The Nonlinear Schrödinger Equation for Orthonormal Functions II: Application to Lieb–Thirring Inequalities

In this paper we disprove part of a conjecture of Lieb and Thirring concerning the best constant in their eponymous inequality. We prove that the best Lieb–Thirring constant when the eigenvalues of a Schrödinger operator -Δ+V(x)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$-\Delta +V(x)$$\end{document} are raised to the power κ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\kappa $$\end{document} is never given by the one-bound state case when κ>max(0,2-d/2)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\kappa >\max (0,2-d/2)$$\end{document} in space dimension d≥1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$d\ge 1$$\end{document}. When in addition κ≥1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\kappa \ge 1$$\end{document} we prove that this best constant is never attained for a potential having finitely many eigenvalues. The method to obtain the first result is to carefully compute the exponentially small interaction between two Gagliardo–Nirenberg optimisers placed far away. For the second result, we study the dual version of the Lieb–Thirring inequality, in the same spirit as in Part I of this work Gontier et al. (The nonlinear Schrödinger equation for orthonormal functions I. Existence of ground states. Arch. Rat. Mech. Anal, 2021. https://doi.org/10.1007/s00205-021-01634-7). In a different but related direction, we also show that the cubic nonlinear Schrödinger equation admits no orthonormal ground state in 1D, for more than one function.


Introduction and Main Results
This paper is a continuation of a previous work [GLN21] where the last two authors together with F.Q. Nazar studied the existence of ground states for the nonlinear Schrödinger equation (NLS) for systems of orthonormal functions. In the present paper, we exhibit a connection between the corresponding minimisation problem and the family of Lieb-Thirring inequalities [LT75,LT76,LS10], which enables us to prove results both for the Lieb-Thirring inequalities and the NLS equation studied in [GLN21].
1.1. Lieb-Thirring inequalities. The Lieb-Thirring inequality [LT75,LT76] is one of the most important inequalities in mathematical physics. It has been used by Lieb and Thirring [LT75] to give a short proof of the stability of matter [DL67,LD68,Lie90,LS10] and it is a fundamental tool for studying large fermionic systems. It is also a source of many interesting mathematical questions.

The finite rank Lieb-Thirring constant
Let d ≥ 1, κ ≥ 0 and N ≥ 1, and let L (N ) κ,d be the best constant in the finite rank Lieb- where a − = max(0, −a) and λ n (− + V ) ≤ 0 denotes the nth min-max level of − + V in L 2 (R d ), which equals the nth negative eigenvalue (counted with multiplicity) when it exists and 0 otherwise. The constant L (1) κ,d is finite by the Gagliardo-Nirenberg inequality, under the assumption that (2) These restrictions on κ are optimal in the sense that L (1) κ,d = ∞ for 0 ≤ κ < 1/2 in d = 1 and for κ = 0 in d = 2. Note that L (N ) κ,d is finite under the same restrictions as for L κ,d . The Lieb-Thirring theorem states that the limit is finite: This was proved by Lieb and Thirring [LT75,LT76] for κ > 1/2 in d = 1 and for κ > 0 in d ≥ 2. The critical cases κ = 0 in d ≥ 3 and κ = 1/2 in d = 1 are respectively due to Cwikel-Lieb-Rozenblum [Cwi77,Lie76,Roz72] and Weidl [Wei96]. An important question is to determine the value of the optimal Lieb-Thirring constant L κ,d . This plays for instance a central role in Density Functional Theory [LLS20]. One possibility is that it is attained for an optimal potential V having N < ∞ bound states, that is, L κ,d = L (N ) κ,d . An opposite scenario is that a sequence V N of optimal potentials for L (N ) κ,d tends to be very spread out and flat as N → ∞ so as to have more and more bound states. In this case L κ,d is equal to the semi-classical constant L sc κ,d := Indeed, recall that if we scale a fixed nice potential V with V − = 0 in the manner V ( x), we obtain in the limit → 0 Lieb and Thirring conjectured in [LT76] that the best constant should be given either by the one bound state case, or by semi-classical analysis: This conjecture has generated a huge interest in mathematical physics. Although the conjecture is still believed to hold in dimension d = 1, it is now understood that the situation is more complicated in dimensions d ≥ 2. In Sect. 1.1.3 below we will give a precise account of what is known and what is not as of today. Most of the previous works have focused on determining when L κ,d equals the semi-classical constant L sc κ,d . Much fewer works have studied the plausibility that L κ,d equals L (1) κ,d or even L (N ) κ,d for some N ≥ 1. In the next section we state our results in this direction.

Results on the non-optimality of the finite rank Lieb-Thirring constant
Our first theorem states that for an appropriate range of κ, the optimal constant in the Lieb-Thirring inequality can never be attained by a potential having finitely many bound states.
In particular, we have L (N ) κ,d < L κ,d for all N ≥ 1.
In addition, for any N ≥ 2 there exist optimisers V N for L (N ) κ,d . When N = N k we have λ N (− + V N ) < 0, that is, − + V N has at least N negative eigenvalues.
As we will discuss below, this result, in particular, disproves the Lieb-Thirring conjecture (5) in dimension d = 2 in the range 1 < κ 1.165 and suggests a new scenario for the optimal Lieb-Thirring constant.
It is unclear whether the passage to a subsequence is really necessary or whether the conclusion holds also for N k = k.
The proof of Theorem 1 proceeds by studying the dual formulation of the Lieb-Thirring inequality (1) in a similar manner as what was done in [GLN21] for the nonlinear Schrödinger equation. This is explained in detail in the next section, where we also collect more properties of V N . This duality argument requires the assumption κ ≥ 1. It is an interesting open question whether Theorem 1 is valid for all κ > max{0, 2−d/2} instead of (6). The value of the critical exponent max{0, 2 − d/2} will be motivated later. In Sect. 4 we provide a direct proof for N = 2 which covers this range of κ, as stated in the following result.
The conclusion L (1) κ,d < L κ,d for the appropriate range of κ is new for all dimensions 2 ≤ d ≤ 7. Let us briefly sketch an alternative way of arriving at this strict inequality for d ≥ 8. It is shown in [GGM78] that the best Cwikel-Lieb-Rozenblum constant satisfies [Fra21]. Here, the constant L (1) 0,d is defined in terms of the Sobolev optimiser. A variation of the monotonicity argument from [AL78] shows that L (1) κ,d /L sc κ,d is strictly decreasing (see Theorem 3 and Lemma 9 below). This implies that L κ,d ≥ L sc κ,d > L (1) κ,d for all κ ≥ 0 and all d ≥ 8, as claimed. In contrast to this argument, our Theorem 2 is not only valid in all dimensions, in the mentioned range of κ, but it gives the additional information that the two-bound states constant The mechanism used in our proof is completely different from [GGM78]. There, the authors increased the coupling constant in front of the potential to reach the semi-classical limit. On the other hand, the proof of Theorem 2 consists of placing two copies of the one-bound state optimiser far away in the appropriate manner, and computing the resulting exponentially small attraction.
Our proof of Theorem 2 does not work for κ = 0 in dimensions d = 5, 6, 7 (where one still has 2 − d/2 < 0). Understanding this case is an open problem.

Discussion
We now discuss in detail the consequences of Theorems 1 and 2 with regard to the Lieb-Thirring conjecture (5).
There are many results on the Lieb-Thirring constants L κ,d . The best estimates currently known are in [FHJN21]. We mention here a selection of results pertinent to our theorem and refer to [Fra21] for a detailed discussion of known results and open problems. We recall the following known properties: 3/2,1 = L sc 3/2,1 ; 1 Only the case N = ∞ is considered in [AL78] but the argument applies the same to any finite N ≥ 1.
For N = 1, we will see in Theorem 3 that κ → L κ,d /L sc κ,d is in fact strictly decreasing.
Next we discuss the one-bound state constant L (1) κ,d . In Sect. 2 we will prove the following result, which provides some new properties of the function κ → L (1) κ,d /L sc κ,d .
In particular, κ 1∩sc (d) is decreasing with the dimension.
That the two curves κ → (L (1) κ,d , L sc κ,d ) cross at a unique point was part of the Lieb-Thirring conjecture [LT76]. In Fig. 1 The inequality is strict because otherwise we would have L κ,d = L sc κ,d = L κ,d at κ = κ 1∩sc (d) which cannot hold by Theorems 1 and 2 . We now discuss some further consequences of our results, mostly in the physical dimensions d ≤ 3.
• In dimension d = 2, we have 1.165 κ 1∩sc (2) < κ sc (2) ≤ 3/2 and this is the best we can say at present. Numerical simulations in [Lev14] did not provide any hint of what is happening in the region 1 ≤ κ 1.165. However, our Theorem 1 in dimension d = 2 shows that L κ,2 > L (N ) κ,2 for all κ > 1 and N ≥ 1. In particular, for 1 < κ 1.165, we disprove the Lieb-Thirring conjecture that the constant is given by the N = 1 optimiser in 2D. It can indeed not be given by any finite rank optimiser.
• In dimension d = 3, a system with 5 bound states was numerically found in [Lev14] to be better than the one bound state for κ 0.855, showing that the one bound state case ceases to be optimal before the critical value 0.8627 in Fig. 1. Our Theorem 2 implies that the one-bound state constant L (1) κ,d can indeed not be optimal for all κ > 1/2. This disproves the Lieb-Thirring conjecture that the constant is given by the N = 1 optimiser for 1/2 < κ 0.8627 in 3D.
• In dimension d ≥ 3, a common belief is that κ sc (d) = 1 for all d ≥ 3. The validity of this conjecture would have some interesting physical consequences, for instance an exact lower bound involving the Thomas-Fermi kinetic energy in Density Functional Theory [LLS20]. Our Theorem 1 does not contradict this belief, since we prove that the optimal Lieb-Thirring potential cannot have a finite number of bound states. But many other situations are still possible, as we now discuss.
Theorem 1 suggests to interpret the Lieb-Thirring inequality within the framework of statistical mechanics. For an optimal potential V N for L (N ) κ,d , we can think of the corresponding N first orthonormal eigenfunctions of − + V N as describing N fermions in R d [GLN21,Rmk. 8]. Theorem 1 says that in the limit N → ∞, the N particles always attract each other, at least along a subsequence N k . We conjecture that for κ > max{2 − d/2, 0} they will form a large cluster of size proportional to N 1/d (if is, for instance, normalised to N ) and that V N will converge in the limit to a bounded, but non-decaying potential V ∞ . There would then be no optimiser for the Lieb-Thirring constant L κ,d . The semi-classical constant L sc κ,d corresponds to the case where the limiting potential V ∞ is constant over R d , that is, the system is translation-invariant. In statistical mechanics, this is called a fluid phase. In principle, the limiting potential V ∞ could also be a non-trivial periodic function, which is then interpreted as a solid phase. We see no obvious physical reasons for discarding this possibility, in particular in low dimensions where periodic systems are ubiquitous [BL15]. This mechanism does not seem to have been considered before in the context of Lieb-Thirring inequalities. In particular, it seems natural to conjecture that the system is in a solid phase for all 2 − d/2 < κ < κ sc (d) in dimensions d = 2, 3. In [FGL21] we shall discuss this new point of view in detail.
Remark 4. In dimension d = 2, some preliminary numerical tests suggest that the difference L κ,2 − L (1) κ,2 might be very small in the region 1 < κ 1.165. This makes the problem difficult to simulate as we need high precision.

Dual Lieb-Thirring inequalities.
Our strategy to prove Theorem 1 is to study the dual version of the Lieb-Thirring inequality (1). This dual version is well known for κ = 1 and it is often used in practical applications. The dual inequality for κ > 1 appears, for instance, in [LP93], but is less known and we briefly recall it in this subsection. There is no known dual problem for κ < 1, except for a certain substitute for κ = 0 in dimensions d ≥ 3 [Fra14].
Let 0 ≤ γ = γ * be a self-adjoint non-negative operator of Rank(γ ) ≤ N , of the form γ = N j=1 n j |u j u j | with u 1 , . . . , u N an orthonormal family in L 2 (R d ). For 1 ≤ q < ∞, we denote by its q-th Schatten norm [Sim05], and use the convention that γ S ∞ = γ is the operator norm. The density of γ is the function ρ γ ∈ L 1 (R d ) defined by and the kinetic energy of γ is with d ≥ 1, and let q := 2 p+d−dp 2+d−dp for 1 ≤ p < 1 + 2 d , +∞ for p = 1 + 2 d .
We denote by K (N ) p,d the best (that is, largest possible) constant in the inequality valid for all 0 ≤ γ = γ * with Rank(γ ) ≤ N . The fact that K p,d > 0 is a consequence of the next result, together with the Lieb-Thirring theorem.
Lemma 5 (Duality). Let 1 ≤ N ≤ ∞, d ≥ 1 and 1 ≤ p ≤ 1 + 2 d , and set Then, The lemma says that the inequality (11) is dual to the finite-rank Lieb-Thirring inequality (1). This is because the density ρ γ is the variable dual to the potential V whereas the density matrix γ can be interpreted as the dual of the Schrödinger operator − + V . Hence p is the dual exponent of κ + d/2 and q the one of κ. The proof of Lemma 5, provided in Appendix A, also shows how to relate the corresponding optimisers, assuming they exist. A similar argument, but without the constraint on the rank, can be found for instance in [LP93].
We denote This constant is related to the constant L κ,d in (3) by and is the best constant in the inequality valid for all 0 ≤ γ = γ * . In Sect. 3, we study the dual problem (11) and prove the following result which, together with Lemma 5, immediately implies Theorem 1.
Theorem 6 (Existence of optimisers and properties). Let d ≥ 1 and 1 ≤ p ≤ 1 + 2/d. After an appropriate normalisation, any optimiser γ for K (N ) p,d has rank 1 ≤ R ≤ N < ∞ and can be written in the form where the corresponding orthonormal system (u 1 , . . . , u R ) solves the nonlinear Schrödinger equation Here μ j are the R first negative eigenvalues of H γ := − − ρ p−1 γ . In particular, this operator has at least R negative eigenvalues. If R < N , then it has exactly R negative eigenvalues. Finally, the potential V = −ρ and any optimiser for K (N k ) p,d must have rank R = N k . In particular, The assertions in (i) and (ii) follow by applying well-known methods from the calculus of variation adapted to the setting of operators; see, for instance, [Sol91,Bac93,FLSS07,Lew11]. For (iii), we use ideas from [GLN21], which consist in evaluating the exponentially small interaction between two copies of an optimiser placed far from each other, in order to show that p,d admits an optimiser of rank N . The proof is provided in Sect. 3 below. This argument inspired our proof of Theorem 2 for κ < 1 and N = 2, provided in Sect. 4. There we use the N = 1 Gagliardo-Nirenberg optimiser to construct a trial state for N = 2 but we do not prove the existence of an optimal potential.

Fermionic nonlinear Schrödinger equation.
The system of coupled nonlinear equations (16) has some similarities with that studied in [GLN21], where one has n j = 1 instead of (15). Here we exhibit a link between the two problems and use this to solve a question left open in [GLN21].
In [GLN21] the authors studied the minimisation problem Under the additional assumption that p < 2, it was also shown that there is an infinite sequence of integers N 1 = 1 < N 2 = 2 < N 3 < · · · such that J (N k ) has a minimiser γ of rank N k . This minimiser is a projector of the form γ = N k j=1 |u j u j |, where u 1 , . . . , u N k form an orthonormal system and solve the fermionic NLS equation The existence of minimisers for J (N k ) therefore proves the existence of solutions of the fermionic NLS equation (18), for all 1 ≤ p < min{2, 1 + 2/d} and N = N k . In dimension d = 1, this does not cover the case p ∈ [2, 3). In the present paper, we prove the following result for the case p = 2, which was announced in [GLN21] and actually also follows from the analysis in [Ld78]. The theorem is reminiscent of a similar result for the true Schrödinger (Lieb-Liniger [LL63]) model in 1D describing N particles interacting with the delta potential. In the attractive case, only two-particle (singlet) bound states exist [McG64,Yan68,Ld78]. The same result in the Hartree-Fock case was proved in [Ld78]. The spatial component of the singlet state coincides with our N = 1 solution.
In the case N = 1 and 1 < p and Q is the unique positive radial solution to the NLS equation When d = 1 and p = 2, we have the explicit formula Our strategy to prove Theorem 7 for d = 1 is to relate J (N ) to the dual Lieb-Thirring constant K (N ) κ,1 for κ = 3/2, and use K (N ) 3/2,1 = K (1) 3/2,1 . The proof is given in Sect. 5.1 below.
Even if J (N ) has no minimiser for N ≥ 2 if d = 1 and p = 2, one may still wonder whether the fermionic NLS equation (18) possesses orthonormal solutions. We believe there are no other solutions than the N = 1 case and are able to prove this for N = 2, using the fundamental fact that the system is completely integrable [Man74]. The following is stronger than Theorem 7 for N = 2.
The proof can probably be generalised to show that there are no solutions for all N ≥ 3 at p = 2 but we only address the simpler case N = 2 here. The proof is given in Sect. 5.2. More comments about the NLS problem (17) can be read in Appendix B.
Structure of the paper. In the next section we recall useful facts about the Lieb-Thirring constant L (1) κ,d and provide the proof of Theorem 3. In Sect. 3, we prove Theorem 6, which implies Theorem 1. Section 4 is devoted to the proof of Theorem 2. We prove Theorem 7 and Theorem 8 in Sects. 5.1 and 5.2 , respectively. The proof of duality (Lemma 5) is given in Appendix A whereas Appendix B contains more comments on the NLS model from [GLN21]. Finally, in Appendix C we compare our results with those in [HKY19].

The One-Bound State Constant L (1) κ,d : Proof of Theorem 3
In this section we discuss some properties of the one-bound state constant L (1) κ,d and provide the proof of Theorem 3. The Gagliardo-Nirenberg inequality states that for all 1 < p < +∞ for d = 1, 2, with the best constant K GN p,d > 0. In dimension d = 1 one can take p → +∞. The constants K GN p,1 and the optimisers are known explicitly in d = 1 [Nag41]. In particular, the optimiser is unique up to translations, dilations and multiplication by a phase factor. As explained, for instance, in [Tao06,Fra13,CFL14], by combining the results on existence [Str77,BL83,Wei83], symmetry [GNN81,ALT86] and uniqueness [Cof72,Kwo89,McL93] one infers that in any d ≥ 2 as well, there is a unique optimiser Q, up to translations, dilations and multiplication by a phase factor. This function can be chosen positive and to satisfy (20). The integral R d Q 2 dx will be a dimension-dependent constant.
For an operator γ of rank one the inequality (11) is equivalent to (23), hence we obtain K (1) The duality argument from Lemma 5 shows that By the implicit function theorem and the non- κ,d is a real-analytic function of κ. In this paper we will only use the continuity of κ → L (1) κ,d , which is more elementary and which we explain now for completeness. We claim that which implies the continuity of L (1) κ,d on the corresponding intervals. To prove this fact, we can notice that By Hölder's inequality, p −1 → log ||u|| L 2 p (R d ) is convex. Hence after minimising over , and log-concave in 1/ p. Log-concavity implies continuity on the interior of the interval of definition and then upper semicontinuity implies continuity up to the endpoints.
Our goal in the rest of this section is to compare L (1) κ,d with the semi-classical constant L sc κ,d . First, the argument from [AL78] can be used to prove that κ → L (1) κ,d /L sc κ,d is nonincreasing. We show here that it is even strictly decreasing, which is (i) in Theorem 3.
Proof. Following [AL78], we use the fact that for any 0 ≤ κ < κ and λ ∈ R, we have for some constant c κ,κ > 0. Let V ∈ L κ+d/2 (R d ). By the variational principle we have for any t ≥ 0 and we can bound, using the definition of L (1) κ ,d , Thus, integrating over t using (27) on both sides, we obtain This shows that that is, κ → L (1) κ,d /L sc κ,d is nonincreasing. As was recalled at the beginning of this section, it is known that for the optimisation problem corresponding to L (1) κ ,d there is an optimiser. This optimiser is a power of the solution of the positive solution of (20) and therefore does not vanish. Since for any V ∈ L κ+d/2 (R d ) and for any t > 0, the function −(V + t) − is supported on a set of finite measure, this function cannot be an optimiser for L (1) κ ,d . Therefore the second inequality in (28) is strict for all t > 0 and, consequently, inequality (29) is strict for any V ∈ L κ+d/2 (R d ). Taking, in particular, V to be an optimiser corresponding to L (1) κ,d , we obtain that inequality (30) is strict, which is the assertion of the lemma.
Next, we prove an inequality relating the constant L (1) κ,d with the ones in lower dimensions, in the spirit of the Laptev-Weidl method of lifting dimensions [LW00].
Lemma 10. For any d ≥ 2 and κ > 0, we have The same inequality holds for Proof. Let V be the optimizer for L (1) κ,d with corresponding ground state u, which can both be expressed in terms of the NLS solution Q in (20). We write and taking the scalar product with u, we find λ 1 (− + V ) > λ 1 − x 1 + λ(x 1 ) . The strict inequality is because u does not solve an eigenvalue equation in x 1 at fixed x 2 . This gives and we obtain inequality (31) for κ > 0.
κ,d and u the corresponding ground state. More precisely, u is an eigenfunction corresponding to the eigenvalue zero if d ≥ 5 and it is a zero energy resonance function (that is, an where S d−n is the optimal Sobolev constant in dimension d − n. We conclude that On the other hand, we have where the strict inequality follows from the fact that for no is algebraically decaying, whereas we know that the optimal potential for L (1) κ+ d−n 2 ,n is exponentially decaying.) Combining the last two inequalities we obtain  HLW00], one sees that for n = d − 1 and κ ∈ {1/2} ∪ [3/2, ∞) the bound (31) with ≤ instead of < holds for the usual Lieb-Thirring constants L κ,d . One might wonder whether this is true more generally.
In [Mar90], Martin used a similar idea but instead of removing he added one dimension by considering the potential W (x, t) := V (x) + λt 2 . This led to the inequality This is larger than 1 in dimensions d ∈ {3, . . . , 7} but smaller than 1 in dimension d ≥ 8, as noted in [LT76,GGM78]. In fact, this is decreasing with the dimension for d ≥ 4 by (10) and the value in dimension d = 8 equals L 1,2 ≥ (8π) −1 = L sc 1,2 . Since the Gaussian does not satisfy the Euler-Lagrange equation for K GN 2,2 , the inequality is, in fact, strict, as claimed.
On the other hand, it also follows from (10) that L (1) κ,d < L sc κ,d for all d ≥ 2 and κ ≥ 3/2. We deduce that in dimensions 2 ≤ d ≤ 7 the two continuous curves L (1) κ,d and L sc κ,d must cross. The crossing point is unique by Lemma 9 and this concludes our proof of Theorem 3.

Finite Rank Lieb-Thirring Inequalities: Proof of Theorem 6
This section contains the proof of Theorem 6 which, for convenience, we split into several intermediate steps. Our goal is to study the optimisation problem corresponding to inequality (11), namely where we recall that q := 2 p+d−dp 2+d−dp Throughout the paper, the constants p, q and κ are linked by the relations (we set p = p/( p − 1) and κ = κ/(κ − 1)) Taking (34) to the power 1 2 ( p − 1), and letting p → 1, so that q → 1 as well, we recover the equality for all 0 ≤ γ = γ * . On the other hand, taking p = 1 + 2/d, so that q = ∞, we recover the better known dual Lieb-Thirring inequality (36) We can think of (11) as a specific interpolation between these two cases. Note that a direct proof of (36) with N = +∞ can be found in [Rum11], see also [LS13,Sab16,Nam18]. The original Lieb-Thirring proof proceeds by proving (1) and then deducing (36) by duality.
3.1. Proof of (i) on the existence of optimisers. Consider a minimising sequence (γ n ) with Rank(γ n ) ≤ N for (34), normalised such that with ρ n := ρ γ n . We have γ n ≤ γ n S q = 1 and hence This proves that ρ n is bounded in L 1 (R d ). On the other hand, the Hoffmann-Ostenhof [HH77] inequality states that all 2 ≤ r < 2 * where 2 * = 2d/(d − 2) in dimension d ≥ 3 and 2 * = +∞ in dimensions d = 1, 2, by the Sobolev inequality. In particular, we can choose r = p. From [Lie83a] or from [Lio84b, Lem. I.1], we know that , a sequence τ k ∈ R d and a subsequence so Due to (37) we know that the first possibility cannot happen and we may assume that √ ρ n √ ρ = 0, after extraction of a subsequence and translation of the whole system by τ n . We may also extract a weak- * limit for γ n in the trace class topology and infer γ n γ where ρ γ = ρ = 0, hence γ = 0. By passing to the limit, we have γ = γ * ≥ 0 and Rank(γ ) ≤ N .
In the third line we used Fatou's lemma in the Schatten space S q . Next, we argue using the method of the missing mass as in [Lie83b], see also [Fra13], noticing that K (N ) p,d can be rewritten as Using Hölder's inequality in the form In the third line we used Rank(η n γ n η n ) ≤ N . Passing to the limit we obtain Tr(− γ ) θ and therefore γ = 0 is an optimiser. The case p = 1 + 2/d is similar. This time, we use γ ≤ lim inf n→∞ γ n = 1 and η n γ n η n ≤ γ n = 1 to bound and arrive at the same conclusion that γ is an optimiser.

Proof of (ii) on the equation. Let γ be an optimiser such that
This normalisation is always possible by scaling and by multiplying γ by a positive constant. Then we have 2+d−dp .
We start with the case q < ∞, that is, p < 1 + 2/d. Assume that we have a smooth curve of operators γ (t) = γ + tδ + o(t) for some δ = δ * , with γ (t) = γ (t) * ≥ 0 and Rank(γ (t)) ≤ N . By expanding we find Hence γ commutes with the mean-field operator H γ := − − pρ p−1 γ /θ . We can therefore write γ = R j=1 n j |u k j u k j | for some eigenvectors u k j of H γ (with eigenvalue μ k j ) and some n j > 0. In particular, H γ admits at least R eigenvalues.
Using now γ (t) = γ + tδ for a δ supported on the range of γ and for t small enough in (39), we find that Evaluating this identity on u k j we infer that This shows that μ k j < 0 and Since γ is assumed to be of rank R, we in particular deduce that H γ has at least R negative eigenvalues.
Next, we show that the μ k j are necessarily the R first eigenvalues. Assume that one eigenvector of H γ with eigenvalue < μ R does not belong to the range of γ , so there is 1 ≤ j ≤ R with u k j = u j with k j > j and u j not in the range of γ . Consider the new operator γ := γ − n j |u k j u k j | + n j |u j u j | := γ + δ, which has the same rank and the same S q norm as γ . We have by convexity , a contradiction. Hence μ k j = μ j . Finally, when R < N and μ R+1 < 0, we can consider the operator with t ≥ 0, which has rank R + 1 ≤ N . From (39) we obtain another contradiction. Hence H γ cannot have more than R negative eigenvalues when R < N . As a conclusion, we have shown that Taking the trace of γ q we find that θ Tr(γ q ) Replacing γ by ( p/θ ) 1 p−1 γ we find the equation mentioned in the statement. The arguments for q = +∞ ( p = 1 + 2/d) are similar. We start with a minimiser normalised so that

The first perturbation γ (t) := e it H γ e −it H = γ + it[H, γ ] + o(t) leaves the operator norm invariant and provides the equation
In order to prove that μ k j < 0, we consider the operator γ := γ − n j |u k j u k j | which has one less eigenvalue and satisfies γ 2/d ≤ γ 2/d = K (N ) 1+2/d,d . We find

Simplifying by K (N )
1+2/d,d > 0, this gives the estimate where the last negative sign is by strict convexity of t → t 1+2/d . Hence γ has its range into the negative spectral subspace of H γ , an operator which thus possesses at least R negative eigenvalues. Next we show that n j = γ for all j = 1, . . . , R. Assume on the contrary that 0 < n j < γ (this can only happen when R ≥ 2). Taking γ (t) = γ + t|u k j u k j | which has the same operator norm for t small enough, we obtain which is a contradiction since μ k j < 0, as we have seen. We conclude that n j = γ for all j = 1, . . . , R. The argument for showing that μ k 1 , . . . , μ k R are the R first eigenvalues is exactly the same as before.

Proof of (iii) on the rank of optimisers.
In this subsection, we prove the following result.

Proposition 11 (Binding
p,d we take N 3 = 3 and otherwise we take N 3 = 4. We then go on by induction to obtain the assertion of (iii). Hence we now concentrate on proving Proposition 11.
Proof of Proposition 11. We follow ideas from [GLN21, Section 2.4]. Let γ := N j=1 n j |u j u j | be a minimiser of rank N for K (N ) p,d , normalised in the manner Tr with n j = c|μ j | 1/(q−1) . Note that the first eigenfunction u 1 is positive, hence the nonlinear potential never vanishes. By usual regularity arguments, this shows that the u j are C ∞ and decay exponentially at infinity. For R > 0, we set u j,R (x) := u j (x − Re 1 ) where e 1 = (1, 0, . . . , 0), and we introduce the Gram matrix Since the functions u i are exponentially decaying, E R goes to 0, and the overlap matrix S R is invertible for R large enough. We then let We have Expanding as in [GLN21] using . for we obtain after a long calculation and by [GLN21,Lemma 21] we have Since p < 2 by assumption we conclude, as we wanted, that K

Binding for κ < 1 and N = 2: Proof of Theorem 2
In this section we provide the proof of Theorem 2. Define p by p = κ + d/2 let Q be the radial Gagliardo-Nirenberg minimiser, solution to (20), and set m := R d Q 2 dx.
4.1. Some properties of Q. First we relate our constants for N = 1 to Q. We have the Pohozaev identity These follow by multiplying the equation (20) by Q and by x · ∇ Q, respectively. This gives the identity m On the other hand, setting V Q := −Q 2( p−1) , we see that Q is an eigenvector of − + V Q (with corresponding eigenvalue −1), and, by optimality of V Q for L Finally, it is well known that there is C > 0 so that

Test potential for L
(2) κ,d . We now construct a test potential to find a lower bound for (1, 0, . . . , 0). Inspired by the dual problem studied in the previous section, we consider the potential It is important here that we add the two densities and not the corresponding potentials. We do not see how to make our proof work if we would take V = −Q We introduce the quantity Due to the inequality (46), A goes (exponentially fast) to 0 as R goes to infinity. Our main result is the following.
Lemma 12. We have, as R → ∞, The proof of Theorem 2 follows as the leading correction is positive.
Proof. First, we bound A from below similarly to (42). Indeed, noting that the integrand of A is nonnegative and bounding it from below using (46) in a neighbourhood of the origin, we find Next, we turn to the denominator appearing in the lemma. We have Together with (45), this gives Finally, we evaluate the numerator. We set E : We have E → 0 as R → ∞, so for R large enough, we have |E| < m, and the two functions ψ (±) defined by By the variational principle, the two lowest eigenvalues of − + V are not larger than the corresponding eigenvalues of H, and therefore We have We have h → −1 and δ → 0 as R → ∞, and therefore It remains to expand h and to bound δ. We begin with h. We find Integrating and using (20) gives Similarly, where A was defined in (47), and where From (46) We see in a similar fashion that δ ≤ C R d e −R hence δ 2 = o(A) as well. Gathering all the estimates gives where the last equality comes from (44).

Non Existence of Minimisers for the Fermionic NLS: Proof of Theorems 7 and 8
In this section, we prove our results concerning the minimisation problem J (N ) which, we recall, is defined by We assume in the whole section After an appropriate scaling, and using the fact that Tr(γ ) = γ S 1 , the optimal inequality valid for all 0 ≤ γ = γ * ≤ 1 with Tr(γ ) = N , and with best constant One can remove the constraint γ ≤ 1 at the expense of a factor γ d/2 , and we obtain the optimal inequality valid for all 0 ≤ γ = γ * with Tr(γ ) = N .

Link between NLS and Lieb-Thirring
Proof. It is shown in [GLN21, Lemma 11] that the minimisation problem J (N ) can be restricted to operators γ which are orthogonal projectors of rank N . For such operators, we have γ = 1 and

This gives
.
Optimising over projectors γ gives K p,d . In the case N = 1, every operator of rank 1 is proportional to a rank 1 projector, so the two problems coincide, and K (1) [GLN21], it is also proved that J (N ) ≤ N J(1). This implies K There is a similarity between the proof of the above proposition and the arguments in [Ld78,FLST11]. In those works also the sharp Lieb-Thirring inequality for κ = 3/2 is used to obtain an inequality about orthonormal functions.
The relation (52) allows us to prove Theorem 7, which states that J (N ) = N J(1) for all N ∈ N, and that J (N ) admits no minimiser for N ≥ 2. To prove that J (N ) has no minimiser for N ≥ 2, we assume by contradiction that γ is one. By [GLN21, Proposition 16], γ is a rank N projector. In addition, since we have equality in (52), γ is also an optimiser for K (N ) 2,1 . But then, by Theorem 6, it is of the form γ = c N j=1 |μ j | 1/2 |u j u j | for some c. We conclude that μ j = −1/c 2 for all j = 1, . . . , N which is impossible since the first eigenvalue μ 1 of a Schrödinger operator is always simple.
If true, this conjecture would imply by the same argument as in the previous proof that for all 2 ≤ p < 3 and all N ≥ 1, in dimension d = 1, and that the corresponding problems do not have minimisers for N ≥ 2. The weaker conjecture (53) appeared in [GLN21].

5.2.
Proof of Theorem 8: triviality of solutions for d = 1, p = 2 and N = 2. In this subsection we prove Theorem 8: we show that the fermionic NLS equation (18) does not have a solution in the one dimensional case with p = 2 and N = 2. We will make use of the integrability of the equations. In the sequel, we study the ODE system We added an extra factor 2 to obtain the same explicit formulas as in the literature. If  (u 1 , u 2 ) is a real-valued ground state solution to (21), then (v 1 , v 2 ) = 1 √ 2 (u 1 , u 2 ) is a real-valued solution to (54), which satisfies in addition v 1 = v 2 = 1 2 . The key step in the proof of Theorem 8 is the following classification result for (54) under an additional vanishing condition for v 2 .
Let us now prove that all square integrable solutions with v 2 (0) = 0 are of this form. In fact, instead of square integrability we will assume that v j and v j tend to zero at infinity for j = 1, 2. It is not hard to deduce this property from the assumption that the solution is square integrable.
For the proof we will assume that v 2 (0) = 0, for otherwise v 2 = 0 everywhere and the result is well-known (and easy to prove by a variation of the arguments that follow, using only (57a) below).
Any solution (v 1 , v 2 ) that decays at infinity has two constants of motion (v 2 1 + v 2 2 ) 2 + |v 1 | 2 + |v 2 | 2 + μ 1 v 2 1 + μ 2 v 2 2 = 0, (57a) To obtain identity (57a) we multiply the first and second equation in (54) by v 1 and v 2 , respectively, add the resulting identities and then integrate using the fact that the solutions and their derivatives vanish at infinity. The fact that there is a second identity (57b) reflects the integrability of the system [Man74]. Evaluating (57) at x = 0 and using v 2 (0) = 0, we deduce that Thus, the value of v 1 (0) is determined, up to a sign, by μ 1 and μ 2 and we have The assumption v 2 (0) = 0 also shows that −μ 2 (μ 2 − μ 1 ) > 0, hence μ 2 = μ 1 and therefore also v 1 (0) = 0. Let (ṽ 1 ,ṽ 2 ) be a solution of the form (55). The absolute value of a 1 is fixed by (56). We will now show that the sign of a 1 as well as the number a 2 can be determined in such a way thatṽ j (0) = v j (0) andṽ j (0) = v j (0) for j = 1, 2. Once we have shown this, ODE uniqueness implies thatṽ j = v j for j = 1, 2, which is what we wanted to prove.

It remains to choose a 2 . A tedious but straightforward computation yields
The last quotient on the right side is a decreasing function of a 2 2 from [0, ∞] to [−1, 1]. Thus, there is an a 2 2 ∈ (0, ∞) such thatṽ 1 (0) = v 1 (0). This determines the absolute value of a 2 . To determine its sign, we note that the identitiesṽ 1 (0) 2 +ṽ 2 (0 Thus, we can choose the sign of a 2 in such a way thatṽ 2 (0) = v 2 (0).
This shows that we can indeed find a 1 and a 2 such thatṽ j (0) = v j (0) andṽ j (0) = v j (0) for j = 1, 2. As explained before, this implies the result.
We will also need the following lemma in the proof of Theorem 8. Proof. With the notation of Lemma 15, a computation reveals that Integrating gives as wanted.
Proof of Theorem 8. As explained before Lemma 15, it is enough to consider solutions (v 1 , v 2 ) of (54) with v 1 = v 2 = 1 2 . The equations (54) mean that the numbers μ 1 and μ 2 are negative eigenvalues of the operator −∂ 2 x x − 2(v 2 1 + v 2 2 ). It is easy to see that the latter operator is bounded from below and its negative spectrum is discrete. Therefore it has a lowest eigenvalue μ 0 . Let v 0 be a corresponding eigenfunction, normalised by v 0 = 1 2 . It is well-known that the eigenvalue μ 0 is non-degenerate and that v 0 can be chosen positive. In particular, if v is a square integrable real valued solution to −v − 2(v 2 1 + v 2 2 )v = μv which never vanishes, then necessarily μ = μ 0 .
It follows from the equality μ 1 = μ 2 = μ 0 , the simplicity of μ 0 and the normalisation that v 2 1 = v 2 2 . In particular, v 1 and v 2 both satisfy v j + 4v 3 j + μ 0 v j = 0. By uniqueness of the solution to the equation up to translations, this gives (22) for some x 0 ∈ R and a sign ±. Since v 2 1 = v 2 2 the x 0 's for the two functions coincide, while the signs are independent. This completes the proof of the theorem. Funding Open Access funding enabled and organized by Projekt DEAL.
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Appendix A. Proof of Lemma 5
The proof of Lemma 5 splits naturally into two parts. We first deduce (11) from (1). We write our operator γ in the form where (u 1 , . . . , u N ) forms an orthonormal system. The inequality (11) which we wish to prove therefore reads For a constant β > 0 to be determined, let For κ ≥ 1 we use Hölder's inequality in Schatten spaces [Sim05] in the form for all B ≥ 0 of rank ≤ N . Applying this with A = − + V and B = γ we obtain, in view of (1), We optimise in β by choosing We now choose κ = p − d/2, which is > 1 since p < 1 + 2/d and which ensures that Therefore, the best constant K (N ) d, p in (58) satisfies Conversely, assume that inequality (58) holds and let V ∈ L κ+d/2 (R d ). We assume that − + V has at least N negative eigenvalues, the other case being handled similarly. Let u 1 , . . . , u N be orthogonal eigenfunctions corresponding to the N lowest eigenvalues of − + V and let γ = N j=1 n j |u j u j |, n j = |λ j (− + V )| κ−1 , so that We have, for p such that p = κ + d 2 , Recall that n κ j = |λ n (− + V )| κ and therefore the above inequality becomes Therefore the best constant L This proves the lemma.
is non-decreasing. This implies that p → K p,d is non-increasing on the interval (1, 1 + 2/d). In particular, since K sc p,d = K sc 1+2/d,d is independent of p (see Sec. B.4), we deduce that if K p,d = K sc p,d for some p = p 0 , then the same equality holds for all 1 < p ≤ p 0 . This generalises the observation in [GLN21] that if the standard Lieb-Thirring conjecture holds for κ = 1 (that is, K p,d = K sc p,d for p = 1 + 2/d), then K p,d = K sc p,d for all 1 < p < 1 + 2/d.

B.2. Dual inequality.
A natural question is to determine the inequality dual to (60). This is the object of the following lemma.
Remark 23 (The semi-classical constant). We show here that the constant L sc κ,d has an interpretation in terms of a semiclassical limit, thereby justifying its name. Because of the second inequality in (66), this argument shows that the considered scenario is in a certain sense dual to that considered in [GLN21]. For any V ∈ L κ+ d 2 (R d ) and any τ > 0, we have On the other hand, by inequality (62), This shows that Taking the supremum in τ shows that According to the optimality statement in the following lemma, we have This proves, once again, the second inequality in (66) and shows how this inequality is related to a semiclassical limit.
Lemma 24. Let X be a measure space, p > r ≥ 0 and f ∈ L p (X ). Then The constant on the right side is best possible.
Proof. We first recall that Together with (65) (note that we may assume r > 0 by continuity) we need to prove that There is equality when f is a characteristic function and τ is chosen appropriately. This proves Lemma 24.
Remark 25. We wonder whether for all d ≥ 1 and all κ ≥ 3 2 , we have the equality L κ,d = L sc κ,d . This would be the analogue of the equality L κ,d = L sc κ,d [LW00]. We have the following rather tight bounds. Thanks to the explicit formulas (67) and (4), one can numerically plot the two curves κ → L sc κ,d and κ → (κ − 1) κ−1 κ κ L sc κ,d . As stated in Lemma 22, the two curves coincide at κ = 1, but for all κ > 1, it appears that In the region κ ≥ 3/2 where L κ,d = L sc κ,d [LW00], we deduce that | L κ,d − L sc κ,d | is smaller than the constants above. . We deal in [GLN21] and this paper with the right edge where α, β > 0. There is no optimiser without an additional trace constraint. Existence of optimisers was proved on the left edge where β = 0 in [HKY19]. The horizontal edge coincides with the Gagliardo-Nirenberg inequality, with α = 0. Minimisers exist and are all rank-one. In dimension d ≥ 3, the Sobolev inequality has a formal rank-one optimiser. For d = 3, 4, however, it is not bounded on L 2 (R d ) since the associated function is not in L 2 (R d ). It is expected that a minimiser exists for the Lieb-Thirring inequality only in dimension d = 1, where it should be rank-one. In dimension d = 1, our study is limited to p < 2 p < 1 + 2/(d − 2). When normalised in the manner γ = 1 and Tr(− γ ) = θ R d ρ