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 \(\gamma \) in the form
$$\begin{aligned} \gamma =\sum _{j=1}^Nn_j|u_j\rangle \langle u_j|, \quad \text {so that} \quad \rho _\gamma (x) = \sum _{j=1}^N n_j | u_j |^2(x), \end{aligned}$$
where \((u_1,\ldots ,u_N)\) forms an orthonormal system. The inequality (11) which we wish to prove therefore reads
$$\begin{aligned} \sum _{j=1}^Nn_j \Vert \nabla u_j\Vert ^2 \ge K_{d,p}^{(N)}\left( \int _{{{\mathbb {R}}}^d} \rho _\gamma ^p {{\mathrm {d}}x} \right) ^{\frac{2}{d(p-1)}} \left( \sum _{j=1}^N n_j^{q} \right) ^{-\frac{2}{d(p-1)}+1}. \end{aligned}$$
(58)
For a constant \(\beta >0\) to be determined, let
$$\begin{aligned} V(x) = - \beta \rho _\gamma ^{p-1}. \end{aligned}$$
For \(\kappa \ge 1\) we use Hölder’s inequality in Schatten spaces [Sim05] in the form
$$\begin{aligned} {\mathrm{Tr}}AB \ge - \left( \sum _{n=1}^N \lambda _n(A)_-^\kappa \right) ^{\frac{1}{\kappa }} \left( {\mathrm{Tr}}B^{\kappa '} \right) ^{\frac{1}{\kappa '}} \end{aligned}$$
for all \(B\ge 0\) of rank \(\le N\). Applying this with \(A=-\Delta +V\) and \(B=\gamma \) we obtain, in view of (1),
$$\begin{aligned}&\sum _{j=1}^N n_j\int _{{{\mathbb {R}}}^d} |\nabla u_j|^2 \,{\mathrm {d}}x - \beta \int _{{{\mathbb {R}}}^d} \left( \sum _{j=1}^N n_j|u_j|^2 \right) ^p {\mathrm {d}}x = \sum _{j=1}^Nn_j \int _{{{\mathbb {R}}}^d} \left( |\nabla u_j|^2 + V |u_j|^2\right) {\mathrm {d}}x \\&\qquad \ge - \left( \sum _{j=1}^N n_j^{\kappa '}\right) ^{\frac{1}{\kappa '}} \left( \sum _{j=1}^N \left| \lambda _j(-\Delta +V) \right| ^\kappa \right) ^{\frac{1}{\kappa }} \\&\qquad \ge - \left| \! \left| \gamma \right| \! \right| _{{\mathfrak {S}}^{\kappa '}}\left( L_{\kappa ,d}^{(N)} \int _{{{\mathbb {R}}}^d} V(x)_-^{\kappa +\frac{d}{2}}\,{\mathrm {d}}x \right) ^{\frac{1}{\kappa }} \\&\qquad = - \left| \! \left| \gamma \right| \! \right| _{{\mathfrak {S}}^{\kappa '}} \left( L_{\kappa ,d}^{(N)} \right) ^{\frac{1}{\kappa }} \beta ^{1+\frac{d}{2\kappa }} \left( \int _{{{\mathbb {R}}}^d} \rho _\gamma ^{(p-1)\big (\kappa +\frac{d}{2}\big )}{\mathrm {d}}x \right) ^{\frac{1}{\kappa }}. \end{aligned}$$
We optimise in \(\beta \) by choosing
$$\begin{aligned} \beta = \left( \frac{2\kappa }{2\kappa +d} \, \frac{\int _{{{\mathbb {R}}}^d} \rho _\gamma ^p {\mathrm {d}}x}{\left| \! \left| \gamma \right| \! \right| _{{\mathfrak {S}}^{\kappa '}} \left( L_{\kappa ,d}^{(N)} \right) ^{\frac{1}{\kappa }} \left( \int _{{{\mathbb {R}}}^d} \rho _\gamma ^{(p-1)(\kappa +d/2)}{\mathrm {d}}x \right) ^{\frac{1}{\kappa }}} \right) ^{\frac{2\kappa }{d}} \end{aligned}$$
and obtain
$$\begin{aligned} \sum _{j=1}^N n_j \int _{{{\mathbb {R}}}^d} |\nabla u_j|^2 \,{\mathrm {d}}x \ge \left( \frac{2\kappa }{2\kappa +d} \right) ^{\frac{2\kappa }{d}} \frac{d}{2\kappa +d} \frac{ \left( \int _{{{\mathbb {R}}}^d} \rho _\gamma ^p\, {\mathrm {d}}x \right) ^{1+\frac{2\kappa }{d}}}{\Vert \gamma \Vert _{{\mathfrak {S}}^{\kappa '}}^{\frac{2\kappa }{d}} \left( L_{\kappa ,d}^{(N)} \right) ^{\frac{2}{d}} \left( \int _{{{\mathbb {R}}}^d} \rho _\gamma ^{(p-1)(\kappa +\frac{d}{2})}{\mathrm {d}}x \right) ^{\frac{2}{d}} } \,. \end{aligned}$$
We now choose \(\kappa =p' - d/2\), which is \(> 1\) since \(p<1+2/d\) and which ensures that \((p-1)(\kappa +d/2)=p\). Thus,
$$\begin{aligned} \sum _{j=1}^N n_j\int _{{{\mathbb {R}}}^d} |\nabla u_j|^2 \,{\mathrm {d}}x \ge \left( \frac{2p'-d}{2p'} \right) ^{\frac{2p'}{d} - 1} \frac{d}{2p'} \, \frac{ \left( \int _{{{\mathbb {R}}}^d} \rho _\gamma ^p\, {\mathrm {d}}x \right) ^{\frac{2}{d(p-1)}}}{\Vert \gamma \Vert _{{\mathfrak {S}}^{\kappa '}}^{\frac{2p'}{d}-1} \left( L_{p'-d/2,d}^{(N)} \right) ^{\frac{2}{d}} } \,. \end{aligned}$$
Therefore, the best constant \(K_{d,p}^{(N)}\) in (58) satisfies
$$\begin{aligned} K_{d,p}^{(N)} \ge \left( \frac{2p'-d}{2p'} \right) ^{\frac{2p'}{d} - 1} \frac{d}{2p'} \, \frac{1}{\left( L_{p'-d/2,d}^{(N)} \right) ^{\frac{2}{d}} } \,. \end{aligned}$$
Conversely, assume that inequality (58) holds and let \(V\in L^{\kappa +d/2}({{\mathbb {R}}}^d)\). We assume that \(-\Delta +V\) has at least N negative eigenvalues, the other case being handled similarly. Let \(u_1,\ldots ,u_N\) be orthogonal eigenfunctions corresponding to the N lowest eigenvalues of \(-\Delta +V\) and let
$$\begin{aligned} \gamma =\sum _{j=1}^Nn_j\,|u_j\rangle \langle u_j|,\qquad n_j = |\lambda _j(-\Delta +V)|^{\kappa -1} \,, \end{aligned}$$
so that
$$\begin{aligned} {\mathrm{Tr}}(-\Delta +V)\gamma = \sum _{j=1}^Nn_j\,\lambda _j(-\Delta +V) = - \sum _{j=1}^N|\lambda _j(-\Delta +V)|^{\kappa } \,. \end{aligned}$$
We have, for p such that \(p' = \kappa + \frac{d}{2}\),
$$\begin{aligned} \sum _{j=1}^N |\lambda _j(-\Delta +V)|^{\kappa }&= - \sum _{j=1}^N n_j\int _{{{\mathbb {R}}}^d} \left( |\nabla u_j|^2 + V |u_j|^2\right) {\mathrm {d}}x \\&\le - K_{d,p}^{(N)} \left( \int _{{{\mathbb {R}}}^d} \rho _\gamma ^p\,{\mathrm {d}}x \right) ^{\frac{2}{d(p-1)}} \left( \sum _{j=1}^N n_j^{\kappa '} \right) ^{-\frac{2}{d(p-1)}+1} + \left\| V_- \right\| _{p'} \left\| \rho _\gamma \right\| _p. \end{aligned}$$
Setting \(x := \Vert \rho \Vert _p\), this is of the form \( - \alpha x^{\frac{2p}{d(p-1)}} + \beta x\), with \(\frac{2p}{d(p-1)} > 1\). So it is bounded from above by
$$\begin{aligned} \sum _{j=1}^N |\lambda _j(-\Delta +V)|^{\kappa }&\le \left( K_{d,p}^{(N)} \right) ^{-\frac{d(p-1)}{2}({d+p-dp)}} \left( \frac{d}{2p'} \right) ^{\frac{d}{2p'-d}} \left( \frac{2p'-d}{2p'} \right) \\&\quad \times \left( \int _{{{\mathbb {R}}}^d} V_-^{p'}\,{\mathrm {d}}x \right) ^{\frac{2}{2p'-d}} \left( \sum _{j=1}^N n_j^{\kappa '} \right) ^{\frac{2-d(p-1)}{2p-d(p-1)}}. \end{aligned}$$
Recall that
$$\begin{aligned} n_j^{\kappa '} = |\lambda _n(-\Delta +V)|^{\kappa } \end{aligned}$$
and therefore the above inequality becomes
$$\begin{aligned} \sum _{j=1}^N |\lambda _j(-\Delta +V)|^\kappa \le \left( K_{d,p}^{(N)} \right) ^{-\frac{d}{2}} \left( \frac{d}{2p'} \right) ^{\frac{d}{2}} \left( \frac{2p'-d}{2p'} \right) ^{\frac{2p'-d}{2}} \int _{{{\mathbb {R}}}^d} V_-^{p'}\,{\mathrm {d}}x \,. \end{aligned}$$
Therefore the best constant \(L_{\kappa ,d}^{(N)}\) in (1) satisfies
$$\begin{aligned} L_{\kappa ,d}^{(N)} \le \left( K_{d,p}^{(N)} \right) ^{-\frac{d}{2}} \left( \frac{d}{2p'} \right) ^{\frac{d}{2}} \left( \frac{2p'-d}{2p'} \right) ^{\frac{2p'-d}{2}}. \end{aligned}$$
This proves the lemma. \(\square \)
Appendix B. Comments on the NLS Model and its Dual
This appendix contains some additional comments on the minimisation problem \(J(\lambda )\) in (49) studied in [GLN21]. The latter was considered for \(\lambda \in {{\mathbb {R}}}_+\) instead of just \(\lambda =N\in {{\mathbb {N}}}\). It is equivalent to the inequality
$$\begin{aligned}&{\widetilde{K}}_{p,d}^{(\lambda )} \left( \int _{{{\mathbb {R}}}^d}\rho _\gamma (x)^p\,{\mathrm {d}}x\right) ^{\frac{2}{d(p-1)}} \le \Big ({\mathrm{Tr}}(\gamma )\Big )^{\frac{d+2-dp}{d(p-1)}}\;\Vert \gamma \Vert ^{\frac{2}{d}}\;{\mathrm{Tr}}(-\Delta \gamma ),\nonumber \\&\quad \hbox { for all}\ 1\le p\le 1+\frac{2}{d} \end{aligned}$$
(59)
with \({\mathrm{Tr}}(\gamma )\le \lambda \), which is a particular interpolation between the trace formula \(\Vert \gamma \Vert _{{\mathfrak {S}}^1} = {\mathrm{Tr}}(\gamma ) =\Vert \rho _\gamma \Vert _1\), and the Lieb–Thirring inequality (36) at \(p=1+2/d\). As discussed in Sect. 1.2, another interpolation involving the Schatten space norm \(\Vert \gamma \Vert _q^{\frac{d+2-dp}{d(p-1)}+\frac{2}{d}}\) instead of \(\left| \! \left| \gamma \right| \! \right| _1^{\frac{d+2-dp}{d(p-1)}} \;\Vert \gamma \Vert ^{\frac{2}{d}}\) is the dual Lieb–Thirring inequality (14).
B.1. An inequality with no optimiser
Optimising (59) over all possible \(\lambda \)’s, we arrive at the inequality without constraints
$$\begin{aligned} {\boxed {{\widetilde{K}}_{p,d} \left( \int _{{{\mathbb {R}}}^d}\rho _\gamma (x)^p\,{\mathrm {d}}x\right) ^{\frac{2}{d(p-1)}} \le \Big ({\mathrm{Tr}}(\gamma )\Big )^{\frac{d+2-dp}{d(p-1)}}\;\Vert \gamma \Vert ^{\frac{2}{d}}\;{\mathrm{Tr}}(-\Delta \gamma ),}} \end{aligned}$$
(60)
for all \(\gamma =\gamma ^*\ge 0\), with the best constant
$$\begin{aligned} \widetilde{K}_{p,d}:=\left( \sup _{\lambda >0}\frac{|J(\lambda )|}{\lambda }\right) ^{-\frac{d+2-dp}{d(p-1)}}\frac{1}{p-1}\left( \frac{d}{2p}\right) ^{\frac{2}{d(p-1)}}\left( 1+\frac{2}{d}-p\right) ^{-\frac{d+2-dp}{d(p-1)}}. \end{aligned}$$
(61)
We recall from [GLN21, Section 1.3] that
$$\begin{aligned} \sup _\lambda \frac{|J(\lambda )|}{\lambda }=\lim _{\lambda \rightarrow \infty }\frac{|J(\lambda )|}{\lambda }<\infty . \end{aligned}$$
From the results in [GLN21] we can deduce that the inequality (60) has no optimiser.
Lemma 17
Let \(d\ge 1\) and \(1<p<\min (2,1+2/d)\). Then \({\widetilde{K}}_{p,d}<{\widetilde{K}}_{p,d}^{(\lambda )}\) for all \(\lambda >0\). In particular, the inequality (60) admits no optimiser.
Proof
It was shown in [GLN21, Corollary 22] that \(J(\lambda )/\lambda \) is always above its limit. Therefore \({\widetilde{K}}_{p,d}<{\widetilde{K}}_{p,d}^{(\lambda )}\) and there cannot be an optimiser with finite trace.\(\quad \square \)
We believe that the optimisers of \({\widetilde{K}}^{(N)}_{p,d}\) converge in the limit \(N\rightarrow \infty \) to periodic or translation-invariant operators, as discussed at the end of Sect. 1.1 and in [GLN21].
Remark 18
(Monotonicity in p) By Hölder’s inequality, for any \(\gamma =\gamma ^*\ge 0\) the function
$$\begin{aligned} p\mapsto \left( \int _{{{\mathbb {R}}}^d} \rho _\gamma (x)^p \,dx \right) ^{\frac{2}{d(p-1)}} \left( \int _{{{\mathbb {R}}}^d} \rho _\gamma (x)\,dx \right) ^{- \frac{2}{d(p-1)}} \end{aligned}$$
is non-decreasing. This implies that \(p\mapsto {\widetilde{K}}_{p,d}\) is non-increasing on the interval \((1,1+2/d)\). In particular, since \({\widetilde{K}}_{p,d}^\mathrm{sc}=K_{1+2/d,d}^{\mathrm{sc}}\) is independent of p (see Sec. B.4), we deduce that if \({\widetilde{K}}_{p,d}={\widetilde{K}}_{p,d}^{\mathrm{sc}}\) for some \(p=p_0\), then the same equality holds for all \(1<p\le p_0\). This generalises the observation in [GLN21] that if the standard Lieb–Thirring conjecture holds for \(\kappa =1\) (that is, \({\widetilde{K}}_{p,d}={\widetilde{K}}_{p,d}^{\mathrm{sc}}\) for \(p=1+2/d\)), then \({\widetilde{K}}_{p,d}={\widetilde{K}}_{p,d}^\mathrm{sc}\) 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.
Lemma 19
(Dual formulation of (60)). Let \(d\ge 1\) and let \(\kappa >1\) and \(p<1+2/d\) be related by \(p'=\kappa +d/2\). Then (60) is equivalent to the inequality
$$\begin{aligned} {\boxed {{\mathrm{Tr}}(-\Delta +V+\tau )_- \le {\widetilde{L}}_{\kappa ,d}\; \tau ^{1-\kappa } \int _{{{\mathbb {R}}}^d} V_-^{\kappa +\frac{d}{2}}\,{\mathrm {d}}x, }} \end{aligned}$$
(62)
valid for all \(\tau >0\) and all \(V\in L^{\kappa +\frac{d}{2}}({{\mathbb {R}}}^d)\), in the sense that the best constants are related by
$$\begin{aligned} {\widetilde{K}}_{p,d} \widetilde{L}_{\kappa , d}^{\frac{2}{d}} = \left( 1 - \frac{d(p-1)}{2}\right) ^{\frac{d + 2 - dp}{d(p-1)}} \frac{d}{2} \frac{(p-1)^{\frac{2 + d}{d}}}{p^{\frac{2p}{d(p-1)}}} = \frac{d}{2}\dfrac{(\kappa - 1)^{\frac{2}{d}(\kappa - 1)}}{(\kappa + \frac{d}{2})^{\frac{2}{d}\kappa + 1}}. \end{aligned}$$
(63)
Proof
Assume that (62) holds and let \(0\le \gamma \le 1\) of finite kinetic energy. Set \(\lambda :={\mathrm{Tr}}(\gamma )\) and \(\rho :=\rho _\gamma \). Then, for all \(\tau >0\) and all \(0\ge V\in L^{\kappa +\frac{d}{2}}({{\mathbb {R}}}^d)\), from (62) with the abbreviation \(L:={\widetilde{L}}_{p'-d/2,d}\) we have
$$\begin{aligned} {\mathrm{Tr}}(-\Delta \gamma )&= {\mathrm{Tr}}(-\Delta +V+\tau )\gamma - \int _{{{\mathbb {R}}}^d} V\rho \,{\mathrm {d}}x - \tau \lambda \\&\ge - L \tau ^{-\kappa +1} \int _{{{\mathbb {R}}}^d} V_-^{\kappa +\frac{d}{2}}\,{\mathrm {d}}x + \int _{{{\mathbb {R}}}^d} V_- \rho \,{\mathrm {d}}x - \tau \lambda \,. \end{aligned}$$
We first optimise in V by taking
$$\begin{aligned} V = - \frac{1}{L^{p-1}} \frac{(p-1)^{p-1}}{p^{p-1}} \tau ^{(\kappa - 1)(p-1)} \rho ^{p-1}, \end{aligned}$$
and obtain
$$\begin{aligned} {\mathrm{Tr}}(-\Delta \gamma ) \ge \frac{(p-1)^{p-1}}{p^{p}} \frac{1}{L^{p-1}} \tau ^{(\kappa -1)(p-1)} \int _{{{\mathbb {R}}}^d} \rho ^p\,{\mathrm {d}}x -\tau \lambda . \end{aligned}$$
We then optimise in \(\tau \) by taking (note that \((\kappa - 1)(p-1) = 1 - \frac{d}{2}(p-1) \in (0, 1)\), so the function is indeed bounded from above)
$$\begin{aligned} \tau = \frac{1}{\lambda ^{\frac{2}{d(p-1)}}} \left( 1 - \frac{d(p-1)}{2} \right) ^{\frac{2}{d(p-1)}} \frac{(p-1)^{\frac{2}{d}}}{p^{\frac{2p}{d(p-1)}}} \frac{1}{L^{\frac{2}{d}}} \left( \int _{{{\mathbb {R}}}^d} \rho ^p\,{\mathrm {d}}x \right) ^{\frac{2}{d(p-1)}}, \end{aligned}$$
and we obtain finally
$$\begin{aligned} {\mathrm{Tr}}( - \Delta \gamma ) \ge \frac{1}{\lambda ^{\frac{d + 2 - dp}{d(p-1)}}} \frac{1}{L^{\frac{2}{d}}} \left( 1 - \frac{d(p-1)}{2}\right) ^{\frac{d + 2 - dp}{d(p-1)}} \frac{d}{2} \frac{(p-1)^{\frac{2 + d}{d}}}{p^{\frac{2p}{d(p-1)}}} \left( \int _{{{\mathbb {R}}}^d} \rho ^p\,{\mathrm {d}}x \right) ^{\frac{2}{d(p-1)}}. \end{aligned}$$
Comparing with (60) shows the first bound
$$\begin{aligned} {\widetilde{K}}_{p,d} L^{\frac{2}{d}} \ge \left( 1 - \frac{d(p-1)}{2}\right) ^{\frac{d + 2 - dp}{d(p-1)}} \frac{d}{2} \frac{(p-1)^{\frac{2 + d}{d}}}{p^{\frac{2p}{d(p-1)}}}. \end{aligned}$$
Conversely, assume that (60) holds and let \(V\in L^{\kappa +\frac{d}{2}}({{\mathbb {R}}}^d)\) and \(\tau >0\). We set \(\gamma ={\mathbb {1}}(-\Delta +V+\tau <0)\), \(\rho =\rho _\gamma \) and \(\lambda ={\mathrm{Tr}}(\gamma )\). We obtain, from (60) with the abbreviation \(K={\tilde{K}}_{p,d}\),
$$\begin{aligned} {\mathrm{Tr}}(-\Delta +V+\tau )_-&= - {\mathrm{Tr}}(-\Delta +V+\tau )\gamma = -{\mathrm{Tr}}(-\Delta \gamma ) - \int _{{{\mathbb {R}}}^d} V\rho \, {\mathrm {d}}x - \tau \lambda \\&\le - K \frac{1}{\lambda ^{\frac{d + 2 - dp}{d(p-1)}}} \left( \int _{{{\mathbb {R}}}^d} \rho ^p\, {\mathrm {d}}x \right) ^{\frac{2}{d(p-1)}} + \int _{{{\mathbb {R}}}^d} V_-\rho \, {\mathrm {d}}x - \tau \lambda \,. \end{aligned}$$
Seen as a function of \(\lambda \), the right-hand side is smaller than its maximum, attained for
$$\begin{aligned} \lambda = \left( \frac{2}{d(p-1)} - 1\right) ^{\frac{d(p-1)}{2}} \left( \frac{K}{\tau } \right) ^{\frac{d(p-1)}{2}} \int _{{{\mathbb {R}}}^d} \rho ^p \,{\mathrm {d}}x \,, \end{aligned}$$
so
$$\begin{aligned} {\mathrm{Tr}}(-\Delta +V+\tau )_-&\le \int _{{{\mathbb {R}}}^d} V_-\rho \,{\mathrm {d}}x \\&\quad - \frac{2}{d(p-1)} \left( \frac{2}{d(p-1)} - 1 \right) ^{\frac{d(p-1)}{2} - 1} K^{\frac{d(p-1)}{2}} \tau ^{1 - \frac{d(p-1)}{2} } \int _{{{\mathbb {R}}}^d} \rho ^p \, {\mathrm {d}}x . \end{aligned}$$
Now, seen as a function of \(\rho \), it is again smaller than its maximum. We deduce that (recall that \(\kappa = \frac{p}{p-1} - \frac{d}{2} = 1 + \frac{1}{p-1} + \frac{d}{2}\))
$$\begin{aligned}&{\mathrm{Tr}}(-\Delta +V+\tau )_-\\&\quad \le \left( \frac{d}{2} \right) ^{\frac{1}{p-1}} \left( \frac{2}{d(p-1)} - 1 \right) ^{\frac{d + 2 - dp}{2(p-1)}} \left( \frac{p-1}{p} \right) ^{\frac{p}{p-1}} \frac{1}{K^{\frac{d}{2}}} \tau ^{1 - \kappa } \int _{{{\mathbb {R}}}^d} V_-^{\kappa + \frac{d}{2}}\, {\mathrm {d}}x \,. \end{aligned}$$
Comparing with (62) shows that
$$\begin{aligned} {\widetilde{L}}_{\kappa ,d} K^{\frac{d}{2}}&\le \left( \frac{d}{2} \right) ^{\frac{1}{p-1}} \left( \frac{2}{d(p-1)} - 1 \right) ^{\frac{d + 2 - dp}{2(p-1)}} \frac{(p-1)^{\frac{p}{p-1}}}{p^{\frac{p}{p-1}}} \\&= \left( \frac{d}{2} \right) ^{\frac{d}{2}} \left( 1 - \frac{d(p-1)}{2} \right) ^{\frac{d + 2 - dp}{2(p-1)}} \frac{(p-1)^{1 + \frac{d}{2}}}{p^{\frac{p}{p-1}}} \,. \end{aligned}$$
This proves the lemma.\(\quad \square \)
B.3. Weak Lieb–Thirring inequalities
The dual inequality (62) provides an estimate on the quantity
$$\begin{aligned} \sup _{\tau>0}\left\{ \tau ^{\kappa -1}{\mathrm{Tr}}(-\Delta +V+\tau )_-\right\} =\sup _{\tau >0}\left\{ \tau ^{\kappa -1}\sum _{n\ge 1}\Big (\lambda _n(-\Delta +V)+\tau \Big )_-\right\} . \end{aligned}$$
(64)
A natural question is to ask how this supremum compares with
$$\begin{aligned} {\mathrm{Tr}}(-\Delta +V)_-^\kappa =\sum _{n\ge 1}|\lambda _n(-\Delta +V)|^\kappa \end{aligned}$$
appearing in the usual Lieb–Thirring inequality. In this subsection we show that (64) is equivalent to the weak \(\ell ^\kappa \) norm of the negative eigenvalues of \(-\Delta +V\). In this sense (62) is weaker than the ordinary Lieb–Thirring inequality for \(\kappa \), which bounds the (strong) \(\ell ^\kappa \) norm of the eigenvalues. The results of this subsection concern the ‘analytic content’ of the inequalities and ignore, at least to some extent, the question of sharp constants.
Let X be a measure space and \(p>r\ge 0\). For a measurable function f we set
$$\begin{aligned}{}[f]_{p,r}' := \sup _{\tau >0} \left\{ \tau ^{1-\frac{r}{p}} \left( \int _X (|f|-\tau )_+^r \,{\mathrm {d}}x \right) ^\frac{1}{p} \right\} . \end{aligned}$$
When \(r = 0\), we get
$$\begin{aligned}{}[f]_{p,0}' = \sup _{\tau>0} \tau |\{|f|>\tau \}|^{1/p} \end{aligned}$$
which is the standard quasinorm in weak \(L^p\). Actually, it turns out that for all \(0 \le r<p\), \([f]_{p,r}'\) is an equivalent quasinorm in this space.
Lemma 20
If \(p>r\ge 0\), then for any measurable f on X,
$$\begin{aligned} \left( \frac{(p-r)^{p-r}\, r^r}{p^p} \right) ^{\frac{1}{p}} [f]_{p,0}' \le [f]_{p,r}' \le \left( \frac{\Gamma (p-r)\,\Gamma (r+1)}{\Gamma (p)} \right) ^\frac{1}{p} [f]_{p,0}' \,. \end{aligned}$$
Proof
We set \(\lambda (\sigma ) := |\{|f|>\sigma \} |\) for brevity. First, for any \(\sigma >\tau \), we have the inequality
$$\begin{aligned} {\mathbb {1}}_{ \{| f |> \sigma \} } \le {\mathbb {1}}_{\{| f |> \sigma \} } \left( \dfrac{| f | - \tau }{\sigma - \tau } \right) ^r \le {\mathbb {1}}_{\{| f | > \tau \} } \left( \dfrac{| f | - \tau }{\sigma - \tau } \right) ^r. \end{aligned}$$
Integrating gives the inequality
$$\begin{aligned} \lambda (\sigma ) \le \dfrac{1}{(\sigma -\tau )^{r}} \int _X (|f|>\tau )_+^r\,{\mathrm {d}}x \le \dfrac{1}{\tau ^{p-r} (\sigma -\tau )^{r}} \left( [f]_{p,r}' \right) ^p \,. \end{aligned}$$
We optimise in \(\tau \) by choosing \(\tau = \left( \frac{p-r}{p} \right) \sigma \), and obtain that
$$\begin{aligned} \sigma ^p \lambda (\sigma ) \le \dfrac{p^p}{(p-r)^{p-r} r^r} \left( [f]_{p,r}' \right) ^p. \end{aligned}$$
which is the first bound. Conversely, we use the identity
$$\begin{aligned} (|f|-\tau )_+^r = r \int _\tau ^\infty {\mathbb {1}}_{\{ |f|>\sigma \}} (\sigma -\tau )^{r-1} \,{\mathrm {d}}\sigma . \end{aligned}$$
Integrating over X gives
$$\begin{aligned} \tau ^{p-r} \int _X (|f|-\tau )_+^r \,{\mathrm {d}}x = r \tau ^{p-r} \int _\tau ^\infty \lambda (\sigma ) (\sigma -\tau )^{r-1} \,{\mathrm {d}}\sigma \,. \end{aligned}$$
(65)
Estimating \(\lambda (\sigma )\le \sigma ^{-p} \left( [f]_{p,0}' \right) ^p\) we obtain
$$\begin{aligned} \tau ^{p-r} \int _X (|f|-\tau )_+^r \,{\mathrm {d}}x \le r \left( [f]_{p,0}' \right) ^p \int _1^\infty \dfrac{(s-1)^{r-1}}{s^p}\,{\mathrm {d}}s = \left( [f]_{p,0}' \right) ^p \frac{r\,\Gamma (p-r)\,\Gamma (r)}{\Gamma (p)} \,, \end{aligned}$$
which is the second bound.\(\quad \square \)
Note that if \(\lambda _n(-\Delta +V)\) denote the negative eigenvalues of \(-\Delta +V\), repeated according to multiplicities, then
$$\begin{aligned} \sup _{\tau >0} \tau ^{\kappa -1} {\mathrm{Tr}}(-\Delta +V+\tau )^\kappa _- = \left( \left[ \lambda _\cdot (-\Delta +V) \right] _{\kappa ,1}' \right) ^{\frac{1}{\kappa }} \,. \end{aligned}$$
Thus, combining Lemmas 19 and 20 , we obtain
Corollary 21
(Weak Lieb–Thirring inequality). Inequalities (62) and (60) are equivalent to the inequality
$$\begin{aligned} {\boxed {\left| \! \left| \Big (\lambda _n (-\Delta +V)\Big )_{n\ge 1} \right| \! \right| _{\ell ^\kappa _{\mathrm{w}}}^\kappa \lesssim \int _{{{\mathbb {R}}}^d} V(x)_-^{\kappa +\frac{d}{2}}\, {\mathrm {d}}x }} \end{aligned}$$
for all \(V\in L^{\kappa +\frac{d}{2}}({{\mathbb {R}}}^d)\).
The equivalence claimed in this corollary is weaker than that in Lemma 19 since the (not displayed) constant depends on the choice of the norm in \(\ell ^\kappa _{\mathrm{w}}\).
B.4. Semiclassical constants
It was proved in [GLN21, Lemma 10] that \({\widetilde{K}}_{p,d}\) is not larger than its semiclassical counterpart, which is independent of p and given by the \(p=1+2/d\) semi-classical constant
$$\begin{aligned} {\widetilde{K}}_{d}^{\mathrm{sc}}= K_{1+2/d,d}^{\mathrm{sc}} = \frac{4 \pi ^2 d}{d+2} \left( \frac{d}{| \mathbb {S}^{d-1} |} \right) ^{\frac{2}{d}}. \end{aligned}$$
Together with Proposition 13, we obtain
$$\begin{aligned} {\boxed { K_{p,d} \le {\widetilde{K}}_{p,d}\le {\widetilde{K}}_{d}^{\mathrm{sc}} .}} \end{aligned}$$
In the dual picture, we have a similar result:
Lemma 22
For all \(\kappa \ge 1\), we have
$$\begin{aligned} {\boxed { \dfrac{(\kappa - 1)^{\kappa - 1}}{\kappa ^\kappa } \ L_{\kappa , d} \ge {\widetilde{L}}_{\kappa ,d}\ge {\widetilde{L}}_{\kappa , d}^{\mathrm{sc}}, }} \end{aligned}$$
(66)
where the semi-classical constant \({\tilde{L}}_{\kappa , d}^{\mathrm{sc}}\) is defined by
$$\begin{aligned} {\widetilde{L}}_{\kappa , d}^{\mathrm{sc}} := \dfrac{(\kappa - 1)^{(\kappa - 1)}(1 + \frac{d}{2})^{1 + \frac{d}{2}}}{(\kappa + \frac{d}{2})^{\kappa + \frac{d}{2}}} L^{\mathrm{sc}}_{1,d} \end{aligned}$$
(67)
with the semiclassical constant \(L^{\mathrm{sc}}_{1,d}\) at \(\kappa =1\) given by (4).
Proof
Both inequalities in (66) follow from the explicit formulas (13) and (63).\(\quad \square \)
Remark 23
(The semi-classical constant). We show here that the constant \({\widetilde{L}}_{\kappa , d}^\mathrm{sc}\) 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 \(V\in L^{\kappa +\frac{d}{2}}({{\mathbb {R}}}^d)\) and any \(\tau >0\), we have
$$\begin{aligned} \tau ^{\kappa -1} {\mathrm{Tr}}(-\hbar ^2\Delta +V+\tau )_- \underset{\hbar \rightarrow 0}{\sim }\tau ^{\kappa -1} \hbar ^{-d} L_{1,d}^{\mathrm{sc}} \int _{{{\mathbb {R}}}^d} (V+\tau )_-^{1+\frac{d}{2}}\,{\mathrm {d}}x \,. \end{aligned}$$
On the other hand, by inequality (62),
$$\begin{aligned} \tau ^{\kappa -1} {\mathrm{Tr}}(-\hbar ^2\Delta +V+\tau )_-&= \hbar ^{2\kappa } ( \hbar ^{-2}\tau )^{\kappa -1} {\mathrm{Tr}}(-\Delta +\hbar ^{-2} V + \hbar ^{-2} \tau )_- \\&\le \hbar ^{2\kappa } {\widetilde{L}}_{\kappa ,d} \int _{{{\mathbb {R}}}^d} \left( \hbar ^{-2} V \right) _-^{\kappa +\frac{d}{2}}\,{\mathrm {d}}x = \hbar ^{-d} {\widetilde{L}}_{\kappa ,d} \int _{{{\mathbb {R}}}^d} V_-^{\kappa +\frac{d}{2}}\,{\mathrm {d}}x \,. \end{aligned}$$
This shows that
$$\begin{aligned} \tau ^{\kappa -1} \int _{{{\mathbb {R}}}^d} (V+\tau )_-^{1+\frac{d}{2}}\,dx \le \frac{{\widetilde{L}}_{\kappa ,d}}{L_{1,d}^{\mathrm{sc}}} \int _{{{\mathbb {R}}}^d} V_-^{\kappa +\frac{d}{2}}\,{\mathrm {d}}x \,. \end{aligned}$$
Taking the supremum in \(\tau \) shows that
$$\begin{aligned} \left[ V_- \right] '_{\kappa +\frac{d}{2},1+\frac{d}{2}} \le \left( \frac{{\widetilde{L}}_{\kappa ,d}}{L_{1,d}^{\mathrm{sc}}} \right) ^{\frac{1}{\kappa +\frac{d}{2}}} \left\| V_- \right\| _{L^{\kappa +\frac{d}{2}}}. \end{aligned}$$
According to the optimality statement in the following lemma, we have
$$\begin{aligned} \left( \frac{{\widetilde{L}}_{\kappa ,d}}{L_{1,d}^{\mathrm{sc}}} \right) ^{\frac{1}{\kappa +\frac{d}{2}}} \ge \left( \frac{(\kappa -1)^{\kappa -1}\, (1+\frac{d}{2})^{1+\frac{d}{2}}}{(\kappa +\frac{d}{2})^{\kappa +\frac{d}{2}}} \right) ^\frac{1}{\kappa +\frac{d}{2}}. \end{aligned}$$
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, \(p>r\ge 0\) and \(f\in L^p(X)\). Then
$$\begin{aligned}{}[f]_{p,r}' \le \left( \frac{(p-r)^{p-r}\,r^r}{p^p} \right) ^\frac{1}{p} \Vert f\Vert _p \,. \end{aligned}$$
The constant on the right side is best possible.
Proof
We first recall that
$$\begin{aligned} \int _X |f|^p\,{\mathrm {d}}x = p \int _0^\infty \lambda (\sigma )\sigma ^{p-1}\,{\mathrm {d}}\sigma . \end{aligned}$$
Together with (65) (note that we may assume \(r > 0\) by continuity) we need to prove that
$$\begin{aligned} r \tau ^{p-r} \int _\tau ^\infty \lambda (\sigma ) (\sigma -\tau )^{r-1} \,{\mathrm {d}}\sigma \le \frac{(p-r)^{p-r}\,r^r}{p^p} p \int _0^\infty \lambda (\sigma )\sigma ^{p-1}\,{\mathrm {d}}\sigma \,. \end{aligned}$$
We write \(\lambda = \int _0^\infty {\mathbb {1}}_{\{\lambda > b\}}\,{\mathrm {d}}b\) and, since \(\lambda \) is non-increasing, for any \(b>0\) the function \({\mathbb {1}}_{\{\lambda > b\}}\) is the characteristic function of an interval with left endpoint at zero. Thus, it suffices to prove the above inequality for such characteristic functions. A computation shows that
$$\begin{aligned} r \tau ^{p-r} \int _\tau ^\infty {\mathbb {1}}_{[0,a)}(\sigma )(\sigma -\tau )^{r-1} \,{\mathrm {d}}\sigma = \tau ^{p-r} (a-\tau )_+^r \end{aligned}$$
and
$$\begin{aligned} p \int _0^\infty {\mathbb {1}}_{[0,a)}(\sigma )\sigma ^{p-1}\,{\mathrm {d}}\sigma = a^p \,. \end{aligned}$$
Thus, the inequality follows from the elementary equality
$$\begin{aligned} \sup _{a>0} \tau ^{p-r} (a-\tau )_+^r = \frac{(p-r)^{p-r}\,r^r}{p^p} a^p \,. \end{aligned}$$
There is equality when f is a characteristic function and \(\tau \) is chosen appropriately. This proves Lemma 24. \(\square \)
Remark 25
We wonder whether for all \(d \ge 1\) and all \(\kappa \ge \frac{3}{2}\), we have the equality \({\widetilde{L}}_{\kappa , d} = {\widetilde{L}}_{\kappa , d}^{\mathrm{sc}}\). This would be the analogue of the equality \(L_{\kappa , d} = L_{\kappa , d}^{\mathrm{sc}}\) [LW00]. We have the following rather tight bounds. Thanks to the explicit formulas (67) and (4), one can numerically plot the two curves \(\kappa \mapsto {\widetilde{L}}_{\kappa , d}^{\mathrm{sc}}\) and \(\kappa \mapsto \dfrac{(\kappa - 1)^{\kappa - 1}}{\kappa ^\kappa } L_{\kappa , d}^\mathrm{sc}\). As stated in Lemma 22, the two curves coincide at \(\kappa = 1\), but for all \(\kappa > 1\), it appears that
$$\begin{aligned} 0< \dfrac{(\kappa - 1)^{\kappa - 1}}{\kappa ^\kappa }\, L_{\kappa , d}^{\mathrm{sc}} - {\widetilde{L}}_{\kappa , d}^{\mathrm{sc}} < {\left\{ \begin{array}{ll} 0.004 &{} \text {for} \quad d = 1, \\ 0.0009 &{} \text {for} \quad d = 2, \\ 0.0002 &{} \text {for} \quad d = 3. \\ \end{array}\right. } \end{aligned}$$
In the region \(\kappa \ge 3/2\) where \(L_{\kappa , d} = L_{\kappa , d}^{\mathrm{sc}}\) [LW00], we deduce that \(| {\widetilde{L}}_{\kappa , d} - {\widetilde{L}}_{\kappa , d}^{\mathrm{sc}}|\) is smaller than the constants above.
Appendix C. An inequality on the Other Side of the Lieb–Thirring Exponent
In this section we would like to compare our inequality (60) with the following related inequality,
$$\begin{aligned}&K'_{p,d} \left| \! \left| \rho _\gamma \right| \! \right| _{L^{p}({{\mathbb {R}}}^d)}^\frac{2p}{d(p-1)}\le \Vert \gamma \Vert ^{\frac{d-(d-2)p}{d(p-1)}}\;{\mathrm{Tr}}(-\Delta \gamma ),\nonumber \\&\quad 1+\frac{2}{d}\le p< 1+\frac{2}{d-2},\quad d\ge 3. \end{aligned}$$
(68)
This inequality remains valid in dimensions \(d=1,2\), with \(1/(d-2)\) replaced by \(+\infty \). Note that the exponent p in (68) lies on the other side of the Lieb–Thirring exponent, compared to the situation considered in this paper. Inequality (68) appears in [LL86, Eq. (3.7)] for \(p=2\) and \(d=3\).
The proof of (68) in dimension \(d\ge 3\) is simple. Indeed, the Hoffmann-Ostenhof [HH77] inequality (38) together with the Sobolev inequality give
$$\begin{aligned} S_{\frac{d}{d-2},d}\left| \! \left| \rho _\gamma \right| \! \right| _{L^{\frac{d}{d-2}}({{\mathbb {R}}}^d)}\le {\mathrm{Tr}}(-\Delta \gamma )\qquad \text {for all } d\ge 3 \text { and all } \gamma =\gamma ^*\ge 0. \end{aligned}$$
(69)
Using Hölder’s inequality and the Lieb–Thirring inequality (36) (with constant \(K_{1+2/d,d}=\inf _N K_{1+2/d,d}^{(N)}>0\)) we obtain (68).
Our inequality (60) interpolates with respect to p between the Lieb–Thirring inequality and the trace equality \(\Vert \gamma \Vert _{{\mathfrak {S}}^1} = {\mathrm{Tr}}(\gamma ) =\Vert \rho _\gamma \Vert _1\). In contrast, inequality (68) interpolates between the Lieb–Thirring inequality (36) and the Sobolev inequality (69). The situation is summarized in Fig. 2.
An interesting difference between (60) and (68) arises when one considers the question of existence of minimizers. Recall from Lemma 17 that (60) never has optimisers. On the other hand, in [HKY19] the existence of optimisers for (68) was proved when \(1+2/d<p< 1+2/(d-2)\). When normalised in the manner \(\Vert \gamma \Vert =1\) and \({\mathrm{Tr}}(-\Delta \gamma )=\theta \int _{{{\mathbb {R}}}^d}\rho _\gamma ^p\), these optimisers were shown in [HKY19, Thm. 2] to solve the equation
$$\begin{aligned} \gamma ={\mathbb {1}}_{(-\infty ,0)}\left( -\Delta -\rho _\gamma ^{p-1}\right) +\delta ,\qquad \text {with}\quad 0\le \delta =\delta ^*\le {\mathbb {1}}_{\{0\}}\left( -\Delta -\rho _\gamma ^{p-1}\right) . \end{aligned}$$
(70)
In other words, \(\gamma \) is the orthogonal projection onto all the negative eigenfunctions, except possibly on the kernel of \(-\Delta -\rho _\gamma ^{p-1}\). If these optimisers \(\gamma \) have a finite rank N (they do for \(d\ge 3\) and p large enough), then they must be NLS ground states in the sense of [GLN21].
We now slightly refine the result in [HKY19] by showing that the operator \(-\Delta -\rho _\gamma ^{p-1}\) has no zero eigenvalues and, in particular, one has \(\delta =0\) in (70).
Proposition 26
Let \(1+\frac{2}{d}< p<\infty \) if \(d=1,2\) and \(1+\frac{2}{d}<p<1+\frac{2}{d-2}\) and let \(\gamma \) be an optimiser for (68), normalised so that \(\Vert \gamma \Vert =1\) and \({\mathrm{Tr}}(-\Delta \gamma )=\theta \int _{{{\mathbb {R}}}^d}\rho _\gamma ^p\). Then
$$\begin{aligned} \ker \left( -\Delta -\rho _\gamma ^{p-1}\right) =\{0\} \,. \end{aligned}$$
Proof
We begin by proving that \(\delta =0\) in (70). We denote by \(u_j\) and \(\mu _j\) the eigenfunctions and eigenvalues of \(-\Delta -\rho _\gamma ^{p-1}\) and by \(n_j\) the corresponding eigenvalues of \(\gamma \). From (70) we know that \(n_j=1\) if \(\mu _j<0\). By arguing as in (40), we have the estimate
$$\begin{aligned} \mu _{j}\le \frac{\theta \int _{{{\mathbb {R}}}^d}\rho _\gamma ^p}{n_j}\left( 1-\frac{n_j}{\theta \int _{{{\mathbb {R}}}^d}\rho _\gamma ^p}\int _{{{\mathbb {R}}}^d}\rho _\gamma ^{p-1}|u_j|^2-\left( \frac{\int _{{{\mathbb {R}}}^d}(\rho _\gamma -n_j|u_j|^2)^p}{\int _{{{\mathbb {R}}}^d}\rho _\gamma ^p}\right) ^{\frac{1}{\theta p}}\right) \end{aligned}$$
(71)
with \(\theta =d/(2p')\in (1/p,1)\). We claim that the right side is negative, which yields \(\mu _j<0\), that is, \(\delta \equiv 0\) in (70). To see this, we remark that for any \(f\ge 0\) and any probability measure \({\mathbb {P}}\), we have by Hölder’s inequality twice
$$\begin{aligned} \int f\,d{\mathbb {P}}\le \left( \int f^p\,d{\mathbb {P}}\right) ^{\frac{1}{p}}\le \theta \left( \int f^p\,d{\mathbb {P}}\right) ^{\frac{1}{\theta p}}+(1-\theta ). \end{aligned}$$
The second inequality is strict when \(\int f^p\,d{{\mathbb {P}}}\ne 1\). This may be rewritten in the form
$$\begin{aligned} 1+\theta ^{-1}\int (f-1)\,d{\mathbb {P}}\le \left( \int f^p\,d{\mathbb {P}}\right) ^{\frac{1}{\theta p}}. \end{aligned}$$
(72)
Choosing \(f=1-n_j|u_{j}|^2/\rho _\gamma \) and \({\mathbb {P}}=\rho _\gamma ^p/\int _{{{\mathbb {R}}}^d}\rho _\gamma ^p\), we obtain \(\mu _j<0\) in (71) since \(f\le 1\) and \(f\ne 1\), hence \(\int _{{{\mathbb {R}}}^d}f^p\,d{{\mathbb {P}}}<1\). We have thus proved that \(\delta \equiv 0\) in (70).
We now show that \(\ker (-\Delta -\rho _\gamma ^{p-1})=\{0\}\). Indeed, assume on the contrary that \(\mu _j=0\) (then \(n_j=0\) by the previous argument). Consider this time the perturbation \(\gamma (t)=\gamma +t|u_j\rangle \langle u_j|\), which cannot be an optimiser for \(t>0\). Taking \(\mu _j = 0\) and \(n_j = -t\) in (71) gives the (strict) inequality
$$\begin{aligned} \left( \frac{ \int _{{{\mathbb {R}}}^d}\left( \rho _\gamma +t|u_j|^2\right) ^p}{ \int _{{{\mathbb {R}}}^d}\rho _\gamma ^p}\right) ^{\frac{1}{\theta p}}<1+\frac{t\int _{{{\mathbb {R}}}^d}\rho _\gamma ^{p-1}|u_j|^2}{\theta \int _{{{\mathbb {R}}}^d}\rho _\gamma ^p} \end{aligned}$$
(73)
for all \(0< t<\Vert \gamma \Vert \). By (72) with \(f=1+t|u_j|^2/\rho _\gamma \), which satisfies \(\int _{{{\mathbb {R}}}^d}f^p\,d{{\mathbb {P}}}>1\), we have
$$\begin{aligned} \left( \frac{ \int _{{{\mathbb {R}}}^d}\left( \rho _\gamma +t|u_j|^2\right) ^p}{ \int _{{{\mathbb {R}}}^d}\rho _\gamma ^p}\right) ^{\frac{1}{\theta p}}>1+\frac{t\int _{{{\mathbb {R}}}^d}\rho _\gamma ^{p-1}|u_j|^2}{\theta \int _{{{\mathbb {R}}}^d}\rho _\gamma ^p} \end{aligned}$$
and we obtain a contradiction. Therefore \(\ker (-\Delta -\rho _\gamma ^{p-1})=\{0\}\), as claimed.\(\quad \square \)