Spectral gaps in Graphene antidot lattices

16 Pages·2017·0.62 MB·English

Spectral gaps in Graphene antidot lattices
Jean-Marie Barbaroux, Horia D. Cornean, Edgardo Stockmeyer
To cite this version:
Jean-Marie Barbaroux, Horia D. Cornean, Edgardo Stockmeyer. Spectral gaps in Graphene antidot
lattices. Integral Equations and Operator Theory, 2017, 89 (4), pp.631-646. ￿10.1007/s00020-017-
2411-9￿. ￿hal-01546056￿
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SPECTRAL GAPS IN GRAPHENE ANTIDOT LATTICES
JEAN-MARIE BARBAROUX, HORIA CORNEAN, AND EDGARDO STOCKMEYER
Abstract. We consider the gap creation problem in an antidot graphene lattice, i.e. a
sheetofgraphenewithperiodicallydistributedobstacles. Weproveseveralspectralresults
concerning the size of the gap and its dependence on different natural parameters related
to the antidot lattice.
1. Introduction
Graphene, a two-dimensional material made of carbon atoms arranged in a honeycomb
structure, has risen a lot of attention due to its many unique properties. Remarkably,
charge carriers close to the Fermi energy behave as massless Dirac fermions. This is due
to its energy band structure which exhibits two bands crossing at the Fermi level making
graphene a gapless semimetal [5]. Many efforts have been carried out for the possibility of
tuning an energy gap in graphene [6].
The main physical motivation of our work is related to the so-called antidot graphene
lattice[13], whichconsistsofaregularsheetofgraphenehavingaperiodicarrayofobstacles
well separated from each other. These obstacles can be thought, for instance, as actual
holes in the graphene layer [13]. More generally, substrate induced obstacles [9] or those
created by doping or by mechanical defects have also been considered in the literature (see
e.g.,[6] and references therein). It has been observed both experimentally and numerically
that such an array causes a band gap to open up around the Fermi level, turning graphene
from a semimetal into a gapped semiconductor (see e.g.,[3], [9], and [6]).
In [7] and [4] there are given several proposals concerning the modelling of this phe-
nomenon. In one of these proposals the authors replace the usual tight-binding lattice
model by a two-dimensional massless Dirac operator, while a hole is modeled with the
help of a periodic mass term. For a large mass term held fixed the authors numerically
analyse how the gap appearing near the zero energy depends on the natural parameters
of the antidot lattice, namely, the area occupied by one mass-insertion versus the area
of the super unit-cell which contains only one such hole. For holes with armchair type of
boundaries, this model is in very good agreement with tight-binding and density functional
ab-initio calculations [4] (see also [3, 11, 12]). Moreover, the Dirac operator with a mass
term varying in a superlattice has also been used to explain the gap appearing when a
layer of graphene is placed on substrate of hexagonal boron nitride [9] (see also [15] for the
inclusion of electronic interaction).
Date: May 16, 2017.
2010 Mathematics Subject Classification. Primary 81Q10; Secondary 46N50, 81Q37, 34L10, 47A10.
Key words and phrases. Dirac operator, graphene.
1
In this article we consider the Dirac model with a periodic mass term and we estimate
the size of the energy gap in terms of the strength and shape of the mass-insertion together
with the natural parameters of the antidot lattice. Let us now formulate more precisely
the problem we want to investigate.
1.1. Setting and main results. Let χ : R2 → R be a bounded function supported on
a compact subset S included in Ω := (−1/2,1/2]2 satisfying
(cid:90) (cid:90)
χ(x)dx := Φ (cid:62) 0 and |χ(x)|2dx = 1. (1.1)
Ω Ω
We use the standard notation for the Pauli matrices
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
0 1 0 −i 1 0
σ = , σ = , σ = .
1 1 0 2 i 0 3 0 −1
We define the massless free Dirac operator in L2(R2,C2) as
(cid:18) (cid:19)
1 0 −i∂ −∂
H := σ ·∇ = 1 2 . (1.2)
0 i −i∂ +∂ 0
1 2
Let α ∈ (0,1] be a dimensionless parameter and let L > 0 have dimensions of length.
In physical units, the operator describing a mass-periodically perturbed graphene sheet is
given by
(cid:88) (cid:16)x−γL(cid:17)
H(cid:101)(α,µ,L) = (cid:126)v H +σ µ χ , (1.3)
F 0 3
αL
γ∈Z2
where (cid:126) is Plank’s constant divided by 2π and v is the Fermi velocity in graphene. Here
F
µ (cid:62) 0 has dimensions of energy and represents the strength of the mass-insertion which is
LZ2-periodic. We note that in order for the continuum Dirac model to hold one needs L
to be much larger than the distance between the carbon atoms constituting graphene.
As it is well known, the spectrum of H covers the whole real line. Our main interest
0
in this paper is finding sufficient conditions that the function χ must satisfy in order to
create a gap around zero in the spectrum of H(cid:101) and to estimate its size in terms of α,µ,
and L. By making the scaling transformation x (cid:55)→ Lx one gets that
(cid:126)v (cid:16) µL(cid:17)
F
H(cid:101)(α,µ,L) = H α, . (1.4)
L (cid:126)v
F
Note that µL is a dimensionless parameter. Here, for β > 0, we define the operator in
(cid:126)vF
L2(R2,C2)
(cid:18) (cid:19)
(cid:88) x−γ
H(α, β) := H +σ β χ . (1.5)
0 3
α
γ∈Z2
This new operator is clearly periodic with respect to Z2 and (just like H(cid:101)) it is self-adjoint
on the first Sobolev space H1(R2,C2) (see [16]).
Given a self-adjoint operator T, we denote by ρ(T) its resolvent set. Here is the first
main result of our paper.
2
Theorem 1.1. Assume Φ (cid:54)= 0. Then there exist two constants C > 0 and δ ∈ (0,1) such
that for all α ∈ (0,1/2] and β > 0 obeying αβ < δ we have
(cid:2) (cid:3)
−α2β(Φ−Cαβ), α2β(Φ−Cαβ) ⊂ ρ(H(α,β)).
Remark. Let us comment on the consequences of this result regarding the energy gap,
E , for the family H(cid:101) defined in (1.3). Let us define the area of the supercell A := L2
g t
and A := α2L2 representing the area supporting one mass perturbation. In view of (1.4),
r
Theorem 1.1 states that for αµL small enough and for Φ in (1.1) positive
(cid:126)vF
A
E (cid:38) µΦα2 = µΦ r. (1.6)
g
A
t
Remarkably this estimate does not depend on the side L of the supercell. This is to be
contrasted with the regime of µ → ∞ considered in [13] and [4] where it was found that,
for α small enough (see e.g., Equation A.8 of [4]),
(cid:126)v (cid:114)A
F r
E ∼ . (1.7)
g
L A
t
This latter regime can be mathematically investigated using the Dirac operator with
infinite mass boundary conditions proposed in [2] (see [1] for its rigorous definition).
In the case Φ = 0 it is still possible to prove the existence of a gap opening at zero. The
next result needs some assumptions on χ in terms of its Fourier coefficients
(cid:90)
χˆ(m) = e−2iπx·mχ(x)dx, m ∈ Z2. (1.8)
Ω
Note that Φ = 0 means that χˆ(0,0) = 0. In terms of the operator H in (1.5), this time we
keep α = 1 and we make β smaller than certain constant times the L∞-norm of χ. Then
the gap can still survive but it scales with β3 instead of β. As a consequence, the gap for
H(cid:101) has the following behaviour, for µL small enough,
(cid:126)vF
(cid:18)µL(cid:19)2
E (cid:38) µ . (1.9)
g (cid:126)v
F
Theorem 1.2. Assume Φ = χˆ(0,0) = 0, and at the same time:
(cid:88) (cid:88) m·m(cid:48)
χˆ(m)χˆ(m(cid:48))χˆ(m−m(cid:48)) (cid:54)= 0. (1.10)
|m|2|m(cid:48)|2
m(cid:54)=0m(cid:48)(cid:54)=0
Then there exist two positive numerical constants β and C such that for every 0 < β <
0
β /(cid:107)χ(cid:107) we have
0 ∞
(cid:2) (cid:3)
−Cβ3, Cβ3 ⊂ ρ(H(1,β)).
3
Let us describe a particular class of potentials where assumption (1.10) holds true.
Assume that χ(x) is of the form
(cid:88)
χ(x) = 2cos(2πm·x), N (cid:29) 1.
N−10(cid:54)|m|(cid:54)N+10
By construction, all the Fourier coefficients χˆ(m) = χˆ(−m) equal either 1 or 0. The
non-zero coefficients are those for which m lies in an annulus with outer radius N + 10
and inner radius N − 10. When N becomes large enough, the triples of vectors m, m(cid:48)
and m−m(cid:48) for which the Fourier coefficients in (1.10) are simultaneously non-zero form
a triangle which “almost” coincides with an equilateral triangle with side-length equal to
N. Here “almost” means that the angle between m and m(cid:48) is close to π/3 when N is large
enough. Thus the scalar product m·m(cid:48) is positive whenever the Fourier coefficients are
non-zero (provided N is large enough) and the double sum in (1.10) is also positive.
In the rest of the paper we give the proof of the two theorems listed above.
2. Proof of Theorem 1.1
Throughout this work we use the notation
(cid:90)
(cid:16) (cid:17)1/p
(cid:107)f(cid:107) = |f(x)|pdx , p ∈ [1,∞).
p
Ω
Note that our conditions on χ imply:
0 < Φ (cid:54) (cid:107)χ(cid:107) (cid:54) (cid:107)χ(cid:107) = 1. (2.11)
1 2
2.1. Bloch-Floquet representation and proof of Theorem 1.1.
In this subsection we start by presenting the main strategy of the proof of Theorem 1.1.
It consists of a suitable application of the Feshbach inversion formula to the Bloch-Floquet
fiber of H(α,β). The main technical ingredients are Lemmas 2.2 and 2.3 whose proof can
be found in the next subsection. At the end of this subsection we present the proof of
Theorem 1.1.
Let S(R2,C) denote the Schwartz space of test functions. Consider the map
U˜ : S(R2,C) ⊂ L2(R2,C) −→ L2(Ω,L2(Ω,C)),
(cid:88)
(U˜Ψ)(x;k) := e2πik·(x+γ) Ψ(x+γ), x,k ∈ Ω.
γ∈Z2
It is well known (see [14]) that U˜ is an isometry in L2(R2,C) that can be extended to a
unitary operator. We denote its unitary extension by the same symbol. We define the
˜
Bloch-Floquet transform as U := U ⊗1 . Then we have that
C2
(cid:90) ⊕
UH(α,β)U∗ = h (α,β)dk, h (α,β) := (−i∇PBC −2πk)·σ +βχ (x)σ ,
k k x α 3
Ω
4
where each fiber Hamiltonian h (α,β) is defined in L2(Ω,C2). Here ∇PBC is the gradient
k x
operator with periodic boundary conditions and
χ (x) := χ(x/α).
α
The spectra of H(α,β) and h (α,β) are related through
k
(cid:91)
σ(H(α,β)) = σ(h (α,β)). (2.12)
k
k∈Ω
We will use the standard eigenbasis of ∇PBC given by
x
ψ (x) := e2πim·x, m ∈ Z2,x ∈ Ω,
m
which is periodic and satisfies
−i∇PBCψ = 2πmψ .
x m m
For m ∈ Z2 define the projections
P := |ψ (cid:105)(cid:104)ψ |⊗1 and Q := Id−P . (2.13)
m m m C2 0 0
Lemma 2.1. Let α ∈ (0,1) and β > 0. Then, for every k ∈ Ω and ψ ∈ P L2(Ω,C2), we
0
have that
(cid:107)P h (α,β)P ψ(cid:107) (cid:62) βα2Φ(cid:107)ψ(cid:107).
0 k 0
Proof. For every k ∈ Ω we have:
P h (α,β)P = (−2πσ ·k+α2βΦσ )P .
0 k 0 3 0
Using the anticommutation relations of the Pauli matrices we get for any ψ ∈ P L2(Ω,C2)
0
that
(cid:107)P h (α,β)P ψ(cid:107)2 = ((2πk)2 +(α2βΦ)2)(cid:107)ψ(cid:107)2 (cid:62) (α2βΦ)2(cid:107)ψ(cid:107)2.
0 k 0
(cid:3)
The previous lemma shows that h (α,β) has a spectral gap of order βα2 on the range
k
of P . In order to investigate whether that is still the case on the full Hilbert space we use
0
the Feshbach inversion formula (see Equations (6.1) and (6.2) in [10]). The latter states in
this case that z ∈ ρ(h (α,β)) if the Feshbach operator
k
F (z) := P (h (α,β)−z)P −β2P χ σ Q (Q (h (α,β)−z)Q )−1Q χ σ P ,
P0 0 k 0 0 α 3 0 0 k 0 0 α 3 0
is invertible on P L2(Ω,C2). Here we used that P h (α,β)Q = βP χ σ Q .
0 0 k 0 0 α 3 0
The next lemma shows that the inverse of Q (h (α,β) − z)Q is well defined on the
0 k 0
range of Q .
0
Lemma 2.2. There exists a constant δ ∈ (0,1) such that for all α ∈ (0,1/2) and β > 0
with αβ < δ, we have that Q (h (α,β) − z)Q is invertible on the range of Q , for any
0 k 0 0
z ∈ [−π/2,π/2] and k ∈ Ω.
5
The following lemma controls the second term of the Feshbach operator F (z)
P0
B (z) := β2P χ σ Q (Q (h (α,β)−z)Q )−1Q χ σ P .
P0 0 α 3 0 0 k 0 0 α 3 0
Lemma 2.3. There exist two constants δ ∈ (0,1) and C > 0 such that for all α ∈ (0,1/2)
and β > 0 with αβ < δ we have
(cid:107)B (z)ψ(cid:107) (cid:54) Cβ2α3(cid:107)ψ(cid:107),
P0
for any z ∈ [−π/2,π/2], k ∈ Ω, and ψ ∈ P L2(Ω,C2).
0
Having stated all the above ingredients we can proceed to the proof of our first main
result.
Proof of Theorem 1.1. In view of (2.12) it is enough to show the invertibility of the Fes-
hbach operator uniformly in k ∈ Ω. Using Lemmas 2.1 and 2.3 we get that for any
ψ ∈ P L2(Ω,C2)
0
(cid:107)F (z)ψ(cid:107) (cid:62) (cid:107)(P (h (α,β)−z)P )ψ(cid:107)−(cid:107)B (z)ψ(cid:107)
P0 0 k 0 P0
(cid:62) (βα2Φ−|z|−cα3β2)(cid:107)ψ(cid:107).
This concludes the proof by picking αβ so small that Φ > Cαβ. (cid:3)
2.2. Analysis of the Feshbach operator.
In this section we provide the proofs of Lemmas 2.2 and 2.3 from the previous section.
For that sake, let us first state some preliminary estimates. Let
h(0) := (−i∇PBC −2πk)·σ (2.14)
k x
Lemma 2.4. There exists a constant C > 0, independent of α ∈ (0,1), such that for all
k ∈ Ω
(cid:112)
(cid:107) |χ |P (cid:107) (cid:54) α (2.15)
α 0
(cid:13) (cid:13) α2
(cid:13)|χ |1/2(h(0) ±i)−1|χ |1/2(cid:13) (cid:54) C(α+ ) (2.16)
(cid:13) α k α (cid:13) 1−α
(cid:13) (cid:13) √ α
(cid:13)|χ |1/2(h(0) ±i)−1(cid:13) (cid:54) C( α+ ). (2.17)
(cid:13) α k (cid:13) 1−α
Proof. In order to show (2.15) we compute for f,g ∈ L2(Ω,C2) (see also (2.11)):
(cid:112) (cid:112)
|(cid:104)f, |χ |P g(cid:105)| (cid:54) |(cid:104)f, |χ |ψ (cid:105)||(cid:104)ψ ,g(cid:105)| (cid:54) (cid:107)χ (cid:107)1/2(cid:107)f(cid:107) (cid:107)g(cid:107) (cid:54) α(cid:107)f(cid:107) (cid:107)g(cid:107) .
α 0 α 0 0 α 1 2 2 2 2
We now turn to the proof of equations (2.16) and (2.17). Denote the integral kernels of
(H −i)−1 and (h(0)±i)−1 by (H −i)−1(x,x(cid:48)) and (h(0)±i)−1(x,x(cid:48)). In the proof we will
0 k 0 k
use the identity for x (cid:54)= x(cid:48)
(cid:88)
(h(0) −i)−1(x,x(cid:48)) = e2πik·(x+γ−x(cid:48))(H −i)−1(x+γ,x(cid:48)), x,x(cid:48) ∈ Ω. (2.18)
k 0
γ∈Z2
6
Let us first estimate the quadratic form of (h(0)−i)−1 for any φ,ψ ∈ L2(Ω,C2). According
k
to Lemma A.1 we have
(cid:88) (cid:90) e−|x−x(cid:48)−γ|
|(cid:104)φ,(h(0) −i)−1ψ(cid:105)| (cid:54) C |φ(x)| |ψ(x(cid:48))|dxdx(cid:48)
k |x−x(cid:48) −γ|
γ∈Z2 Ω×Ω
√ (cid:88) (cid:90) |φ(x)||ψ(x(cid:48))|
(cid:54) Ce 2 e−|γ| dxdx(cid:48),
|x−x(cid:48) −γ|
γ∈Z2 Ω×Ω
where in the last step we bound the exponential using that |x−x(cid:48)−γ| (cid:62) |γ|−|x−x(cid:48)| (cid:62)
√
|γ|− 2.
Now assume that the support of φ lies in Ω := (−α/2,α/2]2, i.e., x ∈ Ω above. Then
α α
it is easy to check that if γ (cid:54)= 0 then |x−x(cid:48) +γ| (cid:62) (1−α). Therefore, we find for such a
case that
√ (cid:16)(cid:90) |φ(x)||ψ(x(cid:48))| 1 (cid:88) (cid:17)
|(cid:104)φ,(h(0) −i)−1ψ(cid:105)| (cid:54) Ce 2 dxdx(cid:48) + ( e−|γ|)(cid:107)φ(cid:107) (cid:107)ψ(cid:107) .
k |x−x(cid:48)| 1−α 1 1
Ω×Ω
γ(cid:54)=0
Using the Hardy-Littlewood-Sobolev inequality for p = r = 4/3 (see Lieb and Loss Theo-
rem 4.3) we get
(cid:90) |φ(x)||ψ(x(cid:48))|
dxdx(cid:48) (cid:54) C(cid:107)φ(cid:107) (cid:107)ψ(cid:107) . (2.19)
|x−x(cid:48)| 4/3 4/3
Ω×Ω
Thus, denoting the universal constants by C we obtain, for any φ,ψ ∈ L2(Ω,C2) with
supp(φ) ⊂ Ω ,
α
|(cid:104)φ,(h(0) −i)−1ψ(cid:105)| (cid:54) C(cid:0)(cid:107)φ(cid:107) (cid:107)ψ(cid:107) +(1−α)−1(cid:107)φ(cid:107) (cid:107)ψ(cid:107) (cid:1). (2.20)
k 4/3 4/3 1 1
For some f ∈ L2(Ω,C2) we observe that (see Remark 2.11)
(cid:107)|χ |1/2f(cid:107) (cid:54) (cid:107)|χ |1/2(cid:107) (cid:107)f(cid:107) (cid:54) α(cid:107)f(cid:107) . (2.21)
α 1 α 2 2 2
Moreover, Ho¨lder’s inequality yields
(cid:107)|χ |1/2f(cid:107) (cid:54) (cid:0)(cid:107)|χ |2/3(cid:107) (cid:107)|f|4/3(cid:107) (cid:1)3/4
α 4/3 α 3 3/2
√ (2.22)
= (cid:107)χ (cid:107)1/2(cid:107)f(cid:107) (cid:54) α(cid:107)f(cid:107) .
α 2 2 2
In order to get the desired bounds we recall that the norm of an operator T is given by
(cid:107)T(cid:107) = sup |(cid:104)f,Tg(cid:105)|/((cid:107)f(cid:107)(cid:107)g(cid:107)). Hence we find (2.16) by using (2.20) with φ = |χ |1/2f
f,g(cid:54)=0 α
and ψ = |χ |1/2g together with the bounds (2.21) and (2.22). Analogously, we obtain
α
(2.17) using again (2.20) with φ = |χ |1/2f and ψ = g. (cid:3)
α
Lemma 2.5. For any f ∈ Q D(h(0)) we have
0 k
(cid:107)h(0)Q f(cid:107) (cid:62) π(cid:107)f(cid:107). (2.23)
k 0
7
Proof. For all m ∈ Z2 and k ∈ Ω, we have the identity
P h(0)P = 2πσ ·(m−k)P .
m k m m
Thus, we obtain for all f ∈ Q D(h(0)),
0 k
(cid:88) (cid:88)
(cid:107)h(0)Q f(cid:107)2 = (2π)2 (cid:107)σ ·(m−k)P f(cid:107)2 (cid:62) π2 (cid:107)P f(cid:107)2.
k 0 m m
m(cid:54)=0 m(cid:54)=0
(cid:3)
Let us introduce some notation: For a self-adjoint operator T and an orthogonal projec-
tion Q, we define
ρ (T) := {z ∈ C such that Q(T −z)Q : RanQ → RanQ is invertible}.
Q
We set
(cid:16) (cid:17)−1
R (z) := Q (h(0) −z)Q (cid:22) , z ∈ ρ (h(0)),
0 0 k 0 RanQ0 Q0 k
R(z) := (Q (h (α,β)−z)Q (cid:22) )−1, z ∈ ρ (h (α,β)).
0 k 0 RanQ0 Q0 k
Let U : L2(Ω,C2) → RanQ and W : RanQ → L2(Ω,C2) be defined as
0 0
(cid:112) (cid:16)(cid:112) (cid:17)
W := β |χ |σ Q ,
α 3 0
(cid:112) (cid:16) (cid:112) (cid:17)
U := βQ sgn(χ ) |χ | .
0 α α
Lemma 2.6. There exists C > 0, independent of α ∈ (0,1/2] and β > 0, such that for all
|z| (cid:54) π/2,
(cid:107)WR (z)U(cid:107) < Cαβ.
0
Proof. Note that due to Lemma 2.5 we have for |z| (cid:54) π/2
2
(cid:107)R (z)(cid:107) (cid:54) . (2.24)
0
π
Using the first resolvent identity
WR (z)U = WR (i)U +(z −i)WR (i)2U +(z −i)2WR (i)R (z)R (i)U. (2.25)
0 0 0 0 0 0
We shall estimate separately each term on the right hand side of (2.25).
Since h(0) commutes with the projections P we have
k m
R (i) = (h(0) −i)−1 −(P (h(0) −i)P (cid:22) )−1. (2.26)
0 k 0 k 0 RanP0
Thus, using (2.15), we obtain the estimate
(cid:112) (cid:112)
(cid:107)W(P (h(0) −i)P (cid:22) )−1U(cid:107) (cid:54) β(cid:107) |χ |P (cid:107)(cid:107)(h(0) −i)−1(cid:107)(cid:107) P |χ |(cid:107)
0 k 0 RanP0 α 0 k 0 α (2.27)
(cid:54) βα2.
8
The identity (2.26) together with the inequalities (2.16) and (2.27), imply that there are
universal constants c,C > 0, such that for |α| < 1/2
(cid:107)WR (i)U(cid:107) (cid:54) (cid:107)W(h(0) −i)−1U(cid:107)+(cid:107)W(P (h(0) −i)P (cid:22) )−1U(cid:107)
0 k 0 k 0 RanP0 (2.28)
(cid:54) cβα+βα2 (cid:54) Cβα.
This bounds the first term on the right hand side of (2.25). To estimate the second one
we first notice that
(h(0) −i)−2 = (Q (h(0) −i)Q (cid:22) )−2 +(P (h(0) −i)P (cid:22) )−2
k 0 k 0 RanQ0 0 k 0 RanP0
Therefore, using the same strategy as in (2.27), there exists C > 0 independent of α and
β such that
(cid:13) (cid:16) (cid:17) (cid:13)
(cid:107)(z −i)WR (i)2U(cid:107) = (cid:13)(z −i)W (h(0) −i)−2 −(P (h(0) −i)P (cid:22) )−2 U(cid:13)
0 (cid:13) k 0 k 0 RanP0 (cid:13)
(cid:54) (cid:107)(z −i)W(h(0) −i)−2U(cid:107)+|z −i|βα2
k (2.29)
(cid:112) (cid:112)
(cid:54) πβ(cid:107) |χ |(h(0) −i)−1(cid:107)(cid:107)(h(0) −i)−1 |χ |(cid:107)+πβα2
α k k α
(cid:54) Cβα,
where we used (2.17) in the last inequality.
Finally, we bound the last term on the right hand side of (2.25). Observe that from
Lemma 2.4 and inequality (2.15), we obtain that there exists c > 0 such that for all
α (cid:54) 1/2
√
(cid:112) (cid:112) (cid:112)
(cid:107) |χ |R (i)(cid:107) (cid:54) (cid:107) |χ |(h(0) −i)−1(cid:107)+(cid:107) |χ |P (P (h −i)P )−1(cid:107) (cid:54) c α.
α 0 α k α 0 0 k 0
Therefore, using (2.24) and (2.17)
(cid:112) (cid:112)
(cid:107)(z −i)2WR (i)R (z)R (i)U(cid:107) (cid:54) β|z −i|2(cid:107) |χ |R (i)(cid:107)(cid:107)R (z)(cid:107)(cid:107)R (i) |χ |(cid:107)
0 0 0 α 0 0 0 α
(cid:54) Cβ|z −i|2α.
In view of (2.25), the latter bound together with (2.28) and (2.29) concludes the proof. (cid:3)
Before stating the next lemma we define the set
S := {z ∈ ρ (h(0)) : (cid:107)WR (z)U(cid:107) < 1}.
Q0 k 0
Lemma 2.7. For any z ∈ S we have that z ∈ ρ (h (α,β)) and
Q0 k
R(z) = R (z)−R (z)U(1+WR (z)U)−1WR (z).
0 0 0 0
Proof. For z ∈ S we define for short
B(z) := R (z)−R (z)U(1+WR (z)U)−1WR (z).
0 0 0 0
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