=1 0

j

f 0 () d;

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

C;0 () = n if n+1   < n

C;0 () = 0 if  < 0 or   1 :

and

(4)

For cases like this where A and B are nite rank, it is known that the spectral shift function is just the dierence of the dimensions of the spectral subspaces | which leads immediately to another way of seeing why (4) is true. In particular, C;0 enjoys the following important properties:  C;0 takes only values in N 0 (or Z if C is not non-negative).  For any non-negative function F on [0; 1) with F (0) = 0, we have Z

F (jC;0 ()j) d =

1 X j=1




?

F (j ) j ? j+1 :

(5)

 In addition, if F is monotone increasing, then Z

F (jC;0 ()j) d =

1 X j=1



F (j ) ? F (j ? 1) j :

(6)

The rst two claims follow immediately from (4). Formally, the last claim follows from the second by summation by parts. However, since N X j=1

?




F (j ) j ? j+1 =

N ? X j=1




F (j ) ? F (j ? 1) j ? F (N )N +1 ;

AN OPTIMAL Lp -BOUND ON THE KREIN SPECTRAL SHIFT FUNCTION

3

this usually poses some growth restriction on F (j ) to do the limit N ! 1. We prefer not to do this but use positivity instead. Notice 1 X j=1

?




F (j ) j ? j+1 = =

j ? 1 X X j=1 n=1


?

F (n) ? F (n ? 1) j ? j+1

X ?

1nj








?

F (n) ? F (n ? 1) j ? j+1 :

By the assumptions on F and 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 1 X j=1

?




F (j ) j ? j+1 = =

1 ? X

F (n) ? F (n ? 1)

n=1 1 ? X n+1

1 ?
X j=n

j ? j+1







F (n) ? F (n ? 1) n ;

where we used the fact that the last sum telescopes and n ! 0 as n ! 1. In particular, the right-hand side of (5) and (6) is nite 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 nite rank operators C = A ? B rst 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; 1) vanishing at zero. Given a non-negative compact operator C with singular values j (C ), we have Z

Z

F (jA;B ()j) d  F (jC;0 ()j) d =

(7)

1 X j=1

F (j ) ? F (j ? 1)]j (C )

1 for all pairs of bounded operators A; B with 1 j=n j (jA ? B j)  j=n j (C ) for all n 2 N . In particular, this is the case if jA ? B j  C . Remark. Moreover, if F is strictly convex, the above inequality is strict if either the modulus of 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 dicult to nd necessary and sucient conditions on A and B alone for the case of equality in (7). Specializing to F (t) = tp for some p  1 we get as a corollary, Corollary 2. Let A;B be the Krein spectral shift function for the pair A; B. In terms of the singular values j of the dierence A ? B , we have the Lp -bound

kA;B kp  kjA?Bj;0 kp =

P

P

1  X n=1



np ? (n ? 1)p n

1=p

:

(8)

4

D. HUNDERTMARK AND B. SIMON

Remarks. i) There are two dierent ways to see that (8) is, indeed, stronger than the bound (3) by Combes et al. First, the direct argument: Rewrite 1  X



np ? (n ? 1)p n

1=p

=

1 X

1=p

np (n ? n+1 )

n=1 p and consider the right-hand side as the l -norm of the function n ! np in the weighted lp -space lp () with measure (n) := n ? n+1 for n 2 N . Write n = 1+(n ? 1) and use Minkowski's inequality for the lp ()-norm to get n=1

1 X

np (n)

n=1

1=p



1 X

(n)

1=p

+

1 X

(n ? 1)p(j )

1=p

n=1 n=2 1 X 1=p = 11=p + (n ? 1)p(n) n=2 X 1=p N 1 X 1 =p p  n + (n ? N ) (n) ; n=1 n=N

where, for the last inequality, we repeated the rst step N times. Using monotone convergence for the limit N ! 1, we conclude 1  X n=1



np ? (n ? 1)p n

1=p



1 X

n=1

1n=p :

(9)

For the soft argument, note that, according to (6), 1  X n=1



np ? (n ? 1)p n

1=p

= kC;0 kp ;

where C is any non-negative compact operator with singular values n . The bound (3) for C;0 immediately implies the inequality (9). Of course, the direct argument is, in some sense, a reformulation inductive proof of Combes et al. P of the p?1 n < 1, then A;B 2 Lp (R). Note that n ii) Our result shows that if 1 n=1 this cannot be improved as far as only conditions on n are used. It is also strictly better than the result by Combes et al. For example, if n = n?p log(n + 2)? , then Combes et al. require  > p to conclude A;B 2 Lp (R), while our result only needs > 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 mf be its distribution function, that is, mf (t) := jf : jf ()j > tgj, with jAj the Lebesgue measure of a Borel set A  R. We will write mA;B for the distribution function of A;B . The following lemma is the core of both proofs: Lemma 3 (Basic Lemma). With C = A ? B, we have for all n 2 N0 Z

1

n

mA;B (t) dt 

1 X

j=n+1

j (C ) =

Z

1

n

mjC j;0(t) dt:

AN OPTIMAL Lp -BOUND ON THE KREIN SPECTRAL SHIFT FUNCTION

Remark. Setting (x ? s)+ := supf0; x ? sg, we have Z 1 Z mf (t) dt = (jf ()j ? s)+ d s for all s  0. Hence Lemma 3 is equivalent to Z

(jA;B ()j ? n)+ d 

1 X

5

(10)

Z

j=n+1

j (C ) = (jjC j;0 ()j ? n)+ d:

Proof of Lemma 3: By the formula (4) for jC j;0 , Z

(jjC j;0 j() ? n)+ d = =

1 X

(j ? n)(j ? j+1 )

j=n+1 1 X 1 X l=n+1 j=l

(j ? j+1 ) =

1 X l=n+1

l ;

proving the second part of the assertion in the lemma. For the inequality, let j , j be two sets of orthonormal vectors such that A?B = P1 ;
i j . Note that j =  j since A and B are self-adjoint. Set C0 := 0 j=1 j hjP and Cn := nj=1 j hj ;
i j for n 2 N . The spectral shift function is transitive. In particular, A;B = A;A+Cn + A+Cn;B : By (2) we know jA;A+Cn j  n. Thus (jA;B ()j ? n)+  jA+Cn ;B ()j and hence, by (10) and (1): Z

1

n

Z

?




mA;B (t) dt = (jA;B ()j ? n)+ d  tr C ? Cn =

1 X

j=n+1

j :

In the following we will write m and  for mA;B , A;B , respectively. First proof P of Theorem 1: For n 2 N 0 , put an := F (n + 1) ? F (n), so we have F (n) = 0l