On the stability of the area law for the entanglement entropy of the Landau Hamiltonian
aa r X i v : . [ m a t h - ph ] F e b ON THE STABILITY OF THE AREA LAW FOR THE ENTANGLEMENTENTROPY OF THE LANDAU HAMILTONIAN
PAUL PFEIFFER
Abstract.
We consider the two-dimensional ideal Fermi gas subject to a magnetic field whichis perpendicular to the Euclidean plane R and whose strength B ( x ) at x ∈ R converges tosome B > k x k → ∞ . Furthermore, we allow for an electric potential V ε which vanishesat infinity. They define the single-particle Landau Hamiltonian of our Fermi gas (up to gaugefixing). Starting from the ground state of this Fermi gas with chemical potential µ ≥ B westudy the asymptotic growth of its bipartite entanglement entropy associated to L Λ as L → ∞ forsome fixed bounded region Λ ⊂ R . We show that its leading order in L does not depend on theperturbations B ε := B − B and V ε if they satisfy some mild decay assumptions. Our result holdsfor all α -R´enyi entropies α > /
3; for α ≤ /
3, we have to assume in addition some differentiabilityof the perturbations B ε and V ε . The case of a constant magnetic field B ε = 0 and with V ε = 0 wastreated recently for general µ by Leschke, Sobolev and Spitzer. Our result thus proves the stabilityof that area law under the same regularity assumptions on the boundary ∂ Λ. Contents
1. Introduction 1Acknowledgment 22. Notations and preliminaries 33. Setting and main result 44. The Ansatz for the proof of Theorem 3.3 75. On the Landau Hamiltonian 126. Proof of Theorem 4.7 20Appendix A. 28Appendix B. 30References 311.
Introduction
Bipartite entanglement entropy is an important quantity that measures correlations of particlesinside a given region with the particles outside that region. These non-trivial correlations are solelydue to the Fermi–Dirac statistics of the particles involved. In recent years there has considerableinterest and progress in quantifying these correlations. Mathematicians and physicists alike realizedfascinating connections between the large scale asymptotics of entanglement entropy and certainsemi-classical asymptotic formulas of traces of certain operators, mostly Toeplitz operators in thediscrete case and Wiener–Hopf operators in the continuous case.In the discrete setting, Jin and Korepin related the Fisher–Hartwig conjecture of Toeplitz matricesto the scaling of the entanglement entropy in the XY -chain in a transverse magnetic field in [7]. Morerelevant to our continuous setting here is the discovery of Gioev and Klich [5] that a conjecture byHarold Widom (proved by Alexander V. Sobolev [20]) gives the precise leading asymptotic growth ofthe bipartite entanglement entropy in ground states of the free Fermi gas. It displays a logarithmicallyenhanced area law of the order L d − ln( L ), where L is a scaling parameter, see below. In [8], this Date : 16th February 2021. was finally proved by Leschke, Sobolev and Spitzer. In [14], M¨uller and Schulte proved that this lawis stable under a perturbation by a compactly supported potential (in dimension d ≥ R with single-particle Hamiltonian H as in ourmodel is given by the (Fermi) spectral projection 1 ≤ µ ( H ), where µ ∈ R . The function 1 ≤ µ is theindicator function of the set ( −∞ , µ ] ⊂ R and the number µ is called the Fermi energy. Let h bethe von Neumann entropy function, see (3.3). For a given bounded region Λ ⊂ R we denote by 1 Λ the (multiplication operator associated to the) indicator function on Λ. Then we define the localentropy (or entanglement entropy) S (Λ) to be the (usual Hilbert space) trace of h applied to thespatially to Λ reduced Fermi projection, that is, S (Λ) := tr h (1 Λ ≤ µ ( H )1 Λ ) . (1.1)At positive temperature a definition of entanglement entropy or mutual information needs to beamended, see [9].For a fixed region Λ, it is generally hard or impossible to calculate the entropy. However, ifwe introduce a scaling parameter L > L Λ for L → ∞ , there are interesting results. They all assume some kindof regularity of the boundary ∂ Λ and assume the Hamiltonian H to be of a certain form. For H = −∇ + V , with some assumptions on V , there are results presented in [4, 8, 9, 13, 14, 16, 17].In this paper, we consider the Hamiltonian H = ( − i ∇ − A ) + V ε , which is a slight perturbation ofthe Landau Hamiltonian H for a constant magnetic field and no electric field, see (3.2) and (3.11).Entanglement entropy of the ground state of the latter Landau Hamiltonian (for the ground statewith chemical potential µ = B ) has been studied in [11, 18, 19] with some additional assumptionson the region Λ. The case of µ = B has been solved by Charles and Estienne in [3] and then forarbitrary µ by Leschke, Sobolev and Spitzer in [10], both under some regularity assumptions on theboundary ∂ Λ. Our main result is Corollary 3.6. It shows that the leading order asymptotic growth ofthe entanglement entropy does not change, if we add such a slight perturbation in both the magneticfield and the electric potential. Hence, we will not need to recalculate the value of the leading term,as we only estimate that this perturbation leads to an error term of smaller order in the scalingparameter L .Our proof is based on a statement by Aleksandrov and Peller in [1], which is Proposition 3.4 inthis paper. With the help of this and approximations of the R´enyi entropy functions h α (see (3.3)),we can reduce our result to some p -Schatten norm estimates, as we prove in Section 3.Proving these p -Schatten norm estimates relies on a result by Birman and Solomyak in [2], whichis our Proposition 4.6, and allows us to estimate the p -Schatten norms of operators with sufficientlydifferentiable kernels. To get a representation of the kernel of the spectral projection of the perturbedHamiltonian, we use the contour integral representation and the resolvent expansion. This hasrecently been done for perturbations of the free case ( H = −∇ + V ) by M¨uller and Schulte in[14], which inspired me to try this approach. In our case ( B > Acknowledgment
I would like to thank Wolfgang Spitzer for introducing me to this topic, proof reading multipleprevious versions, and generally providing advice.
N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 3 Notations and preliminaries
Let N = { , , , . . . } be the natural numbers and Z + be the positive integers.Let n be a positive integer and k be a natural number. For x ∈ R n or x ∈ C n , let k x k be its2-norm. The space of p -integrable, complex valued functions on R n is called L p ( R n ). The Sobolevspace W k,p ( R n ) is the space of complex valued p -integrable functions on R n , such that their first k distributional derivatives are p -integrable functions. We define C kb ( R n , C k ) as the subspace of C n ( R d , C k ), such that all derivatives of order 0 ≤ j ≤ n are bounded.For any non-empty set Λ ⊂ R n and any point x ∈ R n , we define the distance asdist( x, Λ) := inf y ∈ Λ k x − y k , (2.1)and for any r > r -neighbourhood of Λ as D r (Λ) := { y ∈ R n | dist( y, Λ) < r } . (2.2)Furthermore, 1 Λ : R n → { , } ⊂ R is the indicator function of Λ, Λ ∁ := R n \ Λ is the complementof Λ, and if Λ is measurable, let | Λ | be its n -dimensional Lebesgue measure, and if it has Lipschitzboundary ∂ Λ, let | ∂ Λ | be the ( n − ∂ Λ.For any x ∈ R n , we set D r ( x ) = D r ( { x } ). For any x ∈ C n , j ∈ Z + , we inductively define x ⊗ j ∈ ( C n ) ⊗ j ∼ = C n j by setting x ⊗ := x ∈ ( C n ) ⊗ =: C n and x ⊗ ( j +1) = x ⊗ j ⊗ x ∈ ( C n ) ⊗ j ⊗ C n =:( C n ) ⊗ ( j +1) ∼ = C n j +1 . Every appearance of · ⊗ j refers to this tensor product.By J we denote the matrix J := (cid:18) − (cid:19) . (2.3)For a complex number ζ , let ℜ ζ be its real part.For a multiplication operator with a function G : R → C n , we use a slight abuse of notation andcall it G as well. We will regard G as a function, if and only if it is written as G ( · ), where · is eitherleft as a place holder or is a point in R . This is relevant to decide, whether we are applying anoperator to the underlying function or taking the composition of a multiplication operator and anyother operator. C will always refer to a generic constant, that may depend on some, but never on all variables. F will be used similarly, but the dependency on one complex variable will be important, which is whywe write F as a function of that variable. Both may change from line to line.For any compact operator S and any p ∈ R + , we define the p -Schatten von Neumann (quasi-)normby the expression k S k pp = X n ∈ Z + s n ( S ) p , (2.4)where ( s n ( S )) n ∈ Z + is the decreasing sequence of singular values of S counted with multiplicity. Theoperator norm of S is written as k S k ∞ .We recall some properties of the p -Schatten von Neumann (quasi-) norms. In the following, wewill refer to them as p -Schatten norms. Proposition 2.1.
Let < p ≤ q ≤ ∞ and let S, T be operators on a Hilbert space. The p -Schattennorm satisfies the properties Monotonicity I: k S k p ≥ k S k q . Monotonicity II: If S ≥ T ≥ , then k S k p ≥ k T k p . Triangle inequality: If p ≥ , then k S + T k p ≤ k S k p + k T k p . p -triangle inequality: If p ≤ , then k S + T k pp ≤ k S k pp + k T k pp . Powers: If S ≥ , then k S p k qq = k S q k pp . Square: k S k p = k S ∗ S k p/ , where S ∗ denotes the adjoint of S . Adjoint: k S ∗ k p = k S k p . H¨older I:
Let r = p + q with r > . Then k ST k r ≤ k S k p k T k q . PAUL PFEIFFER
H¨older II:
Let r = αp + − αq with < α < and r > . Then k S k r ≤ k S k αp k S k − αq . Hilbert Schmidt kernel: If T : L ( R d ) → L ( R d ) has an integral kernel t , which is squareintegrable, then k T k = k t k L ( R d d ) . Orthogonality: If ST ∗ = 0 or S ∗ T = 0 , then k S k p ≤ k S + T k p . Most of these have for example been proven by McCarthy in [12]. We will now briefly prove theremaining ones.
Proof. “Monotonicity II” follows, as the inequality holds for the ordered sequence of singular values.“H¨older II” is an application of “H¨older I” with the operators | S | α and | S | − α and the properties“Square” and “Powers”. “Hilbert–Schmidt kernel” can be seen as a corollary of Lemma 2.2 in [12].“Orthogonality” is based on the observation, that if S ∗ T = 0, we have ( S + T ) ∗ ( S + T ) = S ∗ S + T ∗ T ,“Monotonicity II”, and “Adjoint” to replace the condition S ∗ T = 0 by the non-equivalent condition ST ∗ = 0. (cid:3) Definition 2.2.
We say a densely defined operator T on L ( R ) has the integral kernel t : R × R → C , if for any f ∈ C c ( R ) , the identity ( T f )( x ) = Z R t ( x, y ) f ( y ) dy (2.5) holds. We say, that t is nice , or respectively, that T is a nice integral operator, if for any fixed x ,the functions t ( x, · ) and t ( · , x ) are in L ( R ) with a norm bounded independently of x . In this case,we define iker T ( x, y ) := t ( x, y ) . (2.6) Lemma 2.3.
Let
S, T be nice integral operators on L ( R ) with integral kernels s, t . Let x, z ∈ R .Then we have the identities iker( S + T )( x, z ) =( s + t )( x, z ) , (2.7)iker( ST )( x, z ) = Z R s ( x, y ) t ( y, z ) dy. (2.8) In particular, S + T and ST are nice integral operators. The first statement is trivial and the second follows by Fubini to interchange the integral over y with the one over z , for any test function f ∈ C c ( R ).3. Setting and main result
We introduce the Landau Hamilton operator H with a constant magnetic field B >
0, definedon (a suitable subspace of) L ( R ), with magnetic gauge A given for any x ∈ R by A ( x ) := B Jx, (3.1) H :=( − i ∇ − A ) . (3.2)The spectrum of H , σ ( H ), equals B (2 N + 1). Let P l be the projection onto the eigenspace witheigenvalue B (2 l + 1) for l ∈ N .Furthermore, for any α >
0, we introduce the α -R´enyi entropy functions h α : [0 , → [0 , ln(2)], h α ( x ) := ( − α ln ( x α + (1 − x ) α ) for α = 1 , − x ln x − (1 − x ) ln(1 − x ) for α = 1 , (3.3)for x ∈ (0 ,
1) and h α (0) = h α (1) = 0. Throughout this paper, let Λ ⊂ R be a bounded open setwith Lipschitz boundary.Let µ ∈ R \ B (2 N + 1). We define 1 ≤ µ ( H ) as the spectral projection associated to H and µ .We are interested in how the leading order asymptotic expansion of the local entropy, S α ( L Λ) := tr h α (1 L Λ ≤ µ ( H )1 L Λ ) , (3.4) N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 5 as L → ∞ changes under slight perturbations of H . The trace is defined as the usual Hilbert spacetrace of trace class operators on L ( R ). This quantity is the local entropy or entanglement entropyof the ground state restricted to L Λ. Under the assumption that Λ has C boundary, the leadingterm of order L for the operator H = H has been calculated by Leschke, Sobolev and Spitzer in[10]. This allows us to focus on bounding the error term that arises, as we introduce a perturbationto H . Our main result is Corollary 3.6 and relies on the exact calculations of the leading term for H = H , see [10], and the estimates we will prove in this paper.The following condition is needed to state our main results and a lot of results along the way.Throughout this paper, we fix 0 < ε < Definition 3.1.
Let λ, C ∈ (0 , ∞ ) . We say a function f : R n → C has ( λ, C ) decay, if it ismeasurable and for all x ∈ R n , we have k f ( x ) k ≤ C (1 + k x k ) λ . (3.5) Let γ ≥ be a natural number. We call a magnetic field B ε : R → R and a potential V ε : R → R ( γ, C d ) tame, if all directional derivatives of order ≤ m ≤ γ − of B ε have (1 + ε, C d ) decay andthe same differentials of V ε have ( ε, C d ) decay. We say that they are (0 , C d ) tame, if they are (1 , C d ) tame.Remark. All the following estimates will depend on B ε , V ε only through ε, γ, C d . Maybe somewhatcounterintuitively, small values of ε correspond to slowly decaying B ε , V ε .To define the perturbed Hamiltonian H , we need to choose a gauge A ε of the magnetic field B ε .We choose the convolution, which is given by A ε ( x ) := (cid:18) B ε ∗ J · π k·k (cid:19) ( x ) = Z R B ε ( x − y ) Jy π k y k dy (3.6)for any x ∈ R . Its relevant properties are summed up in the following Lemma. Lemma 3.2.
For any x ∈ R , m , m ∈ N with m + m ≤ γ − , the gauge A ε satisfies ∇ x × A ε ( x ) = B ε ( x ) , (3.7) ∇ x · A ε ( x ) = 0 , (3.8) k ∂ m x ∂ m x A ε ( x ) k ≤ C (1 + k x k ) ε . (3.9) Remark.
A gauge satisfying (3.8) is commonly referred to as a Coulomb gauge. The restriction to ε < ε > k x k ) − decayin A ε .The proof can be found in Appendix B.Now we define the perturbed gauge A and the perturbed Hamiltonian H by A := A − A ε , (3.10) H := ( − i ∇ − A ) + V ε . (3.11)As we can see, this gauge corresponds to the magnetic field B − B ε , that is, ∇ x × A ( x ) = B − B ε ( x ).The operator H is self-adjoint and its domain agrees with the domain of H , which we will see inCorollary 4.2.We need the following p -Schatten quasi norm estimate, which will be proven in the next section. Theorem 3.3.
Let l ∈ N , γ ∈ Z + . Let B ε , V ε be ( γ, C d ) tame and let ≥ p > γ +2 . Let a, b ∈ R \ B (2 N + 1) with a < b . Then we have the estimates (cid:13)(cid:13) L Λ [ a,b ] ( H )1 L Λ ∁ (cid:13)(cid:13) pp ≤ CL, (3.12) (cid:13)(cid:13) L Λ (cid:0) [ a,b ] ( H ) − [ a,b ] ( H ) (cid:1) L Λ ∁ (cid:13)(cid:13) pp ≤ CL − pε . (3.13) The constants C depend on γ, C d , a, b, Λ , p, ε . PAUL PFEIFFER
Finally, we need the following statement due to Aleksandrov and Peller, which is a Corollary ofTheorem 5.11 in [1] and the inclusion C ∞ c ( R ) ⊂ B ∞ , ( R ), where the latter refers to the Besov spaceas used by Aleksandrov and Peller. Proposition 3.4 (based on Theorem 5.11 in [1]) . Let f ∈ C ∞ c ( R ) . Then there is a constant C < ∞ ,such that for any self-adjoint bounded operators A, B , such that A − B is trace class, we have theestimate k f ( A ) − f ( B ) k ≤ C k A − B k . (3.14)Now we can prove the main results of this paper. Theorem 3.5.
Let α > and choose β = min(0 . , α ) . Define γ as the smallest positive integer,such that γ > β − . Let B ε , V ε be ( γ, C d ) tame. Let a, b ∈ R \ B (2 N + 1) , a < b and I := [ a, b ] .Then we have tr ( h α (1 L Λ I ( H )1 L Λ ) − h α (1 L Λ I ( H )1 L Λ )) = o ( L ) , (3.15) as L → ∞ .Remark. The choice of β = 0 . α ≥ . γ , namely 1. For α > ,we can get away with a non-differentiable B ε , V ε .The assumption that a, b B (2 N + 1) cannot be dropped, as the following counter exampleillustrates. Let B ε = 0 , a = 0 and b = B . By Corollary 4.2, the spectrum of H has an accumulationpoint at B . If we assume V ε > H are strictly larger than B and hence 1 I ( H ) = 0. But Theorem 8 in [10], which we will elaborate on shortly, states, that theleading order asymptotic expansion of tr h α (1 L Λ I ( H )1 L Λ ) for large L is of order O ( L ) and does notvanish. On the other hand, if we assume that V ε < B , B + δ ] in the spectrum of H . Hence, we can move b to B + δ without changing the operators.Now we can apply our Theorem 3.5. Hence under our general assumptions, it is possible to get bothone-sided limits, when b = B . We expect similar results, whenever a or b is in the spectrum of H .It is, however, a little more complicated to see, whether the leading order expansion for H changes,when we add or remove a single Landau level from the interval I .The following corollary is our main result. It combines Theorem 8 in [10], which can be stated asthe corollary for the case B ε = V ε = 0, with our Theorem 3.5. Corollary 3.6.
Let α > and choose β = min( α, . . Define γ as the smallest positive integer,such that γ > β − . Let B ε , V ε be ( γ, C d ) tame. Let µ σ ( H ) and define ν as the largest integer,such that B (2 ν + 1) < µ . Assume that the boundary ∂ Λ is C -smooth. Then S α ( L Λ) = tr( h α (1 L Λ ≤ µ ( H )1 L Λ ) = L p B | ∂ Λ | M ≤ ν ( h α ) + o ( L ) , (3.16) as L → ∞ with < M ≤ ν ( h α ) < ∞ as described in [10]. In the case µ < B , the projection is finite dimensional and the entropy has an order at most O (1) in L as L → ∞ . Proof of Theorem 3.5.
We define the function g α : [0 , → [0 , ln(2)] by the identity g α (4 x (1 − x )) = h α ( x ) . (3.17)The symmetry of h α guarantees the existence of g α . We have g α ( t ) = h α (cid:18) − √ − t (cid:19) . (3.18)Let ε >
0. We choose a smooth cut-off function ϕ : [0 , → [0 ,
1] with ϕ ( x ) = 1, if x ≤ ε , and ϕ ( x ) = 0 , if x ≥ ε . Now we write g α ( t ) = (1 − ϕ ( t )) g α ( t ) + ϕ ( t ) g α ( t ) . (3.19) N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 7
The advantage of this decomposition is that the first summand is smooth, and the second summandis small. The second summand can be bounded using the fact, that h α is β -H¨older continuous on[0 ,
1] and smooth on (0 , h α is symmetric around t = and analytic on (0 , t − ). Thus, we see that g α is analytic at t = 1. Hence g α ∈ C ∞ ((0 , β -H¨older continuous on [0 , β = min( α, . β ′ < β ≤ , such that γ > β ′ −
2. Hence, we have ϕ ( t ) g α ( t ) ≤ Cε β − β ′ t β ′ . (3.20)We define P, P ′ as the spectral projections, P :=1 I ( H ) , (3.21) P ′ :=1 I ( H ) . (3.22)We observe h α (1 L Λ P ( ′ ) L Λ ) = g α (4 | L Λ ∁ P ( ′ ) L Λ | ) . (3.23)We can now apply Proposition 3.4. Thus, (cid:13)(cid:13) ((1 − ϕ ) g α ) (cid:0) | L Λ ∁ P ′ L Λ | (cid:1) − ((1 − ϕ ) g α ) (cid:0) | L Λ ∁ P L Λ | (cid:1)(cid:13)(cid:13) (3.24) ≤ C (cid:13)(cid:13) | L Λ ∁ P ′ L Λ | − | L Λ ∁ P L Λ | (cid:13)(cid:13) (3.25) ≤ C k L Λ ∁ ( P ′ − P )1 L Λ k (3.26) ≤ CL − ε . (3.27)Note that the last constant C depends on ε , but not on L . In the second step we used the identity | A | − | B | = A ∗ ( A − B ) + ( A ∗ − B ∗ ) B . In the last step, we used Theorem 3.3 with p = 1.We can also apply Theorem 3.3 for the remaining term, after using (3.20), 1 ≥ β ′ > γ +2 andthat H = H is admissible for Theorem 3.3. (cid:13)(cid:13)(cid:13) ( ϕg α ) (cid:16) | L Λ ∁ P ( ′ ) L Λ | (cid:17)(cid:13)(cid:13)(cid:13) ≤ Cε β − β ′ (cid:13)(cid:13)(cid:13) | L Λ ∁ P ( ′ ) L Λ | (cid:13)(cid:13)(cid:13) β ′ β ′ ≤ Cε β − β ′ L. (3.28)Hence, | tr h α (1 L Λ P ′ L Λ ) − tr h α (1 L Λ P L Λ ) | ≤ C ( ε ) L − ε + Cε β − β ′ L. (3.29)Note that the first constant C ( ε ) depends on ε while the second one does not. This term is in o ( L ),as for any ε > L large enough to let the first term be less than ε L . This provesthat the leading term expansion of the α -R´enyi entropy for the perturbed Landau Hamiltonian H agrees with the main term in the same expansion for the Landau Hamiltonian H . This finishes theproof. (cid:3) Remark.
We can actually pick ε dependent on L , which does lead to a smaller error term, if webound the constant C ( ε ) more precisely. This does however not lead to an improved error term inCorollary 3.6, as the known error term for the constant magnetic field is too large. Hence, I did notinclude the details here.4. The Ansatz for the proof of Theorem 3.3
The goal of this section is to explain how to prove Theorem 3.3 and, to reduce it to two moretechnical statements. The general approach has been inspired by [14].We define H ε := H − H , (4.1)where H and H were defined in (3.11) and (3.2).We expand H ε as H ε = H − H (4.2) PAUL PFEIFFER = ( − i ∇ − A ) − ( − i ∇ − A ) + V ε (4.3)= ( A − A ) · ( − i ∇ − A + A − A ) + ( − i ∇ − A ) · ( A − A ) + V ε (4.4)= 2 A ε · ( − i ∇ − A ) + A ε + V ε . (4.5)We used (3.8), which is equivalent to ∇ · A ε = A ε · ∇ , in the last step. We now introduce the pseudopotential W ε := A ε + V ε . (4.6)We introduce a few more operators. Let I ⊂ N be cofinite, ζ ∈ C and ζ B (2 I + 1). Then wedefine the bounded, self-adjoint operator M I,ζ := X l ∈ I P l B (2 l + 1) − ζ . (4.7)For ζ σ ( H ), we have the identity M N ,ζ = 1 H − ζ . (4.8)This resolvent operator has been studied before, in particular we also need the special case T l := M N \{ l } ,B (2 l +1) = X k = l P k B ( k − l ) . (4.9)Hence it is more convenient to deal with the operator M I,ζ in this generality.We define n as the smallest integer such that n > ε . (4.10)The following lemma will be proved in a Section 5 after some preparations. Lemma 4.1.
Let B ε and V ε be (1 , C d ) tame. Then for any I ⊂ N cofinite and any ζ ∈ C \ B (2 I +1) ,the operator H ε M I,ζ is in the n -Schatten class, and the n -Schatten norm is in L ∞ loc ( C \ σ ( H )) as a function of ζ . The upper bound for the norm depends on B . As p -Schatten class operators are compact, we now know that H ε is H compact. This impliesthe corollary Corollary 4.2.
The essential spectrum of H agrees with the essential spectrum of H which is B (2 N + 1) .Remark. The statement is also true if V = 0 and B is smooth and converges to B as k x k → ∞ (at any rate), see [6]. They state smoothness of B as a condition, but I think it is not required.However, their algebraic proof does not imply that the eigenspaces of H and H are at all related.As σ ( H ) is discrete, this implies, that σ ( H ) = σ p ( H ) and that the continuous part of the spectrumof H vanishes. We continue with the Riesz integral representation. Fact 4.3.
For any path Γ in C that intersects R in exactly two points λ < λ , does not intersect σ ( H ) ⊂ R and has winding number +1 around ( λ + λ ) / , we have the identity − πi Z Γ dζH − ζ = 1 λ Corollary 4.4. For any n ∈ Z + , ζ σ ( H ) ∪ σ ( H ) , we have H − ζ = n − X k =0 ( − k H − ζ (cid:18) H ε H − ζ (cid:19) k + (cid:18) H − ζ H ε (cid:19) n H − ζ (cid:18) H ε H − ζ (cid:19) n , (4.14) where H ε = H − H , as in (4.1) . With the use of Lemma 4.1 we will be able to bound the final term. For the summands inCorollary 4.4 except the last summand, we can resolve the path integral over some paths. Lemma 4.5. Let l, k ∈ N and Γ be the path along the circle ∂D B ( B (2 l + 1)) that rotates in positivedirection. Then we have − πi Z Γ H − ζ (cid:18) H ε H − ζ (cid:19) k dζ = k X m =0 ( T l H ε ) m P l ( H ε T l ) k − m , (4.15) where H ε = H − H , as in (4.1) .Proof. Let N > l and either I = N and ζ ∈ Γ or I = N \ { l } and ζ = B (2 l + 1). We introduce P ≤ N := P n ≤ N P n and P >N := 1 − P ≤ N . We continue with the identity P ≤ N M I,ζ = X j ∈ I,j ≤ N P j B (2 j + 1) − ζ . (4.16)There is a constant C , independent of N and ζ , such that the estimate k P >N M I,ζ k ∞ ≤ CN holds.Furthermore, by Lemma 4.1 and as the 4 n -Schatten norm is an upper bound for the operatornorm, we have the estimate k H ε M I,ζ k ∞ < C with a constant C independent of N and ζ . We usethe telescope sum b ( ab ) k − c ( ac ) k = P kk ′ =0 ( ba ) k ′ ( b − c )( ac ) k − k ′ , which holds in any ring, and thetriangle inequality to get (cid:13)(cid:13)(cid:13) M I,ζ ( H ε M I,ζ ) k − P ≤ N M I,ζ ( H ε P ≤ N M I,ζ ) k (cid:13)(cid:13)(cid:13) ∞ (4.17) ≤ k X k ′ =0 (cid:13)(cid:13)(cid:13) ( M I,ζ H ε ) k ′ P >N M I,ζ ( H ε P ≤ N M I,ζ ) k − k ′ (cid:13)(cid:13)(cid:13) ∞ ≤ CN , (4.18)where C is independent of N and ζ . The second step relies on the submultiplicativity of the norm,and the identity M I,ζ P ≤ N = P ≤ N M I,ζ . Thus, we have − πi Z Γ H − ζ (cid:18) H ε H − ζ (cid:19) k dζ (4.19)= − πi lim N →∞ Z Γ P ≤ N M N ,ζ ( H ε P ≤ N M N ,ζ ) k dζ (4.20)= − πi lim N →∞ Z Γ X σ ∈{ ,...,N } k +1 P σ Q kj =1 H ε P σ j Q kj =0 ( B (2 σ j + 1) − ζ ) dζ (4.21)= lim N →∞ X σ ∈{ ,...,N } k +1 P σ k Y j =1 H ε P σ j (Q σ j = l B ( σ j − l ) if { j | σ j = l } = 10 else (4.22)= lim N →∞ k X m =0 ( P ≤ N T l H ε ) m P l ( H ε P ≤ N T l ) k − m (4.23)= k X m =0 ( T l H ε ) m P l ( H ε T l ) k − m . (4.24) In the first step, we used that (4.18) holds uniformly in ζ ∈ Γ for I = N . In the second step, weinserted (4.16) k +1 times and multiplied out all terms in order to get a finite sum. We then exchangedthis finite sum with the complex path integral and resolvent this complex-valued integral. The fourthstep uses (4.16) in reverse. The final step follows by (4.18) for I = N \ { l } and ζ = B (2 l + 1). Thisfinishes the proof. (cid:3) We intend to use an estimate on the singular values due to Birman and Solomyak. It is based onProposition 2.1 in [2]. We will use the case p = 2 , a = γ + ν > , X = [0 , , Y = D ∁ R (0) ⊂ R andwe will use the Lebesgue measures on both sides. Now we can write the statement as Proposition 4.6. Let T : L ([0 , ) → L ( D ∁ R (0)) be an operator with integral kernel t . Let γ ∈ N , γ ≥ and < ν < . We define N ( γ, ν, T ) := Z D ∁ R (0) dy Z [0 , dx ′ Z [0 , dx ′′ (cid:13)(cid:13) ∇ ⊗ γx ′ t ( x ′ , y ) − ∇ ⊗ γx ′′ t ( x ′′ , y ) (cid:13)(cid:13) k x ′ − x ′′ k ν (4.25)+ Z D ∁ R (0) dy Z [0 , | t ( x, y ) | dx. (4.26) Then there is a constant C , depending only on γ and ν , such that the singular values of T, s n ( T ) , n ∈ Z + satisfy the upper bounds s n ( T ) ≤ Cn − γ + ν N ( γ, ν, T ) . (4.27) Remark. For any p > / (1 + γ + ν ), we have k T k p ≤ C N ( γ, ν, T ) , (4.28)where C depends on p, γ, ν , but not on T or R .With this proposition and the integral kernel estimates we still have to establish, we will provethe following theorem in Section 6. Theorem 4.7. Let k, l ∈ N , γ ∈ Z + with k ≥ m . Let B ε , V ε be ( γ, C d ) tame and let ≥ p > γ +2 .Then there is a constant C > and a λ > , such that for any R ≥ , we have the upper bound forany x ∈ R (cid:13)(cid:13)(cid:13) [0 , + x ( T l H ε ) m P l ( H ε T l ) k − m D ∁ R ( x ) (cid:13)(cid:13)(cid:13) p ≤ C exp (cid:0) − λR (cid:1) (1 + k x k ) kε . (4.29) The constant C depends on B , l, k, m, γ, p, ε, C d , but is independent of R and x .Remark. For k = 0, this is Lemma 12 in [10].We will now follow Theorem 13 in [10]. But we go a slightly different direction with the proof . Theorem 4.8. Let k, l ∈ N , γ ∈ Z + with k ≥ m , let B ε , V ε be ( γ, C d ) tame and let ≥ p > γ +2 .Then for any L > we have (cid:13)(cid:13) L Λ ( T l H ε ) m P l ( H ε T l ) k − m L Λ ∁ (cid:13)(cid:13) pp ≤ CL − pkε . (4.30) The constant C depends on Λ , B , l, k, m, γ, p, ε, C d .Proof. We define T := ( T l H ε ) m P l ( H ε T l ) k − m . (4.31)We choose an h ∈ [0 , . We will now use the p -Schatten norm property we called orthogonalityin the first and forth step, and the p -triangle inequality in the second step. Hence, k L Λ T L Λ ∁ k pp (4.32) We replace a sum by an integral. N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 11 ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X z ∈ Z ,z + h ∈ D √ ( L Λ) [0 , + z + h T L Λ ∁ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pp (4.33) ≤ X z ∈ Z ,z + h ∈ D √ ( L Λ) (cid:13)(cid:13) [0 , + z + h T L Λ ∁ (cid:13)(cid:13) pp (4.34) ≤ X z ∈ Z ,z + h ∈ D √ ( L Λ) (cid:13)(cid:13) [0 , + z + h T L Λ ∁ (cid:13)(cid:13) pp (4.35) ≤ X z ∈ Z ,z + h ∈ D √ ( L Λ) (cid:13)(cid:13)(cid:13)(cid:13) [0 , + z + h T D ∁ dist( z + h ,L Λ ∁ ) ( z + h ) (cid:13)(cid:13)(cid:13)(cid:13) pp (4.36) ≤ X z ∈ Z ,z + h ∈ D √ ( L Λ) C exp (cid:16) − pλ dist( z + h , L Λ ∁ ) (cid:17) (1 + k z + h k ) pkε . (4.37)In the third step we used, that L bounded operators do not care about 0-sets. The last step followsby Theorem 4.7. The constant C is independent of z, h . Now we can integrate this upper boundover h ∈ [0 , . This integral can be resolved by Lemma A.2. Hence, we have k L Λ T L Λ ∁ k pp ≤ Z [0 , dh X z ∈ Z ,z + h ∈ D √ ( L Λ) C exp (cid:16) − pλ dist( z + h , L Λ ∁ ) (cid:17) (1 + k z + h k ) pkε (4.38)= Z D √ ( L Λ) exp (cid:16) − pλ dist( x, L Λ ∁ ) (cid:17) (1 + k x k ) pkε dx (4.39)= L Z D √ L (Λ) exp (cid:16) − pλL dist( x ′ , Λ ∁ ) (cid:17) (1 + k x ′ k ) pkε dx ′ (4.40) ≤ CL Z Λ exp (cid:16) − pλL dist( x ′ , Λ ∁ ) (cid:17) (1 + L k x ′ k ) kε dx ′ + L − pkε (cid:12)(cid:12)(cid:12) D √ L (Λ) \ Λ (cid:12)(cid:12)(cid:12) . (4.41)The constant C does not depend on L . We are left to show, that the term behind CL is boundedby CL − − pkε .As L ≥ 1, by (A.27), we have (cid:12)(cid:12)(cid:12) D √ L (Λ) \ Λ (cid:12)(cid:12)(cid:12) ≤ CL , (4.42)because we can ignore the L part. The constant depends on Λ and this is the desired estimate.In order to estimate the remaining integral, we apply Lemma A.3 and then once more Lemma A.4to estimate the following integral. Thus, Z Λ exp (cid:16) − pλL dist( x ′ , Λ ∁ ) (cid:17) dx ′ (4.43)= Z R pλL h exp (cid:0) − pλL h (cid:1) (cid:12)(cid:12)(cid:12) { x ′ ∈ Λ | dist( x ′ , Λ ∁ ) ≤ h } (cid:12)(cid:12)(cid:12) dh (4.44) ≤ Z ∞ pλL h exp (cid:0) − pλL h (cid:1) Chdh (4.45)= Z ∞ C exp (cid:0) − ( h ′ ) (cid:1) ( h ′ ) dh ′ L (4.46)= CL . (4.47) In the second to last step, we used the substitution ( h ′ ) = pλL h . The constant C depends on p, λ and in turn on p, l, k, m, γ, B and the decay of B ε , V ε .To deal with the denominator in (4.41), we use 0 ∈ Λ. Hence there is an r > 0, such that B r (0) ⊂ Λ. For the integral over Λ ∩ D r (0) ∁ , we can estimate the denominator by CL − pkε anduse the integral estimate above for the enumerator. For the integral over D r (0) we estimate theenumerator by Ce − L and the denominator by 1. This finishes the proof. (cid:3) We can now conclude the proof of Theorem 3.3. Proof of Theorem 3.3. We begin with a fixed Landau level, meaning we assume B (2 l − < a n n . Now we use Corollary 4.4. Hence, for any ζ ∈ im Γ, we have1 H − ζ = n − X k =0 ( − k H − ζ (cid:18) H ε H − ζ (cid:19) k + (cid:18) H − ζ H ε (cid:19) n H − ζ (cid:18) H ε H − ζ (cid:19) n . (4.48)The path integral over every summand for 0 ≤ k ≤ n − (cid:13)(cid:13)(cid:13) H ε H − ζ (cid:13)(cid:13)(cid:13) n ≤ F ( ζ ) by Lemma 4.1, and note k H − ζ k ∞ =1 / dist(Γ , σ ( H )) and k L Λ ( ∁ ) k = 1. Then we use H¨older type inequalities and the monotonicity ofthe Schatten norm so that (cid:13)(cid:13)(cid:13)(cid:13) L Λ ( ∁ ) (cid:18) H − ζ H ε (cid:19) n H − ζ (cid:18) H ε H − ζ (cid:19) n L Λ (cid:13)(cid:13)(cid:13)(cid:13) p (4.49) ≤ (cid:13)(cid:13)(cid:13)(cid:13) L Λ ( ∁ ) (cid:18) H − ζ H ε (cid:19) n H − ζ (cid:18) H ε H − ζ (cid:19) n L Λ (cid:13)(cid:13)(cid:13)(cid:13) n n (4.50) ≤ (cid:13)(cid:13)(cid:13)(cid:13) H − ζ H ε (cid:13)(cid:13)(cid:13)(cid:13) n n (cid:13)(cid:13)(cid:13)(cid:13) H − ζ (cid:13)(cid:13)(cid:13)(cid:13) ∞ (cid:13)(cid:13)(cid:13)(cid:13) H ε H − ζ (cid:13)(cid:13)(cid:13)(cid:13) n n (4.51) ≤ F ( ζ ) . (4.52)The upper bound function F may have changed from above. This finishes the upper bound for asingle Landau level, as the function F is integrable over the path Γ.For every l ∈ N , such that a < B (2 l + 1) < b , we choose a circle path, such that the last onehits R at b , each two neighbouring paths hit R at one common point not in σ ( H ), the first path hits R at a and every circle has a real-valued centre. Then we apply the estimate for a single Landaulevel and the p -triangle inequality. If there is no Landau eigenvalue between a and b , the associatedprojections are finite dimensional and will lead to an O (1) term with respect to L . (cid:3) On the Landau Hamiltonian In this section we establish several properties of the Landau Hamilton operator H and theoperators P l , M I,ζ and in particular, their integral kernels. At the end of this section, we will alsoinclude an important integral bound.We introduce the Laguerre polynomials and their generating function. For any l ∈ N , the Laguerrepolynomials L l is given by L l : [0 , ∞ ) → R , t l X k =0 (cid:18) lk (cid:19) ( − k k ! t k . (5.1)For any s ∈ [0 , ∞ ), − < t < 1, their generating function is given by X l ∈ N t l L l ( s ) = 11 − t exp (cid:18) − ts − t (cid:19) . (5.2) N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 13 Let x, y ∈ R . For l ∈ N , we define p l as the integral kernel of P l , p l ( x, y ) := B π exp (cid:18) − B k x − y k + i B h x | Jy i (cid:19) L l (cid:0) B k x − y k / (cid:1) . (5.3)Furthermore, for 0 < t < 1, we define the operator Q t := P l t l P l . Its integral kernel is given by q t ( x, y ) := X l t l p l ( x, y ) (5.4)= B π (1 − t ) exp (cid:18) − B k x − y k + i B h x | Jy i − B t − t k x − y k (cid:19) (5.5)= B π (1 − t ) exp (cid:18) − B (1 + t )4(1 − t ) k x − y k + i B h x | Jy i (cid:19) . (5.6)We easily calculate (cid:18) − i ∇ x − B Jx (cid:19) q t ( x, y ) = (cid:18) iB (1 + t )2(1 − t ) ( x − y ) − B J ( x − y ) (cid:19) q t ( x, y ) , (5.7)and (cid:18) − i ∇ x − B Jx (cid:19) ⊗ q t ( x, y ) (5.8)= (cid:18) iB (1 + t )2(1 − t ) ( x − y ) − B J ( x − y ) (cid:19) ⊗ + (cid:18) B (1 + t )2(1 − t ) (cid:18) (cid:19) − B J (cid:19)! q t ( x, y ) . (5.9) Lemma 5.1. For any j ∈ N , there are C, a > , independent of l, B , such that for any x, y ∈ R (cid:13)(cid:13) ( − i ∇ x − A ( x )) ⊗ j p l ( x, y ) (cid:13)(cid:13) ≤ B . j Ca l exp (cid:18) − B k x − y k (cid:19) . (5.10)The norm on the left-hand side is the 2-norm on C j . Proof. Using the explicit formula for the Laguerre polynomials, for any t ≥ , j ′ ∈ N , < δ < 1, webound the j ′ th differential as follows: kL ( j ′ ) l ( t ) k ≤ l − j ′ X k =0 l t k k ! (5.11)= l − j ′ X k =0 (cid:18) δ (cid:19) l ( δt ) k k ! (5.12) ≤ (cid:18) δ (cid:19) l exp( δt ) . (5.13)Each of the j differential operators have to be resolved with the product rule, where we apply the − i ∇ x to the polynomial, which is resolved by chain rule, and − i ∇ x − A ( x ) to the exponential. Thiswill always be the exponential times a polynomial expression in x − y , taking values in C j . Thisleads to the first bound, with a constant C depending only on j , as the dependency on l is encodedentirely in the polynomial L l and its differentials. Thus, we have (cid:13)(cid:13) ( − i ∇ x − A ( x )) ⊗ j p l ( x, y ) (cid:13)(cid:13) (5.14) ≤ C j X j ′ =0 (cid:13)(cid:13)(cid:13)(cid:13) L ( j ′ ) l (cid:18) B k x − y k (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) (cid:16) p B k x − y k (cid:17) j B . j exp (cid:18) − B k x − y k (cid:19) . (5.15)By setting t = B k x − y k / δ = in (5.13), we can finally estimate (cid:13)(cid:13) ( − i ∇ x − A ( x )) ⊗ j p l ( x, y ) (cid:13)(cid:13) (5.16) ≤ CB . j j X j ′ =0 (cid:13)(cid:13)(cid:13)(cid:13) L ( j ′ ) l (cid:18) B k x − y k (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) (cid:16) p B k x − y k (cid:17) j exp (cid:18) − B k x − y k (cid:19) (5.17) ≤ CB . j l exp (cid:18) B k x − y k (cid:19) (cid:16) p B k x − y k (cid:17) j exp (cid:18) − B k x − y k (cid:19) (5.18) ≤ CB . j l exp (cid:18) B k x − y k (cid:19) exp (cid:18) − B k x − y k (cid:19) (5.19) ≤ CB . j l exp (cid:18) − B k x − y k (cid:19) . (5.20)In the second to last step, we used that polynomials can be bounded by exponentials. The constant C changed, but still only depends on j . (cid:3) Lemma 5.2. Let I ⊂ N be cofinite, ζ ∈ C , ζ B (2 I + 1) and l ∈ N , such that l ≥ max( I ∁ ∪{ℜ ζ + B B } ) . Then we have the identity B M I,ζ = Z t − ζ/B Q t − X l ≤ l t l P l dt + X l ∈ I,l ≤ l P l (2 l + 1 − ζ/B ) . (5.21) Proof. The idea of this proof is the formal identity Z X l ∈ I t l − ζ/B P l dt = X l ∈ I 11 + 2 l − ζ/B P l . (5.22)Now we need to establish the precise meaning of this identity. First, we note that t − ζ/B =exp( − ζ/B ln( t )) is well defined, as t > 0. If ℜ ( ζ ) /B ≥ l + 1, then the integral of the summandsfor l will not exist, which is the reason we introduced l . We bounded the real part of ζ a littlestronger then necessary to make the proof easier. Hence, we have Z X l>l t l − ζ/B P l dt = X l>l 11 + 2 l − ζ/B P l . (5.23)For any single l ≥ l , the integrals exists as a Bochner integral with respect to the operator norm.As these operators are scalar multiples of a set of orthogonal projections corresponding to pairwiseorthogonal subspaces, we can exchange the integral and the sum by dominated convergence, as wecan replace the triangle inequality by a maximum. (cid:3) We will deal with a few integral kernels that have a singularity at the diagonal. To describe sucha singularity, for any s ∈ R , we introduce b s : R → [0 , ∞ ) , ( x, y ) − D √ B (0) ( x − y ) ln( √ B k x − y k ) s = 0 , D √ B (0) ( x − y ) k x − y k s s = 0 . (5.24) Lemma 5.3. Let I ⊂ N be cofinite. Then there is a function F ∈ L ∞ loc ( C \ (2 I + 1)) , such that thefollowing pointwise upper bounds hold for all x, y ∈ R , x = y and ζ ∈ C \ B (2 I + 1) : | iker M I,ζ ( x, y ) | ≤ F (cid:18) ζB (cid:19) (cid:18) b ( x, y ) + exp (cid:18) − B k x − y k (cid:19)(cid:19) , (5.25) k iker( − i ∇ − A ) M I,ζ ( x, y ) k ≤ F (cid:18) ζB (cid:19) (cid:18) b ( x, y ) + p B exp (cid:18) − B k x − y k (cid:19)(cid:19) , (5.26) k ( − i ∇ x − A ( x )) ⊗ iker M I,ζ ( x, y ) k ≤ F (cid:18) ζB (cid:19) (cid:18) b ( x, y ) + B exp (cid:18) − B k x − y k (cid:19)(cid:19) . (5.27) N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 15 Remark. The last inequality is structurally different, because the implied operator ( − i ∇− A ) ⊗ M I,ζ does not have a nice integral kernel (see Definition 2.2). The differential of the integral kernel canstill be considered but is not L with respect to y for any fixed x and hence not a nice integral kernel.It is a singular integral kernel. While there is a lot of theory developed on such a kernel, we canavoid dealing with such a kernel in this paper. Proof. The set I ⊂ N is fixed throughout the proof.For any t ∈ [0 , , l ∈ N , j ∈ { , , } , we define q t,j ( x, y ) := ( − i ∇ x − A ( x )) ⊗ j q t ( x, y ) , (5.28) p l,j ( x, y ) := ( − i ∇ x − A ( x )) ⊗ j p l ( x, y ) . (5.29)As q t,j , p l,j are nice integral kernels, we can apply dominated convergence and see that q t,j ( x, y ) = iker (cid:0) ( − i ∇ − A ) ⊗ j Q t (cid:1) ( x, y ) , (5.30) p l,j ( x, y ) = iker (cid:0) ( − i ∇ − A ) ⊗ j P l (cid:1) ( x, y ) . (5.31)We choose l ∈ N minimal, such that (2 l − B > ℜ ζ and l ≥ max( I ∁ ). Now we use therepresentation established in Lemma 5.2. To prove, that for j ∈ { , } , the operators have integralkernels, we want to use Lemma A.6. Hence, we only need to show, that the following inequalityholds, in order to finish the proof for j = 0 , Z (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) t − ζ/B q t ,j ( x, y ) − X l ≤ l t l p l,j ( x, y ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) dt + X l ∈ I,l ≤ l k p l,j ( x, y ) k (2 l + 1 − ζ/B ) (5.32) ≤ B F (cid:18) ζB (cid:19) (cid:18) b ( x, y ) + B exp (cid:18) − B k x − y k (cid:19)(cid:19) . (5.33)For j = 2, however, we need to consider, that as the integrand is smooth on (0 , 1) and the summandsat the end are smooth, we can try to exchange the integral with the differential operator ( − i ∇ − A ).This will work, if the absolute value of the differential is integrable, by dominated convergence. Hencethe above integral bound also covers the case j = 2 and we will now proceed to bound all terms atthe same time by choosing j ∈ { , , } . We want to use Lemma 5.1 to bound the first integral onthe interval (0 , t ) and the sums. Hence, Z t (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) t − ζ/B q t ,j ( x, y ) − X l ≤ l t l p l,j ( x, y ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) dt (5.34) ≤ Z t X l>l t l −ℜ ( ζ ) /B | p l,j ( x, y ) | dt (5.35) ≤ Z t X l>l t l −ℜ ( ζ ) /B CB . j a l exp (cid:18) − B k x − y k (cid:19) dt (5.36)= CB . j X l>l ( t a ) l t (2 l + 1 − ℜ ( ζ ) /B ) t ℜ ( ζ ) /B exp (cid:18) − B k x − y k (cid:19) (5.37) ≤ F ( ζ/B ) B . j exp (cid:18) − B k x − y k (cid:19) . (5.38)The last step holds, if t a < 1, so we fix such a t now . The function F is in L ∞ loc ( C \ B (2 I + 1)),as l is chosen locally bounded in ζ/B . For fixed l the function F is continuous. The next step is Actually a = 16, so we could choose for example t = 0 . 1, but the value is not relevant. bounding the remaining finite sum terms. Here, we will use, that l ≤ l and hence a l ≤ C . Thus, Z t (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)X l ≤ l t l − ζ/B p l,j ( x, y ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) dt + X l ∈ I,l ≤ l k p l,j ( x, y ) k (2 l + 1 − ζ/B ) (5.39) ≤ CB . j X l ≤ l (cid:18)Z t t l −ℜ ( ζ ) /B dt (cid:19) + X l ∈ I,l ≤ l (cid:18) k l + 1 − ζ/B k (cid:19) exp (cid:18) − B k x − y k (cid:19) (5.40) ≤ F ( ζ/B ) B . j exp (cid:18) − B k x − y k (cid:19) . (5.41)The function F is L ∞ loc ( C \ B (2 I + 1)) by the same argumentation as F . We will now turn ourattention to the last remaining term. It is given by Z t (cid:13)(cid:13)(cid:13) q t ,j ( x, y ) t − ζ/B (cid:13)(cid:13)(cid:13) dt. (5.42)The integrand is given by (5.7) for j = 1 and by (5.9) for j = 2. Only in the following lines, wedenote by j δ ( j, 2) the function, that is 1, if j = 2 and 0 otherwise. We introduce the parameter h := √ B k x − y k and estimate Z t (cid:13)(cid:13)(cid:13) t − ζ/B q t ,j ( x, y ) (cid:13)(cid:13)(cid:13) dt (5.43) ≤ (cid:16) t −ℜ ( ζ ) /B + 1 (cid:17) Z t CB − t (cid:18) √ B h − t ) (cid:19) j + δ ( j, B − t ) ! exp (cid:18) − t − t ) h (cid:19) dt (5.44) ≤ Z F ( ζ/B ) B − t (cid:18) √ B h − t ) (cid:19) j + δ ( j, B − t ) ! exp (cid:18) − − t ) h (cid:19) dt. (5.45)In the last step, we used the fact, that t t ≥ √ − > to bound the factor in the exponential.The function F is just continuous on C .We change variables to s := h − t ) . The interval is changed to ( h / , ∞ ) and the determinant is h / (5 s ). In total we have Z t (cid:13)(cid:13)(cid:13) t − ζ/B q t ,j ( x, y ) (cid:13)(cid:13)(cid:13) dt (5.46) ≤ F (cid:18) ζB (cid:19) B . j Z ∞ h / sh (cid:18)(cid:16) sh (cid:17) j + δ ( j, sh (cid:19) exp ( − s ) h s ds (5.47) ≤ F (cid:18) ζB (cid:19) B . j Z ∞ h / s (cid:18)(cid:16) sh (cid:17) j + δ ( j, sh (cid:19) exp ( − s ) ds =: Θ . (5.48)If h > 1, we can bound the integrand by C exp( − s ). The reduction in the exponent takes care ofthe factor s , that appears in the case j = 2. Negative powers of h can be bounded by one. Theintegral can then be resolved and we haveΘ ≤ CF (cid:18) ζB (cid:19) B . j exp (cid:18) − h (cid:19) (5.49)= CF (cid:18) ζB (cid:19) B . j exp (cid:18) − B k x − y k (cid:19) . (5.50)This is the desired upper bound.If h ≤ j > 0, we can set the lower interval limit to 0 and get an integrable function in s multiplied by h − j . This gives usΘ ≤ CF (cid:18) ζB (cid:19) B . j h − j ≤ CF (cid:18) ζB (cid:19) B b j ( x, y ) , (5.51) N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 17 which is the desired upper bound.Finally, if h ≤ j = 0, we get a constant from the integral starting at . For the integral up to , we can bound the integrand by s . Hence, the remaining integral is bounded by C (1 − ln( h )) = C (1 + b ( x, y )). Once again, this is the desired result. (cid:3) We need one very important bound, which will have multiple uses later. Lemma 5.4. Let u , u , u : R → R + be functions, such that ln ◦ u j is Lipschitz with Lipschitzconstant C lip > . Let ≤ s , s < and λ > be real numbers. Then there is a constant C > ,depending only on B , s , s , λ and C lip , such that for all x, y ∈ R , x = y we have the estimate Z R (cid:0) b s ( x, y ) + exp( − B λ k x − y k ) (cid:1) (cid:0) b s ( y, z ) + exp( − B λ k y − z k ) (cid:1) u ( x ) u ( y ) u ( z ) dy (5.52) ≤ Cb s + s − ( x, z ) + C exp (cid:0) − B λ k x − z k (cid:1) u ( x ) u ( x ) u ( x ) . (5.53) If / ( u u u ) ∈ L ( R ) and s + s < , then the integral kernel is Hilbert–Schmidt. This is to be used together with Lemma 5.3 with λ = . The general λ is included to be able tochain more resolvents inductively. As all summands in the integral are positive, we may assume thatthey have the same constants in front. Proof. We first need two minor results. Let a, b ∈ R , j ∈ { , , } . Then for any δ > 0, we have u j ( a ) u j ( b ) = exp (ln ◦ u j ( a ) − ln ◦ u j ( b )) (5.54) ≤ exp ( C lip k a − b k ) (5.55) ≤ exp δ k a − b k + C lip δ ! (5.56)= C ( C lip , δ ) exp (cid:0) δ k a − b k (cid:1) . (5.57)We used the Young inequality. Furthermore (for any x, y, z ∈ R ) we have the identity k x − y k + k y − z k = 12 k x − z k + 2 (cid:13)(cid:13)(cid:13)(cid:13) y − x + z (cid:13)(cid:13)(cid:13)(cid:13) . (5.58)We write R := √ B . Let us begin with the left-hand side of (5.53) and just write out most of theH¨older estimates. Hence, LHS ≤ Cu ( x ) u ( z ) (5.59) Z D R ( x ) b s ( x, y ) b s ( y, z ) dy (cid:13)(cid:13)(cid:13)(cid:13) u ( · ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ ( D R ( x )) (5.60)+ exp (cid:18) − B λ k x − z k (cid:19) Z R exp (cid:16) − B (cid:13)(cid:13) y − x + z (cid:13)(cid:13) (cid:17) u ( y ) dy (5.61)+ k b s ( x, · ) k L ( D R ( x )) (cid:13)(cid:13) exp (cid:0) − B λ k· − z k (cid:1)(cid:13)(cid:13) L ∞ ( D R ( x )) (cid:13)(cid:13)(cid:13)(cid:13) u ( · ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ ( D R ( x )) (5.62)+ k b s ( · , z ) k L ( D R ( z )) (cid:13)(cid:13) exp (cid:0) − B λ k x − ·k (cid:1)(cid:13)(cid:13) L ∞ ( D R ( z )) (cid:13)(cid:13)(cid:13)(cid:13) u ( · ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ ( D R ( z )) ! . (5.63)The L ∞ norms of the non-exponential terms can be bounded by a constant times the functionevaluated at the centre, where the constant is given by (5.55), using a as the centre of the ball and b as any point in the ball. For the L ∞ norms of the exponential terms, we use Lemma A.5 with x := y − z . We are left to estimate the four L norms, some of which are written as integrals. Thelast two L norms can be bounded by a constant and that is sufficient. For the exponential integral,we first use (5.57) with δ = B to replace the u ( y ) in the denominator by u (( x + z ) / LHS ≤ Cu ( x ) u ( z ) × (5.64) Z D R ( x ) b s ( x, y ) b s ( y, z ) dy u ( x ) + exp (cid:0) − B λ/ k x − z k (cid:1) u (( x + z ) / 2) (5.65)+ exp (cid:18) − B λ k x − z k (cid:19) u ( x ) + exp (cid:18) − B λ k x − z k (cid:19) u ( z ) ! . (5.66)If we apply (5.57) again, we can get the desired bound for the last three summands. So, we onlyneed to get the same bound for the first summand. If k x − z k > R , the first summand vanishes.Otherwise, the term 1 /u ( z ) can be bounded by C/u ( x ) by (5.57). In the case 2 R ≥ k x − z k ≥ R/ R , which can then be bounded by a constanttimes the Gaussian. We are left to consider the case k x − z k < R/ 2. So, we are left to bound theintegral Z D R ( x ) b s ( x, y ) b s ( y, z ) dy. (5.67)We have b s ( x, · ) ∈ L p for any 1 ≤ p < /s and b s is symmetric in x, y . Hence, if s + s < 2, wecan bound this by a constant (independent of x, z ) using H¨older. This can then by bounded by theGaussian, as k x − z k ≤ R . We are left with the case s + s ≥ 2, where we want to bound theintegral by b s + s − ( x, z ) + C . As s , s < 2, we have s , s > 0. Let e ∈ R be the standard unitvector and let D r ,r (0) be the annulus between the two radii r ≤ r . Then we have Z D R ( x ) b s ( x, y ) b s ( y, z ) dy (5.68) ≤ Z D R ( x ) k x − y k s k y − z k s dy (5.69)= Z D R (0) k y k s k y − ( z − x ) k s dy (5.70)= Z D R k x − z k− (0) k x − z k − s − s k y k s k y − e k s dy (5.71) ≤ R D (0) k y k − s k y − e k − s dy k x − z k s + s − + Z D ,R k x − z k− (0) k x − z k − s − s k y k s k y − e k s dy (5.72) ≤ C k x − z k s + s − + Z D ,R k x − z k− (0) C k x − z k − s − s k y k s + s dy (5.73) ≤ Cb s + s − ( x, z ) . (5.74)In the final step, we have to consider the case s + s = 2 separately. In this case, the integral at theend yields the term b ( x, z ) up to a constant. In the case s + s > 2, the integral over k y k − s − s can be bounded by a constant, independent of x, z and we are left with the correct singularity at thediagonal. This finishes the proof of the upper bound.If 1 / ( u u u ) ∈ L and s + s < 3, we get C Z R dx Z R dz b s + s − ( x, z ) + exp (cid:0) − B λ k x − z k (cid:1) u ( x ) u ( x ) u ( x ) ! (5.75) N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 19 = C Z R dx Z R d ( x − z ) b s + s − ( x, z ) + exp (cid:0) − B λ k x − z k (cid:1) u ( x ) u ( x ) u ( x ) ! (5.76) ≤ Z R dx C ( u ( x ) u ( x ) u ( x )) ! ≤ C. (5.77)Hence the integral kernel is Hilbert–Schmidt. (cid:3) Corollary 5.5. Let n ∈ Z + and for any ≤ i ≤ n , let there be an operator K i with integral kernel k i on L ( R ) , log-Lipschitz functions u i , v i : R → R + , λ i > , and ≤ s i < . Assume the integralkernels k i satisfy the upper bound u i ( x ) | k i ( x, y ) | v i ( y ) ≤ Cb s i ( x, y ) + C exp (cid:0) − B λ i k x − y k (cid:1) , (5.78) for any x = y . Define K := Q ni =0 K i and let s = − n + n X i =0 s i . (5.79) Then K has an integral kernel k and there are λ > , C > , such that for any x = y , we have theinequality | k ( x, y ) | ≤ Cb s ( x, y ) + C exp (cid:0) − B λ k x − y k (cid:1)Q ni =0 u i ( x ) v i ( x ) . (5.80)For s < 0, we can replace b s by 0 in (5.80), as b s is bounded and can be absorbed in the Gaussian.The proof follows by induction on n , as nice integral operators (see Definition 2.2) are closed undercomposition and K i is by assumption always a nice integral operatorNow we have established everything necessary to prove Lemma 4.1. Proof of Lemma 4.1. We recall that I ⊂ N is cofinite and ζ ∈ C \ B (2 I + 1). By the triangleinequality, we have k H ε M I,ζ k n ≤ k A ε · ( − i ∇ − A ) M I,ζ k n + k W ε M I,ζ k n . (5.81)We can estimate pointwise for x = y , using first the assumption that B ε , V ε are (1 , C d ) tame andLemma 3.2 and then Lemma 5.3. Thus, we have k iker A ε ( − i ∇ − A ) M I,ζ ( x, y ) k (5.82) ≤ C (1 + k x k ) ε k iker( − i ∇ − A ) M I,ζ ( x, y ) k (5.83) ≤ F (cid:18) ζB (cid:19) b ( x, y ) + √ B exp (cid:0) − B k x − y k (cid:1) (1 + k x k ) ε . (5.84)And now for the other part, we get k iker W ε M I,ζ ( x, y ) k (5.85) ≤ C (1 + k x k ) ε k iker M I,ζ ( x, y ) k (5.86) ≤ F (cid:18) ζB (cid:19) b ( x, y ) + exp (cid:0) − B k x − y k (cid:1) (1 + k x k ) ε (5.87) ≤ F (cid:18) ζB (cid:19) b ( x, y ) + √ B exp (cid:0) − B k x − y k (cid:1) (1 + k x k ) ε . (5.88)In the last step we used b ≤ Cb and 1 = C √ B .We use properties we denoted as powers and Hilbert–Schmidt kernel of the p -Schatten norms from ?? . Hence the 4 n -Schatten norm of T can be calculated as the 4 n th root of the square integral of the integral kernel of ( T T ∗ ) n We note, that u ( x ) := (1 + k x k ) ε is log Lipschitz. We want to useCorollary 5.5. Hence, we define for 0 ≤ i ≤ n − K i := ( H ε M I,ζ i even , ( H ε M I,ζ ) ∗ i odd . (5.89)For even i , we choose u i ( x ) = u ( x ) , v i ( x ) = 1 and for odd i , we choose v i ( x ) = u ( x ) , u i ( x ) = 1. Wealways have s i = 1. Now we can apply Corollary 5.5 and get for any x = y that (cid:12)(cid:12) iker (cid:0) ( H ε M I,ζ ) ( H ε M I,ζ ) ∗ (cid:1) n ( x, y ) (cid:12)(cid:12) (5.90) ≤ F (cid:18) ζB (cid:19) b ( x, y ) + exp (cid:0) − λB k x − y k (cid:1) (1 + k x k ) n + ε . (5.91)The function F is in L ∞ loc ( C \ σ ( H )). This integral kernel is in L , as 2 n ε > 1. The b term onlyappears for n = 1, as for n > 1, we get s < 0, which corresponds to a bounded b s . (cid:3) Proof of Theorem 4.7 In order to apply Proposition 4.6, we need to apply some differentials to the integral kernel of( T l H ε ) m P l ( H ε T l ) k − m . We denote by t l the integral kernel of T l . By Lemma 5.3, we can only applyone full differential in x or y to t l , before we get a function, that is not a nice integral kernel (asdefined in Definition 2.2 anymore. However, the operator P l has a smooth integral kernel, whichis why we would like to move differentials over to it. For this purpose, we have to study some(anti-)commutators. Lemma 6.1. For any l ∈ N , x, y ∈ R with x = y , we have the identity ∇ x t l ( x, y ) = (cid:18) − i B J ( x − y ) − ∇ y (cid:19) t l ( x, y ) . (6.1) Proof. We use the fact, that the adjoint of an operator with integral kernel ( x, y ) k ( x, y ) has theintegral kernel ( x, y ) k ( y, x ) and that Q t is self-adjoint. We recall (5.7). Hence, ∇ x q t ( x, y ) = (cid:18) − B (1 + t )2(1 − t ) ( x − y ) + i B Jy (cid:19) q t ( x, y ) , (6.2) ∇ y q t ( x, y ) = ∇ y q t ( y, x ) (6.3)= (cid:18) − B (1 + t )2(1 − t ) ( y − x ) − i B Jx (cid:19) q t ( x, y ) . (6.4)If we add these up, we get ∇ x q t ( x, y ) + ∇ y q t ( x, y ) = − i B J ( x − y ) q t ( x, y ) . (6.5)As this identity is independent of t , the same identity is satisfied, if we replace q t by p l by a simplecoefficient comparison, in view of (5.4) and as q t is real analytic for | t | < 1. We define the function k : (0 , × R × R by k t ( x, y ) := t − (2 l +1) q t ( x, y ) − l X k =0 t k p k ( x, y ) ! + l − X k =0 p k ( x, y )2( k − l ) . (6.6)Hence, we have for any t ∈ (0 , ∇ x k t ( x, y ) + ∇ y k t ( x, y ) = − i B J ( x − y ) k t ( x, y ) . (6.7)By Lemma 5.2, Lemma 5.3 and Lemma A.6, for any l ∈ N , j = 0 , 1, we have the representation B ( − i ∇ x − A ( x )) ⊗ j t l ( x, y ) = Z ( − i ∇ x − A ( x )) ⊗ j k t ( x, y ) dt. (6.8) N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 21 Hence, the equality holds for t l . (cid:3) Lemma 6.2. For any k , k , m , m ∈ N , there are polynomial functions s k k m m : R → C ofdegree at most m + m , such that for any l ∈ N and any Ψ ∈ W m + m , ( R ) , the identity ∂ m x ∂ m x Z R t l ( x, y )Ψ( y ) dy = Z R X k ≤ m ,k ≤ m s k k m m ( x − y ) t l ( x, y ) ∂ k y ∂ k y Ψ( y ) dy (6.9) holds for any x ∈ R .Proof. We use induction on ( m , m ). The case (0 , 0) is tautological. By Lemma 5.3, ∇ x t l ( x, · ) ∈ L . ( R ). The cases (0 , 1) and (1 , 0) follow from Lemma 6.1, dominated convergence to exchangeintegral and differentiation, and integration by parts. We need the steps ( m , m ) → ( m + 1 , m )and ( m , m ) → ( m , m + 1). As they work analogously, we will only consider the first case. Thus, ∂ m +1 x ∂ m x Z R t l ( x, y )Ψ( y ) dy (6.10)= ∂ x Z R X k ≤ m ,k ≤ m s k k m m ( x − y ) t l ( x, y ) ∂ k y ∂ k y Ψ( y ) dy (6.11)= Z R X k ≤ m ,k ≤ m ( ∂ x s k k m m ) ( x − y ) t l ( x, y ) ∂ k y ∂ k y Ψ( y ) dy (6.12) − Z R X k ≤ m ,k ≤ m s k k m m ( x − y ) (cid:18) i B J ( x − y ) (cid:19) t l ( x, y ) ∂ k y ∂ k y Ψ( y ) dy (6.13)+ Z R X k ≤ m ,k ≤ m s k k m m ( x − y ) t l ( x, y ) ∂ k +1 y ∂ k y Ψ( y ) dy. (6.14)We used the product rule, the case (1 , s k k m m ,but we are only interested in their degree and independence of l . The recursive formula lets usconclude, that the degree of s k k m m is at most m + m − k − k . This finishes the proof. (cid:3) The next step is to move the differentials past H ε . For this, we easily see the identity[ ∇ j , H ε ] = [ ∇ j , W ε + 2 A ε · ( − i ∇ − A )] (6.15)= 2( ∂ x j A ε ) · ( − i ∇ − A ) − A ε · B Je j + ( ∂ x j W ε ) . (6.16)Here, ∂ x j A ε and ∂ x j W ε refer to the multiplication operators of the partial derivatives of the functions A ε ( · ) , W ε ( · ).For any x ∈ R , we introduce B ε,x ( x ) := B ε ( x + x ) , x ∈ R (6.17)The operators A ε,x , V ε,x , W ε,x , H ε,x are defined accordingly. For any x ∈ R , we define A ε,x ( x ) := B ε ( · + x ) ∗ J · π k·k ( x ) (6.18) V ε,x ( x ) := V ε ( x + x ) (6.19) H ε,x := ( − i ∇ − A + A ε,x ) + V ε,x − H (6.20) W ε,x ( x ) := W ε ( x + x ) . (6.21)We have not defined A x and H x , as this may lead to confusion with A and H , if we set x = 0.The idea behind this is, that by conjugating with the unitary operator U x : L ( R ) → L ( R ), asdefined in Lemma A.1, we observe the identity (cid:13)(cid:13)(cid:13) [0 , + x ( T l H ε ) m P l ( H ε T l ) k − m D ∁ R ( x ) (cid:13)(cid:13)(cid:13) p = (cid:13)(cid:13)(cid:13) [0 , ( T l H ε,x ) m P l ( H ε,x T l ) k − m D ∁ R (0) (cid:13)(cid:13)(cid:13) p , (6.22) as the p -Schatten norm is unitarily invariant.Now, we can prove our next major step. Theorem 6.3. Let k, l, m, m , m ∈ N with k ≥ m . If m > , let B ε , V ε be ( m + m , C d ) tame. If m = 0 let B ε , V ε be (1 , C d ) tame. Then there are C, λ > , α ≥ , α ≥ kε ∈ R , such that for any x , x, z ∈ R , we have the upper bound (cid:13)(cid:13) ∂ m x ∂ m x iker( T l H ε,x ) m P l ( H ε,x T l ) k − m (cid:13)(cid:13) ( x, z ) ≤ C (1 + k x k ) α (1 + k x + x k ) α exp (cid:0) − λ k x − z k (cid:1) . (6.23)The idea of the proof is to move the differentials forward, until they get applied to P l , which hasa smooth integral kernel. Then we can use Lemma 5.4 or Corollary 5.5 and are done. Remark. For 0 < m < k , it may be possible to get an even stronger decay under slightly strongerassumptions on the decay of the first differentials of B ε , V ε , by estimating the kernel of P l H ε,x T l =[ P l , H ε,x ] T l . In the case, where B ε = 0 and ∇ V ε has (1 + ε, C d ) decay, we can show α ≥ kε using this approach. For B ε = 0, there is the problem of moving the stronger decay of the differentialof B ε over to the differential of A ε and the problem, that − i ∇ − A does not commute with H . Proof. First, we note that H ε,x T l and P l are nice integral operators, as defined in Definition 2.2.This allows us to write compositions of these operators as the related integrals over the kernels.We do an induction on m and k over the following statement:The theorem holds for any m ′ , m ′ ∈ N , such that m ′ + m ′ ≤ m + m . The values of α , λ maydepend on m ′ , m ′ . The value of α is given by m + m + 2 m . We observe (cid:12)(cid:12)(cid:12) ∂ m ′ x ∂ m ′ x p l ( x, y ) (cid:12)(cid:12)(cid:12) ≤ C (1 + k y k ) m ′ + m ′ (1 + k x − y k ) l + m ′ + m ′ exp (cid:18) − B k x − y k (cid:19) (6.24) ≤ C (1 + k x k ) m ′ + m ′ exp (cid:18) − B k x − y k (cid:19) . (6.25)In the last step, we used (5.57) and the fact, that polynomials can be bounded by exponentials. Weconsider the case m = 0. For k = 0, we are done by (6.25). For k > θ ( y, z ) := iker ( H ε,x T l ) k ( y, z ) (6.26)and note iker P l ( H ε,x T l ) k ( x, y ) = Z R p l ( x, y ) θ ( y, z ) dy. (6.27)For k = 1, θ ( y, z ) is equal to (cid:18) W ε ( y + x ) + 2 A ε ( y + x ) · (cid:18) − i ∇ y − B Jy (cid:19)(cid:19) t l ( y, z ) . (6.28)Now we can use the ( ε, C d ) decay of A ε and V ε , Lemma 5.3 and b ≤ Cb to arrive at | θ ( y, z ) | ≤ C (1 + k y + x k ) ε (cid:18) b ( y, z ) + p B exp (cid:18) − B k y − z k (cid:19)(cid:19) . (6.29)for k = 1. For general k ≥ 1, we apply Corollary 5.5 to θ and have | θ ( y, z ) | ≤ C (1 + k y + x k ) kε (cid:0) b − k ( y, z ) + exp (cid:0) − λB k y − z k (cid:1)(cid:1) (6.30) ≤ C (1 + k y + x k ) kε (cid:0) b ( y, z ) + exp (cid:0) − λB k y − z k (cid:1)(cid:1) . (6.31)for some C, λ > 0. In the last step, we used b − k ≤ Cb , which holds as k ≥ It may be possible to remove the 2 m in here, but it is not relevant for this paper. N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 23 Now we use (6.25), Lemma 5.4 and interchange integration and differentiation by dominatedconvergence. Hence, (cid:12)(cid:12)(cid:12)(cid:12) ∂ m ′ x ∂ m ′ x Z R p l ( x, y ) θ ( y, z ) dy (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (1 + k x k ) m ′ + m ′ (1 + k x + x k ) kε exp (cid:0) − λ k x − z k (cid:1) . (6.32)The precise value of λ is irrelevant and we can choose α = m ′ + m ′ and α = kε .Now we do the step ( m, k ) → ( m + 1 , k + 1). For any y, z ∈ R , we define ϕ ( y, z ) := iker( T l H ε,x ) m P l ( H ε,x T l ) k − m ( y, z ) . (6.33)We know that for any m ′ + m ′ ≤ m + m , (cid:12)(cid:12)(cid:12) ∂ m ′ x ∂ m ′ x ϕ ( y, z ) (cid:12)(cid:12)(cid:12) ≤ C (1 + k y k ) α (1 + k y + x k ) α exp (cid:0) − λ k y − z k (cid:1) . (6.34)By (5.57), we can get a similar bound with y in front instead of x , if we slightly decrease λ andincrease C . Hence ϕ ( · , y ) ∈ W m + m , ( R ).We need to bound the integral kernel ψ ( x, z ) := ∂ m ′ x ∂ m ′ x Z R t l ( x, y ) ( H ε,x ϕ ( · , z )) ( y ) dy (6.35)in a similar manner.We will first consider the case m ′ = m ′ = 0. Here, we just use (6.31) to bound the integral kernelof ( T l H ε,x ) m and then Lemma 5.4 and the induction hypothesis to prove the statement for m + 1.In the remaining case, we assume w.l.o.g. that m ′ > 0. As A ε , W ε ∈ C m + m − b ( R ) and ϕ ( · , z ) ∈ W m + m , ( R ) for any z ∈ R , we know that H ε,x ϕ ( · , z ) ∈ W m + m − , ( R ) for any z ∈ R . Hence, we can apply Lemma 6.2 with the pair ( m ′ − , m ′ ) and get ψ ( x, z ) = ∂ x Z R dy X k ≤ m ′ − ,k ≤ m ′ (cid:0) s k k ( m ′ − m ′ ( x − y ) t l ( x, y ) (cid:1) (6.36) × ∂ k y ∂ k y W ε ( y + x ) + X j =1 A ε ( y + x ) j (cid:18) − i∂ y j − B Jy ) j (cid:19) ϕ ( y, z ) . (6.37)As the second factor is in L ( R ) and the first differential of the first factor is in L . ( R ), we canexchange integration and differentiation by dominated convergence. We now resolve every differentialby the product rule. In the next equation, each differential is only applied to the immediatelyfollowing function. We have ψ ( x, z ) = Z R X k ≤ m ′ − ,k ≤ m ′ X α + α + α = k ,β + β + β = k X j =1 (6.38) (cid:0) ∂ x s k k ( m ′ − m ′ ( x − y ) t l ( x, y ) + s k k ( m ′ − m ′ ( x − y ) ∂ x t l ( x, y ) (cid:1) (6.39) × M (cid:0) ∂ α + α y ∂ β + β y W ε ( y + x ) (cid:1) (cid:0) ∂ α y ∂ β y ϕ ( y, z ) (cid:1) (6.40) − iM (cid:0) ∂ α + α y ∂ β + β y A ε ( y + x ) j (cid:1) (cid:0) ∂ y j ∂ α y ∂ β y ϕ ( y, z ) (cid:1) (6.41) − M (cid:0) ∂ α y ∂ β y A ε ( y + x ) j (cid:1) (cid:0) ∂ α y ∂ β y ( Jy ) j (cid:1) (cid:0) ∂ α y ∂ β y ϕ ( y, z ) (cid:1) ! dy. (6.42)The rational numbers M , M , M are combinatorial expressions in the summation indices. Theirprecise values are irrelevant. Now we can bound every single term and combine this to bound theintegral.There are at most m + m differentials applied to ϕ , hence we can use the induction hypothesisto bound this. Here, we will get an α ≥ kε and α = m + m + 2 m . The values of λ and C arenot relevant. At most m + m − A ε , W ε . Any of these terms will be bounded bythe decay assumptions on B ε , V ε . Specifically, it will have at least ( C, ε ) decay, for some constant C .The function ( Jy ) j is linear. Its differentials are all bounded by 1 + k y k . Most of them evenvanish.We shift our focus to the first term. We start by bounding the polynomial s − . Its degree is atmost m + m − 1. We want to use Lemma 5.3. On the first summand, we can immediately do so.On the second summand, we need a zero addition. Thus, (cid:12)(cid:12) ∂ s k k ( m ′ − m ′ ( x − y ) t l ( x, y ) + s k k ( m ′ − m ′ ( x − y ) ∂ x t l ( x, y ) (cid:12)(cid:12) (6.43) ≤ C (1 + k x − y k ) m + m − (cid:18) | t l ( x, y ) | + (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) ∂ x + i B Jy ) − i B Jy ) (cid:19) t l ( x, y ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (6.44) ≤ C (1 + k x − y k ) m + m − (6.45) × (cid:18) (1 + k y k ) (cid:18) b ( x, y ) + exp (cid:18) − B k x − y k (cid:19)(cid:19) + b ( x, y ) + exp (cid:18) − B k x − y k (cid:19)(cid:19) (6.46) ≤ C (1 + k x − y k ) m + m − (1 + k y k ) (cid:18) b ( x, y ) + exp (cid:18) − B k x − y k (cid:19)(cid:19) (6.47) ≤ C (1 + k y k ) (cid:18) b ( x, y ) + exp (cid:18) − B k x − y k (cid:19)(cid:19) . (6.48)We used, that b ≤ Cb and, that the exponential is larger than a polynomial. Now we applyLemma 5.4. The value of α increases by 2, as we have two potential (1 + k y k ) factors (one from( Jy ) and one from the last estimate). The value of α increases by ε , as every summand containssome A ε or W ε term. The value of λ could be bounded more precisely, but we just divide it by threeagain. This finishes the proof. (cid:3) To estimate the fractional derivative in Proposition 4.6, we need the following geometric fact. Fact 6.4. Let x ′ , x ′′ , y ∈ R be pairwise distinct and k x ′ − y k ≤ k x ′′ − y k . Then there is a C path Γ from x ′ to x ′′ of length at most π k x ′ − x ′′ k , such that for any point x on the path, k x − y k ≥ k x ′ − y k . This allows us to show the following estimate. Lemma 6.5. For each y ∈ R , let ϕ y ∈ C ( R , R ) . Let there be a constant C > , independent of x, y , such that for all x, y , we have k ϕ y ( x ) k ≤ C exp ( k x − y k ) and k∇ ϕ y ( x ) k ≤ C exp ( k x − y k ) .Let < ν < , h ∈ { , } and l ∈ N . Then there is a constant C , depending only on C , ν, B , l , suchthat for any pairwise distinct x ′ , x ′′ , y ∈ R , k ϕ y ( x ′ ) ∇ hx ′ t l ( x ′ , y ) − ϕ y ( x ′′ ) ∇ hx ′′ t l ( x ′′ , y ) kk x ′ − x ′′ k ν (6.49) ≤ C (1 + k x ′ k + k x ′′ k ) k x ′ − x ′′ k ν max x ∈{ x ′ ,x ′′ } (cid:18) b ν ( x, y ) + exp (cid:18) − B k x − y k (cid:19)(cid:19) . (6.50) The expression ∇ hx refers to ∇ x , if h = 1 and the identity otherwise.Proof. We can bound the left-hand side either with the triangle inequality or with the gradienttheorem. We will need both and begin with the second one.We can write the difference as a path integral of the gradient along a path Γ as described inFact 6.4. Hence, we begin with estimating the gradient. We use Lemma 5.3 for j = 0 , , ϕ y . The A in Lemma 5.3 is moved to the other side. The dependency on h isbounded against the worst case after the first upper bound. Hence, (cid:13)(cid:13) ∇ x (cid:0) ϕ y ( x ) ∇ hx t l ( x, y ) (cid:1)(cid:13)(cid:13) (6.51) ≤ X j ∈{ , , } C (1 + k x k ) j exp ( k x − y k ) C (cid:18) b j ( x, y ) + exp (cid:18) − B k x − y k (cid:19)(cid:19) (6.52) N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 25 ≤ C (1 + k x k ) exp ( k x − y k ) (cid:18) b ( x, y ) + exp (cid:18) − B k x − y k (cid:19)(cid:19) (6.53) ≤ C (1 + k x k ) (cid:18) b ( x, y ) + exp (cid:18) − B k x − y k (cid:19)(cid:19) . (6.54)W.l.o.g. we assume that k x ′ − y k ≤ k x ′′ − y k . Along the path Γ, the second half of the upper boundfor the gradient maximizes at ( x ′ , y ). For any x on the path Γ, k x k ≤ C ( k x ′ k + k x ′′ k ). Bounding thepath integral by the supremum times the path length, we get k ϕ y ( x ′ ) ∇ hx ′ t l ( x ′ , y ) − ϕ y ( x ′′ ) ∇ hx ′′ t l ( x ′′ , y ) k (1 + k x ′ k + k x ′′ k ) (6.55) ≤ C (cid:18) b ( x ′ , y ) + exp (cid:18) − B k x ′ − y k (cid:19)(cid:19) π k x ′ − x ′′ k (6.56) ≤ C k x ′ − x ′′ k ( k x ′ − y k − if k x ′ − y k ≤ / √ B , exp (cid:0) − B k x ′ − y k (cid:1) if k x ′ − y k > / √ B . (6.57)The constant C depends on C , B , l .Now we use the triangle inequality to bound the difference. Then we use Lemma 5.3, the upperbound for ϕ and, that k x ′ − y k ≤ k x ′′ − y k . We also use, that b ≤ Cb . The denominator on theleft-hand side appears, if we have h = 1, as we move A to the other side in Lemma 5.3. We squareit for convenience. Hence, we get k ϕ y ( x ′ ) ∇ hx ′ t l ( x ′ , y ) − ϕ y ( x ′′ ) ∇ hx ′′ t l ( x ′′ , y ) k (1 + k x ′ k + k x ′′ k ) (6.58) ≤ X x ∈{ x ′ ,x ′′ } C exp ( k x − y k ) C (cid:18) b ( x, y ) + exp (cid:18) − B k x − y k (cid:19)(cid:19) (6.59) ≤ C ( k x ′ − y k − if k x ′ − y k ≤ / √ B , exp (cid:0) − B k x ′ − y k (cid:1) if k x ′ − y k > / √ B . (6.60)Once again, the constant C depends on C , B , l .As we have these two upper bounds (6.57) and (6.60), we can use the fact min( a, b ) ≤ a ν b − ν for any a, b > 0. Hence, we know k ϕ y ( x ′ ) ∇ hx ′ t l ( x ′ , y ) − ϕ y ( x ′′ ) ∇ hx ′′ t l ( x ′′ , y ) k (1 + k x ′ k + k x ′′ k ) (6.61) ≤ C k x ′ − x ′′ k ν ( k x ′ − y k − ν if k x ′ − y k ≤ / √ B , exp (cid:0) − B k x ′ − y k (cid:1) if k x ′ − y k > / √ B , (6.62) ≤ C k x ′ − x ′′ k ν (cid:18) b ν ( x ′ , y ) + exp (cid:18) − B k x ′ − y k (cid:19)(cid:19) . (6.63)The constant C depends only on C , B , l, ν . If we divide both sides by k x ′ − x ′′ k ν , multiply by(1 + k x ′ k + k x ′′ k ) and write the maximum over x ′ , x ′′ on the right-hand side, we get the desiredresult. It is a maximum over x ′ , x ′′ , as we assumed that k x ′ − y k ≤ k x ′′ − y k . (cid:3) We can now finally present the proof of Theorem 4.7. Proof of Theorem 4.7. We begin by conjugating with the unitary operator U x : L ( R ) → L ( R ),as defined in Lemma A.1. Hence, we have (cid:13)(cid:13)(cid:13) [0 , + x ( T l H ε ) m P l ( H ε T l ) k − m D ∁ R ( x ) (cid:13)(cid:13)(cid:13) p = (cid:13)(cid:13)(cid:13) [0 , ( T l H ε,x ) m P l ( H ε,x T l ) k − m D ∁ R (0) (cid:13)(cid:13)(cid:13) p . (6.64) We have already done this exact step in (6.22), but as it is a pivotal part of this proof, we wanted to repeat ithere for the reader’s convenience. For x ∈ [0 , , z ∈ D ∁ r (0) we define θ ( x, z ) := iker( T l H ε,x ) m P l ( H ε,x T l ) k − m ( x, z ) . (6.65)We choose 0 < ν < 1, such that p > γ + ν +1 . We want to use Proposition 4.6 and the remarkbelow it. Hence, we have N ( γ, ν, ( T l H ε,x ) m P l ( H ε,x T l ) k − m ) (6.66)= Z D ∁ R (0) dz Z [0 , dx ′ Z [0 , dx ′′ (cid:13)(cid:13) ∇ ⊗ γx ′ θ ( x ′ , z ) − ∇ ⊗ γx ′′ θ ( x ′′ , z ) (cid:13)(cid:13) k x ′ − x ′′ k ν (6.67)+ Z D ∁ R (0) dz Z [0 , dx k θ ( x, z ) k . (6.68)The second summand can be bounded easily. We use Theorem 6.3, noting α = 2 m, α ≥ kε , tobound θ . Hence, Z D ∁ R (0) dz Z [0 , dx k θ ( x, z ) k (6.69) ≤ C Z D ∁ R (0) dz Z [0 , dx exp (cid:0) − λ k x − z k (cid:1) (1 + k x + x k ) kε ≤ C exp (cid:0) − λR (cid:1) (1 + k x k ) kε . (6.70)In the last step we used Lemma A.5. The constant C depends on B , l, k and m .We will now assume m > 0. The case m = 0 is the easier special case. We define θ ( y, z ) := iker (cid:2) ( T l H ε,x ) m − P l ( H ε,x T l ) k − m (cid:3) ( y, z ) . (6.71)We can now write ∇ ⊗ γx θ ( x, z ) = ∇ ⊗ γx Z R t l ( x, y ) ( W ε ( y + x ) + 2 A ε ( y + x ) · ( − i ∇ y − A ( y ))) θ ( y, z ) dy. (6.72)The differential is resolved by the product rule. We will now deal with a single component ofthis element of C γ . We assume without loss of generality, that we differentiate at least once inthe direction corresponding to ∂ . Let m + m = γ − m + 1 times ∇ and m times ∇ has the form ∂ m +11 ∂ m Z R t l ( x, y ) H ε ( y + x ) θ ( y, z ) dy. (6.73)We know, that A ε , W ε ∈ C γ − b by the assumptions on B ε , V ε and θ ( · , z ) ∈ W γ, ( R ) by Theorem 6.3.Hence, we can apply Lemma 6.2 with the pair ( m , m ). Hence, we have(6.73) = ∂ Z R X k ≤ m ,k ≤ m s k k m m ( x − y ) t l ( x, y ) ∂ k y ∂ k y ( H ε,x θ ( · , z )) ( y ) dy. (6.74)As the second factor is in L ( R ) and the differential of the first term is in L . ( R ), we can exchangeintegration and differentiation by dominated convergence. Hence, we finally have(6.73) = Z R X k ≤ m ,k ≤ m ∂ x ( s k k m m ( x − y ) t l ( x, y )) ∂ k y ∂ k y ( H ε,x θ ( · , z )) ( y ) dy. (6.75)We define ϕ yhk k as the unique polynomials in x − y , such that ∂ x ( s k k m m ( x − y ) t l ( x, y )) = ϕ y k k ( x ) t l ( x, y ) + ϕ y k k ( x ) ∂ x t l ( x, y ) . (6.76)As they are polynomials, they are exponentially bounded and admissible for Lemma 6.5. In thelast few steps, we have established the identity ∂ m +1 x ∂ m x θ ( x, z ) = Z R X k ≤ m ,k ≤ m X h =0 ϕ yhk k ( x ) ∂ hx t l ( x, y ) ∂ k y ∂ k y ( H ε,x θ ( · , z )) ( y ) dy. (6.77) N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 27 Now we bound the difference of this term at x = x ′ and x = x ′′ , (cid:12)(cid:12)(cid:12) ∂ m +1 x ′ ∂ m x ′ θ ( x ′ , z ) − ∂ m +1 x ′′ ∂ m x ′′ θ ( x ′′ , z ) (cid:12)(cid:12)(cid:12) k x ′ − x ′′ k ν (6.78) ≤ P k ≤ m ,k ≤ m k x ′ − x ′′ k ν (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Z R X h =0 ϕ yhk k ( x ′ ) ∂ hx ′ t l ( x ′ , y ) − ϕ yhk k ( x ′′ ) ∂ hx ′′ t l ( x ′′ , y ) ! (6.79) × ∂ k y ∂ k y (( W ε ( y + x ) + 2 A ε ( y + x ) · ( − i ∇ y − A ( y ))) θ ( y, z )) dy (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (6.80) ≤ C Z R dy max x ∈{ x ′ ,x ′′ } (cid:16) b ν ( x, y ) + exp (cid:0) − B k x − y k (cid:1)(cid:17) (1 + k y k ) α exp (cid:0) − λ k y − z k (cid:1) k x ′ − x ′′ k ν (1 + k y + x k ) (1+( k − ε (6.81) ≤ C max x ∈{ x ′ ,x ′′ } (1 + k x k ) α exp (cid:0) − λ ′ k x − z k (cid:1) k x ′ − x ′′ k ν (1 + k x k ) kε (6.82) ≤ C exp (cid:16) − λ ′ k− z k (cid:17) k x ′ − x ′′ k ν (1 + k x k ) kε . (6.83)In the second step we pulled the absolute value inside, used Lemma 6.5 on the first term, used, that A ε , V ε are ( γ, C d ) tame to bound the first γ − A ε and W ε and used Theorem 6.3 tobound the first γ differentials of the integral kernel θ . As the resulting upper bound is independentof h, k , k , the sums can be absorbed in the constant C .Then we used Lemma 5.4 in the third step. We also used, that x ′ , x ′′ ∈ [0 , to get the secondfactor in the denominator independent of x ′ , x ′′ . The constant C depends on B , l, m, k, γ, ν, C d , ε but not on x , x ′ , x ′′ , z . In the last step, we used x ′ , x ′′ ∈ [0 , and Lemma A.5.We will now switch to the case m = 0.We want to bound the integrand in (6.67). The first step is to write the difference as the pathintegral of the gradient, where the path is chosen as in Fact 6.4. Then we bound this by the supremumtimes the path length. Hence, we get (cid:13)(cid:13) ∇ ⊗ γx ′ θ ( x ′ , z ) − ∇ ⊗ γx ′′ θ ( x ′′ , z ) (cid:13)(cid:13) k x ′ − x ′′ k ν ≤ π max x ∈ im (Γ) (cid:13)(cid:13)(cid:13) ∇ ⊗ ( γ +1) x θ ( x, z ) (cid:13)(cid:13)(cid:13) k x ′ − x ′′ k ν . (6.84)We apply Theorem 6.3 to all components of the right-hand side. As m = 0, we can apply thiswithout requiring differentiability of A ε , V ε . We get α = γ + 1 and α = kε . Thus, we have (cid:13)(cid:13) ∇ ⊗ γx ′ θ ( x ′ , z ) − ∇ ⊗ γx ′′ θ ( x ′′ , z ) (cid:13)(cid:13) k x ′ − x ′′ k ν (6.85) ≤ C max x ∈ im (Γ) (cid:0) (1 + k x k ) γ +1 exp (cid:0) − λ k x − z k (cid:1)(cid:1) k x ′ − x ′′ k ν (1 + k x k ) kε (6.86) ≤ C exp (cid:0) − λ k− z k (cid:1) k x ′ − x ′′ k ν (1 + k x k ) kε , (6.87)(6.88)for some C, λ > 0. In the last step, we used x ′ , x ′′ ∈ [0 , and hence im (Γ) ⊂ [ − , , Lemma A.5and ν ≤ ν . This is essentially the same upper bound, as we have for the case m > N for general m .We can resolve the integral over x ′′ in (6.67), as 2 × ν < 2. The integral over x ′ is then anintegral over a bounded function. The final step is Lemma A.5. Hence, we get Z D ∁ R (0) dz Z [0 , dx ′ Z [0 , dx ′′ (cid:13)(cid:13) ∇ ⊗ γx ′ ( θ ( x ′ , z )) − ∇ ⊗ γx ′′ ( θ ( x ′′ , z )) (cid:13)(cid:13) k x ′ − x ′′ k ν (6.89) ≤ Z D ∁ R (0) dz Z [0 , dx ′ Z [0 , dx ′′ C exp (cid:0) − λ ′ k z k (cid:1) k x ′ − x ′′ k ν (1 + k x k ) kε (6.90) ≤ Z D ∁ R (0) dzC exp (cid:0) − λ ′ k z k (cid:1) (1 + k x k ) kε (6.91)= Cπ exp( − λ ′ R )(1 + k x k ) kε . (6.92)This finishes the proof. (cid:3) Appendix A.Recall, the operator H ε,x has been defined in (6.20) Lemma A.1. Let x ∈ R . Then there is a unitary operator U x : L ( R ) → L ( R ) , such that thefollowing identities hold for any f ∈ L ∞ ( R ) : U x f ( X ) U − x = f ( X + x ) , (A.1) U x ( − i ∇ − A ) U − x = ( − i ∇ − A ) , (A.2) U x H ε U − x = H ε,x , (A.3) U x T l U − x = T l . (A.4) Here, X refers to the multiplication operator with the identity on R and f ( X ) is defined by functionalcalculus and hence the multiplication operator with the function f .Proof. In this proof ψ will always refer to a C ∞ c ( R ) function, that we regard as the test function forour densely defined operators. For any x ∈ R , we define the three unitary operators U x , U x , U x by ∀ x ∈ R : ( U x ψ ) ( x ):= ψ ( x + x ) , (A.5) ∀ x ∈ R : ( U x ψ ) ( x ):= ψ ( x ) exp (cid:18) − i B h x | Jx i (cid:19) , (A.6) U x := U x U x . (A.7)We conjugate our operator with the unitary operator U x . This does not change the Schatten norm.We have U x U x f ( X ) U − x U − x = U x f ( X ) U − x (A.8)= f ( X + x ) . (A.9)Now, we need to check how ( − i ∇ − B JX ) behaves under conjugation with U x . Hence, we get (cid:18) U x (cid:18) − i ∇ − B JX (cid:19) U − x U − x ψ (cid:19) ( x ) (A.10)= exp (cid:18) − i B h x | Jx i (cid:19) (cid:18) − i ∇ x − B Jx (cid:19) exp (cid:18) i B h x | Jx i (cid:19) ψ ( x − x ) (A.11)= (cid:18) − i ∇ x − B Jx (cid:19) ψ ( x − x ) + ψ ( x − x ) ( − i ∇ x ) (cid:18) i B h x | Jx i (cid:19) (A.12)= (cid:18) − i ∇ x − B J ( x − x ) (cid:19) ψ ( x − x ) (A.13)= (cid:18) U − x (cid:18) − i ∇ − B JX (cid:19) ψ (cid:19) ( x ) . (A.14)In the second step, we used the product and chain rule and the exponentials cancel. The interiorderivative is then resolved in the next step. N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 29 In conclusion, we have U x ( − i ∇ − A ) U − x = ( − i ∇ − A ) . (A.15)This implies U x T l U − x = T l . (A.16)Together with (A.9), this implies the identity U x H ε U − x = H ε,x . (A.17)This finishes the proof. (cid:3) Lemma A.2. Let Ω ⊂ R n be measurable and let f : Ω → C be integrable. Then we have the identity Z [0 , n X z ∈ Z n ,z + h ∈ Ω f ( z + h ) dh = Z Ω f ( x ) dx. (A.18) Proof. We observe Z [0 , n X z ∈ Z n ,z + h ∈ Ω f ( z + h ) dh = Z [0 , n X z ∈ Z n Ω ( z + h ) f ( z + h ) dh (A.19)= Z R n Ω ( x ) f ( x ) dx (A.20)= Z Ω f ( x ) dx. (A.21)In the second step we used Fubini with [0 , n × Z n = R n . (cid:3) Lemma A.3. Let f ∈ C ( R ) with f ′ ≤ and lim t →∞ f ( t ) = 0 , h ∈ C ( R , R ) and Λ ⊂ R bemeasurable. Then we have Z Λ f ( h ( x )) dx = Z R − f ′ ( t ) |{ x ∈ Λ | h ( x ) ≤ t }| dt. (A.22) As both integrands are positive, we do not need to require the existence of the integral, both sidesbeing ∞ is an option.Proof. We use the fundamental theorem of calculus and Fubini. As everything is positive, we canapply both theorems. Thus, Z Λ f ( h ( x )) dx = Z Λ dx Z ∞ h ( x ) ( − f ′ ( t )) dt (A.23)= Z R dx Z R dt Λ ( x )1 ( h ( x ) , ∞ ) ( t )( − f ′ ( t )) (A.24)= Z R dt Z R dx Λ ( x )1 ( h ( x ) , ∞ ) ( t )( − f ′ ( t )) (A.25)= Z R − f ′ ( t ) |{ x ∈ Λ | h ( x ) ≤ t }| dt. (cid:3) Lemma A.4. Let Λ ⊂ R be a bounded Lipschitz region. Then there is a constant C > , such thatfor any r > (cid:12)(cid:12)(cid:12) { x ∈ Λ | dist( x, Λ ∁ ) ≤ r } (cid:12)(cid:12)(cid:12) ≤ Cr, (A.26) (cid:12)(cid:12)(cid:12) { x ∈ Λ ∁ | dist( x, Λ) ≤ r } (cid:12)(cid:12)(cid:12) ≤ C ( r + r ) . (A.27)In both cases, for small r we have an approximately linear dependency. In the first case, it isbounded by | Λ | < ∞ and in the second case it is contained in a ball of radius r + r , which explainsthe r term. Lemma A.5. Let R, λ > be real numbers and x , x ∈ R with k x − x k ≤ R . Then we have exp( − λ k x k ) ≤ e λR exp( − λ k x k ) , (A.28) Z D ∁ R (0) exp( − λ k x ′ k ) dx ′ = πλ exp( − λR ) . (A.29)For k x k ≤ R , the estimate is trivial. Otherwise, the proof follows by taking the ln, dividing by λ and then completing the square. Lemma A.6. For every t ∈ (0 , , let K t : L ∞ ( R ) → L ∞ ( R ) be an operator with a nice integralkernel k t : R × R → C , as defined in Definition 2.2. Assume, that for every x ∈ R , the function [0 , × R → C : ( t, y ) k t ( x, y ) is integrable, its integral is bounded independently of x , and thesame holds for x and y reversed. Then we have iker (cid:18)Z K t dt (cid:19) ( x, y ) = Z k t ( x, y ) dt. (A.30) Proof. The integral R K t dt exists as a Bochner integral with respect to the operator norm fromL ∞ ( R ) to L ∞ ( R ) by the integrability assumptions on the kernel. Let f ∈ C c ( R ). Then, for every x ∈ R , we have (cid:18)(cid:18)Z K t dt (cid:19) f (cid:19) ( x ) = (cid:18)Z K t f dt (cid:19) ( x ) (A.31)= Z (cid:18)Z R k t ( x, y ) f ( y ) dy (cid:19) dt (A.32)= Z R (cid:18)Z k t ( x, y ) dt (cid:19) f ( y ) dy. (A.33)The first step holds, as the Bochner integral commutes with the (linear, bounded) evaluation op-erator. The second step is the definition of k t and the last step is Fubini, as f is bounded and weassumed k · ( x, · ) to be integrable for any x ∈ R . The same holds, if x and y are reversed, hencethis is a nice integral kernel again. (cid:3) Appendix B. Proof of Lemma 3.2. We define A ε ( x ) := Z R Jy π k y k B ε ( x − y ) dy. (B.1)The last property will be seen by bounding this integral. C ε will be a constant depending onlyon ε , that may change from line to line. To begin with we have the bound k A ε ( x ) k ≤ Z R k y k C d (1 + k y − x k ) ε dy (B.2) ≤ Z D k x k (0) k y k C d (1 + k y − x k ) ε dy (B.3)+ Z D ∁ k x k (0) k y k C d (1 + k y − x k ) ε dy (B.4) ≤ Z D (0) k y k C d ( k x k + k y − e k ) ε k x k − − − ε dy (B.5)+ Z D ∁ k x k (0) k y k C d (1 + k y/ k ) ε dy (B.6) ≤ min (cid:8) C ε C d k x k − ε , C ε C d k x k (cid:9) (B.7) N THE STABILITY OF THE AREA LAW FOR THE E.E. OF THE LANDAU HAMILTONIAN 31 + C ε C d max {k x k , } − ε + C ε C d D (0) ( x ) (B.8) ≤ C ε C d (1 + k x k ) ε . (B.9)In the second to last step, we got the first minimum by ignoring either of the summands in thedenominator of the bounded domain integral and for the second part we just did a different boundon the annulus from k x k to 1, if k x k < F ( f )( ξ ) := 12 π Z R f ( x ) exp( − ix · ξ ) dx, ξ ∈ R (B.10)for any n ∈ N and f ∈ L ∩ L ( R , C n ). It can be expanded to tempered distributions and has thefollowing properties for any ξ ∈ R : F ( · f ( · ))( ξ ) = i ∇F ( f )( ξ ) , (B.11) F ( ∇ f ( · ))( ξ ) = − iξ F ( f )( ξ ) , (B.12) F (1)( ξ ) = 2 πδ ( ξ ) , (B.13) F ( f ∗ g )( ξ ) = 2 π F ( f )( ξ ) F ( g )( ξ ) . (B.14)Here δ refers to the δ -distribution at 0. Furthermore, the Fourier transform is linear and invertible.As A ε and B ε are bounded, they are both tempered distributions. Now we can apply the Fouriertransform to our first two claimed equations and are left to show − πiJξ · F (cid:18) J · π k·k (cid:19) ( ξ ) F ( B ε )( ξ ) = F ( B ε )( ξ ) , (B.15) − πiξ · F (cid:18) J · π k·k (cid:19) ( ξ ) F ( B ε )( ξ ) = 0 . (B.16)Basically, this equation does not depend on B ε . Now we have to compute the Fourier transform of Jx π k x k , F (cid:18) J · π k·k (cid:19) ( ξ ) = 12 π (cid:0) Ji ∇ ( − ∆) − F (1) (cid:1) ( ξ ) (B.17)= iJ ∇ ( − ∆) − δ ( ξ ) (B.18)= iJ ∇ π ln( k ξ k ) (B.19)= iJ ξ π k ξ k . 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