c√ 2
ξ2H+D(ξ )2
H
1
2, ξ∈ dom(D), which means that A is D-bounded.
The equivalence of (ii) and (iii) follows from the fact that the operators
(±i + D)
1+ D2−1
2 and
1+ D21
2(±i + D)−1 are unitary. 2
Remark 24. It follows from the proof of Lemma 23 that
√1
2ADA
1+ D2−1
2 √2AD.
Observe that, according to Lemma 23 and Remark 24, a path t→ Dt is Γ -differentiable at the point t= 0 (as defined in Section 4) if and only if dom(D0)⊆ dom(Dt)in some neighbourhood of t= 0 and
tlim→0
Dt− D0
t − ˙D0
D0
= 0,
in other words, if and only if the path t→ Dt is differentiable with respect to the graph norm of the operator D0at the point t= 0. This observation further extends to CΓ1-paths.
6.2. The subspace L(D)
Recall that an operator A : dom(A)→ H is called symmetric if and only if
A(ξ ), η
= ξ, A(η)
, ξ, η∈ dom(A).
Let D : dom(D)→ H be a self adjoint linear operator and let L(D) be the real linear space consisting of all symmetric A : dom(A)→ H where dom(D) ⊆ dom(A) and such that12 A(1+ D2)−12 ∈ B(H).
Next, observe that, since A is symmetric and (1+ D2)−12 is self adjoint, we have
1+ D2−12
A(ξ ), η
= A(ξ ),
1+ D2−12 (η)
= ξ, A
1+ D2−12 (η)
. Consequently, we have the implication
A
1+ D2−12
∈ B(H) ⇒
1+ D2−12
A∈ B(H). (27)
More precisely, if the operator A(1+ D2)−12 is bounded, then the operator (1+ D2)−12Ais closable and its closure is also bounded.
Lemma 25. L(D) is a closed subspace of B(H).
Proof. We prove that the subspace L(D) is closed with respect to the weak operator topology.
Let Bn∈ L(D), n 1, and
wo- lim
n→∞Bn= B ∈ B(H).
This means that there is a sequence Anof symmetric operators such that
Bn= An
1+ D2−1
2 and dom(D)⊆ dom(An).
We need to show that B∈ L(D). Introduce the operator A : dom(A) → H by setting
dom(A)= dom(D) and A = B
1+ D21
2.
We clearly have that the operator A(1+ D2)−12 is closable and its closure coincides with B i.e.,
B= A
1+ D2−12 .
Consequently, we have only to verify that the operator A is symmetric. To this end, observe that
12 Here A(1+ D2)−12 ∈ B(H) means that the operator A(1 + D2)−12 is closable and the closure belongs to B(H).
A(ξ ), η
= B
1+ D21
2(ξ ), η
= lim
n→∞
Bn
1+ D212 (ξ ), η
= lim
n→∞
An(ξ ), η
= lim
n→∞
ξ, An(η)
= lim
n→∞
ξ, Bn
1+ D21
2(η)
= ξ, B
1+ D21
2(η)
= ξ, A(η)
, ξ, η∈ dom(A).
Thus, A is symmetric and therefore B∈ L(D). 2 For a closely related argument to the following see [29].
Lemma 26. Let Dj: dom(Dj)→ H, j = 0, 1 be a self adjoint linear operator and let B ∈ L(D0). If D1− D0 is D0-bounded and D1− D0D0 <1, then the operator Bθ = (1 + D21)−θ2B(1+ D20)θ2 is bounded and
Bθ c0B,
for some constant c0>0 and every 0 θ 1.
Proof. Let A : dom(A)→ H be a symmetric linear operator such that B= A
1+ D20
−1
2 ∈ B(H). (28)
In particular,
dom(D0)⊆ dom(A).
The operator A is D0-bounded (see Lemma 23). According to (26), the operator A is also D1 -bounded. This further means (using (27) and Lemma 23) that
1+ D12−1
2A∈ B(H). (29)
Let Ejn = EDj(−n, n) be the spectral projection of the operator Dj, j = 0, 1. The operator E1nAEn0is bounded since
En1AEn0= En1B 1+ D02
1
2En0. Let
Cθ,n=
1+ D12−θ
2En1BEn0
1+ D02θ
2.
We clearly have that
nlim→∞
Cθ,n(ξ ), η
=
Bθ(ξ ), η
, ξ ∈ dom(D0), η∈ dom(D1).
Thus, it is sufficient to show that the operators Cn,θ are uniformly bounded with respect to n.
This follows from (28), (29) and Lemma 27 below. 2
Lemma 27. Let A, Bj, j= 0, 1, be bounded linear operators. If Bj, j= 0, 1, are positive, then the operator B11−θAB0θ is bounded and
B11−θAB0θ B1A1−θAB0θ, 0 θ 1. (30)
Proof. The lemma is a straightforward application of the three lines lemma (see [7, Lem-ma 1.1.2]) to the holomorphic function
fξ,η(z)= B1A−1+zAB0−z
B11−zAB0z(ξ ), η
, ξ, η∈ H, considered in the strip S= {z ∈ C: 0 < z < 1}. 2
6.3. The proof of Theorems 21 and 22
The proof of Theorems 21 and 22 rests on the properties of the operator Tφ(D1, D0)where the function φ is given by
φ(λ, μ)=ϑ (λ)− ϑ(μ) λ− μ
1+ μ21
2. (31)
The difficulty about this operator is the fact that it is not bounded on B(H). Thus, a direct application of the methods of double operator integrals and harmonic analysis is not feasible.
Nevertheless, we shall show that this operator, when considered on the subspace L(D0), is bounded and possesses all the properties needed to prove Theorems 21 and 22.
Let Dj: dom(Dj)→ H, j = 0, 1, be a self adjoint linear operator such that D1− D0is D0 -bounded andD1− D0D012. In order to introduce the operator Tφ= Tφ(D1, D0): L(D0)→
B(H) (φ is given in (31)), let us consider another function
ψ (λ, μ)=
1+ λ21
4ϑ (λ)− ϑ(μ) λ− μ
1+ μ21
4. (32)
The operator Tψ= Tψ(D1, D0)is bounded on B(H) (Tψ is equal to the operator Tθwith θ=12 introduced in the proof of [37, Theorem 14]). Observe also that the bound of operator Tψ does not depend of the operators Dj, j= 0, 1.
The fact that the operator Tψ is bounded on B(H) and Lemma 26 imply that the mapping
B∈ L(D0)→ Tψ
1+ D12−1
4B
1+ D201
4
∈ B(H) (33)
is a bounded linear operator L(D0)→ B(H) whose norm depends only on the quantity
D1− D0D0. Furthermore, it is known (see [36], see also [35] for a more complete and de-tailed exposition) that the operators Tφand Tψ are bounded fromL2→ L2, whereL2⊆ B(H) is the Hilbert–Schmidt ideal and the following identity holds
Tφ(B)
1+ D02−1
4 =
1+ D21−1
4Tψ(B), B∈ L2.
The latter identity suggests that the mapping (33) is a (unique) bounded extension of the operator Tφ fromL2∩ L(D0)to the space L(D0). Motivated by this observation, we shall write Tφ= Tφ(D1, D0)for the mapping (33).
Proof of Theorem 21. Let Tφ,t = Tφ(Dt, D0)and let H= Tφ,0(G)where G= ˙D0(1+ D02)−12 (observe that the subspace L(D0)is closed in B(H) and therefore G ∈ L(D0)). It now follows from Theorem 17 that
When t → 0, the first term vanishes due to assumptions of the theorem and the fact that the operators Tφ,t are uniformly bounded. To finish the proof of the theorem, we need to justify that
tlim→0Tφ,t(G)− Tφ,0(G) =0. (34) Observing that Tψ,t are uniformly bounded, we see that it is sufficient to show
tlim→0Ct− C0 = 0 and lim
t→0Tψ,t(C0)− Tψ,0(C0) =0. (35) The first limit in (35) is due to the following identity
Ct− C0=
and the following estimate (see Lemma 28.(i) below)
1+ D2t−1
For the second limit in (35), observe that the operator Tψ,t is precisely the operator Tψ,t intro-duced in the proof of [37, Theorem 21]. Thus, the required limit follows from [37, Formula (6.8)]
and the estimate (see Lemma 28.(ii) below)
1+ Dt2
is
2 −
1+ D20is
2 c0(Dt− D0)(1+ D0)−12, |s| s0, (37) for some constant c0>0 which may depend on s0. The latter estimate is an improvement of [37, Formula (6.16)]. The proof of the theorem is finished. 2
Proof of Theorem 22. The proof is similar to the proof of Theorem 21. Indeed, letting Tφ,t,s= Tφ(Dt, Ds)(φ is given in (31)), we obtain
Ht− H0= Tφ,t,t(Gt)− Tφ,0,0(G0)
= Tφ,t,t(Gt)− Tφ,t,t(G0)+ Tφ,t,t(G0)− Tφ,t,0(G0)+ Tφ,t,0(G0)− Tφ,0,0(G0).
The first term vanishes since the operators Tφ,t,t are uniformly bounded and the assumption of the theorem; the last one does due to (34). To finish the proof, we need to show that
tlim→0Tφ,t,t(G0)− Tφ,t,0(G0) =0.
Let Tψ,t,s= Tψ(Dt, Ds), where the function ψ is given in (32) and let Ct,s=
1+ D2t
−14 G0
1+ D2s
14
∈ B(H).
It then follows from the definition of the operator Tφ,t,s that Tφ,t,t(G0)− Tφ,t,0(G0)
= Tψ,t,t(Ct,t)− Tψ,t,0(Ct,0)
= Tψ,t,0(Ct,0− C0,0)+ Tψ,t,t(Ct,t− C0,0)+ Tψ,t,t(C0,0)− Tψ,t,0(C0,0). (38) The first term vanishes due to the fact that the operators Tψ,t,s are uniformly bounded and (35).
For the last term in (38), observe that the operator Tψ,t,s is the operator ¯Tt,s introduced in the proof of [37, Theorem 21]. Therefore, the last term in (38) vanishes due to [37, Formula (6.11)]
and the estimate (37). Thus, to finish the proof of the theorem, we need only justify that
tlim→0Ct,t− C0,0 = 0. (39)
For the latter, observe that Ct,t− C0,0=
1+ D2t
−1
4G0 1+ Dt2
1
4 −
1+ D02−1
4G0
1+ D021
4
= Ct,t
1−
1+ D2t−1
4
1+ D021
4 +
1+ D2t−1
4
1+ D021
4 − 1 C0,0. It is now clear that (39) follows from (36) and the fact that the operators Ct,t are uniformly bounded. The proof of the theorem is finished. 2
Lemma 28. Let Dj: dom(Dj)→ H, j = 0, 1, be a linear self adjoint operator. If D1− D0 is Thus, it is sufficient to show that
G−112G
Suppose that η∈ Φ(B(H)). The required estimate then follows from Theorem 17 which guaran-tees the identity is justified by [37, Lemmas 7 and 9] and the following representation of the function η given in (40)
where the function f is given by
f (t )= (1 + t)−1
t14 + t−14−1
, t >0.
(ii) We keep the notations of the proof above. Let s0be fixed. Referring to Lemma 26 again, we need only show that
Gis1 − Gis0 c0G−112(D1− D0)G−
1 2
0 .
Let
ζ (λ1, λ0)= γ112γ1is− γ0is λ1− λ0
γ
1 2
0. If ζ ∈ Φ(B(H)), then we have the identity (see Theorem 17)
Gis1 − Gis0 = Tζ
G−
1 2
1 (D1− D0)G−
1 2
0
,
where Tζ= Tζ(D1, D0)and (ii) follows. Thus, we need to establish that ζ∈ Φ(B(H)). To this end, note that the latter function admits the representation
ζ (λ1, λ0)= γ1is2γ
is 2
0
λ1 γ1f
γ0 γ1
+λ0
γ0f γ1
γ0
,
where the function f is given by
f (t )= tis2 − t−is2 (1+ t)(t14− t−14)
.
This, together with [37, Lemmas 7 and 9], implies that ζ ∈ Φ(B(H)) (with the norm depending on s). The proof of the lemma is finished. 2
Acknowledgments
This research was supported by the Australian Research Council and the Hausdorff Institute for Mathematics. We also thank the referee and Matthias Lesch for valuable comments on an earlier draft.
Appendix A
Homotopical equivalence betweenF∗andF∗±1
We regard the setsF∗andF∗±1as topological spaces endowed with norm topology.
Theorem 29. The spaceF∗±1is a deformation retract of the spaceF∗i.e., there is a continuous mapping r :[0, 1] × F∗→ F∗such that
(i) r(0, F )= F , F ∈ F∗; (ii) r(1, F )∈ F∗±1, F∈ F∗; (iii) r(1, F )= F , F ∈ F∗±1.
Proof. Recall thatK is the two-sided ideal of all τ -compact operators of M and π is the homo-morphism
π:M → M/K.
Also recall that if F is a self adjoint τ -Fredholm operator and δF= π(F )−1−1, then, for every 0 δ < δF, the spectral projection EF(−δ, δ) is τ -finite (see Lemma 2) i.e.,
τ
EF(−δ, δ)
<+∞, 0 δ < δF.
Consider the intermediate spaceF∗±1⊆ F∗ ⊆ F∗of all τ -Fredholm operators F such that the projection EF(−δ, δ) is τ -finite, for every 0 δ < 1. Clearly, it is sufficient to show that F∗ is a deformation retract ofF∗and thatF∗±1is a deformation retract ofF∗ .
Observing that the function F→ π(F )−1 is continuous, we see that the deformation retract betweenF∗andF∗ is given by the mapping
r1(t, F )= F
1− t + tπ(F )−1−1∈ F∗, 0 t 1.
To construct the deformation retract betweenF∗ and F∗±1, let us consider the continuous function χ (t) which is constantly 1 for t 1, constantly −1 for t −1 and linear for −1 t 1.
The function χ is given by
χ (t )=1
2|t + 1| −1 2|t − 1|.
Observe that the mapping F → χ(F ) is continuous in the norm topology (see [8, Theo-rem X.2.1]). Observe also that χ (F )∈ F∗±1, for every F∈ F∗ . Indeed, let us fix F∈ F∗ and let F1= χ(F ). Clearly, F1 1. To see that 1 − F12is τ -compact, consider the points δn= 1 −1n, n= 1, 2, . . . , and the projections
En= EF(−δn+1,−δn] + EF[δn, δn+1).
Observe that, since F∈ F∗ , the projection Enis τ -finite, for every n= 1, 2, . . . . Noting that EF(−1, 1) = EF1(−1, 1) and
1− F12
EF1{±1} = 0, we obtain that
1− F12= 1− F12
EF(−1, 1) = 1− F12
∞
n=1
En=∞
n=1
1− F12
En∞
n=1
2 nEn.
The latter means that the operator 1− F12is τ -compact. Thus, we may define the deformation retract betweenF∗ andF∗±1by setting
r2(t, F )= (1 − t)F + tχ(F ) ∈ F∗ , 0 t 1.
The theorem is proved. 2
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