% % AMS-LATEX % % \documentstyle[12pt,amssymb,amstex]{amsart} \sloppy \renewcommand{\baselinestretch}{1.3} \newcommand{\bl}{{\pmb{\pmb |}}} \newtheorem{tm}{\\ $\ \ \ \$ Theorem} \newtheorem{lm}{\\ $\ \ \ \$ Lemma} \newtheorem{cl}{\\ $\ \ \ \$ Corollary} \newtheorem{pr}{\\ $\ \ \ \$ Proposition} \newtheorem{df}{\\ $\ \ \ \$ Definition} \newcounter{rem} \newcommand{\rema}{{\bf Remark \thesection.\refstepcounter{rem}\arabic{rem}.\ }} \begin{document} \title{On the Lieb-Thirring constants $L_{\gamma,1}$ for $\gamma\geq 1/2.$} \author{Timo Weidl} \date{April 2, 1995} \address{Timo Weidl\\Royal Institute of Technology\\Department of Mathematics\\ S-10044 Stockholm\\Sweden\\int+46-8-7906196, fax: int+46-8-7231788} \email{weidl@@math.kth.se} \maketitle \begin{abstract} Let $E_i(H)$ denote the negative eigenvalues of the one-dimensional Schr\"odinger operator $Hu:=-u^{\prime\prime}-Vu,\ V\geq 0,$ on $L_2({\Bbb R}).$ We prove the inequality % $$\label{Wei} \sum_i|E_i(H)|^\gamma\leq L_{\gamma,1}\int_{\Bbb R} V^{\gamma+1/2}(x)dx,$$ % for the "limit" case $\gamma=1/2.$ This will imply improved estimates for the best constants $L_{\gamma,1}$ in \eqref{Wei} as $1/2<\gamma<3/2.$\end{abstract} {\bf 0.} Let $H=-\Delta-V$ denote the Schr\"odinger operator in $L_2({\Bbb R}^d).$ If the potential $V\geq 0$ decreases sufficiently fast at infinity, the negative part of the spectrum of $H$ is discrete. Let $\{E_i(H)\}$ be the corresponding increasing sequence of negative eigenvalues, each eigenvalue occurs with its multiplicity. This sequence is either finite or tends to zero. Estimates on the behavior of the sequence of eigenvalues in terms of the potential have been in the focus of research for many years. In the earlier papers the main attention was paid to bounds on the number of negative eigenvalues ( \cite{Barg},\cite{B},\cite{Schwinger},\cite{Rbl}, \cite{C},\cite{[L]}, \cite{LiYau},\cite{Blanch} ). In \cite{LT} Lieb and Thirring proved inequalities of the type % $$\label{LiTh} \sum_i|E_i(H)|^\gamma\leq L_{\gamma,d}\int_{{\Bbb R}^d} V^{\gamma+\varkappa}(x)dx, \ \ \varkappa=d/2.$$ % Since that these estimates and the corresponding constants $L_{\gamma,d}$ have been studied intensively (e.g. \cite{Lieb},\cite{GGM},\cite{RoHe}). Up to now it was known, that \eqref{LiTh} holds for all $\gamma\geq 0$ if $d\geq 3,$ for $\gamma>0$ if $d=2,$ and for $\gamma>1/2$ if $d=1.$ On the contrary \eqref{LiTh} fails for $\gamma=0, d=2$ and for $\gamma<1/2, d=1.$ In this paper we prove \eqref{LiTh} for the remaining case $d=1,\gamma=1/2,$ which does not seem to have been settled so far. This result will imply an essential improvement for the estimates on the constants $L_{\gamma,1},\ 1/2<\gamma<3/2.$ Moreover we deduce a new integral bound on the transmission coefficient of the corresponding scattering problem. In conclusion the author expresses his gratefulness to M. Sh. Birman, who introduced him to the topic of negative bound states of Schr\"odinger operators. Moreover I am grateful to A. Laptev, under whose intensive supervision this paper was written. \vspace{3mm} {\bf 1.} In this subsection we provide some auxiliary results on the negative spectrum of the Neumann problem for the Sturm-Liouville-operator % $(L_I^Nu)(x)=-u^{\prime\prime}(x)-V(x)u(x),$ % $x\in I=[0,l], \ \ u^\prime(0)=u^\prime(l)=0,\ \ 0\leq V(x)\in L_1(I).$ % Let $N_I(V,E)$ be the number of eigenvalues $E_i(L_I^N)$ of $L_I^N$ below $E<0.$ According to the Birman-Schwinger principle (\cite{B},\cite{Schwinger}), the value of $N_I(V,E)$ does not exceed the square of the Hilbert-Schmidt norm of the integral operator % $(Q_E u)(x)=\sqrt{V(x)}\int_0^l G(x,y,E)\sqrt{V(y)}u(y),\ \ x\in I.$ % Here % $G(x,y,E)=\Biggl\{ \begin{array}{cc} \frac{\cosh(\lambda x)\cosh(\lambda(y-l))}{\lambda\sinh(\lambda l)}\ \ x\leq y \\ \frac{\cosh(\lambda y)\cosh(\lambda(x-l))}{\lambda\sinh(\lambda l)}\ \ y\leq x \end{array}\Biggr . ,\ \ \lambda=\sqrt{|E|},\ E<0,\ x,y\in I,$ % denotes the Green function of the problem $-u^{\prime\prime}-Eu,\ u^\prime(0)=u^\prime(l)=0$ on $I.$ In view of % $|G(x,y,E)|\leq \frac{\coth(\lambda l)}{\lambda}$ % one obtains the inequality % $$\label{N} N_I(V,E)\leq \frac{\coth^2(\lambda l)}{\lambda^2} \biggl(\int_I V(x)dx\biggr)^2,\ \ \lambda=\sqrt{|E|},\ E<0.$$ % We apply \eqref{N} to the lowest eigenvalue $E_1(L_I^N),$ and find % $$\label{N1} \vartheta (\lambda_1 l) \leq l \int_I V(x)dx, \ \ \lambda_1=\sqrt{ |E_1(L_I^N)| }>0, \ \ \vartheta(x):=x\tanh x.$$ % The function $\vartheta(x)=x\tanh x$ is strongly increasing in $x\geq 0.$ Let $\varsigma(y)$ be the inverse function of $\vartheta(x)=y,\ x,y>0.$ From \eqref{N1} we immediately conclude \begin{lm}\label{lm1} Let $E_1(L_I^N)$ be the lowest eigenvalue of the Neumann problem $L_I^N$ on $I=[0,l].$ Assume $0\leq V\in L_1(I).$ Then the estimate % $$\label{N1lm} \lambda_1\leq \varsigma(l \int_I V(x)dx)/l,\ \ \ \ \ \lambda_1=\sqrt{ |E_1(L_I^N)| }\geq 0,$$ % holds. \end{lm} \vspace{3mm} Next we recall a criteria, providing the existence of not more then one negative eigenvalue of the operator $L_I^N.$ First notice, that for functions $u\in C^\infty(I),$ satisfying the orthogonality condition $\int_I udx=0,$ the inequality % $$\label{3} |u(x)|^2\leq \frac{l}{3}\int_I|u^\prime|^2dx,\ \ x\in I,$$ % holds. Indeed, we have % $lu(x_0)=\int_0^{x_0}u^\prime (x)xdx-\int_{x_0}^lu^\prime (x)(l-x)dx.$ % This gives % $|u(x_0)|^2\leq \frac{(x_0^{3/2}+(l-x_0)^{3/2})^2}{3l^2} \int_0^l |u^\prime (x)|^2dx .$ % Passing to the upper bound in $x_0\in I$ we find \eqref{3}. The constant $l/3$ in \eqref{3} is sharp. \begin{lm}\label{lm2} Assume, that for the non-trivial potential $0\leq V$ the estimate % $$\label{NN} l\int_I V(x)dx \leq 3$$ % holds. Then the Neumann problem $L_I^N$ on $I=[0,l]$ has exactly one negative eigenvalue. \end{lm} {\bf Proof.} The existence of the eigenvalue is obvious. By \eqref{3} we find % $$\label{ininin} \int_I |u^\prime|^2dx-\int_I V(x)|u|^2dx \geq 0, \ \ u\in C^\infty([0,l]),\ \int_I udx=0.$$ % The inequality \eqref{ininin} holds on a set of functions of codimension one with respect to the domain of the quadratic form of the Neumann problem $L_I^N.$ Thus $L_I^N$ itself has not more than one negative eigenvalue. $\Box$ \vspace{3mm} {\bf 2.} We turn now our attention to the one-dimensional Schr\"odinger operator % $Hu=-u^{\prime\prime}-V(x)u,\ \ x\in{\Bbb R}, \ 0\leq V\in L_1({\Bbb R}),$ % realized as a self-adjoint operator on $L_2({\Bbb R})$ in the form sum sense. Let $H_+$ and $H_-$ denote selfadjoint operators on $L_2({\Bbb R}_\pm),$ corresponding to the Neumann problem on the positive and negative semi-axes respectively. Assume $V\not\equiv 0$ on ${\Bbb R}_+.$ Fix the point $l_0=0,$ and by iteration construct the sequence $l_k, k\in{\Bbb K}\subset{\Bbb N},$ % $$\label{NNcon} l^{(k)}\int_{l_k}^{l_{k+1}}V(x)dx=3,\ \ l^{(k)}:=l_{k+1}-l_k.$$ % If it occurs that $\int_{l_n}^\infty V(x)dx=0,$ we formally choose $l_{n+1}=+\infty.$ For the elements of the sequence $l^{(k)}$ we have the bound $l^{(k)}\geq 3/\int V(x)dx>0.$ Hence the intervals $I_k:=[l_k,l_{k+1}],\ k\geq 0,$ cover ${\Bbb R}_+.$ On each interval we consider the Neumann problem $L_{I_k}^Nu=-u^{\prime\prime}-V(x)u,\ u^\prime(l_k)=u^\prime(l_{k+1})=0.$ Let $H_+^N=\oplus_{k\in{\Bbb K}} L_{I_k}^N$ denote the orthogonal sum of these operators. We have $H_+^N\leq H_+.$ For the ordered sequence of the respective negative eigenvalues this implies % $$\label{specpro} E_i(H_+^N)\leq E_i(H_+).$$ % In case of a semi-infinite interval the potential is identically zero on this interval, the respective Neumann problem has no negative spectrum. Therefore it will not play any role in our considerations. By Lemma \ref{lm2} the Neumann problem $L_{I_k}^N$ on the finite intervals $I_k$ has exactly one negative eigenvalue. Because of \eqref{NNcon} the bound \eqref{N1lm} for $\lambda_{1}(I_k):=\sqrt{|E_{1}(L_{I_k}^N)|}$ turns into $\lambda_1(I_k)\leq \varsigma(3)/l^{(k)},$ or equivalently \footnote{On the other hand for $u(x)=1/\sqrt{l}$ one has $E_1(L_{I_k}^N)\leq (L_{I_k}^Nu,u)_{L_2(I_k)}=-l^{-1}\int_{I_k}V(x)dx,$ and $\lambda_1(I_k)\geq \sqrt{1/3} \int_{I_k} V(x)dx.$} % $$\label{victoria} \lambda_1(I_k)\leq \frac{\varsigma(3)}{3} \int_{I_k} V(x)dx.$$ % Since $V\in L_1({\Bbb R}_+),$ the sequence $\int_{I_k} V ( x )dx$ tends to zero as $k\to\infty.$ Thus both operators $H_+^N$ and $H_+$ are semibounded and their negative spectra are discrete. The negative spectrum of $H_+^N$ coincides (as set and in its multiplicity) with the sequence of eigenvalues $\{E_1(L_{I_k}^N)\}=\{-\lambda_1^2(I_k)\}.$ By \eqref{specpro} we have $|E_i(H_+)|\leq |E_i(H_+^N)|.$ Together with $0\leq V\in L_1({\Bbb R}_+)$ this implies % $\sum_i \sqrt{|E_i(H_+)|}\leq\sum_i \sqrt{|E_i(H_+^N)|}=\sum_{k} \lambda_{1}(I_k)\leq$ % $\leq \frac{\varsigma(3)}{3}\sum_k \int_{I_k} V(x)dx= \frac{\varsigma(3)}{3}\int_0^\infty V(x)dx,$ % and we find the claimed result for the negative eigenvalues of the Neumann operator on the semi-axes % $$\label{upp+-} \sum_i \sqrt{|E_i(H_+)|}\leq L_{\frac{1}{2},1}^+ \int_{{\Bbb R}_+} V(x)dx, \ \ L^+_{\frac{1}{2},1}\leq \varsigma(3)/3<1.005.$$ % Naturally the analogous bound with the same constant holds for the operator $H_-.$ Because of $H_-\oplus H_+\leq H$ we obtain the analog estimate on the negative eigenvalues of the Schr\"odinger operator $H$ on ${\Bbb R}$ % $$\label{upp} \sum_i \sqrt{|E_i(H)|}\leq L_{\frac{1}{2},1} \int_{{\Bbb R}} V(x)dx, \ \ L_{\frac{1}{2},1}\leq \varsigma(3)/3<1.005.$$ % We recall the reverse estimate for the operator $H$ (see \cite{LT} and \cite{GGM}). The first sum rule of Faddeev-Zakharov \cite{FadZ} states % $$\label{FadZ} \int V(x)dx = 4\sum_i \sqrt{|E_i(H)|}+\pi^{-1}\int \ln(1-|R(k)|^2)dk,$$ % for (not necessary sign-defined) potentials $V\in C_0^\infty({\Bbb R}).$ In this $R(k)\in [0,1]$ is the reflection coefficient of the operator $H.$ The integrand on the right hand side is negative, hence % $$\label{FadZ1} \sum_i \sqrt{|E_i(H)|}\geq\frac{1}{4}\int V(x)dx.$$ % This bound can be closed to all potentials $V\in L_1({\Bbb R})$. The estimate from below on the constant $L_{1/2,1}$ can be improved. For a potential $0\leq V\in L_1({\Bbb R})$ the number $N(V,E)$ of eigenvalues $E_i(H)0,$ the operator $H_\alpha u=-u^{\prime\prime}-\alpha Vu$ has exactly one negative eigenvalue $E_1(H_\alpha).$ This eigenvalue obeys the asymptotics (see \cite{Si}) % $\sqrt{|E_1(H_\alpha)|}= ( \alpha/2 )\int Vdx+o ( \alpha ),\ \ \alpha\to 0.$ % We conclude $L_{1/2,1}\geq 1/2.$ The previous arguments can be adapted to the problem on the semi-axes. Assume, that $0\leq V$ is continuous on ${\Bbb R}_+$ up to the point zero, and has compact support. We supply this potential with a small coupling constant $\alpha>0,$ and consider the lowest eigenvalue $E_1(H_{+,\alpha})$ of the respective Neumann problem on ${\Bbb R}_+.$ Let $u_\alpha(x)$ denote the corresponding eigenfunction. The even extension $u_\alpha(x)=u_\alpha(-x)$ is an eigenfunction of the operator $H_\alpha$ with the extended potential $V(x)=V(-x)$ on ${\Bbb R}.$ The corresponding eigenvalue is $E_1(H_\alpha)=E_1(H_{+,\alpha}).$ Since the operators $H_\alpha$ and $H_{+,\alpha}$ have only one negative eigenvalue for sufficiently small $\alpha>0,$ we find % $\sqrt{|E_1(H_+)|}=\alpha\int_0^\infty Vdx+o ( \alpha ),\ \ \alpha\to 0.$ % We obtain $1\leq L_{1/2,1}^+<1.005,$ our bound on the constant for the Neumann problem on the semi-axes is almost sharp! Finally we remark the analog of \eqref{FadZ1} for the operator $H_+.$ For a sumable potential $V(x)=V(-x)$ it holds % $$\label{FadZ2} \int_{0}^\infty V(x)dx\leq 2\sum_i\sqrt{|E_i(H)|}\leq 2 \sum_i\sqrt{|E_i(H_-\oplus H_+)|}=4\sum_i\sqrt{|E_i(H_+)|}.$$ % The results of this subsection we summarize in \begin{tm}\label{thethe} 1.) The inclusion $0\leq V\in L_1({\Bbb R}_+)$ implies the inequality % $$\label{upp0} \sum_i \sqrt{|E_i(H_+)|}\leq L^+_{\frac{1}{2},1}\int_0^\infty V(x)dx.$$ % For the best constant $L^+_{1/2,1}$ in \eqref{upp0} we have the estimate $1\leq L_{1/2,1}^+\leq \varsigma(3)/3<1.005.$ Reversely, a priori assuming $0\leq V\in L_1^{loc}({\Bbb R}_+),$ the discreteness of the negative spectrum together with the convergence of the sum in \eqref{upp0} imply $V\in L_1$ and \eqref{FadZ2}. 2.) The inclusion $0\leq V\in L_1({\Bbb R})$ implies the inequality % $$\label{upp1} \sum_i \sqrt{|E_i(H)|}\leq L_{\frac{1}{2},1}\int V(x)dx.$$ % For the best constant $L_{1/2,1}$ in \eqref{upp1} we have the estimate $1/2\leq L_{1/2,1}\leq \varsigma(3)/3<1.005.$ Reversely, a priori assuming $0\leq V\in L_1^{loc}({\Bbb R}),$ the discreteness of the negative spectrum together with the convergence of the sum in \eqref{upp1} imply $V\in L_1$ and \eqref{FadZ1}. \end{tm} {\bf Remark.} As usually one can drop the assumption $V\geq 0.$ One has to ensure, that the corresponding operators $H,H_+$ are defined in the form sum sense, and the integrant in \eqref{upp0} and \eqref{upp1} has to be replaced by $V_+(x):=\max\{0,V(x)\}.$ Notice, that \eqref{upp1} and \eqref{FadZ} together with $1/2\leq L_{1/2,1}<\infty$ imply \begin{tm} Assume $V\in C_0^\infty({\Bbb R}), 2V_\pm=|V|\pm V,$ and let $R(k)$ be the reflection coefficient for the corresponding one-dimensional Schr\"odinger operator $Hu=-u^{\prime\prime}-Vu$ on $L_2({\Bbb R}).$ Then the integral estimate % $\frac{1}{\pi} \int |\ln (1-|R(k)|^2)| dk \leq \int V_-dx+(4L_{1/2,1}-1)\int V_+dx \leq(4L_{1/2,1}-1)\|V\|_{L_1({\Bbb R})}$ % holds. \end{tm} \vspace{3mm} {\bf 3.} We turn now to the case $\gamma>1/2.$ We restrict our considerations to the operator $H$ on $L_2({\Bbb R}).$ Here the inequalities % $$\label{Wei12} \sum_i|E_i(H)|^\gamma\leq L_{\gamma,1}\int_{\Bbb R} V^{\gamma+1/2}(x)dx,$$ % are well established, but we will give an essential improvement of the estimates for the corresponding constants $L_{\gamma,1}.$ For $\gamma\geq 3/2$ in \cite{AiL} has been proven, that $L_{\gamma,1}=L_{\gamma,1}^{cl}.$ The last notation stands for the classical constant % $L_{\gamma,1}^{cl}=\frac{\Gamma(\gamma+1)}{2\sqrt{\pi} \Gamma(\gamma+\frac{3}{2})}.$ % Hence we will stress on the case $1/2<\gamma<3/2.$ We shall compare our results with the bounds of Lieb and Thirring % $$\label{LiThCo} L_{\gamma,1}\leq L_{\gamma,1}^{LT} :=\frac{\gamma^{\gamma+1}} {\sqrt{2}(\gamma-1/2)^{\gamma+1/2}(\gamma+1/2)},$$ % and their improvements by Glaser, Grosse and Martin (\cite{GGM}) % \label{GGM1} L_{\gamma,1}\leq L_{\gamma,1}^{GGM}:=\inf_{10,\] % we can find a better constant by "interpolating" between the cases $\gamma=1/2$ and $\gamma=3/2.$ Indeed, the ratio % $\psi(\gamma,\tilde{V}):=\frac{\sum_i |E_i(H)|^\gamma} {\int \tilde{V}^{\gamma+1/2}dx}$ % is analytic and continuous up to the boundary for complex $\gamma$ in the strip $1/2<\Re\gamma<3/2.$ On the boundary we have the estimates % $|\psi(\gamma,\tilde{V})| \leq L_{1/2,1}\leq \varsigma(3)/3,\ \ \mbox{as}\ \Re\gamma=1/2, \ \ |\psi(\gamma,\tilde{V})| \leq L_{3/2,1}=3/16,\ \ \mbox{as}\ \Re\gamma=3/2.$ % By the Hadamard Lemma we obtain % $$\label{sharr}\psi(\gamma,\tilde{V})\leq \tilde{L}_{\gamma,1}^{**} :=\bigl(\frac{\varsigma(3)}{3}\bigr)^{\frac{3}{2}-\gamma} \bigl(\frac{3}{16}\bigr)^{\gamma-\frac{1}{2}},\ \ 1/2<\gamma<3/2.$$ % In particular, $\tilde{L}_{1,1}^{**}<0.4341.$ We notice, that \eqref{sharr} is sharper than the results for characteristic functions by A. Laptev in \cite{La} for the case of dimension one . For completeness we recall the estimate from below on the constants $L_{\gamma,1},$ obtained in \cite{LT}. To do so we consider the best constants $L_{\gamma,1}^1$ in the inequalities % $$\label{1//2} |E_1(H)|^\gamma\leq L_{\gamma,1}^1 \int V^{\gamma+1/2}dx,\ \ \gamma\geq 1/2.$$ % Obviously $L_{\gamma,1}\geq L_{\gamma,1}^1.$ For $\gamma>1/2$ the corresponding variational equation can be solved analytically and one obtains \footnote{In particular this gives $0.2451L_{\gamma,1}^{cl},$ this implies $L_{\gamma,1}>L_{\gamma,1}^{cl}$ as $1/2\leq \gamma<3/2$ (see \cite{LT} and also \cite{RoHe}). We proved \begin{tm} For the numerical values of the best possible constants $L_{\gamma,1},\ 1/2\leq\gamma\leq 3/2$ in \eqref{Wei12} the estimate % $2L_{\gamma,1}^{cl}\biggl(\frac{\gamma-1/2}{\gamma+1/2}\biggr)^{\gamma-1/2} \leq L_{\gamma,1}\leq L_{\gamma,1}^{*}= 4\varsigma(3)L_{\gamma,1}^{cl}/3= \frac{2\varsigma(3)\Gamma(\gamma+1)}{3\sqrt{\pi} \Gamma(\gamma+\frac{3}{2})}, \ \ \frac{1}{2}\leq\gamma\leq\frac{3}{2},$ % holds. For potentials $\tilde{V}$ proportional to characteristic functions, the constant $L_{\gamma,1}$ in the Lieb-Thirring inequality can be replaced by $\tilde{L}_{\gamma,1}^{**}$ from \eqref{sharr}. \end{tm} Notice, that the bound $L_{\gamma,1}^*$ on $L_{\gamma,1}$ does not tend to $L_{\frac{3}{2},1}=L_{\frac{3}{2},1}^{cl}=3/16$ as $\gamma\to 3/2-0.$ For $\gamma$ near $3/2$ the estimate on $L_{\gamma,1}$ can be improved. To do so we shall recall some auxiliary material from real interpolation theory. \vspace{3mm} {\bf 4.} Let $\ell_p$ denote the ideal of $p$-sumable sequences $\{u_n\}_{n\in{\Bbb N}},$ equipped by the standard quasi-norm % $\|\{u_n\}\|_{\ell_p}^p:=\sum_n |u_n|^p\ ,\ \ p>0.$ % For a sequence $\{u_n\}_{n\in{\Bbb N}}\in \ell_{p_0}+\ell_{p_1}$ one can define the $(p_0,p_1)-K$-function % $K(\{u_n\},t,p_0,p_1):=\inf_{ \begin{array}{cc}u_n=u_n^{(0)}+ u_n^{(1)}\\u_n^{(i)}\in\ell_{p_i} \end{array}}\Bigl(\|u_n^{(0)}\|_{\ell_{p_0}}^{p_0} +t\|u_n^{(1)}\|_{\ell_{p_1}}^{p_1} \Bigr),\ \ t>0.$ % For a function $f\in L_{p_0}+L_{p_1}$ one may use the analogous definition % $K(f,t,p_0,p_1):=\inf_{ \begin{array}{cc}f=f_0+ f_1\\f_i\in L_{p_i} \end{array}}\Bigl(\|f_0\|_{L_{p_0}}^{p_0} +t\|f_1\|_{L_{p_1}}^{p_1} \Bigr),\ \ t>0.$ % On functions $h:(0,\infty)\to [0,\infty)$ we define the functionals % $\Phi_{\eta,q}[h]= \Bigl(\int_0^\infty(t^{-\eta}h(t))^q\frac{dt}{t}\Bigr)^{1/q}, \ \ \eta\in(0,1),\ 00}t^{-\eta}h(t), \ \ \eta\in(0,1).$ % Notice, that $h_1(t)\leq h_2(t)$ implies $\Phi_{\eta,q}[h_1]\leq \Phi_{\eta,q}[h_2].$ % According to the power theorem'' of real interpolation theory, see \cite{BL}, it holds % $$\label{h1} \Phi_{\eta,q}[K(\{u_n\},\cdot,p_0,p_1)]\asymp\|\{u_n\}\|_{\ell^{p,r}}^p,$$ % $$\label{h2} \Phi_{\eta,q}[K(f,\cdot,p_0,p_1)]\asymp\|f\|_{L^{p,r}}^p,$$ % $p=(1-\eta)p_0+\eta p_1,\ \ r=pq,\ \eta\in(0,1),\ 0From \eqref{hilfe1} and \eqref{LiTh} it follows, that % \[K(\{E_k(H)\},t,p_0,p_1)\leq \|\{a_k\}\|_{\ell_{p_0}}^{p_0} +t\|\{b_k\}\|_{\ell_{p_1}}^{p_1}\leq$ % $\leq (1+N)\sum_n |E_n(H_0)|^{p_0}+ t(1+N^{-1})\sum_m|E_m(H_1)|^{p_1}$ % $\leq (1+N)\theta^{-\varkappa}L_{p_0,d} \|V_0\|_{L_{p_0+\varkappa}({\Bbb R}^d)}^{p_0+\varkappa} +t(1+N^{-1})(1-\theta)^{-\varkappa}L_{p_1,d} \|V_1\|_{L_{p_1+\varkappa}({\Bbb R}^d)}^{p_1+\varkappa}.$ % Interchanging the definitions of the sequences $\{a_k\}$ and $\{b_k\}$ one can see, that in the previous expression the role of $N$ and $1/N$ can be interchanged. Thus we can assume, that $N$ is of the form $k$ or $1/k,\ k\in{\Bbb N}.$ Passing to the infimum over all suitable decompositions $V=V_0+V_1$ one finds % $$K(\{E_k(H)\},t,p_0,p_1)\leq \frac{(1+N)L_{p_0,d}}{\theta^\varkappa} K(V,t\frac{(1+N^{-1})(1-\theta)^{-\varkappa}L_{p_1,d}} {(1+N)\theta^{-\varkappa}L_{p_0,d}},p_0+\varkappa,p_1+\varkappa).$$ % $N=..,\frac{1}{3},\frac{1}{2},1,2,3,..\ \ \varkappa=d/2,$ % with $01.14 .$ \begin{tm} For the constant $L_{\gamma,1}$ in \eqref{upp1} the bound $L_{\gamma,1}\leq\min\{L_{\gamma,1}^*,L_{\gamma,1}^{**}\}, \ 1/2<\gamma<3/2,$ holds. \end{tm} \vspace{3mm} {\bf 7.} Let $\{\phi_i\}$ be some $L_2({\Bbb R}^d)-$orthonormal system, $\phi_i\in W_2^1({\Bbb R}^d).$ Then \eqref{LiTh} implies (\cite{LT},\cite{Lieb}) % $$\label{ll1} \sum_{i=1}^n\int|\nabla\phi_i|^2dx\geq K_{p,d} \biggl(\int\rho_\phi^{p/(p-1)}dx\biggr)^{2(p-1)/d},$$ % $\rho_\phi(x):=\sum_{i=1}^n|\phi_i(x)|^2,$ % % $\max\{d/2,1\}\leq p<1+d/2,\ \ \mbox{excluding}\ p=1\ \mbox{for}\ d=2,$ % with suitable constants $K_{p,d}.$ In case of $d=1$ and $p=3/2$ this turns into % $$\label{ll} \sum_{i=1}^n\int|\phi_i^\prime|^2dx\geq K_{3/2,1}\int\rho_\phi^{3}dx.$$ The constant $K_{3/2,1}$ is related to $L_{1,1}$ via the formula % $L_{1,1}=2/\sqrt{27 K_{3/2,1}}.$ % Our improved estimate on $L_{1,1}$ implies $K_1\geq 0.203,$ in compare with $K_1\geq 1/12$ in \cite{Lieb}. 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