Third Order Trace Formula

Third Order Trace Formula Arup Chattopadhyay 1 and

Kalyan B. Sinha

2

arXiv:1207.3500v1 [math.FA] 15 Jul 2012

Abstract In [5], Dykema and Skripka showed the existence of higher order spectral shift functions when the unperturbed self-adjoint operator is bounded and the perturbations is Hilbert-Schmidt. In this article, we give a different proof for the existence of spectral shift function for the third order when the unperturbed operator is self-adjoint (bounded or unbounded, but bounded below).

Keywords. Trace formula, spectral shift function , perturbations of self-adjoint operators.

1

Introduction.

Notations: Here, H will denote the separable Hilbert space we work in; B(H), Bp (H) [p ≥ 1], the set of bounded, Schatten p- class operators in H respectively with k.k, k.kp as the associated norms. In particular B1 (H) and B2 (H) are known as the set of trace class and Hilbert-Schmidt class operators in H. Let A be a self-adjoint operator in H with σ(A) as the spectra and EA (λ) the spectral family. The symbols Dom(A), Ker(A), Ran(A) and TrA denote the domain, kernel, range and trace of the operator A respectively. Let A (possibly unbounded) and V be two self-adjoint operators in H such that V ∈ B1 (H), then Krein [9] proved that there exists a unique real-valued L1 (R)- function ξ with support in the interval [a, b] ( where a = min{inf σ(A+V ), inf σ(A)} and b = max{sup σ(A+ V ), sup σ(A)} ) such that Z b Tr [φ (A + V ) − φ (A)] = φ′ (λ)ξ(λ)dλ, (1.1) a

for a large class of functions φ . The function ξ is known as Krein’s spectral shift function and the relation (1.1) is called Krein’s trace formula. In 1985, Voiculescu approached the trace formula (1.1) from a different direction. Later Voiculescu [13], and Sinha and Mohapatra ([11], [12]) proved that Z Tr [φ (A + V ) − φ (A)] = lim Trn [φ ((A + V )n ) − φ (An )] = φ′ (λ)ξ(λ)dλ, (1.2) n−→∞

1

J.N.Centre for Advanced Scientific Research, Bangalore-560064, INDIA; [email protected] J.N.Centre for Advanced Scientific Research and Indian Institute of Science, Bangalore-560064, INDIA; [email protected] 2

1

by adapting the Weyl-von Neumann’s theorem (where φ(.) is a suitable function and (A + V )n , An are finite dimensional approximations of (A + V ) and A respectively and Trn is the associated finite dimensional trace).In [8], Koplienko considers instead φ(A + V ) − φ(A) − D (1) φ(A)(V ), where D (1) φ(A) denotes the first order Frechet derivative of φ at A [1] and finds a trace formula for this expression. If V ∈ B2 (H), then Koplienko’s formula asserts that there exists a unique function η ∈ L1 (R) such that Z ∞ (1) Tr{φ(A + V ) − φ(A) − D φ(A)(V )} = φ′′ (λ)η(λ)dλ (1.3) −∞

for rational functions φ with poles off R. In [4], Koplienko trace formula was derived using finite dimensional approximation method, while Dykema and Skripka [5] obtained the formula (1.3) in the semi-finite von Neumann algebra setting and also studied the existence of higher order spectral shift function. In ([5], Theorem 5.1), Dykema and Skripka showed that for a self-adjoint operator A (possibly unbounded) and a self-adjoint operator V ∈ B2 (H), the following assertions hold: (i) There is a unique finite real-valued measure ν3 on R such that the trace formula Z ∞ 1 (2) (1) Tr{φ(A + V ) − φ(A) − D φ(A)(V ) − D φ(A)(V, V )} = φ′′′ (λ)dν3 (λ), (1.4) 2 −∞ holds for suitable functions φ, where D (2) φ(A) is the second order Frechet derivative of φ at A [1]. The total variation of ν3 is bounded by 3!1 kV k32 . (ii) If, in addition, A is bounded, then ν3 is absolutely continuous. It is noted that there has been a more recent preprint by Potapov , Skripka and Sukochev [10] in which similar and further results have been announced. This paper is organized as follows. In section 2 , we establish the formula (1.4) for bounded self-adjoint case and section 3 is devoted to the unbounded self-adjoint case.

2

Bounded Case

The next three lemmas are preparatory for the proof of the main theorem of this section, theorem 2.5. n−1 Lemma 2.1. Let, for a given n ∈ N, {ak }k=0 be a sequence of complex numbers such that an−k−1 = ak . Then j−1 n−1 n n−1 n−j−1 X X X X X ak = (n + 1) ak . ak + j=0

k=0

j=1

k=0

2

k=0

Proof. By changing the indices of summation and using the fact an−k−1 = ak , we get that n−1 X j=0

n−j−1 X

ak +

k=0

j−1 n X X j=1

ak =

n−1 n−1 X X j=0

k=0

=

an−k−1 +

j=0

k=j

n−1 X

aj +

j=0

j n−1 X X

n−1 X j=0

n−1 X

ak =

ak =

j=0

k=0

n−1 X

aj + n

j=0

k=0

n−1 n−1 X X n−1 X

ak +

k=j

j n−1 X X j=0

ak = (n + 1)

k=0

ak

k=0

n−1 X

ak .

k=0

✷ Lemma 2.2. Let A and V be two bounded self-adjoint operators in an infinite dimensional Hilbert space H such that V ∈ B3 (H). Let p(λ) = λr (r ≥ 0).Then   1 (2) r r r (1) r Tr (A + V ) − A − D (A )(V ) − D (A )(V, V ) 2 Z Z (2.1) r−2 1 s X   r−k−2 k r−k−2 k ds dτ Tr V Aτ V Aτ − V A VA , =r k=0

0

0

where Aτ = A + τ V and 0 ≤ τ ≤ 1. Proof. For X ∈ B(H), p(A + X) − p(A) =

r−1 P

(A + X)r−j−1XAj and hence

j=0

r−2 r−j−2 r−1

X

X X

r−j−1 j kA + Xkr−j−k−2kXkkAkk kXkkAkj , A XA ≤

p(A + X) − p(A) −

j=0

j=0

proving that D (1) (Ar )(X) =

r−1 P

k=0

Ar−j−1 XAj .

j=0

Again for X, Y ∈ B(H),

D (1) ((A + X)r )(Y ) − D (1) (Ar )(Y ) =

r−1 X

(A + X)

r−j−1

j

Y (A + X) −

j=0

=

r−1 X j=0

=

Ar−j−1 Y Aj

j=0

r−1 X     r−j−1 r−j−1 j (A + X) −A Y (A + X) + Ar−j−1 Y (A + X)j − Aj j=0

r−2 r−j−2 X X

(A + X)

j=0

r−1 X

r−j−k−2

k

j

XA Y (A + X) +

j−1 r−1 X X j=1

k=0

3

k=0

Ar−j−1 Y (A + X)k XAj−k−1,

leading to the estimate kD (1) ((A + X)r )(Y ) − D (1) (Ar )(Y ) r−2 r−j−2 X X

−

j=0

Ar−j−k−2XAk Y Aj +

k=0

j−1 r−1 X X j=1

k=0

k

j

!

Ar−j−1 Y Ak XAj−k−1 k = (kXk2 )

for kXk ≤ 1, proving that (2)

r

D (A )(X, Y ) =

r−2 r−j−2 X X j=0

r−j−k−2

A

XA Y A +

j−1 r−1 X X j=1

k=0

Ar−j−1 Y Ak XAj−k−1.

(2.2)

k=0

Recall that As = A + sV ∈ Bs.a. (H) (0 ≤ s ≤ 1), and a similar calculation shows that the map [0, 1] ∋ s 7−→ Ars is continuously differentiable in norm-topology and r−1

r−1

X X d r (As ) = Ar−j−1 V Ajs = Ajs V Ar−j−1 . s s ds j=0 j=0 Hence Z

1

d r (As ) − D (1) (Ar )(V ) ds 0 Z 1 X Z 1 X r−1 r−1 Z s

d r−j−1 j r−j−1 j Ar−j−1 V Ajτ , = ds As V As − A VA = ds dτ τ dτ 0 0 0 j=0 j=0 r

(1)

r

r

(A + V ) − A − D (A )(V ) =

ds

which by an application of Leibnitz’s rule reduces to Z

1

ds

0

Z

s

dτ

0

r−2 r−j−2 X X j=0

Ar−j−k−2 V Akτ V Ajτ + τ

j−1 r−1 X X j=1

k=0

Ar−j−1 V Akτ V Aj−k−1 τ τ

k=0

!

and using (2.2), we get 1 (A + V )r − Ar − D (1) (Ar )(V ) − D (2) (Ar )(V, V ) 2 Z 1 Z s j−1 r−j−2 r−1 X r−2 X X X k j Ar−j−1 V Akτ V Aj−k−1 Ar−j−k−2 V A V A + = ds dτ { τ τ τ τ τ 0

0

j=0

−

j=0

j=1

k=0

r−2 r−j−2 X X

r−j−k−2

A

k

j

VA VA −

r−1 X j=1

k=0

4

k=0

j−1 X

Ar−j−1 V Ak V Aj−k−1 }.

k=0

(2.3)

Let us denote the sum of the first and third term inside the integral in (2.3) to be I1 ≡

r−2 r−j−2 X  X j=0

=

Ar−j−k−2 V Akτ V Ajτ − Ar−j−k−2V Ak V Aj τ

k=0

r−2 r−j−2 X X j=0

k=0

+



r−2 r−j−2 X X  r−j−k−2    r−j−k−2 k j Aτ −A V Aτ V Aτ + Ar−j−k−2V Akτ − Ak V Ajτ j=0

r−2 X j=0

r−j−2

X k=0

k=0

  Ar−j−k−2V Ak V Ajτ − Aj ∈ B1 (H),

since V ∈ B3 (H) and Akτ − Ak ∈ B3 (H) ∀τ ∈ [0, 1] and k ∈ {0, 1, 2, 3, ......}. Thus by the cyclicity of trace , we have that Tr(I1 ) =

r−2 r−j−2 X X j=0

k=0

  Tr Aτr−k−2 V Akτ V − Ar−k−2V Ak V .

Again if we set the sum of the second and fourth term inside the integral in (2.3) to be j−1 r−1 X X  r−j−1  V Akτ V Aj−k−1 Aτ I2 ≡ − Ar−j−1V Ak V Aj−k−1 ∈ B1 (H), τ j=1

k=0

a similar calculation shows that Tr(I2 ) =

j−1 r−1 X X j=1

k=0

  Tr Aτr−k−2V Akτ V − Ar−k−2 V Ak V .

  By applying Lemma 2.1 with n = r − 1 and ak = Tr Aτr−k−2 V Akτ V − Ar−k−2 V Ak V and using the cyclicity of trace, we conclude that Tr(I1 ) + Tr(I2 ) = r

r−2 X

k=0 r−2 X

=r

k=0

  Tr Aτr−k−2 V Akτ V − Ar−k−2 V Ak V

(2.4)



 Tr V Aτr−k−2 V Akτ − V Ar−k−2 V Ak .

Hence combining (2.3) and (2.4), we get the required expression (2.1).

✷

Lemma 2.3. Let B be a bounded operator in an infinite dimensional Hilbert space H(i.e. e as a Hilbert space B ∈ B(H)). Define MB : B2 (H) 7−→ B2 (H) ( looking upon B2 (H) ≡ H 5

with inner product given by trace i.e.hX, Y i2 = Tr{X ∗ Y } for X, Y ∈ B2 (H)) by MB (X) = BX − XB ; X ∈ B2 (H). Then e (i.e. MB ∈ B(H)) e with M∗ = MB∗ . (i) MB is a bounded operator on H B

e are left invariant by left (ii) Ker (MB ) and its orthogonal complement Ran (MB∗ ) in H n ∗ n and right multiplication by B and (B ) (n = 1, 2, 3, ...) respectively. L e = Ker (MB ) Ran (MB∗ ) ; B2 (H) ∋ X = X1 ⊕ X2 , where X1 ∈ Ker (MB ) and (iii) H X2 ∈ Ran (MB∗ ). (iv) If Ker (MB ) = Ker (MB∗ ), then Ker (MB ) and Ran (MB ) are generated by their e we have (X ∗ ) = X ∗ and (X ∗ ) = X ∗ , where X = self-adjoint elements and for X ∈ H, 1 2 1 2 e X1 ⊕ X2 and X ∗ = (X ∗ )1 ⊕ (X ∗ )2 are the respective decompositions of X and X ∗ in H. e X = X1 ⊕ X2 with X1 and X2 (v) If Ker (MB ) = Ker (MB∗ ), then for X = X ∗ ∈ H, both self-adjoint.

e and for X = X ∗ ∈ H, e we have (vi) (a) For B = B ∗ ∈ B(H), MB is self-adjoint in H ∗ ∗ X1 = X1 and X2 = X2 .

(b) For B = (A + i)−1 ( where A is an unbounded self-adjoint operator in H), MB e and for X = X ∗ ∈ H, e we have X1 = X ∗ and X2 = X ∗ , where is bounded normal in H 1 2 e X = X1 ⊕ X2 is the decomposition of X in H.

(vii) (a) Let [0, 1] ∋ τ −→ Aτ ∈ Bs.a (H)( set of bounded self-adjoint operators in H) be e ∋ X ≡ X1τ ⊕ X2τ be the self-adjoint decomposition continuous in operator norm , and let H e are continuous. with respect to Aτ . Then τ −→ X1τ , X2τ ∈ H

(b) Let {Aτ }τ ∈[0,1] be a family of unbounded self-adjoint operators in H such that [0, 1] ∋ τ −→ (Aτ + i)−1 is continuous in operator norm. Then the conclusions of (vii)(a) is valid e with respect to Bτ ≡ (Aτ + i)−1 . for the decomposition of H

Proof. The proofs of (i) to (iii) are standard and for (iv), we note that since Ker (MB ) = if X ∗
∈ Ker (MB ) and hence for any X ∈ Ker (MB ) Ker (MB∗ ), X ∈ Ker (MB ) if and
only X−X ∗ X+X ∗ can be written as X = +i , proving that Ker (MB ) is generated by its 2 2i self-adjoint elements. Similarly, by a similar argument we conclude that Ran (MB ) is also generated by its self-adjoint elements. e and X = X1 ⊕ X2 and X ∗ = (X ∗ ) ⊕ (X ∗ ) be the corresponding decompoLet X ∈ H, 1 2 e Then for any Y1 = Y ∗ ∈ Ker (MB ), sitions of X and X ∗ in H. 1 hX, Y1 i2 = hX1 , Y1i2 = Tr{X1∗ Y1 } = Tr{Y1 X1∗ } = hY1 , X1∗ i2 = hX1∗ , Y1i2 .

But on the other hand, hX, Y1 i2 = Tr{X ∗ Y1 } = Tr{(Y1 X)∗ } = Tr{Y1 X} = hX ∗ , Y1 i2 = h(X ∗ )1 , Y1i2 6

and hence h(X ∗ )1 − X1∗ , Y1 i2 = 0 ∀ Y1 = Y1∗ ∈ Ker (MB ), which implies that h(X ∗ )1 − X1∗ , Y i2 = 0 ∀ Y ∈ Ker (MB ), proving that (X ∗ )1 = X1∗ . Similarly, by the same argument we conclude that (X ∗ )2 = X2∗ . The result (v) and (vi(a)) follows from (iv) and (v) respectively. For (vi(b)), it suffices to note that any X ∈ B2 (H) commuting with (A + i)−1 commutes with the spectral family EA (.) of A. For (vii(a)), since the map [0, 1] ∋ τ −→ MAτ is holomorphic, then ( using Theorem 1.8, page 370, [7] ) we conclude that the map [0, 1] ∋ τ −→ P0 (τ ) (where P0 (τ ) is the projection onto Ker(MAτ )) is continuous and since X1τ ≡ P0 (τ )X we get that the map [0, 1] ∋ τ −→ X1τ is continuous. Similarly, since the map [0, 1] ∋ τ −→ I − P0 (τ ) is continuous and X2τ = (I − P0 (τ ))X then we conclude that the map [0, 1] ∋ τ −→ X2τ is also continuous. Conclusions of (vii(b)) follows immediately from (vii(a)) since the map [0, 1] ∋ τ −→ M(Aτ +i)−1 is holomorphic, and since M(Aτ +i)−1 is normal for each τ . Remark 2.4. Let A and V be two bounded self-adjoint operators in an infinite dimensional Hilbert space H such that V ∈ B2 (H) and Aτ = A + τ V (0 ≤ τ ≤ 1). Apply Lemma 2.3 with B = A and Aτ respectively to get V = V1 ⊕ V2 = V1τ ⊕ V2τ , with Vj and Vjτ (j = 1, 2) self-adjoint and therefore kV k22 = kV1 k22 + kV2 k22 = kV1τ k22 + kV2τ k22 ∀ 0 ≤ τ ≤ 1. Theorem 2.5. Let A and V be two bounded self-adjoint operators in an infinite dimensional Hilbert space H such that V ∈ B2 (H). Then there exist a unique real-valued function η ∈ L1 ([a, b]) such that   Z b 1 (2) (1) Tr p(A + V ) − p(A) − D p(A)(V ) − D p(A)(V, V ) = p′′′ (λ)η(λ)dλ, (2.5) 2 a where p(.) is a polynomial in [a, b], Rb η(λ)dλ = 61 T r(V 3 ).

a = [inf σ(A)] − kV k,

b = [sup σ(A)] + kV k and

a

Proof. It will be sufficient to prove the theorem for p(λ) = λr (r ≥ 0). Note that for r = 0, 1 or 2, both sides of (2.5) are identically zero. We set Aτ = A + τ V and 0 ≤ τ ≤ 1. Then by lemma 2.2, we have that   1 (2) r r r (1) r Tr (A + V ) − A − D (A )(V ) − D (A )(V, V ) 2 Z Z r−2 1 s X   ds dτ Tr V Aτr−k−2 V Akτ − V Ar−k−2V Ak =r k=0

0

0

7

= r(r − 1)

Z

1

ds

0

+ r

Z

s

0

r−2 X k=0

Z

  dτ Tr V1τ2 Aτr−2 − V12 Ar−2

1

ds

0

Z

s

dτ Tr

0



V2τ Aτr−k−2 V2τ Akτ

r−k−2

− V2 A

k

V2 A



(2.6) ,

where we have also noted the invariance, orthogonality and continuity properties in Lemma 2.3 (ii) − (vii) and set V = V1 ⊕ V2 = V1τ ⊕ V2τ ∈ B2 (H) as in Remark 2.4. Using the spectral families Eτ (.) and E(.) of the self-adjoint operators Aτ and A respectively and integrating by-parts, the first term of the expression (2.6) is equal to Z 1 Z s Z b   r(r − 1) ds dτ λr−2 Tr V1τ2 Eτ (dλ) − V12 E(dλ) 0 0 Z 1 a Z s   = r(r − 1) ds dτ {λr−2 Tr V1τ2 Eτ (λ) − V12 E(λ) |bλ=a 0 0 Z b   − (r − 2)λr−3 Tr V1τ2 Eτ (λ) − V12 E(λ) dλ} Z

1

Z

a

s

  dτ Tr V1τ2 − V12 0 0 Z 1 Z s Z b   + r(r − 1)(r − 2) ds dτ λr−3 Tr V12 E(λ) − V1τ2 Eτ (λ) dλ.

= r(r − 1)br−2

ds

0

0

(2.7)

a

(n)

Since V2 ∈ Ran(MA ), then there exists a sequence {V2 } ⊆ Ran(MA ) such that (n) (n) (n) (n) (n) kV2 − V2 k2 −→ 0 as n −→ ∞ and V2 = AY0 − Y0 A, for a sequence {Y0 } ⊆ (n) B2 (H). Similarly, for every τ ∈ (0, 1], there exists a sequence {V2τ } ⊆ Ran(MAτ ) such that (n) (n) kV2τ − V2τ k2 −→ 0 point-wise as n −→ ∞ and V2τ = Aτ Y (n) − Y (n) Aτ , for some sequence (n) (n) {Y (n) } ⊆ B2 (H). Observe that Y0 and Y (n) must be skew-adjoint for each n, since V2 (n) and V2τ can be chosen to be self-adjoint. Furthermore, by lemma 2.3 (vii)(a), the map [0, 1] ∋ τ −→ V1τ , V2τ are continuous. Hence the second term of the expression (2.6) is equal to Z 1 Z s r−2 X (n) (n) Tr{V2τ Aτr−k−2 V2τ Akτ − V2 Ar−k−2V2 Ak } r ds dτ lim 0

=r =r

n→∞

0

Z

Z

1

ds 0

s

dτ lim

ds

Z

0

k=0 r−2 Z b X

n→∞

0

1 0

Z

s

dτ lim

n→∞

Z

b

(n)

(n)

λr−k−2 µk Tr{V2τ Eτ (dλ)V2τ Eτ (dµ) − V2 E(dλ)V2 E(dµ)}

a k=0 a Z bZ b

(n)

(n)

φ(λ, µ) Tr{V2τ Eτ (dλ)V2τ Eτ (dµ) − V2 E(dλ)V2 E(dµ)},

a

a

λr−1 −µr−1 λ−µ

if λ 6= µ ; = (r − 1)λr−2 if λ = µ, and where the interchange where φ(λ, µ) = of the limit and the integration is justified by an application of the bounded convergence 8

(n)

theorem. Furthermore using the representation of V2τ ∈ Ran(MAτ ), the above reduces to Z 1 Z s Z bZ b   r ds dτ lim φ(λ, µ) Tr{V2τ Eτ (dλ) Aτ Y (n) − Y (n) Aτ Eτ (dµ) n→∞ 0 0 a a h i (n) (n) − V2 E(dλ) AY0 − Y0 A E(dµ)} Z 1 Z s Z bZ b
(n) =r ds dτ lim λr−1 − µr−1 Tr{V2τ Eτ (dλ)Y (n) Eτ (dµ) − V2 E(dλ)Y0 E(dµ)} 0

=r

Z

n→∞

0

1

0

ds

Z

a

Z

s

dτ lim

n→∞

0

a

a b

h i   (n) λr−1 Tr{V2τ Eτ (dλ), Y (n) − V2 E(dλ), Y0 }.

(2.8)

Again by twice integrating by-parts, the expression in (2.8) is equal to Z 1 Z s  i h   (n) r ds dτ lim {λr−1 Tr V2τ Eτ (λ), Y (n) − V2 E(λ), Y0 |bλ=a n→∞ 0 0 Z b  i h   (n) r−2 (n) − (r − 1)λ Tr V2τ Eτ (λ), Y dλ} − V2 E(λ), Y0 a Z 1 Z s Z b h i   (n) = −r(r − 1) ds dτ lim λr−2 Tr{V2τ Eτ (λ), Y (n) − V2 E(λ), Y0 } dλ 0 1

n→∞

0 s

a

Z

 i  h   (n) (n) ds dτ lim {λ Tr V2τ Eτ (µ), Y dµ }|bλ=a − V2 E(µ), Y0 = −r(r − 1) n→∞ 0 0 a Z λ Z 1 Z s Z b  i  h   (n) r−3 (n) + r(r − 1) ds dτ lim Tr V2τ Eτ (µ), Y dµ dλ (r − 2)λ − V2 E(µ), Y0 n→∞ a 0 0 a Z 1 Z s Z b  h i   (n) r−2 = −r(r − 1)b ds dτ lim Tr V2τ Eτ (µ), Y (n) − V2 E(µ), Y0 dµ n→∞ 0 0 a Z λ  Z 1 Z s Z b i  h   (n) r−3 (n) + r(r − 1)(r − 2) ds dτ lim λ Tr V2τ Eτ (µ), Y dµ dλ. − V2 E(µ), Y0 Z

Z

r−2

0

0

n→∞

λ

a

a

(2.9)

Next we note that by an integration by-parts, i    h
  (n) (n) (n) (n) 2 2 = lim Tr V2τ Aτ , Y − V2 A, Y0 Tr V2τ − V2 = lim Tr V2τ V2τ − V2 V2 n→∞ n→∞ Z b  i h   (n) = lim µTr V2τ Eτ (dµ), Y (n) − V2 E(dµ), Y0 n→∞ a  Z b   i h i  h     (n) (n) b (n) (n) |µ=a − = lim µTr V2τ Eτ (µ), Y − V2 E(µ), Y0 Tr V2τ Eτ (µ), Y dµ . − V2 E(µ), Y0 n→∞

a

The boundary term above vanishes and substituting the above in the first expression in (2.9), we get that the right hand side of (2.9) Z 1 Z s Z 1 Z s Z b
(n) r−2 2 2 = r(r − 1)b ds dτ Tr V2τ − V2 + r(r − 1)(r − 2) ds dτ lim λr−3 η2τ (λ)dλ, 0

0

0

9

0

n→∞

a

(n)

where η2τ (λ) =

a

Hence r−2 Z X

r

k=0

Rλ

1

ds 0

Z

s

0

r−2

= r(r − 1)b

 h i   (n) Tr V2τ Eτ (µ), Y (n) − V2 E(µ), Y0 dµ.

  dτ Tr V2τ Aτr−k−2 V2τ Akτ − V2 Ar−k−2V2 Ak Z

1

ds

0

Z

s

dτ Tr

0

V2τ2

−

V22




+ r(r − 1)(r − 2)

Z

1

ds

0

Z

s

dτ lim

n→∞

0

Z

b

(n)

λr−3 η2τ (λ)dλ. a

(2.10)

Combining (2.7) and (2.10) and since kV k22 = Tr (V1τ2 + V2τ2 ) = Tr (V12 + V22 ), we conclude that   1 (2) r r r (1) r Tr (A + V ) − A − D (A )(V ) − D (A )(V, V ) 2 Z 1 Z s Z b   = r(r − 1)(r − 2) ds dτ λr−3 Tr V12 E(λ) − V1τ2 Eτ (λ) dλ 0 0 a Z 1 Z s Z b (n) + r(r − 1)(r − 2) ds dτ lim λr−3 η2τ (λ)dλ n→∞ 0 0 a Z 1 Z s Z b = r(r − 1)(r − 2) ds dτ lim λr−3 ητ(n) (λ)dλ n→∞ 0 0 a Z b = lim (λr )′′′ η (n) (λ)dλ , where n→∞

η

(n)

(λ) ≡

R1

Rs

ds dτ

0

a

(n) ητ (λ)

0

and

(n) ητ (λ)

=

h

Tr{V12 E(λ)

−

V1τ2 Eτ (λ)}

+

(n) η2τ (λ)

i , the inter-

change of limit and the τ - and s- integral is justified by an easy application of bounded convergence theorem. Note that η (n) is a real-valued function ∀ n. Next we want to show that {η (n) } is cauchy in L1 ([a, b]) and we follow the idea from [6]. Rλ Rλ For that let f ∈ L∞ ([a, b]). Define g(λ) = f (t)dt and h(λ) = g(µ)dµ, then g ′ (λ) = f (λ) a

a.e. and h′ (λ) = g(λ). Now consider the expression Z

b a

a

Z b h i  (n)  (n) (m) (m) f (λ) ητ (λ) − ητ (λ) dλ = f (λ) η2τ (λ) − η2τ (λ) dλ Z λ  a Z b i  h   (n) (m) ′′ (n) (m) = h (λ)dλ dµ , Tr V2τ Eτ (µ), Y − Y − V2 E(µ), Y0 − Y0 a

a

10

which on integration by-parts twice and on observing that the boundary term for λ = a vanishes, leads to Z b  i h   (n) (m) ′ h (b) Tr V2τ Eτ (µ), Y (n) − Y (m) − V2 E(µ), Y0 − Y0 dµ a  i h   (n) (m) − {h(λ)Tr V2τ Eτ (λ), Y (n) − Y (m) − V2 E(λ), Y0 − Y0 |bλ=a Z b  h i   (n) (m) − h(λ)Tr V2τ Eτ (dλ), Y (n) − Y (m) − V2 E(dλ), Y0 − Y0 } a Z b  i h   (n) (m) ′ = h (b) Tr V2τ Eτ (µ), Y (n) − Y (m) − V2 E(µ), Y0 − Y0 dµ a Z b  h i   (n) (m) (n) (m) + h(λ)Tr V2τ Eτ (dλ), Y − Y − V2 E(dλ), Y0 − Y0 . a

(2.11)

Next we use the identity Tr



(n) V2τ V2τ

−

(n) V2 V2



=−

Z

a

b

 i h   (n) Tr V2τ Eτ (µ), Y (n) − V2 E(µ), Y0 dµ

to reduce the the above expression in (2.11) to i i  h i h  h   (n) (m) (n) (m) (n) (m) (n) (m) . +Tr V2τ h(Aτ ), Y − Y − V2 h(A), Y0 − Y0 − V2τ V2τ − V2τ g(b)Tr V2 V2 − V2 (2.12) But on the other hand, Z bZ b   h(λ) − h(µ) (n) (n) E(dλ)V2 E(dµ), and h(A), Y = λ−µ a a  h i Z b Z b h(λ) − h(µ)  h i  (n) (m) (n) (m) Tr V2 h(A), Y0 − Y0 = Tr V2 E(dλ) V2 − V2 E(dµ) λ−µ a a (2.13) and hence as in Birman-Solomyak ([2],[3]) and in [4] ,  h

h

h i i i

(n)

(n) (n) (m) (m) (m) Tr V h(A), Y − Y ≤ khk kV k V − V ≤ (b−a)kf k kV k V − V

2 Lip 2 2 ∞ 2 0 0 2 2 2 2 2

and hence R h i b (n) (m) a f (λ) ητ (λ) − ητ (λ) dλ kf k∞

i.e.

kητ(n) − ητ(m) kL1

2

h i   h i

(n)

(n) (m) (m) ≤ 2(b − a)kV k2 V2 − V2

+ V2τ − V2τ 2

2

h  h i i 

(n)

(n) (m) (m) ≤ 2(b − a)kV k2 V2 − V2

+ V2τ − V2τ , 2

11

2

which converges to 0 as n, m → ∞ and ∀ τ ∈ [0, 1]. A similar computation also shows that R1 Rs (n) (n) kητ kL1 ≤ 2(b − a)kV k22 . Therefore { ds dτ ητ (λ) ≡ η (n) (λ)} is also cauchy in L1 ([a, b]) 0

0

and thus there exists a function η ∈ L1 ([a, b]) such that kη (n) − ηkL1 → 0 as n −→ ∞, by the bounded convergence theorem and hence also kηkL1 ≤ (b − a)kV k22 . Therefore, Z b Z b r−3 (n) lim λ η (λ)dλ = λr−3 η(λ)dλ and hence n−→∞

a

a

  Z b 1 (2) r r r (1) r λr−3 η(λ)dλ. Tr (A + V ) − A − D (A )(V ) − D (A )(V, V ) = r(r − 1)(r − 2) 2 a

For uniqueness, let us assume that there exists η1 , η2 ∈ L1 ([a, b]) such that  Z b  1 (2) (1) p′′′ (λ)ηj (λ)dλ, Tr p(A + V ) − p(A) − D p(A)(V ) − D p(A)(V, V ) = 2 a where p(.) is a polynomial and j = 1, 2. Therefore Z b p′′′ (λ) η(λ) dλ = 0 ∀ polynomials p(.) and η ≡ η1 − η2 ∈ L1 ([a, b]), a

Rb Rb which together with the fact that a η1 (λ) dλ = a η2 (λ) dλ = 61 Tr(V 3 ) (which one can easily arrive at by setting p(λ) = λ3 in the above formula), implies that Z b λr η(λ)dλ = 0 ∀ r ≥ 0. Hence by an application of Fubini’s theorem, we get that a

Z

∞ −itλ

e

η(λ) dλ =

−∞

Hence

1

Z

Z ∞ ∞ X 1 (−itλ)n η(λ) dλ = 0. n! −∞ n=0

∞

e−itλ η(λ) dλ = 0 ∀ t ∈ R. −∞

Therefore η is an L ([a, b])- function whose Fourier transform ηˆ(t) vanishes identically, implying that η = 0 or η1 = η2 a.e. ✷ Corollary 2.6. Let A and V be two bounded self-adjoint operators in an infinite dimensional Hilbert space H such that V ∈ B2 (H). Then the function η ∈ L1 ([a, b]) obtained as in theorem 2.5 satisfies the following equation Z b Z 1 Z s Z bZ b h(λ) − h(µ) Tr [V Eτ (dλ)V Eτ (dµ) − V E(dλ)V E(dµ)] , f (λ)η(λ)dλ = ds dτ λ−µ a 0 0 a a where f (λ) , g(λ) and h(λ) are as in the proof of the theorem 2.5. 12

Proof. By Fubini’s theorem we have that Z b Z 1 Z s Z b h i (n) (n) f (λ)η (λ)dλ = ds dτ f (λ) Tr{V12 E(λ) − V1τ2 Eτ (λ)} + η2τ (λ) dλ. 0

a

But

Z

0

a

b

f (λ)

Tr{V12 E(λ)

−

V1τ2 Eτ (λ)}dλ

=

a

Z

a

b

  g ′(λ)Tr V12 E(λ) − V1τ2 Eτ (λ) dλ,

which by integrating by-parts leads to Z b  2    2 g(b) Tr V1 − V1τ + g(λ)Tr V1τ2 Eτ (dλ) − V12 E(dλ)  2  a   2 = g(b) Tr V1 − V1τ + Tr V1τ2 h′ (Aτ ) − V12 h′ (A) Z bZ b  2  h(λ) − h(µ) 2 = g(b) Tr V1 − V1τ + Tr [V1τ Eτ (dλ)V1τ Eτ (dµ) − V1 E(dλ)V1 E(dµ)] . λ−µ a a (2.14) Again by repeating the same above calculations to get (2.12) and (2.13) as in the proof of the theorem 2.5, we conclude that Z b h i (n) (n) (n) f (λ)η2τ (λ)dλ = g(b) Tr V2 V2 − V2τ V2τ a Z bZ b h i h(λ) − h(µ) (n) (n) Tr V2τ Eτ (dλ)V2τ Eτ (dµ) − V2 E(dλ)V2 E(dµ) . + λ−µ a a (2.15) Combining (2.14) and (2.15) we have, Z

b

f (λ)η (n) (λ)dλ a h   i (n) (n) = Tr V12 + V2 V2 − V2τ2 + V2τ V2τ Z 1 Z s Z bZ b  i   h  h(λ) − h(µ) (n) (n) E(dµ) . Tr V Eτ (dλ) V1τ ⊕ V2τ Eτ (dµ) − V E(dλ) V1 ⊕ V2 + ds dτ λ−µ 0 0 a a (2.16) (n)

(n)

But by definition V2 , V2τ converges to V2 , V2τ respectively in k.k2 and we have already proved that η (n) converges to η in L1 ([a, b]). Hence by taking limit on both sides of (2.16) we get that Z b Z 1 Z s Z bZ b h(λ) − h(µ) f (λ)η(λ)dλ = ds dτ Tr [V Eτ (dλ)V Eτ (dµ) − V E(dλ)V E(dµ)] . λ−µ a 0 0 a a (2.17) 13

In the we have used the fact that  right hand  side of (2.17)   (n) (n) (n) V ar G2 − G2 ≤ kV k2 kV2τ − V2τ k2 + kV2 − V2 k −→ 0 as n −→ ∞, where h     i (n) (n) (n) G2 (∆ × δ) = Tr V Eτ (∆) V1τ ⊕ V2τ Eτ (δ) − V E(∆) V1 ⊕ V2 E(δ) and

G2 (∆ × δ) = Tr[V Eτ (∆)V Eτ (δ) − V E(∆)V E(δ)] are complex measures on R2 and  (n) (n) V ar G2 − G2 is the variation of G2 − G2 and also noted that kgkLip ≤ (b − a)kf k∞ . ✷

3

Unbounded Case

Theorem 3.1. Let A be Ran unbounded self-adjoint operator in a Hilbert space H and let ∞ ˆ φ : R −→ C be such that −∞ |φ(t)| (1 + |t|)3 dt < ∞, where φˆ is the Fourier transform of φ. Then φ(A), D (1) φ(A), D (2) φ(A) exist and Z ∞ Z t  (1)  ˆ D φ(A) (X) = i φ(t)dt dβ eiβA Xei(t−β)A and 0

−∞

 (2)  D φ(A) (X, Y ) = i2

Z

∞

ˆ φ(t)dt

Z

0

−∞

t

dβ {

Z

β

dν eiνA Xei(β−ν)A Y ei(t−β)A

0

+

where X, Y ∈ B2 (H).

Z

t−β

dν eiβA Y eiνA Xei(t−β−ν)A }, 0

Proof. That φ(A) and the expressions on the right hand side above exist in B(H) are conseˆ Next quences of the functional calculus and the assumption on φ. Z ∞ Z ∞ Z t  it(A+X)  itA ˆ ˆ φ(A + X) − φ(A) = φ(t) e −e dt = φ(t)dt dβ eiβ(A+X) iXei(t−β)A . −∞

Therefore

Z

−∞

0

Z t ˆ φ(A + X) − φ(A) − i φ(t)dt dβ eiβA Xei(t−β)A −∞ 0 Z ∞ Z t   ˆ =i φ(t)dt dβ eiβ(A+X) Xei(t−β)A − eiβA Xei(t−β)A . ∞

−∞

0

Using the interpolation inequality

iβ(A+X)

e − eiβA ≤ 2(1−ǫ) (βkXk)ǫ (0 ≤ ǫ ≤ 1), we get that

Z ∞ Z t

iβA i(t−β)A ˆ

φ(A + X) − φ(A) − i

φ(t)dt dβ e Xe

−∞ 0  (1−ǫ)  Z ∞ 2 ǫ+1 ˆ kXk |φ(t)| (1 + |t|)ǫ+2 dt, ≤ ǫ+1 −∞ 14

which by virtue of the assumption on φˆ implies that D (1) φ(A) exist and that Z ∞ Z t  (1)  ˆ D φ(A) (X) = i φ(t)dt dβ eiβA Xei(t−β)A . Similarly for X, Y ∈ B(H), 0

−∞

 (1)    D φ(A + X) (Y ) − D (1) φ(A) (Y ) Z ∞ Z t Z 2 ˆ =i φ(t)dt dβ { 0

−∞

β

dν eiν(A+X) Xei(β−ν)A Y ei(t−β)(A+X) 0

+

Z

t−β

dν eiβA Y eiν(A+X) Xei(t−β−ν)A }

0

and one can verify as before that     k D (1) φ(A + X) (Y ) − D (1) φ(A) (Y ) Z ∞ Z t Z 2 ˆ −i φ(t)dt dβ { 0

−∞

β

dν eiνA Xei(β−ν)A Y ei(t−β)A

0

+ Z

Z

t−β

dν eiβA Y eiνA Xei(t−β−ν)A }k

0

∞

ˆ |φ(t)| (1 + |t|)ǫ+2 dt (for some ǫ > 0 and some constant K ≡ K(ǫ)), −∞   proving the expression for D (2) φ(A) (X, Y ). ✷ ≤ KkXkǫ+1kY k

Theorem 3.2. Let A be an unbounded self-adjoint operator in a Hilbert space H, V be a self-adjoint operator such that V ∈ B3 (H) and furthermore let φ ∈ S(R) (the Schwartz class of smooth functions of rapid decrease). Then    1  (2) φ(A + V ) − φ(A) − D (1) φ(A) (V ) − D φ(A) (V, V ) ∈ B1 (H) and 2     (1)  1  (2) D φ(A) (V, V ) Tr φ(A + V ) − φ(A) − D φ(A) (V ) − 2 Z ∞ Z t Z 1 Z s   2 ˆ = i tφ(t) dt dν ds dτ Tr V ei(t−ν)Aτ V eiνAτ − V ei(t−ν)A V eiνA , 0

−∞

0

0

where Aτ = A + τ V and 0 ≤ τ ≤ 1.

Proof. Since [0, 1] ∋ s −→ eitAs is B3 (H)-continuously differentiable, uniformly in t, Z ∞  (1)      ˆ dt eit(A+v) − eitA − D (1) φ(A) (V ) φ(A + V ) − φ(A) − D φ(A) (V ) = φ(t) −∞ Z ∞ Z t Z ∞ Z 1 d itA s ˆ dt ˆ dt φ(t) dβ eiβA V ei(t−β)A = φ(t) ds (e ) − i ds −∞ 0 −∞ 0 Z 1 Z t  Z ∞ Z 1 Z t iβAs i(t−β)As iβA i(t−β)A ˆ = φ(t) dt ds dβ e iV e −i ds dβ e V e −∞

0

0

0

15

0

Z

∞

Z

1

Z

t

  dβ eiβAs V ei(t−β)As − eiβA V ei(t−β)A −∞ 0 0 Z ∞ Z 1 Z t   ˆ = iφ(t) Q(t) dt, where Q(t) ≡ ds dβ eiβAs V ei(t−β)As − eiβA V ei(t−β)A .

=

ˆ dt iφ(t)

ds

0

−∞

0

(3.1)

As before, τ ∈ [0, 1] −→ eiβAτ V ei(t−β)Aτ ∈ B3 (H) is B 3 (H)− continuously differentiable, 2 uniformly with respect to β. Then Z 1 Z t Z s d iβAτ i(t−β)Aτ
Q(t) = ds dβ dτ e Ve dτ 0 0 0 Z 1 Z s Z t Z β Z t−β iνAτ i(β−ν)Aτ i(t−β)Aτ =i ds dτ dβ { dνe Ve Ve + dνeiβAτ V eiνAτ V ei(t−β−ν)Aτ }, 0

0

0

0

0

where Fubini’s theorem has been used to interchange the order of integration. Hence    1 φ(A + V ) − φ(A) − D (1) φ(A) (V ) − D (2) φ(A) (V, V ) 2 Z ∞ Z 1 Z s Z t Z β   ˆ = i2 φ(t)dt ds dτ dβ{ dν eiνAτ V ei(β−ν)Aτ V ei(t−β)Aτ − eiνA V ei(β−ν)A V ei(t−β)A −∞

0

0

0

+

Z

0

0 t−β

  dν eiβAτ V eiνAτ V ei(t−β−ν)Aτ − eiβA V eiνA V ei(t−β−ν)A }

(3.2)

Though each of the four individual terms in the integral in (3.2) belong to B 3 (H), each of 2 the differences in parenthesis [.] belong to B1 (H), e.g.  iνAτ i(β−ν)Aτ i(t−β)Aτ  e Ve Ve − eiνA V ei(β−ν)A V ei(t−β)A     = eiνAτ − eiνA V ei(β−ν)Aτ V ei(t−β)Aτ + eiνA V ei(β−ν)Aτ − ei(β−ν)Aτ V ei(t−β)Aτ   + ei(β−ν)Aτ V ei(t−β)Aτ − ei(t−β)Aτ ∈ B1 (H), and k[.]k1 ≤ |τ | kV k33 {|ν| + |β − ν| + |t − β|}, where we have used the estimate:

iνAits norm

ˆ

e τ − eiνA ≤ |ντ |kV k3 . Hence by the hypothesis on φ, 3    1 φ(A + V ) − φ(A) − D (1) φ(A) (V ) − D (2) φ(A) (V, V ) ∈ B1 (H) and 2

16



  1  (2) Z ≡ Tr φ(A + V ) − φ(A) − D φ(A) (V ) − D φ(A) (V, V ) 2 Z ∞ Z 1 Z s Z t Z β   2ˆ = i φ(t)dt ds dτ dβ { dνTr eiνAτ V ei(β−ν)Aτ V ei(t−β)Aτ − eiνA V ei(β−ν)A V ei(t−β)A 

0

−∞

0



(1)

0

+

Z

0 t−β

0

  dνTr eiβAτ V eiνAτ V ei(t−β−ν)Aτ − eiβA V eiνA V ei(t−β−ν)A }. (3.3)

Again by the cyclicity of trace and a change of variable, the first integral in {.} in (3.3) is equal to Z β   dν Tr ei(t−β+ν)Aτ V ei(β−ν)Aτ V − ei(t−β+ν)A V ei(β−ν)A V 0 (3.4) Z β  i(t−ν)Aτ iνAτ  i(t−ν)A iνA = dν Tr V e Ve −Ve Ve . 0

Similarly, the second integral in {.} in (3.3) is equal to Z t−β   dν Tr V ei(t−ν)Aτ V eiνAτ − V ei(t−ν)A V eiνA .

(3.5)

0

Combining (3.4) and (3.5), we conclude that Z ∞ Z 1 Z s Z t Z β   2ˆ Z= i φ(t)dt ds dτ dβ { dν Tr V ei(t−ν)Aτ V eiνAτ − V ei(t−ν)A V eiνA 0

−∞

0

0

0

+

Z

t−β

0

  dν Tr V ei(t−ν)Aτ V eiνAτ − V ei(t−ν)A V eiνA }.

(3.6)

By a change of variable and the cyclicity of trace, we get that Z t   t dν Tr V ei(t−ν)Aτ V eiνAτ − V ei(t−ν)A V eiνA 0 Z t Z β   = dβ { dν Tr V ei(t−ν)Aτ V eiνAτ − V ei(t−ν)A V eiνA 0 0 Z t−β   + dν Tr V ei(t−ν)Aτ V eiνAτ − V ei(t−ν)A V eiνA }, 0

using which in (3.6) we are lead to the equation Z ∞ Z t Z 1 Z s   2 ˆ Z= i tφ(t)dt dν ds dτ Tr V ei(t−ν)Aτ V eiνAτ − V ei(t−ν)A V eiνA , −∞

0

0

(3.7)

0

by an application of Fubini’s theorem.

17

✷

Theorem 3.3. Let A be an unbounded self-adjoint operator in a Hilbert space H with σ(A) ⊆ [b, ∞) for some b ∈ R and V be a self-adjoint operator   such that V ∈ B2 (H). Then there exist a unique real-valued function η ∈ L1 R, (1+λdλ2 )1+ǫ (for some ǫ > 0) such that for every φ ∈ S(R) (the Schwartz class of smooth functions of rapid decrease)  Z ∞    (1)  1  (2) D φ(A) (V, V ) = φ′′′ (λ)η(λ)dλ. Tr φ(A + V ) − φ(A) − D φ(A) (V ) − 2 −∞ Proof. Equation (3.7), after an application of Fubini’s theprem, yields that    (1)   1  (2) Z ≡ Tr φ(A + V ) − φ(A) − D φ(A) (V ) − D φ(A) (V, V ) 2 Z 1 Z s Z ∞ Z t   2 ˆ = ds dτ i tφ(t)dt dν Tr V ei(t−ν)Aτ V eiνAτ − V ei(t−ν)A V eiνA . 0

0

0

−∞

(3.8)

Now Z t 0

  dν Tr V ei(t−ν)Aτ V eiνAτ − V ei(t−ν)A V eiνA Z t Z ∞Z ∞ = dν ei(t−ν)λ eiνµ Tr [V Eτ (dλ)V Eτ (dµ) − V E(dλ)V E(dµ)] , 0

a

a

(3.9)

where a = min{b, inf σ(Aτ ) [0 < τ ≤ 1]} and Eτ (.) and E(.) are the spectral families of the operator Aτ and A respectively and the measure G : ∆ × δ ⊆ Borel(R2 ) −→ Tr [V Eτ (∆)V Eτ (δ) − V E(∆)V E(δ)] is a complex measure with total variation ≤ 2kV k22 and hence by Fubini’s theorem the right hand side expression in (3.9) is equal to Z ∞Z ∞ Z t dν ei(t−ν)λ eiνµ Tr [V Eτ (dλ)V Eτ (dµ) − V E(dλ)V E(dµ)] a Z ∞aZ ∞ 0 itλ e − eitµ Tr [V Eτ (dλ)V Eτ (dµ) − V E(dλ)V E(dµ)] = i(λ − µ) a a Z 1 Z s Z ∞ Z ∞   2 ˆ = ds dτ i tφ(t)dt teitλ Tr V1τ2 Eτ (dλ) − V12 E(dλ) 0 0 −∞ a Z 1 Z s Z ∞ Z ∞ Z ∞ itλ e − eitµ 2 ˆ Tr [V2τ Eτ (dλ)V2τ Eτ (dµ) − V2 E(dλ)V2 E(dµ)] , + ds dτ i tφ(t)dt i(λ − µ) 0 0 −∞ a a (3.10) where we have set V = V1 ⊕ V2 = V1τ ⊕ V2τ ∈ B2 (H) as in Lemma 2.3 vi(b). Applying 18

Fubini’s theorem in the first expression in right hand side of (3.10), we conclude that Z 1 Z s Z ∞ Z ∞   2 ˆ ds dτ i tφ(t)dt teitλ Tr V1τ2 Eτ (dλ) − V12 E(dλ) 0 0 −∞ a Z 1 Z s Z ∞   = ds dτ φ′′ (λ) Tr V1τ2 Eτ (dλ) − V12 E(dλ) 0 0 a Z 1 Z s Z ∞   = ds dτ φ′′′ (λ) Tr V12 E(λ) − V1τ2 Eτ (λ) dλ, 0

0

a

where we have integrated by-parts and observed that the boundary term vanishes. Thus the first term in (3.10) is equal to Z 1 Z s Z ∞   ds dτ φ′′′ (λ) η1τ (λ) dλ, where η1τ (λ) = Tr V12 E(λ) − V1τ2 Eτ (λ) . (3.11) 0

0

a

Now consider the second expression in the right hand side of (3.10) : Z 1 Z s Z ∞ Z ∞ Z ∞ itλ e − eitµ 2 ˆ ds dτ Tr [V2τ Eτ (dλ)V2τ Eτ (dµ) − V2 E(dλ)V2 E(dµ)] i tφ(t)dt i(λ − µ) 0 0 −∞ a a (3.12)     i⊥ h  = Ran M∗(A +i)−1 = Ran M(Aτ −i)−1 , there exists Since V2τ ∈ Ker M(Aτ +i)−1 τ   (n) (n) a sequence {V2τ } ⊆ Ran M(Aτ −i)−1 i.e. V2τ = (Aτ − i)−1 Y (n) − Y (n) (Aτ − i)−1 for (n)

some Y (n) ∈  B2 (H) such  that V2τ −→ V2τ in k.k2 . Similarly, there exists a sequence (n) (n) (n) (n) (n) {V2 } ⊆ Ran M(A−i)−1 i.e. V2 = (A − i)−1 Y0 − Y0 (A − i)−1 for Y0 ∈ B2 (H) such (n)

that V2 −→ V2 in k.k2 . Furthermore, by lemma 2.3 (vii)(b), the map [0, 1] ∋ τ −→ V1τ , V2τ are continuous. Thus the expression in (3.12) is equal to Z 1 Z s Z ∞ Z ∞ Z ∞ itλ h i e − eitµ (n) (n) 2 ˆ ds dτ i tφ(t)dt lim Tr V2τ Eτ (dλ)V2τ Eτ (dµ) − V2 E(dλ)V2 E(dµ) i(λ − µ) n→∞ 0 0 −∞ a a Z 1 Z s Z ∞ Z ∞ Z ∞ itλ h i e − eitµ (n) (n) ˆ Tr V2τ Eτ (dλ)V2τ Eτ (dµ) − V2 E(dλ)V2 E(dµ) , = ds dτ i2 tφ(t)dt lim n→∞ a i(λ − µ) 0 0 −∞ a (3.13)   (n) (n) (n) since V ar G2 − G2 ≤ kV2τ k2 kV2τ − V2τ k2 + kV k2 kV2 − V2 k −→ 0 as n −→ ∞, (n)

where G2 (∆ × δ) = Tr [V2τ Eτ (∆)V2τ Eτ (δ) − V2 E(∆)V2 E(δ)] and G2 (∆ × δ) is the same (n) expression withsecond V2 -terms replaced by V2 . These are complex measures on R2 and   (n) (n) V ar G2 − G2 is the variation of G2 − G2 . Note that      (n) Tr V2τ Eτ (dλ)V2τ Eτ (dµ) = Tr V2τ Eτ (dλ) (Aτ − i)−1 Y (n) − Y (n) (Aτ − i)−1 Eτ (dµ) =


−(λ − µ) Tr V2τ Eτ (dλ)Y (n) Eτ (dµ) (λ − i)(µ − i) 19

R∞ ˆ and since −∞ |tφ(t)|dt < ∞ and the other functions are bounded, the right hand side expression in (3.13) is equal to   Z 1 Z s Z ∞ Z ∞ Z ∞ itλ −(λ − µ) e − eitµ 2 ˆ ds dτ lim i tφ(t)dt Tr{V2τ Eτ (dλ)Y (n) Eτ (dµ) n→∞ i(λ − µ) (λ − i)(µ − i) 0 0 −∞ a a (n)

=

=

Z Z

1

ds 0

Z

s

dτ lim

n→∞

0

1

ds 0

Z

s

dτ lim

n→∞

0

Z

∞

ˆ −itφ(t)dt

∞

a

−∞

Z

Z

∞

ˆ −itφ(t)dt

Z

Z

− V2 E(dλ)Y0 E(dµ)}

∞ a

∞

eitλ

a

−∞



 eitλ − eitµ Tr{V2τ Eτ (dλ) (Aτ − i)−1 Y (n) (Aτ − i)−1 Eτ (dµ) (n)

− V2 E(dλ) (A − i)−1 Y0 (A − i)−1 E(dµ)} i i h  h (n) (n) ˜ ˜ , − V2 E(dλ), Y0 Tr V2τ Eτ (dλ), Y (3.14)

(n) (n) where Y˜ (n) = (Aτ − i)−1 Y (n) (Aτ − i)−1 and Y˜0 = (A − i)−1 Y0 (A − i)−1 . Again by applying Fubini’s theorem the right hand side expression in (3.14) is equal to Z 1 Z s Z ∞  h i h i (n) ′ (n) ˜ ˜ ds dτ lim −φ (λ) Tr V2τ Eτ (dλ), Y − V2 E(dλ), Y0 , 0

n→∞

0

a

and by integrating by-parts twice and on observing that the boundary term vanishes, this reduces to Z 1 Z s (n) ds dτ lim − {φ′ (λ) Tr(V2τ [Eτ (λ), Y˜ (n) ] − V2 [E(λ), Y˜0 ]) |∞ λ=a n→∞ 0 0 Z ∞ i i h  h (n) dλ} − −φ′′ (λ) Tr V2τ Eτ (λ), Y˜ (n) − V2 E(λ), Y˜0 a Z 1 Z s Z ∞ i i h  h (n) = ds dτ lim dλ φ′′ (λ) Tr V2τ Eτ (λ), Y˜ (n) − V2 E(λ), Y˜0 n→∞ 0 0 a Z 1 Z s Z λ (n) ′′ = ds dτ lim {φ (λ) Tr(V2τ [Eτ (µ), Y˜ (n) ] − V2 [E(µ), Y˜0 ]) dµ |∞ λ=a n→∞ 0 0 a Z λ Z ∞ i  i h  h (n) ′′′ (n) ˜ ˜ − dµ dλ} φ (λ) − V2 E(µ), Y0 Tr V2τ Eτ (µ), Y a a Z λ Z 1 Z s Z ∞  h i h i  (n) ′′′ (n) = ds dτ lim φ (λ) Tr V2 E(µ), Y˜0 − V2τ Eτ (µ), Y˜ dµ dλ 0

=

Z

0

1

ds 0

where

Z

n→∞

a

dτ lim

Z

s

0

(n) η2τ (λ)

n→∞

=

Z

a

a

∞ (n)

φ′′′ (λ) η2τ (λ) dλ,

(3.15)

a

λ

i i h  h (n) − V2τ Eτ (µ), Y˜ (n) dµ. Tr V2 E(µ), Y˜0 20

Here it is worth observing that the hypothesis that A is bounded below is used for the first time here, only for performing the second integration-by-parts. Combining (3.11) and (3.15), we conclude that    (1)   1  (2) Tr φ(A + V ) − φ(A) − D φ(A) (V ) − D φ(A) (V, V ) 2 (3.16) Z 1 Z s Z ∞ ′′′ (n) = ds dτ lim φ (λ) ητ (λ) dλ, 0

(n)

0

(n)

n→∞

a

(n)

where  ητ (λ) =  η1τ (λ) + η2τ (λ). We claim that {ητ } is a cauchy sequence in dλ 1 L R, (1+λ2 )1+ǫ (ǫ > 0) and we follow the idea from [6]. First note that

L∞ (R, dλ) = L∞ (R, ψ(λ)dλ) [ where ψ(λ) = (1+λ12 )1+ǫ ] since the two measures are equivalent. Next, let f ∈ L∞ (R, dλ) and define Z ∞ g(λ) = f (t)ψ(t)dt for λ ∈ R, then g is absolutely continuous with λ

g ′ (λ) = −f (λ)ψ(λ) a.e. and that |g(λ)| ≤

Const.

1 1

′

(1+λ2 ) 2 +ǫ

(for some ǫ′ > 0) for λ → ∞

and bounded. Next consider the expression Z ∞ Z ∞ h i  (n)  (n) (m) (m) f (λ)ψ(λ) ητ (λ) − ητ (λ) dλ = f (λ)ψ(λ) η2τ (λ) − η2τ (λ) dλ −∞ a Z λ Z ∞ i  i h  h (n) (m) (n) (m) ′ ˜ ˜ ˜ ˜ = dµ , − V2τ Eτ (µ), Y − Y −g (λ) dλ Tr V2 E(µ), Y0 − Y0 a

a

which on integration by-parts and on observing that the boundary terms vanishes, leads to Z ∞ i i h  h (n) (m) − V2τ Eτ (λ), Y˜ (n) − Y˜ (m) dλ. g(λ)Tr V2 E(λ), Y˜0 − Y˜0 (3.17) a

Define h(λ) =

Z

λ

g(t)dt for λ ∈ [a, ∞),

then h is bounded, differentiable

[a, ∞)

a

with h′ (λ) = g(λ) ∀ λ ∈ [a, ∞). Hence by integrating by-parts and observing that the

21

boundary term vanishes, the right hand side expression in (3.17) is equal to Z ∞ i i h  h (n) (m) (n) (m) ′ ˜ ˜ ˜ ˜ dλ − V2τ Eτ (λ), Y − Y h (λ)Tr V2 E(λ), Y0 − Y0 a i i h  h (m) (n) (n) (m) ˜ ˜ ˜ ˜ |∞ − V2τ Eτ (λ), Y − Y = h(λ) Tr V2 E(λ), Y0 − Y0 λ=a Z ∞ i i h  h (n) (m) (n) (m) ˜ ˜ ˜ ˜ − − V2τ Eτ (dλ), Y − Y h(λ)Tr V2 E(dλ), Y0 − Y0 a Z ∞ i i h  h (n) (m) (n) (m) ˜ ˜ ˜ ˜ = − V2 E(dλ), Y0 − Y0 h(λ) Tr V2τ Eτ (dλ), Y − Y a i i h  h (n) (m) (n) (m) ˜ ˜ ˜ ˜ . − V2 h(A), Y0 − Y0 = Tr V2τ h(Aτ ), Y − Y

(3.18)

But on the other hand, i Z ∞Z ∞ h (n) (n) ˜ = h(A), Y0 [h(λ) − h(µ)]E(dλ)Y˜0 E(dµ) Za ∞ Za ∞ (n) = [h(λ) − h(µ)](λ − i)−1 (µ − i)−1 E(dλ)Y0 E(dµ) Za ∞ Za ∞ h(λ) − h(µ) (n) = (λ − i)−1 (µ − i)−1 E(dλ)V2 E(dµ) −1 −1 (λ − i) − (µ − i) a a and hence

Z h i (n) h(A), Y˜0 =−

Therefore Z i  h (n) (n) ˜ ˜ =− Tr V2 h(A), Y0 − Y0

∞ a

Z

Z

∞ a

h(λ) − h(µ) (n) E(dλ)V2 E(dµ). λ−µ

i   h h(λ) − h(µ) (n) (m) E(dµ) Tr V2 E(dλ) V2 − V2 λ−µ a a (3.19) and hence as in Birman-Solomyak ([2],[3]) and in [4] ,

h  h i i (n) (n)

(n) (m) Tr V2 h(A), Y˜0 − Y˜0

≤ kf k∞ kψkL1 kV k2 V2 − V2 ∞

∞

2

and hence R h i ∞ (n) (m) f (λ)ψ(λ) η (λ) − η (λ) dλ −∞ τ τ kf k∞

i.e.

kητ(n)

−

ητ(m) kL1 (R,ψ(λ)dλ)

≤ kψkL1

≤ kψk

L1

h i   h i

(n)

(n) (m) (m) kV k2 V2 − V2

+ V2τ − V2τ 2

2

h  h i i 

(n)

(n) (m) (m) kV k2 V2 − V2

+ V2τ − V2τ , 2

22

2

which converges to 0 as m, n −→ ∞ and ∀ τ ∈ [0, 1]. A similar computation also shows that R1 Rs (n) (n) kητ kL1 (R,ψ(λ)dλ) ≤ 2 kψkL1 kV k22 , independent of τ and n. Therefore { 0 ds 0 dτ ητ (λ) ≡ η (n) (λ)} is also cauchy in L1 (R, ψ(λ)dλ) and thus there exists a function η ∈ L1 (R, ψ(λ)dλ) such that kη (n) − ηkL1 (R,ψ(λ)dλ) −→ 0 as n −→ ∞, by the bounded convergence theorem and hence also kηkL1 (R,ψ(λ)dλ) ≤ kψkL1 kV k22 . Therefore by using the Dominated Convergence theorem as well as Fubini’s theorem, from (3.16) we conclude that  Z ∞    (1)  1  (2) D φ(A) (V, V ) = Tr φ(A + V ) − φ(A) − D φ(A) (V ) − φ′′′ (λ)η(λ)dλ. 2 −∞ ✷

The proof of the uniqueness and the real-valued nature of η is postponed till after the corollary 3.4. Corollary 3.4. Let A be an unbounded self-adjoint operator in a Hilbert space H with σ(A) ⊆ [b, ∞) for some b ∈ R and V be a self-adjoint operator such that V ∈ B2 (H). Then the function η ∈ L1 (R, ψ(λ)dλ) obtained as in theorem 3.3 satisfies the following equation Z ∞ f (λ)ψ(λ)η(λ)dλ −∞ (3.20) Z 1 Z s Z ∞Z ∞ h(λ) − h(µ) = ds dτ Tr [V E(dλ)V E(dµ) − V Eτ (dλ)V Eτ (dµ)] , λ−µ 0 0 a a where f (λ), g(λ), h(λ) and ψ(λ) are as in the proof of the theorem 3.3. Proof. By Fubini’s theorem we have that Z ∞ Z 1 Z s Z (n) f (λ)ψ(λ)η (λ)dλ = ds dτ 0

−∞

0

a

∞

h i (n) f (λ)ψ(λ) η1τ (λ) + η2τ (λ) dλ.

But Z

∞

f (λ)ψ(λ)η1τ (λ)dλ = a

Z

a

∞

  −g ′ (λ)Tr V12 E(λ) − V1τ2 Eτ (λ) dλ,

which by integrating by-parts and observing that the boundary terms vanishes, leads to Z ∞     g(λ)Tr V12 E(dλ) − V1τ2 Eτ (dλ) = Tr V12 h′ (A) − V1τ2 h′ (Aτ ) , which a

by lemma 2.3 (ii) is equal to Z ∞Z ∞ h(λ) − h(µ) Tr [V1 E(dλ)V1 E(dµ) − V1τ Eτ (dλ)V1τ Eτ (dµ)] . λ−µ a a 23

(3.21)

Again by repeating the same calculations as in the proof of the theorem 3.3, we conclude that Z ∞ Z ∞Z ∞ h i h(λ) − h(µ) (n) (n) (n) f (λ)ψ(λ)η2τ (λ)dλ = Tr V2 E(dλ)V2 E(dµ) − V2τ Eτ (dλ)V2τ Eτ (dµ) . λ−µ a a a (3.22) Combining (3.21) and (3.22) we have, Z ∞ f (λ)ψ(λ)η (n) (λ)dλ −∞ Z 1 Z s Z ∞Z ∞   h(λ) − h(µ) (n) Tr[(V1 ⊕ V2 )E(dλ) V1 ⊕ V2 E(dµ) = ds dτ λ−µ 0 0 a a   (n) − (V1τ ⊕ V2τ )Eτ (dλ) V1τ ⊕ V2τ Eτ (dµ)].

(3.23)

(n)

(n)

But by definition V2 , V2τ converges to V2 , V2τ respectively in k.k2 and we have already proved that η (n) converges to η in L1 (R, ψ(λ)dλ) .Hence by taking limit on both sides of (3.23) we get that Z ∞ f (λ)ψ(λ)η(λ)dλ −∞ (3.24) Z 1 Z s Z ∞Z ∞ h(λ) − h(µ) Tr [V E(dλ)V E(dµ) − V Eτ (dλ)V Eτ (dµ)] , = ds dτ λ−µ 0 0 a a where we have used the fact that     (n) (n) (n) V ar G2 − G2 ≤ kV k2 kV2τ − V2τ k2 + kV2 − V2 k −→ 0, and that khkLip ≤ kgk∞ ≤ kf k∞ kψkL1 . ✷ Proof of uniqueness and real-valued property of η: For uniqueness in theorem 3.3, let us assume that there exists η1 , η2 ∈ L1 (R, ψ(λ)dλ) such that   Z ∞  (1)   1  (2) Tr φ(A + V ) − φ(A) − D φ(A) (V ) − D φ(A) (V, V ) = φ′′′ (λ)ηj (λ)dλ 2 −∞

for j = 1, 2. Then using corollary 3.4 we conclude that Z Z 1 Z s Z ∞Z ∞ h(λ) − h(µ) Tr [V E(dλ)V E(dµ) − V Eτ (dλ)V Eτ (dµ)] , f (λ)ψ(λ)ηj (λ)dλ = ds dτ λ−µ R 0 0 a a 24

for j = 1, 2 and for all f ∈ L∞ (R). Hence Z f (λ) ψ(λ) η(λ) dλ = 0 ∀ f

∈ L∞ (R),

(3.25)

R

where η(λ) ≡ η1 (λ) − η2 (λ) ∈ L1 (R, ψ(λ)dλ). Since (3.25) is true for all f ∈ L∞ (R), in particular it is true for all real valued f ∈ L∞ (R) i.e. Z f (λ) ψ(λ) η(λ) dλ = 0 ∀ real valued f ∈ L∞ (R). (3.26) R

Let η(λ) = ηRel (λ) + i ηImg (λ), where ηRel (λ) and ηImg (λ) are real valued L1 (R, ψ(λ)dλ)function. Hence from (3.26) , we conclude that Z Z f (λ) ψ(λ) ηRel (λ) dλ = 0 = f (λ) ψ(λ) ηImg (λ) dλ ∀ real valued f ∈ L∞ (R). R

R

(3.27) In particular if we consider f (λ) = sgn ηRel (λ), where sgn ηRel (λ) = 0 ∀ λ such that ηRel (λ) = 0 ; sgn ηRel (λ) = 1 ∀ λ such that ηRel (λ) > 0 ; sgn ηRel (λ) = −1 ∀ λ such that ηRel (λ) < 0 . Then f = sgn ηRel ∈ L∞ (R) and hence Z Z |ηRel (λ)| |ψ(λ)| dλ = sgn ηRel (λ) ηRel (λ) ψ(λ)dλ = 0, R

R

which implies that |ηRel (λ)| |ψ(λ)| = 0 a.e. and hence ηRel (λ) = 0 a.e.. Similarly by the same above argument we conclude that ηImg (λ) = 0 a.e. and hence η(λ) = 0 a.e.. Therefore η1 (λ) = η2 (λ)a.e.. Again, since the right hand side of (3.20) is real for all real valued f ∈ L∞ (R), by a similar argument as above, it follows that η is real valued. Acknowledgment: The authors would like to thank Council of Scientific and Industrial Research (CSIR), Government of India for a research and Bhatnagar Fellowship respectively.

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[5] Dykema, K. and Skripka, A.: Higher order spectral shift. J. Funct. Anal. 257 , 1092-1132 (2009) [6] Gesztesy, F. Pushnitski, A. and Simon, B. On the Kopleinko spectral shift function. I.Basics Z.Mat.Fiz.Anal.Geom.,4(1) 63-107 (2008). [7] Kato, T.: Perturbation Theory of Linear Operators, 2nd edn. Springer , New York (1976) [8] Koplienko, L. S.: Trace formula for nontrace-class perturbations. Sibirsk. Mat.Zh. 25(5), 6-21 (1984) (Russian).English Translation: Siberian Math.J.25(5) ,735-743 (1984) [9] Krein, M. G.: On certain new studies in the perturbation theory for self-adjoint operators. In: Gohberg, I.(ed.) Topics in Differential and Integral equations and Operator theory, OT 7 , pp.107-172. Birkhauser, Basel (1983) [10] Potapov, D. Skripka, A. and Sukochev, F.: arXiv:0912.3056v1.

Spectral shift function of higher order,

[11] Sinha, K. B. and Mohapatra, A. N.: Spectral shift function and trace formula. Proc.Indian Acad.Sci. (Math.Sci.),104(4), 819-853 (1994) [12] Sinha, K. B. and Mohapatra, M. N.: Spectral shift function and trace formula for unitariesa new proof. Integral Equ. Oper. Theory 24 , 285-297 (1996) [13] Voiculescu, D. On a Trace Formula of M.G.Krein. Operator Theory: Advances and Applications,Vol.24, pp.329-332. Birkhauser, Basel (1987)

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