Lieb-Thirring inequalities for wave functions vanishing on the diagonal set

. — We propose a general strategy to derive Lieb–Thirring inequalities for scale-covariant quantum many-body systems. As an application, we obtain a generalization of the Lieb–Thirring inequality to wave functions vanishing on the diagonal set of the conﬁguration space, without any statistical assumption on the particles.

This is Pauli's exclusion principle for fermions (1) .Replacing the minus sign in (1.2) by a plus sign defines bosonic particles, while if the particles are non-identical, i.e. distinguishable, no exchange symmetry may be imposed.The inequality (1.1) was first proved by Lieb and Thirring in 1975 for the case s = 1 relevant to non-relativistic particles [LT75,LT76], and extended by Daubechies in 1983 to general s > 0, thus also including the relativistic case s = 1/2 [Dau83].The constant K = K(d, s) > 0 is independent of N and Ψ N (see [FHJN18] for the best known value of K).
The Lieb-Thirring inequality is a beautiful combination of the uncertainty and exclusion principles of quantum mechanics, and has also been very actively studied in the mathematical literature from the dual perspective of estimation of eigenvalues of one-body Schrödinger operators (see e.g.[Lap12,LS09] for reviews).Historically, the Lieb-Thirring inequality was invented to give a short, elegant proof of the stability of ordinary non-relativistic matter with Coulomb forces [LT75].In that context it is well known that stability of the first kind, i.e. that the ground state energy of the Coulomb system is finite, follows easily from some sort of the uncertainty principle (e.g.Sobolev's inequality).On the other hand, the stability of the second kind, that the ground state energy does not diverge faster than the number of particles, is much more subtle: for this the fermionic nature of particles is crucial.In fact, the stability of the second kind fails for bosonic (or distinguishable) charged systems [Dys67].
Without the anti-symmetry condition (1.2), the Lieb-Thirring inequality (1.1) fails and the best one can get is the Gagliardo-Nirenberg-Sobolev inequality Ψ N (x) 1+2s/d dx (see e.g.[LNP16]).The emergence of the factor N −2s/d can be seen by considering the bosonic trial state Ψ N = u ⊗N (whose density is ).This factor is small when N becomes large, making (1.3) not very useful in applications.
Note that Pauli's exclusion principle (1.2) implies that the wave function Ψ N vanishes on the diagonal set (1.4) := (x 1 , . . ., x N ) ∈ (R d ) N : x i = x j for some i = j , namely there is zero probability for two quantum particles to occupy a common single position in the configuration space.
In this paper, we want to address the following Question 1.1.-Does the Lieb-Thirring inequality (1.1) remain valid if the anti-symmetry assumption (1.2) is replaced by the weaker condition Ψ N | = 0 ?
We will show that the answer is yes if and only if 2s > d.In fact, 2s > d is the optimal condition for the vanishing assumption Ψ N | = 0 to be non-trivial (heuristically this follows from Sobolev's embedding H s (R d ) ⊆ C(R d ) for 2s > d).
The precise statement of our result and its consequences will be presented in the next section.

Main results
Recall that for every s > 0 (not necessarily an integer) the operator (−∆) s on L 2 (R d ) is defined as the multiplication operator |p| 2s in Fourier space, namely The associated space H s (R d ) is a Hilbert space with norm The N -particle space H s (R dN ) is defined in the same way.Let us denote the subspace of functions vanishing on the diagonal set in (1.4) by .
Here C = C(d, s) > 0 is a universal constant independent of N and Ψ N .
We have some immediate remarks.
(1) The condition 2s > d in Theorem 2.1 is optimal.If 2s d, then by the relatively small size, i.e. the large codimensionality, of the diagonal set (see Appendix B) and thus the Lieb-Thirring inequality fails.(2) For d = 1 and s = 1, it is well known that a symmetric wave function which vanishes on the diagonal set is equal to an anti-symmetric wave function up to multiplication by an appropriate sign function [Gir60], and hence (2.1) reduces to the usual Lieb-Thirring inequality [LT76] in this case.However, when d > 1 this boson-fermion correspondence is no longer available and our result is new.Furthermore, one may consider hard-core bosons defined by the higher-order vanishing around diagonals , and subject to symmetry.For large enough order 2s > d there is even for d = 1 a non-trivial difference between these spaces, and our result assumes only the weaker vanishing conditions imposed by H s,N (R d ) (see Appendix B for some further remarks).(3) Theorem 2.1 verifies a conjecture in [LNP16, p. 1362] that the Lieb-Thirring inequality (2.1) holds for all wave functions in the form domain of the interaction potential In fact, we have (again, see Appendix B for details) by the singular nature of the potential at the diagonals.We may think of the potential W s as defining (by Friedrichs extension) a one-parameter family of non-negative and scale-covariant (scaling homogeneously to degree −2s) interacting N -body Hamiltonian operators ANNALES HENRI LEBESGUE (In the case β = 0, then H β is still defined on the quadratic form domain of W s .)One may then ask about the best constant C(β) 0 in the bound i.e. a Lieb-Thirring inequality (generalized uncertainty principle) for H β .The case β > 0 was treated in [LPS15, LNP16], while our setting here concerns the limit β → 0 of zero-range/contact interaction.A crucial difference is the strength of the interaction term, which is of order βN 2 and thus provides a large repulsive energy for fixed β > 0, while for β 1/N it ought to be much weaker than the kinetic term.Nevertheless, for 2s > d the potential W s is singular enough to impose the vanishing condition at , and Theorem 2.1 thus implies that the limiting constant is positive C(0) > 0. Furthermore, for 2s = d we have C(0) = 0 which follows from a straightforward argument through contradiction using the density of C ∞ c (R dN \ ) in H s (R dN ); again see our first remark and Appendix B. Together with the case 2s < d that was treated in [LNP16] and for which also C(0) = 0, this settles the question concerning which H β satisfy the Lieb-Thirring inequality (2.4) with C(β) > 0.
(4) The original proof of the Lieb-Thirring inequality [LT75, LT76] is based on the following operator bound which is a consequence of Pauli's exclusion principle (1.2).Here γ (1) Ψ N is the one-body density matrix of Ψ N , a trace-class operator on L 2 (R d ) with kernel γ (1) However, unlike the full anti-symmetry condition (1.2), the vanishing condition Ψ N | = 0 alone is not known to be sufficient to ensure the operator inequality (2.5), and therefore the original proof in [LT75,LT76] as well as subsequent proofs based on (2.5) (e.g.Rumin's method [Rum11]) do not apply.Our result is in fact more general than as previously formulated.More precisely, define for any k 2 the diagonal set of k-particle coincidences and the corresponding space of N -particle wave functions with a vanishing condition on k .
We have Theorem 2.2 (Lieb-Thirring inequality for wave functions vanishing on k-diagonals).-Let d 1, k 2 and 2s > d(k − 1).Then for every N 1 and every Here The proof of Theorem 2.2 occupies the rest of the paper.Our proof is based on a general strategy of deriving Lieb-Thirring inequalities for wave functions satisfying some partial exclusion properties, which was proposed by Lundholm and Solovej in [LS13a] and developed further in [FS12, LL18, LNP16, LPS15, LS13b, LS14, LS18, Lun18, Nam18].We will quickly review this strategy in Section 3 for the reader's convenience, following the simplification by Lundholm, Nam and Portmann [LNP16].
The main new ingredient is a local version of the exclusion principle using the vanishing condition on the diagonal set.In Section 4, we will discuss a very useful reduction of the desired local exclusion to simply the positivity of a local energy using the scale-covariance of the kinetic operator (−∆) s .This step refines and generalizes a recent bootstrap argument for the energy of ideal anyons by Lundholm and Seiringer [LS18].In Section 5, the remaining crucial fact that the local energy eventually becomes positive with increasing particle number will be settled by means of a new many-particle Poincaré inequality.Some standard and non-standard results on relevant function spaces are collected in the appendices for completeness.
We stress that our method will also work for any other deformations of the Laplacian which retain similar positivity and scale-covariance properties, including other types of point interactions as well as particles subject to intermediate statistics (ideal anyons) in one and two dimensions.

General strategy of deriving Lieb-Thirring inequalities
In the following we will summarize the proof of the usual Lieb-Thirring inequality (1.1) for fermionic wave functions, mainly following the simplified representation in [LNP16].The starting point is the following obvious localization formula: if {Ω} is a collection of disjoint subsets of R d , then where the Neumann localization (−∆) s |Ω is defined via the quadratic form (Sobolev seminorm) Consequently, for any N -body wave function Ψ N ∈ H s (R dN ) we have where the expected local energy on Ω is Hereafter, C = C(d, s) > 0 denotes a universal constant (independent of N , Ψ N and Q).
Lemma 3.1 can be interpreted as a local version of the lower bound (1.3) (the negative term appears due to the lack of Dirichlet boundary condition).
Lemma 3.2 (Local exclusion for fermions).-Let d 1 and s > 0. Let Ψ N be a fermionic wave function in H s (R dN ) satisfying (1.2) for N 2 and let Q be an arbitrary cube in R d .Then where q := #{multi-indices α ∈ N d 0 : 0 |α| < s}.In the non-relativistic case s = 1, Lemma 3.2 simply states that as soon as there is more than one particle on Q the energy must be strictly positive, and furthermore that it grows at least linearly with the number of particles.Such a weak formulation of the exclusion principle was used by Dyson and Lenard in their first proof of the stability of matter [DL67], while its general applicability in the above format was noted by Lundholm and Solovej in [LS13a,LS13b].
Then the support of f can be covered by a collection of disjoint cubes {Q} in R d such that for all α > 0 and 0 q < Λ2 −d , where Conclusion of (1.1).
Let q be as in Lemma 3.2 and let Λ = 2 d q + 1.If N Λ, then (1.1) follows immediately from (1.3), whose proof is similar to (indeed simpler than) that of Lemma 3.1.If N > Λ, then we can apply Lemma 3.3 with f = Ψ N (by standard approximation we may reduce to compact support), α = 2s/d, and obtain a collection of disjoint cubes {Q}.Combining with (3.2), (3.4) and (3.5) we obtain As we can see from the above strategy, the only place where the anti-symmetry (1.2) plays a role is the local exclusion bound in Lemma 3.2.Extending this result to the weaker condition Ψ N | = 0 is the main task of our proof below.

Reduction of local exclusion
In this section, we prove a very useful observation, that allows to reduce the local exclusion (3.5) to the positivity of the local energy, using the scale-covariance of the kinetic energy.This step is inspired by the recent work of Lundholm and Seiringer [LS18] on the energy of ideal anyons.We formulate it abstractly as follows: Lemma 4.1 (Covariant energy bound).-Assume that to any n ∈ N 0 and any cube Q ⊂ R d there is associated a non-negative number ('energy') E n (Q) satisfying the following properties, for some constant s > 0: • (a priori positivity) There exists q 0 such that E n (Q) > 0 for all n q.Then there exists a constant C > 0 independent of n and Q such that Proof.-Note that for q n N , (4.1) holds for some C = C N > 0 by the a priori positivity.The main point here is to remove the N -dependence of the constant. Denote with a uniform constant C > 0 and consider n = N .Split Q 0 into 2 d subcubes of half side length and obtain by the superadditivity, translation-invariance and scale-covariance Consider a configuration {n j } ⊂ N 2 d 0 such that the minimum in (4.3) is attained.The a priori positivity E N > 0 ensures that none of the n j can be N (in the same way we deduce that E 0 = 0).Assume that there exist exactly M numbers n j < q with 0 M 2 d .Then Therefore, from (4.3), (4.2) and Hölder's inequality we deduce that with the same constant C as in (4.2).If we take so that also qM N −1 1, then by Bernoulli's inequality and hence (4.4) reduces to with the same constant C as in (4.2).By induction we obtain (4.6) for all N q, with a constant C independent of N .This is the desired bound (4.1) for the unit cube Q 0 .The result for the general cube follows from scale-covariance and translation-invariance.
Remark 4.2.-It is in fact also possible to allow for E n < 0 for finitely many n > 0 in Lemma 4.1, under a small refinement of the assumption of a priori positivity.It is sufficient that there exists q > 0 and c > 1 such that for all n q (4.7) Namely, with this assumption, the bound in (4.4) may again be used for all n j q, and one obtains , where the function We will apply the above general bound to the local ground-state energy among wave functions satisfying the vanishing condition on k-particle diagonals (4.8) , where we have introduced the "completely localized" kinetic functional (4.9)Ψ N 2 Ḣs, N (Ω) 3), and its properties will be crucial to deduce the desired local exclusion for E Ω [Ψ N ].The seminorm • Ḣs,N (Ω) in general contains only some of the terms of the standard homogeneous Sobolev seminorm • Ḣs (Ω N ) ; however, the corresponding norms (i.e. the seminorms plus the L 2 -norm) are actually equivalent modulo N -dependent constants, not only globally on R dN but also locally on Q N (see Appendix A).
The superadditivity of the energy E N (Ω) follows from the partitioning of the many-body space and by locality respectively non-negativity of any non-local part of the kinetic energy, i.e. (3.1).The method was also used in [LS18, Lemma 4.2] for anyons.

ANNALES HENRI LEBESGUE
Proof.-For any partition A = {A j } J j=1 of {1, 2, . . ., N } (i.e. the A j are disjoint subsets of {1, 2, . . ., N } such that j |A j | = N ), we denote by 1 A the characteristic function of the set Using the operator bound similar to (3.1) the partition of unity and the fact that 1 A commutes with (−∆ x i ) s |Ω j , we can write for any Here we have introduced the shorthand notation Thus in summary Here in the last identity we have used the partition of unity (4.11) again.This implies the desired estimate (4.10).Now we are ready to prove the reduction of the local exclusion.
Lemma 4.4 (Energy positivity implies local exclusion).-Assume that there exists a constant q > 0 such that for any cube Then for all N 1 and for all wave functions Here C > 0 is a constant independent of N , Ψ N and Q.
Proof.-Given (4.12), the energy functional E n (Q) defined in (4.8) verifies all conditions in Lemma 4.1.Therefore, there exists a constant C > 0 independent of n and Q such that Now we adapt the localization method in the proof of Lemma 4.3 to treat the functional . To be precise, for any subset B of {1, . . ., N } we denote by 1 B the characteristic function of the set Here we have used the fact that 1 B commutes with (−∆ x i ) s |Q and the shorthand notation (x 1 , . . ., On the other hand, the partition of unity (4.15) implies that Thus from (4.16) and (4.14) we conclude that by Jensen's inequality and the convexity of the function t → [t] + .

Many-body Poincaré inequality
The crucial fact that the local energy E n (Ω) in (4.8) eventually becomes positive with increasing particle number is the content of the following Poincaré inequality: for all u ∈ C ∞ (Ω N ) whose restriction to k is zero.
Since Theorem 5.1 is of independent interest, we state the result for more general domains although the result for cubes is sufficient for our application.
Conclusion of Theorem 2.2.-From Theorem 5.1 and Lemma 4.4 we obtain the local exclusion bound (4.13).Theorem 2.2 then immediately follows from the proof strategy in Section 3.
It remains to prove Theorem 5.1.The central fact used in the proof is that a function for which the left-hand side of (5.1) vanishes must be a polynomial, and that if a polynomial vanishes on too many diagonals it must be zero.
We prove the other cases by induction.
Step 2. -Consider d = 1 and k > 2. Then f (x 1 , . . ., x N ) = 0 if at least k points x i 's coincide.Then if x k , . . ., x N are mutually different, the one-variable polynomial for all x 1 , . . ., x N .Similarly, by a renumbering, we can show that f (x 1 , x 2 , . . ., x N ) = 0 if at least (k − 1) points x i 's coincide.By induction in k, we conclude that f ≡ 0.
Step 3. -Now consider d > 1 and k 2. Let us denote satisfies that deg y i g S and g = 0 if (at least) k points y i 's coincide.By the result in the 1D case (with the choice n = (S + 1)k) we conclude that g ≡ 0. Similarly, we obtain that f (x 1 , . . ., x N ) = f ((y 1 , z 1 ), . . ., (y N , z N )) = 0 if at least n points z i 's coincide.By induction in d (i.e. using the induction hypothesis with d − 1 and k = n, N (S + 1) d−1 n) we conclude that f ≡ 0.
We will also need the following technical Lemma 5.3, which essentially states that if a multivariable function is a polynomial in each variable separately, then it is a multivariable polynomial.The proof of this seemingly obvious fact is indeed nontrivial; see Carroll [Car61] for an elegant proof in the two variables case.Here we provide an alternative proof for n variables.
) satisfy that for any j = 1, 2, . . ., n and for a.e.(x 1 , . . ., x j−1 , x j+1 , . . ., x n ) ∈ R n−1 the mapping x j → f (x 1 , . . ., x j , . . ., x n ) is a polynomial of degree at most M j .Then f is a polynomial of n variables (x 1 , . . ., x n ) of degree at most M = n j=1 M j .From the proof below, it is clear that we can replace R n by a subdomain (e.g. a cube). Proof.
By assumption, for a.e.x j ∈ R n−1 , the mapping x j → f (x j ; x j ) is a polynomial of degree at most M j .Therefore, for any we have α j > M j for some j, and hence for any test function ϕ ∈ C ∞ c (R n ) using Fubini's theorem and (5.2) we can write Step 2. -Thus it remains to prove that if D α f = 0 as distribution in R n for any |α| > M , then f is a polynomial of n variables.We prove this statement by induction in M .
If M = 0, then D x j f = 0 as distribution for any j = 1, 2, . . ., n, and hence f is constant by [LL01, Theorem 6.1].Now we prove the statement for M 1 using the induction hypothesis for M − 1.From D α f = 0, ∀ |α| > M we have for any j = 1, 2, . . ., n, Thus by the induction hypothesis for M − 1, D x j f is a polynomial of n variables for any j = 1, 2, . . ., n.Since D x j f ∈ C(R n ) for all j = 1, 2, . . ., n, we obtain that f ∈ C 1 (R n ) by [LL01, Theorem 6.10] and we have the formula [LL01, Theorem 6.9] The latter formula and the fact that D x j f is a polynomial of n variables for any j = 1, 2, . . ., n imply that f is a polynomial of n variables.This ends the proof of the Lemma 5.3.
In particular, u n is bounded in the Sobolev space H ν (Ω N ) with ν = min{s, 1}.Indeed, for d = 1 this follows from Lemma A.1 and Sobolev's embedding theorem.If d 2 then s > 1 and the claim follows from Sobolev's embedding theorem combined with that for any Ω the Ḣ1 (Ω N ) and Ḣ1, N (Ω) seminorms are equivalent.By compactness of the embedding H ν (Ω N ) ⊂ L 2 (Ω N ), up to a subsequence, u n converges strongly to a function P in L 2 (Ω N ).Since u n L 2 (Ω N ) = 1 we have that P L 2 (Ω N ) = 1.On the other hand, by Poincaré's inequality for Ḣs (Ω) (combining [LL01, Theorem 8.11] and [HSV13, Lemma 2.2]) where P (n) j (x) is a polynomial in x j of degree s − 1 .In fact, the polynomial can be written explicitly as (5.4) for universal functions ϕ β ∈ C ∞ (Ω).Since u n converges strongly in L 2 (Ω N ), we can conclude that P (n) j (x) → P j (x) strongly in L 2 (Ω N ) and the limit is again a polynomial in x j of degree s − 1 .The assumption (5.3) allows us to identify the limiting functions and we find that Thus the function P (x) is a polynomial in each variable x j (of degree s−1 ).By Lemma 5.3, P (x) is a multivariate polynomial whose degree in each x j is s − 1 .We now want to use that u n = 0 on k to prove that P = 0 on k .Once this is done, then Lemma 5.2 implies that P ≡ 0 if N s d k.This contradicts that P L 2 (Ω N ) = 1 and hence completes our proof.Note that if we can prove that P ≡ 0 in some open subset this is sufficient, in particular we can find some open cube Q ⊆ Ω and consider instead u n and P restricted to Q N .
We consider the diagonal x 1 = x 2 = . . .= x k ; the other cases are treated identically.By Lebesgue's differentiation theorem it suffices to prove that (5.5) lim By Fatou's lemma we have for any δ > 0 that (5.6) Since u n = 0 on k it holds that (5.7)

ANNALES HENRI LEBESGUE
By Lemma A.1, any u ∈ L 2 (Q l ) with u Ḣs, l (Q) < ∞ satisfies that u ∈ H s (Q l ) and moreover there is a constant C depending only on Q, l, s such that If 2s > dl, by Sobolev's embedding theorem (see for instance [DNPV12, Theorem 8.2]), there is for any γ ∈ (0, min{1, 2s−dl 2 }) a constant C so that By assumption 2s > d(k − 1), and hence we can apply this result to the function (x 2 , . . ., x k ) → u n (x 1 , . . ., x k ; x ) (whose Ḣs, k−1 (Q)-seminorm is bounded for a.e.(x 1 , x )).Equation (5.7) then implies that where we set x = (x 2 , . . ., x k ) and x 1 = (x 1 , . . ., x 1 ).Applying (5.8) and Hölder's inequality yields Since u n L 2 (Q N ) + u n Ḣs, N (Q) C and γ > 0, we arrive at (5.5) which completes the proof of Theorem 5.1.We finally note that the many-body nature of the wave functions is crucial for Theorem 5.1 to hold.The following example shows that the requirement that the particle number N is large, in fact typically strictly larger than k, is necessary.

Replacing the condition u|
, does not help.
TOME 4 (2021) Proof.-If N < k there is no diagonal set k and we may take the constant function as a counterexample.For N = k we consider the polynomial u(x 1 , . . ., x k ) := for which, by the arithmetic mean-geometric mean inequality and the triangle inequality, where R 0 may serve as a radial coordinate on R d(k−1) relative to x 1 .Hence, we have that . Thus (analogously to Lemma B.2, and by extension) Thus it is a slightly subtle question of what happens to these spaces in the manybody limit.An even more subtle question is what happens to the local versions of these spaces, i.e. when R d is replaced by Ω R d .For us, the following equivalence of the spaces in the case of cubes will suffice: There exist positive constants c, C depending only on d, s, N so that Lemma A.1 is an immediate consequence of the equivalence (A.1) of the two seminorms on R dN and the following extension Lemma A.2: , where C is a constant depending only on s, d and N .
Proof.-We shall prove the Lemma A.2 by using higher-order reflection through one side of the hypercube Q at a time.To this end we recall that if v ∈ C n ([0, 1]), for some n 0, we can construct an explicit extension where ϕ ∈ C ∞ ((−∞, 0]) such that ϕ(x) ≡ 0 for x < −δ and ϕ(x) ≡ 1 in [−δ/2, 0].What remains is to verify that we can choose the λ j 's so that v ∈ C n .But if we differentiate v for x away from zero we see that the system of equations that we need the λ j to satisfy to get continuity of the derivatives across x = 0 is But the determinant of this matrix is non-zero (it is a Vandermonde matrix) and hence there exists a unique solution (λ 1 , . . ., λ n+1 ).
We shall now prove that we can use this one-dimensional extension repeatedly to construct an extension of u to R dN .The idea is to use the one-dimensional result one coordinate at a time and show that the new function in each step has the quantity corresponding to the Ḣs, N -seminorm controlled by that of u.
Without loss we can assume that u ∈ C n (Q N ) (the construction is stable under approximation), where we take n = s .Consider u(x 1 ; x ), x 1 ∈ [0, 1] and x ∈ [0, 1] dN −1 .And apply the above Lemma A.2 for each fixed x , that is, we define v 1 by It is a simple calculation to use Sobolev's embedding theorem to prove that we can bound the L p -norm of l th order derivatives of v 1 by the corresponding one for u if l n.We need to prove that also the fractional order seminorm is preserved.That is, we wish to show that, with s = m + σ and for all multi-indices |α| = m.If we can prove this inequality, then by repeating the procedure to extend v 1 to x 1 > 1 the same proof gives that we can bound the corresponding Ḣs, N quantity in terms of that of v 1 , and hence u.By repeating the procedure for each coordinate at a time we, after 2dN reflections, find a function u ∈ L 2 (R dN ) satisfying the claims of the lemma.Thus all that remains is to prove (A.2).We start with the first term which is also the most difficult: Clearly the integral over Q×Q is bounded by u Ḣs,N (Q) .We treat the two remaining terms separately.In order to bound the integral over Q × (Q \ Q) we write

ANNALES HENRI LEBESGUE
Thus we can bound the second integral in (A.3) as follows: Using the triangle inequality and Sobolev's embedding theorem one finds that the second term is Ḣs,N (Q) .Since j λ j (−j) −α 1 = 1 for any α 1 m + 1, one obtains for the first integral In the last step we used the inequality (x + jy) 2 (x − y) 2 for x, y 0 and j 1.
TOME 4 (2021) For the last integral in (A.3) we have where we set α as the multi-index α but with α 1 exchanged for β.By the triangle inequality and the fact that j λ j (−j) −β = 1 the integral is smaller than where we used that ψu To show that the remaining terms in (A.2) are u 2 L 2 + u 2 Ḣs, N one can proceed in an almost identical manner.The main difference is that in these terms the differentiation is with respect other variables than the variable in which the extension has been made, and the splitting of the integrals is slightly different.However, in the end this only simplifies each step of the proof of Lemma A.2.

Appendix B. Spaces of contact interaction
We consider in the following only 2-particle diagonals , for simplicity, however analogous statements can be made for the case of k-particle diagonals.Define for N 2 the restricted N -particle spaces .
Proof.-For α = 0 there are in D α x j χ ε a total of |α| derivatives of functions ϕ ε (x j − x k ), k = j, and remaining factors involving the other particles.These factors are uniformly bounded while each derivative yields an additional factor 1/ε, while reducing the support in x j to B 2ε (x k ) \ B ε (x k ).Furthermore, we thus have and ε −2|α|+d , and for any 0 < σ < 1 In χ * ε this could involve different points x k but the worst case is if they are the same, This covers the even-dimensional critical case d = 2m, m ∈ N 1 .
In the odd-dimensional critical case d = 2m + 2σ, σ = 1/2, we observe that which is not enough for 2|α| = d − 1.Instead we shall use χ * ε .For the case 2|α| = d − 1 things are a bit less straightforward.We start with the case d = 1 which is the easiest.Here our approach differs slightly due to the fact that in this case |α| = 0. Let We estimate the seminorm χ * ε Ḣs x j (Ω) .By construction of χ * ε we have that Moreover, the difference is zero whenever (x, y) ∈ U 2 1 ∪ U 2 2 .
For x and y close we need to estimate this quantity more precisely.By Taylor's theorem we can estimate By symmetry in x, y we find The latter term is fairly easy to estimate: We return to the remaining term of (B.2): The inner integral is convergent and hence we are left with When 2|α| = d − 1 and d > 1 the estimates for the difference quotient are a bit more technical.Similarly to above, Taylor's theorem combined with (B.1) yields We estimate the integral Choosing coordinates in a plane containing x k , x and y such that x k = (0, 0), x = (r 1 , 0) and y = (r 2 cos(θ), r 2 sin(θ)) with θ ∈ [0, π) we can write this integral as The integral g(θ) tends to infinity in the limit θ → π.However, this corresponds to x and y being far apart relative to their distance to the x k .When θ is far from 0 we shall instead use the following bound which follows directly from the supremum bound in (B.1) where θ is the angle between the vectors y − x k and x − x k .Note that the bound in (B.3) does not capture the continuity of D α χ * ε and hence cannot be sufficiently accurate for our purposes when |x − y| is small.
We are now ready to start estimating the H s -seminorm of χ * ε .Using the same notation as in the d = 1 case where we used that |D α To bound the integral we use the estimates derived earlier.Recalling that in the case under consideration |α| = d−1 2 the derived bounds tells us that here θ k denotes the angle between the vectors x − x k and y − x k .For each fixed x we rewrite the integral over Ω in spherical coordinates around x k , oriented so that x is located at the south pole.With R = |x − x k |, r = |y − x k | and θ k as before, the integral becomes For θ ∈ [π/2, π] we use the bounds in (B.3), (B.4): Thus this part of the integral is O(ε −1 ).What remains is to bound the integral when r 0 and θ k ∈ [0, π/2).To accomplish this we shall use the bound for the difference of the derivatives derived earlier.Note that since θ k < π/2 we can replace the factor g(θ k ) by a constant without any loss.Using that |x − y| 2 = R 2 + r 2 − 2rR cos θ max{(R − r) 2 , 2rR(1 − cos θ k )} we for any fixed µ ∈ (0, 1) find Consequently, also this part of the integral is O(ε −1 ) which completes the proof.
Lemma B.2. -For all s > 0 it holds that H s, N W (R d ) ⊆ H s, N 0 (R d ).
For I α we have that It suffices to prove that C ∞ c (R dN \ ) is dense in H s , and moreover, using that Moreover, by Lemma B.1 and arguing as in the proof of Lemma B.2 The above generalizes the case d = 2 and s = 1 where it is well known that hard-core bosons have non-extensive energy in the dilute limit [LY01] and thus that a Lieb-Thirring inequality of the type (2.1) cannot hold.See also [Sve81] for generalizations with integer s.

dx
Next, we have the following three key tools [LNP16, Lemmas 8, 11, 12].Lemma 3.1 (Local uncertainty).-Let d 1 and s > 0. Let Ψ N be a wave function in H s (R dN ) for arbitrary N 1 and let Q be an arbitrary cube in R d .Then Theorem 5.1 (Poincaré inequality for functions vanishing on diagonals).-Fix an integer k 2 and a bounded connected Lipschitz domain Ω ⊂ R d .Assume that 2s > d(k − 1).For N ∈ N large enough (N s d k is sufficient) there exists a positive constant C depending only on s, k, N, Ω so that (5.1) u Ḣs, N (Ω) C u L 2 (Ω N ) Lemma 5.2 (Low-degree polynomials vanishing on diagonals are trivial).-Given d, k, S ∈ N 1 and N Proposition 5.4 (Counterexample to the k-body case).-Theorem 5.1 cannot hold for N < k, or for N = k if s is integer and max{d, 2}(k − 1) < 2s < (d + k)(k − 1).
by inserting the partition of unity 1 R dN = due to the non-integrability of W s .For 2s > d and s − d/2 / ∈ Z it again holds by the Hardy-Rellich inequality that H s, N W