The Berezin inequality on domains of infinite measure

TheBerezin inequality gives an upper bound on theRieszmeans of themagnetic Schrödinger operator on a set of finite volume.Wefindan analogous inequality for the magnetic operator with homogeneous magnetic field on sets whose complement in R 2 has finite measure. Similar bounds are obtained for the Heisenberg sub-Laplacian.

and is sharp, which follows from an asymptotic result by Weyl [17]. For the inequality to hold it is essential that the Laplace operator is considered on the set c of finite volume. This guarantees that H only has discrete spectrum consisting of eigenvalues converging to infinity, showing that the left-hand-side of (1) exists. In [5] a similar result to the Berezin inequality (1) has been established for the Dirichlet Laplace operator on the set of infinite measure. To this end one introduces the orthogonal projection P : L 2 (R d ) → L 2 ( ), i.e. the multiplication with the characteristic function χ . The operator P (− )P corresponds to the Laplacian on the set with Dirichlet boundary conditions. Since the continuous spectrum of P (− )P contains the positive real axis, the operator (P (− )P − λ) − is not trace-class on L 2 (R d ). However, it can be compared to a suitable operator to achieve similar results to (1). The authors of [5] considered the difference (− − λ) − − (P (− )P − λ) − and proved that which can be seen as an analogue of the Berezin inequality for perturbations of the continuous spectrum of the Laplace operator. Bounds on traces for these types of problems are a fairly recent research area and we point to [5] for a generalisation of Lieb-Thirring inequalities to this setting. In our paper we aim to find an analogous inequality to (2) for the magnetic operator H B = (−i∇ + A(x)) 2 . Similar to the case of the Laplacian, problems stem from the fact that (P H B P − λ) − is not trace-class. Thus we consider the difference (H B −λ) γ − −(P H B P −λ) γ − and establish lower bounds on the trace of this operator. We also prove a similar inequality for the sub-Laplacian L on the first Heisenberg group H 1 .A key observation for our results is that, for any self-adjoint operator H , a formal computation involving the Berezin-Lieb inequality for convex functions (see [1] and [14]) yields the result It is the object of this work to give correct mathematical meaning to this observation and to explicitly calculate the right-hand-side for the two special choices of H . The Berezin inequality (1) on domains of finite measure has inspired a number of authors and is related to the Li-Yau inequality [13]. In their paper the authors showed that the sum over the first k eigenvalues λ 1 , . . . , λ k of − D c can be bounded from below as This was later proven to be a corollary of (1) via the Legendre transformation, see [11].
In [9] comparable inequalities were established for various classes of differential and pseudo-differential operators including ((− ) α ) D c with α > 0. A similar inequality to (1) can be found for Schrödinger operators with magnetic fields in the case d = 2. The operator H D B, c := (−i∇ + A(x)) 2 on L 2 ( c ) with Dirichlet boundary conditions and arbitrary vector field A satisfies for all γ ≥ 3 2 , which follows from a result by Laptev and Weidl in [12] (see also [6]). In [4] this was generalised to γ ≥ 1 under the restriction that the magnetic field B = d A is constant. In this case the upper bound in (4) can be improved by allowing it to depend on B as shown in [6]. In their paper the authors also proved that, under the assumption that c is a tiling domain, this inequality also holds if 0 ≤ γ < 1, where it is sharp. For γ = 1 the right-hand-side of (5) can be adapted to magnetic operators with additional external potentials V , see [7]. The Berezin inequality was furthermore extended to the sub-Laplacian L on the Heisenberg group H 1 . In [8] (see also [16]), it was proven that the Dirichlet realisation L D c of L on a domain c ⊂ H 1 of finite measure satisfies In our paper we obtain lower bounds on the traces of the differences γ − which are of the same form as the upper bounds in (5) and (6), respectively.
The paper is organised as follows. In Sect. 2 we discuss (3) in the general setting of H being a self-adjoint operator on L 2 (R d ). We then state our main results for the magnetic operator H B with constant magnetic field and the sub-Laplacian L in Theorems 2 and 3, respectively. The complete proofs of these results are given in the subsequent sections.

Statement of the main results
Let H be a self-adjoint operator on the space L 2 (R d ) and let ϕ : R → R be a convex function such that ϕ(H ) − ϕ(P H P ) and ϕ(H ) − P ϕ(H )P are both trace-class. Under these assumptions a generalisation of the Berezin-Lieb inequality as derived in [10] states that tr (P ϕ(H )P − ϕ(P H P )) ≥ 0.
As a consequence, we obtain the inequality by making use of the additivity of the trace. We now simplify the right-hand-side of (7) as follows. To shorten notation denote the trace-class operator Q := ϕ(H ) − P ϕ(H )P and let P c = I − P be the complementary projection of P . Clearly Q can be written as the sum of four operators corresponding to the decomposition of L 2 (R d ) into ran P and ran P c , i.e.
In [15][Theorem VI.25] it is shown that, if T is trace-class and S is bounded, then tr(ST ) = tr(T S). As a result tr(P Q P c ) = 0 as well as tr(P c Q P ) = 0. Thus the trace of Q consists only of the diagonal terms These results are summarised in the following theorem.

Theorem 1 Let H be a self-adjoint operator on L
Then the Berezin-Lieb type inequality While this result is true for arbitrary self-adjoint operators H , we shall now apply it to two special choices of H to obtain the main results of this work. First, consider Schrödinger operators with magnetic fields. Let the magnetic field B(x) be a two-form on R d and the magnetic vector potential A(x) a one-form satisfying B(x) = d A(x). We shall restrict ourselves to the case d = 2 and in the remainder of this work, we furthermore assume that B is constant and positive. Consider the magnetic operator H B = (−i∇ + A(x)) 2 , which is defined as the closure of the form , the set of smooth functions with compact support. The obtained operator is found to be self-adjoint and we can state the first main result.
holds and the right-hand-side can be calculated explicitly as The proof of Theorem 2 is provided in Sect. 3. The lower bound (9) coincides with the upper bound (5) for the magnetic operator on the set c of finite volume. In essence the proof is the same.
Similar results can also be obtained on the first Heisenberg group H 1 . Here, H 1 is considered to be the three-dimensional space R 3 equipped with the non-commutative multiplication On the Heisenberg group, we introduce the two left-invariant vector fields Using these definitions, we consider the quadratic form and note that the closure of this form gives the self-adjoint sub-Laplacian L = −X 2 1 − X 2 2 on H 1 . For a detailed background we refer to the literature, e.g. [3]. The sub-Laplacian L is found to satisfy the following analogue of the Berezin inequality.
Theorem 3 Assume λ ≥ 0, γ ≥ 1 and let be an open subset of R 3 such that R 3 \ has finite measure. Then the inequality holds for the sub-Laplacian L on H 1 and the right-hand-side can be calculated explicitly as Similarly to the previous application, the lower bound (11) coincides with the upper bound in the case of the Heisenberg sub-Laplacian being defined on the domain c of finite measure with Dirichlet boundary conditions, see (6). The proof of Theorem 3 is basically the same as in the case of finite measure [8] and can be found in Sect. 4. Note that this result can easily be generalised to the N -th Heisenberg group H N .
Remark 1 Using Theorem 1 we can also reproduce the results of Frank, Lewin, Lieb and Seiringer [5] and show that for the Laplacian on a set ⊂ R d with complement of finite measure.

The proof of Theorem 2
Let ϕ λ,γ : R → R be the convex function defined as Applying Theorem 1 to this function and the operator H B yields (8) and it only remains to prove (9). This can be done in complete analogy to calculations by Frank, Loss and Weidl [6]. The spectrum of H B is entirely discrete and can be calculated to be (2k + 1)B for k ∈ N ∪ {0}. The projection onto the k-th Landau level is denoted by B,k . The spectral theorem implies that the operator ϕ λ,γ (H B ) can then be written as We multiply this identity from both sides with the projection P c and consider the trace of the obtained expression, that is To explicitly calculate the summands on the right-hand-side of (13), we observe that by the cyclicity of the trace where · σ 2 denotes the Hilbert-Schmidt norm. This norm can be calculated explicitly by using the integral kernel of the operator P c B,k . Let B,k (x, y) be the integral kernel of B,k such that B,k ψ(x) = R 2 B,k (x, y)ψ(y) dy. The integral kernel of the composition P c B,k is then given by k (x, y).
We can calculate the Hilbert-Schmidt norm on the right-hand-side of (14) by double integration of the square of the modulus of this integral kernel, that is To explicitly solve this integral, we point out some important properties of the function B,k (x, y). As the orthogonal projection B,k is self-adjoint it must hold that k (y, x). By evaluating B,k at the delta distribution δ(x − x 0 ) and using the defining property of projections, B,k = B,k B,k , it can easily be concluded that It is furthermore a remarkable fact that the diagonal of the integral kernel of B,k is given by the constant B,k (x, x) = B 2π for all k ∈ N ∪ {0}. Using these properties, identity (15) can be continued as Inserting this equation back into (13) yields the final result which finishes the proof.

The proof of Theorem 3
Let the convex function ϕ λ,γ : R → R be defined as in (12). Theorem 1 applied to L yields (10) and it only remains to show (11) which can be proven following calculations by Hansson and Laptev [8]. Firstly, we introduce the Fourier transformation F x 3 with respect to the variable x 3 , A simple calculation shows that the Heisenberg sub-Laplacian L satisfies the identity For fixed x 3 , the right-hand-side of (17) can be identified with a two-dimensional Schrödinger operator with vector field A = x 3 2 (−x 2 , x 1 ). The corresponding magnetic field B = d A is constant and can be calculated to be B = x 3 . The eigenvalues of this magnetic operator H x 3 are (2k + 1)|x 3 | for k ∈ N ∪ {0}. Note that they depend on the variable x 3 . Similar to the previous section we shall use the Landau projections x 3 ,k to prove (11). With respect to the variable x 3 , the operator F x 3 LF * x 3 simply acts as a multiplication operator and consequently the spectral theorem allows us to write where we have used the tensor product x 3 ,k = x 3 ,k ⊗ I L 2 (R) with I L 2 (R) denoting the identity on L 2 (R). For convenience we introduce the notation μ k (x 3 ) = ϕ λ,γ ((2k + 1)|x 3 |) swallowing the dependence on λ and γ for the moment. As a consequence of (18) we obtain the identity This result can be compared to the analogous equation in the case of a magnetic operator (13). While in this setting the magnetic field was a given constant, we now consider a magnetic field that changes with the variable x 3 . In addition, the Fourier transformation F x 3 has to be dealt with. The summands on the right-hand-side of (19) can be written as Hilbert-Schmidt norms of certain operators, that is for every k ∈ N∪{0}. Here we have used that the multiplication operator μ k (x 3 ) and the projection x 3 ,k commute. The investigation of these Hilbert-Schmidt norms requires us to calculate the integral kernels of the operators involved. Let x = (x 1 , x 2 , x 3 ) and y = (y 1 , y 2 , y 3 ) be two vectors in R 3 . The integral kernel of P c F * can then be computed to be χ c (x) 1 √ 2π e ix 3 y 3 y 3 ,k (x 1 , x 2 , y 1 , y 2 )μ k (y 3 ) 1 2 . As a consequence we obtain the identity | y 3 ,k (x 1 , x 2 , y 1 , y 2 )| 2 dy 1 dy 2 dx dy 3 .
To calculate this integral, we recall (16) and stress again that the diagonal of the integral kernel y 3 ,k (x 1 , x 2 , y 1 , y 2 ) is is known to be the constant |y 3 | 2π . This results in where we have used the definition of μ k (y 3 ) to obtain the last equality. We insert this identity back into (19) and substitute p = (2k + 1)y 3 to conclude that tr P c ϕ λ,γ (L)P c = 1 Here, the last equality follows from the well-known fact that ∞ k=0 1 (2k+1) 2 = π 2 8 . The remaining integral can be easily calculated using partial integration