The Lieb-Thirring inequality revisited

We provide new estimates on the best constant of the Lieb-Thirring inequality for the sum of the negative eigenvalues of Schr\"odinger operators, which significantly improve the so far existing bounds.


Introduction
In 1975, Lieb and Thirring [19,20] proved that the sum of all negative eigenvalues of Schrödinger operators −∆ + V in L 2 (R d ), with a real-valued potential V : R d → R, admits the bound for a finite constant L 1,d > 0 depending only on the dimension, for all d ≥ 1.Here we use the convention that t ± = max{±t, 0}.Inequality (1) should be compared with Weyl's law [18,Theorem 12.12] where with |B 1 | the volume of the unit ball in R d .While (2) is only correct in the semiclassical limit h → 0, the Lieb-Thirring inequality (1) is a universal bound for all finite parameters.A simpler version of (1) is the following bound for a single eigenvalue, which is a consequence of Sobolev's inequality, namely some sort of the uncertainty principle.This inequality is essentially due to Keller [14]; see also [4] for a stability analysis.The Lieb-Thirring inequality (1) extends Sobolev's inequality (3) by taking the exclusion principle into account.The Lieb-Thirring conjecture [20] concerns the best constant in (1) and states that this is given by with L So 1,d being the best constant in (3).While the lower bound L 1,d ≥ max{L cl 1,d , L So 1,d } is obvious, proving the matching upper bound is a major challenge in mathematical physics.
The original proof of Lieb and Thirring [19] gave L 1,d /L cl 1,d ≤ 4π in d = 3.Since then, there have been many contributions devoted to improving the upper bound on L 1,d Date: August 27, 2018.
Our new result is Theorem 1.For all d ≥ 1, the best constant in the Lieb-Thirring inequality (1) satisfies Our estimate is a significant improvement over (5), but in one-dimension is still about 26% bigger than the expected value [20].Historically, the Lieb-Thirring inequality was invented to prove the stability of matter [19].In this context, it can be stated as a lower bound on the fermionic kinetic energy, Here γ is an arbitrary one-body density matrix on L 2 (R d ), i.e. 0 ≤ γ ≤ 1 with Tr γ < ∞, and γ(x, x) is the diagonal part of the kernel of γ (which can be defined properly by the spectral decomposition).By a standard duality argument, (1) is equivalent to (6), and the corresponding best constants are related by In particular, K d should be compared with the semiclassical constant which emerges naturally from the lowest kinetic energy of the Fermi gas in a finite volume.
In 2011, Rumin [23] found a direct proof of (6), without using the dual form (1). His method has been used to derive several new estimates, e.g. a positive density analogue of (6) in [10], and it will be also the starting point of our analysis.Note that Rumin's original proof [23] gives ), and hence namely L 1,1 /L cl 1,1 ≤ √ 5 = 2.236... when d = 1 and and worse estimates in higher dimensions.Therefore, new ideas are needed to push forward the bound.
Our proof of Theorem 1 contains several main ingredients: • First, we will modify Rumin's proof by introducing an optimal momentum decomposition.This gives L 1,1 /L cl 1,1 ≤ 1.618... in d = 1, which is already an improvement over the best-known result (5) in d = 1.
• Second, we use the Laptev-Weidl lifting argument to extend the bound L 1,d /L cl 1,d ≤ 1.618... to arbitrary dimension d, which is an improvement over the best-known result (5).The idea of lifting with respect to dimension is by now classical [16,13,6], but its combination with Rumin's method is not completely obvious and requires an improvement of the bound in [9].
• Third, we take into account a low momentum averaging.This improves further the bound to L 1,1 /L cl 1,1 ≤ 1.456 in d = 1 (and worse estimates in higher dimensions).This is one of our key ideas and deviates substantially from Rumin's original argument.
• Finally, we transfer the one-dimensional bound in the last step to higher dimensions by the lifting argument again.
These steps will be explained in the next four sections.For the proof of Theorem 1 only the last two sections are relevant, but we feel that a slow presentation of the various new ideas might be useful.
As a by-product of our method we obtain Lieb-Thirring inequalities for fractional Schrödinger operators.The inequalities we are interested in have the form and Again, a duality argument shows that the optimal constants in these two inequalities satisfy the relation Finally, the semi-classical constants are given by The main ingredients of the proof of Theorem 1, except the lifting argument, apply equally well to the fractional case.This gives Theorem 2. For all d ≥ 1 and σ > 0, the best constant in the Lieb-Thirring inequality (10) satisfies with the infimum taken over all functions f, ϕ : R In particular, when σ = 1/2 and d = 3, we have C 3,1/2 ≤ 0.046737 and hence The proof of Theorem 2 is presented in the last section; see also Remark 7 in Section 3.For σ = 1 and d > 1, the bound from Theorem 2 is not as good as the lower bound in Theorem 1.For all other cases, Theorem 2 yields the best known constants.In particular in the physically relevant case σ = 1/2 and d = 3, i.e., the ultra-relativistic Schrödinger operator in three dimensions, where K cl 3,1/2 = 3 4 (6π 2 ) 1/3 = 2.923..., our result improves significantly the bounds K 3,1/2 /K cl 3,1/2 ≥ 0.6 in [23, p. 586] and K 3,1/2 K cl d,σ ≥ 0.558 in [5,Eq.(3.4)].
An immediate consequence of Theorem 2 is Corollary 3.For every fixed σ > 0, in the limit of large dimensions we have Indeed, from (11) we have . So ( 14) follows from the first lower bound in Theorem 2 and the fact that (sin(t)/t) 1/t → 1 as t → 0. Note that Rumin's original proof gives a bound similar to (14) but with e replaced by e 2 (see (8)).
As a consequence of ( 14) , we also have The lower bound lim inf d→∞ K d,σ /K cl d,σ ≥ 1 follows from ( 14) and the upper bound [9].Finally, we note that in 2013, Lundholm and Solovej [21] found another direct proof of the kinetic estimate (6).Their approach is based on a local version of the exclusion principle, which is inspired by the first proof of the stability of matter by Dyson and Lenard [7].Recently, the ideas in [21] have been developed further in [22] to show that for all d ≥ 1 and ε > 0 (the gradient error term is always smaller than the kinetic term [11]).Note that from ( 16), as well as from all existing proofs of the Lieb-Thirring inequality (including the present paper), the real difference between dimensions is not visible.Therefore, new ideas are certainly needed to attack the full conjecture (4).

Optimal momentum decomposition
In this section, we use a modified version of Rumin's proof in [23] to prove Proposition 4. For d ≥ 1, the best constant in the Lieb-Thirring inequality (6) satisfies In particular, when d = 1 we get Proof.Let γ be an operator on L 2 (R d ) with 0 ≤ γ ≤ 1.By a density argument, it suffices to consider the case when γ is a finite-rank operator with smooth eigenfunctions.For any function f : R + → R + with ∞ 0 f 2 = 1, using the momentum decomposition and Fubini's theorem we can write Next, we estimate the kernel of f (s/p 2 )γf (s/p 2 ).Using Cauchy-Schwarz and 0 ≤ γ ≤ 1, for every ε > 0 we have the operator inequalities This inequality implies for any x ∈ R d the kernel bound Optimizing over ε > 0 we obtain Moreover, it is straightforward to see that where Consequently, we deduce from ( 20) that Next, inserting ( 23) into ( 17) and integrating over s > 0 lead to Thus, Finally, it remains to minimize A f under the constraint ∞ 0 f 2 = 1.We note that the proof in [23] corresponds to f (t) = 1(t ≤ 1) (although the representation there is rather different), which gives A f = 1 but this is not optimal.From Lemma 5 below we have Inserting this into (25) we conclude the proof of Proposition 4.
In the previous proof we needed the following solution of a minimization problem.
Proof.Heuristically, the optimizer can be found by solving the Euler-Lagrange equation, but to make this rigorous one would have to prove that a minimizer exists.This can be easily done by setting h Since C is closed, which follows easily from Fatou's lemma, and does not contain the zero function, it contains a functions h * of minimal length.Necessarily h * ∈ ∂C, otherwise h * would be in the interior of C and we could shrink it, thus reducing its length a little bit, which is impossible.So h Hence f * is a minimizer which must obey the Euler-Lagrange equation.
A more direct solution is as follows We see that for any f : Here we used t Alternatively one can use a keyhole type contour encircling the positive real axis and the residue theorem, see [2, Section 11.1.III], to directly evaluate The functional equations Γ(1 + z) = zΓ(z) and Γ(z)Γ(1 − z) = π sin(πz) , the last one again valid for −1 < Re z < 1, yield This proves the claimed formula.

Lifting to higher dimensions. I
In dimension d = 1 Proposition 4 yields L 1,1 /L cl 1,1 ≤ 1.618435, which is better than for instance the bound in dimension d = 3, namely L 1,3 /L cl 1,3 ≤ 1.994584.In this section we use a procedure of Laptev and Weidl [15,16] to show that the higher-dimensional fraction L 1,d /L cl 1,d is at least as good as the low-dimensional one.The idea is to consider potentials V on R d that take values in the self-adjoint operators on some separable Hilbert space H.We are looking for an inequality of the form where tr denotes the trace in H, Tr the trace in , and where, by definition, the constant L op 1,d is independent of H. Taking H one-dimensional we see that (26) coincides with (1) and therefore It is not known whether L 1,d and L op 1,d coincide, but in this section we will show that the upper bound on L 1,d from Proposition 4 is, in fact, also an upper bound on L op 1,d .We show this by using the classical duality argument.This shows the analogue of (7), that is, where K op d denote the best constant in the inequality for all operators γ on L 2 (R d ; H) satisfying 0 ≤ γ ≤ 1, where H is an arbitrary (separable) Hilbert space.For such γ, one can consider γ(x, x) as a non-negative operator in H.
The following proof improves an argument from [9].Proposition 6.For d ≥ 1, the best constant in the Lieb-Thirring inequality (29) satisfies In particular, when d = 1 we get K op 1 /K The operator inequality (18) implies that for any x ∈ R d one has (19), understood as an operator inequality in H. Denoting by λ n (T ) the nth eigenvalue, in decreasing order and taking multiplicities into account, of a non-negative operator T , we infer from (19), the variational principle and the computation ( 21) that for any n ∈ N, At this stage we can optimize over ε > 0 and obtain (31) Thus, For fixed n (and x) we obtain after integration over s, .
Summing over n and integrating with respect to x we obtain by ( 30) .
The proposition now follows in the same way as Proposition 4.
Remark 7. The same proof yields the operator-valued analogue of Theorem 2. Since there seems to be no analogue of the following proposition for (−∆) σ with σ = 1, we do not write this out.
In order to obtain good constants in higher dimensions we recall the following bound which is essentially due to Laptev and Weidl [16].The extension to d 1 ≥ 2, which is not needed here, but is interesting in its own right, is due to [12].
In particular, taking d 1 = 1 and using the bound from Proposition 6 together with (27) we obtain the following bound.

Corollary 9. For any
The proof of Proposition 8 is by now standard, but we sketch it for the sake of completeness.We need the following more general family of Lieb-Thirring inequalities, as well as the semi-classical constant • where again V takes now values in the self-adjoint operators on some auxiliary separable Hilbert space H and its negative part V (x) − is in the α + d 2 von Neumann-Schatten ideal, tr denotes the trace over H, and Tr the trace over The celebrated result by Laptev and Weidl [16] says that L op α,d = L cl α,d for any α ≥ 3/2 and any d ≥ 1. (For d = 1, α = 3/2 and in the scalar case, this was shown in the original paper of Lieb and Thirring [20].) Proof of Proposition 8. We follow the argument in [12] closely: Let V be a function on R d taking values in the self-adjoint operators in some Hilbert space H.For any 2 the bound from [16] implies, for any Combining the last two inequalities and observing that (see [12] for a non-computational proof of this identity), we obtain the claimed inequality.

Low momentum averaging
Our main idea to improve the estimate in Proposition 4 is to average over low momenta s ≤ E before using the Cauchy-Schwarz inequality (18).We will actually push forward this idea by adding a weight function.This leads to Proposition 10.For d ≥ 1, the best constant in the Lieb-Thirring inequality (6) satisfies where with the infimum taken over all functions f, ϕ : R Recall the momentum decomposition (17).We have for any and therefore, for every E > 0, This implies that we have the operator inequality with Next, by the Cauchy-Schwarz estimate similarly to (18) (thanks to 0 ≤ γ ≤ 1) we have for every ε > 0. Combining (36) and (38) we get Transferring (39) to a kernel bound, using the same computation as in ( 21)- (22), and then optimizing over ε > 0 we obtain Then optimizing over E > 0 leads to Inserting this into (17) we conclude that namely the best constant in (6) satisfies Optimizing over f, ϕ leads to (34).
Therefore, when d = 1 we get the desired lower bound The upper bound on C 1 requires an explicit choice of (f, ϕ).The analysis from Section 2 suggests the following choice which gives C 1 ≤ 0.381378.We can do slightly better by taking with µ 0 and c 0 determined by

Lifting to higher dimensions. II
In this section we proceed analogously to Section 3 to extend Proposition 10 to the operator-valued case.
Proposition 12.For d ≥ 1, the best constant in the Lieb-Thirring inequality (29) satisfies and denote g as in (37).We follow the proof of Proposition 10 to arrive at the operator inequality (39).As in the proof of Proposition 6 this implies for any x ∈ R d and n ∈ N, Finally, optimizing over E > 0 leads to Inserting this into (17) we conclude that Finally, it remains to optimize over f, ϕ to obtain (43).The numerical values when d = 1 are obtained from the upper bound on C 1 in Lemma 11.

Bounds with fractional operators
The proof of Theorem 2 is essentially the same as that of Theorem 1 (except we do not use the lifting argument) and we only sketch the major steps.Transfering the latter to a kernel bound, using the same computation as in ( 46)-(47), and optimizing over ε > 0 and then E > 0 we obtain the following analogue of (41),

β
and equality is achieved if and only if
the last one.This shows that the infimum is attained if and only if f = f * .It remains to compute the infimum and µ * .Both follow from the formula [1, Abramowitz-Stegun, 6.2.1 and 6.2.2] Proof.Let γ be an operator on L 2 (R d ; H) with 0 ≤ γ ≤ 1.By a density argument we may assume that H is finite-dimensional and that γ is finite rank and with smooth eigenfunctions.The analogue of (17) is cl 1 ≥ 0.381777 and L op 1,d /L cl 1,d ≤ 1.618435.