==== TABLE OF CONTENTS ====
Sharp-One-Dimensional-Lieb-Thirring-Constants//main.tex
Sharp-One-Dimensional-Lieb-Thirring-Constants//preamble.tex
Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/action-framework.tex
Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/action-limit.tex
Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/connecting-path.tex
Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/introduction.tex
Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/matrix-field.tex
Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/rank-one.tex
Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/spectral.tex
Sharp-One-Dimensional-Lieb-Thirring-Constants//references.bib
===========================
==== BEGIN Sharp-One-Dimensional-Lieb-Thirring-Constants//main.tex ====
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\documentclass[11pt]{article}
\input{preamble}
\title{Sharp one-dimensional Lieb--Thirring constants}
\author{OpenAI}
\date{September 23, 2026}
\begin{document}
\maketitle
\begin{abstract}
We resolve affirmatively the remaining cases of the scalar one-dimensional
Lieb--Thirring conjecture: for every $\frac12<\gamma<\frac32$,
the optimal constant is the one-bound-state constant.
The estimate holds for every nonnegative potential in
$L^{\gamma+1/2}(\mathbb R)$, with all negative eigenvalues included.
\end{abstract}
\tableofcontents
\clearpage
\input{sections/introduction}
\input{sections/action-framework}
\input{sections/matrix-field}
\input{sections/rank-one}
\input{sections/connecting-path}
\input{sections/action-limit}
\input{sections/spectral}
\bibliographystyle{plain}
\bibliography{references}
\end{document}
==== END Sharp-One-Dimensional-Lieb-Thirring-Constants//main.tex ====
==== BEGIN Sharp-One-Dimensional-Lieb-Thirring-Constants//preamble.tex ====
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==== END Sharp-One-Dimensional-Lieb-Thirring-Constants//preamble.tex ====
==== BEGIN Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/action-framework.tex ====
\section{The finite-interval action inequality}\label{sec:action}
Fix
\[
0<\sigma<1,\qquad p=1+\sigma,\qquad q=1+\frac1\sigma.
\]
All matrices below are finite dimensional. Write $\Sym_N$ for the real
symmetric $N\times N$ matrices, with inner product $\tr(AB)$ and norm
$\|A\|_{\HS}=(\tr(A^2))^{1/2}$. The notation $A\le B$ means that $B-A$
is positive semidefinite. Vectors are columns, and $|a|$ is their Euclidean norm.
Let $I=(x_-,x_+)$ be a bounded interval. For a positive diagonal matrix
$K=\diag(k_1,\ldots,k_N)$ and $a\in H^1_0(I;\R^N)$, define
\begin{equation}\label{eq:action}
\mathcal E_K(a)=\int_I\bigl(|a'|^2+a^{\transpose}K^2a-|a|^{2q}\bigr)\,dx.
\end{equation}
In one dimension $H^1_0(I)$ embeds into the continuous functions, so every
term in this integral is finite.
\begin{theorem}[Action inequality]\label{thm:action}
Let $u=(u_1,\ldots,u_N)^{\transpose}\in H^1_0(I;\R^N)$ satisfy
$\int_I uu^{\transpose}\,dx=I_N$, and let $k_i>0$. Then
\begin{equation}\label{eq:action-bound}
\sup_{C\text{ lower triangular}}\mathcal E_K(Cu)
\ge \sum_{i=1}^N\int_{-k_i}^{k_i}(k_i^2-s^2)^\sigma\,ds.
\end{equation}
The supremum is over all real lower triangular $N\times N$ matrices.
\end{theorem}
The restriction to lower triangular matrices is what makes this inequality
useful spectrally. If the $u_i$ are eigenfunctions ordered from the lowest
eigenvalue upward, the $i$th component of $Cu$ stays in the span of the first
$i$ eigenfunctions. Section~\ref{sec:spectral} will use this observation to
bound the left side from above by an integral of the potential.
Here is the construction behind the lower bound. For each $\delta>0$,
Section~\ref{sec:field} builds a locally Lipschitz map
$M:\Sym_N\to\Sym_N$ and a $C^1$ function $J:\Sym_N\to\R$ such that
\begin{equation}\label{eq:framework-gradient}
dJ(B)[H]=-\tr(M(B)H).
\end{equation}
They have two decisive properties. First, Section~\ref{sec:rankone} proves
\begin{equation}\label{eq:framework-rankone}
M(B)=aa^{\transpose}\quad\Longrightarrow\quad
a^{\transpose}(K^2-B^2)a\ge |a|^{2q}.
\end{equation}
Second, Section~\ref{sec:path} finds, for every $\varepsilon>0$, a lower
triangular $C$ and a path $B$ satisfying
\begin{equation}\label{eq:framework-path}
\varepsilon B'=M(B)-(Cu)(Cu)^{\transpose},\qquad
B(x_-)=K,\quad B(x_+)=-K.
\end{equation}
The small parameter $\varepsilon$ penalizes failure of the rank-one constraint.
For $a=Cu$, differentiating $J(B)+a^{\transpose}Ba$ along this path produces
the negative term $-\varepsilon^{-1}\|M(B)-aa^{\transpose}\|_{\HS}^2$.
Uniform bounds on the path and on $C$ then give
\[
\sup_C\mathcal E_K(Cu)\ge J(-K)-J(K).
\]
Finally, the boundary difference tends to the right side of
\eqref{eq:action-bound} as $\delta\downarrow0$. We keep $\delta$ fixed until
after the limit $\varepsilon\downarrow0$.
==== END Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/action-framework.tex ====
==== BEGIN Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/action-limit.tex ====
\section{From the matrix path to the action}\label{sec:action-limit}
We now prove Theorem~\ref{thm:action}. The matrix comparison controls points
where $M(B)=aa^{\transpose}$ exactly. The differential equation supplies a
penalty for the remaining points, and compactness makes that penalty
effective without requiring convergence of the chosen coefficients $C$.
\begin{proof}[Proof of Theorem~\ref{thm:action}]
Fix $\delta>0$, and use the field $M$ and primitive $J$ from
Section~\ref{sec:field}. For each $0<\varepsilon\le1$, the connecting-path
result of Section~\ref{sec:path} provides a lower triangular matrix
$C_\varepsilon$ and a path $B_\varepsilon$ satisfying
\eqref{eq:framework-path}. Write $a_\varepsilon=C_\varepsilon u$ and temporarily
suppress the subscripts. The path $B$ is $C^1$, and $a\in H^1_0(I;\R^N)$
is absolutely continuous. The chain and product rules give, almost everywhere,
\begin{align}
(J(B)+a^{\transpose}Ba)'
&=2(a')^{\transpose}Ba
-\tr\bigl((M(B)-aa^{\transpose})B'\bigr)\notag\\
&=2(a')^{\transpose}Ba
-\varepsilon^{-1}\|M(B)-aa^{\transpose}\|_{\HS}^2.
\label{eq:action-identity}
\end{align}
Since $a$ vanishes at the endpoints, integration and
$2(a')^{\transpose}Ba\le |a'|^2+a^{\transpose}B^2a$ yield
\begin{equation}\label{eq:penalty-bound}
J(-K)-J(K)\le\mathcal E_K(a)
+\int_I F_\varepsilon(B(x),a(x))\,dx,
\end{equation}
where
\[
F_\varepsilon(B,a)
=|a|^{2q}-a^{\transpose}(K^2-B^2)a
-\varepsilon^{-1}\|M(B)-aa^{\transpose}\|_{\HS}^2.
\]
The bounds from Section~\ref{sec:path} place every pair
$(B_\varepsilon(x),a_\varepsilon(x))$, for $x\in\overline I$ and
$0<\varepsilon\le1$, in a single compact set $\mathcal K$.
This uses the continuous representative of $u$, the bound on $C_\varepsilon$,
and fixed $\delta$. We claim that
\begin{equation}\label{eq:uniform-penalty}
\eta_\varepsilon:=\max_{(B,a)\in\mathcal K}
\max\{F_\varepsilon(B,a),0\}\longrightarrow0.
\end{equation}
Otherwise there would be $\eta>0$, a sequence $\varepsilon_n\downarrow0$,
and a convergent sequence $(B_n,a_n)\in\mathcal K$ with
$F_{\varepsilon_n}(B_n,a_n)\ge\eta$.
The first two terms of $F_\varepsilon$ are uniformly bounded on
$\mathcal K$, so $M(B_n)-a_na_n^{\transpose}\to0$.
At the limit $(B_*,a_*)$ we have $M(B_*)=a_*a_*^{\transpose}$ and
\[
|a_*|^{2q}-a_*^{\transpose}(K^2-B_*^2)a_*\ge\eta,
\]
contrary to Proposition~\ref{prop:rankone}. This proves
\eqref{eq:uniform-penalty}.
Equation~\eqref{eq:penalty-bound} now implies
\[
J(-K)-J(K)\le
\sup_{C\text{ lower triangular}}\mathcal E_K(Cu)+|I|\eta_\varepsilon.
\]
Let $\varepsilon\downarrow0$ at this fixed $\delta$. By
Lemma~\ref{lem:jensen},
\[
\sup_C\mathcal E_K(Cu)
\ge\sum_i\int_{-k_i}^{k_i}
\bigl((k_i^2-s^2+\delta)^\sigma-\delta^\sigma\bigr)\,ds.
\]
Finally let $\delta\downarrow0$. On each integration interval the integrand
converges to $(k_i^2-s^2)^\sigma$ and lies between zero and this same function,
by subadditivity of the power $\sigma$. Dominated convergence gives
\eqref{eq:action-bound}. No convergence of $C_\varepsilon$, and no bound
uniform in $\delta$, has been used.
\end{proof}
==== END Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/action-limit.tex ====
==== BEGIN Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/connecting-path.tex ====
\section{A matrix path with prescribed endpoints}\label{sec:path}
The action argument needs a symmetric matrix path from $K$ to $-K$.
Its differential equation also contains a lower triangular coefficient
matrix $C$, which we are free to choose. We choose $C$ by matching the
terminal endpoint. At the start of a deformation, the matching problem
has exactly one solution after identifying each row of $C$ with its
negative. A compactness argument shows that this solution count cannot
disappear modulo two.
Fix $\delta>0$ throughout the remaining construction. We use the
matrix field $M=M_\delta$ and scalar function $\mu=\mu_\delta$
constructed above. Recall the properties needed here. The field
$M$ is locally Lipschitz on $\Sym_N$, the space
of real symmetric $N\times N$ matrices. The function $\mu$ is locally
Lipschitz and strictly increasing, with $\mu(0)=0$. For
$K=\operatorname{diag}(k_1,\ldots,k_N)$ and every diagonal sign matrix
$S$,
\begin{equation}\label{path:equivariance}
M(SBS)=SM(B)S.
\end{equation}
For every real unit vector $e$,
\begin{equation}\label{path:comparison}
e^{\mathsf T}M(B)e
\leq \mu\bigl(e^{\mathsf T}K^2e-(e^{\mathsf T}Be)^2\bigr).
\end{equation}
Finally, writing $e_i$ for the $i$th standard basis vector,
\begin{equation}\label{path:coordinate}
Be_i=t e_i
\quad\Longrightarrow\quad
M(B)e_i=\mu(k_i^2-t^2)e_i.
\end{equation}
The equivariance in \eqref{path:equivariance} only concerns sign
matrices, which commute with the fixed matrix $K$.
\begin{proposition}[Connecting path]\label{path:existence}
Let $I=(x_-,x_+)$ be a bounded interval, let
$u\in H^1_0(I;\mathbb R^N)$ satisfy
\[
\int_I u(x)u(x)^{\mathsf T}\,dx=I_N,
\]
and let $K=\operatorname{diag}(k_1,\ldots,k_N)$ with every $k_i>0$.
Suppose $M$ and $\mu$ have the regularity and properties
\eqref{path:equivariance}--\eqref{path:coordinate} just stated.
For every $\varepsilon>0$, there are a real lower triangular matrix
$C$ and a path $B\in C^1(\overline I;\Sym_N)$ such that,
with $a=Cu$,
\begin{equation}\label{path:equation}
\varepsilon B'=M(B)-aa^{\mathsf T},
\qquad B(x_-)=K,\qquad B(x_+)=-K.
\end{equation}
These choices satisfy
\begin{equation}\label{path:operator-bound}
\|B(x)\|_{\mathrm{op}}\leq\|K\|_{\mathrm{op}}
\qquad(x\in I).
\end{equation}
For fixed $M$, $K$, and $I$, the matrices $C$ can be chosen bounded
uniformly for $0<\varepsilon\leq1$.
\end{proposition}
The parity principle is the compact-one-manifold argument underlying
mod-two degree; see \cite[Section~4]{Milnor1965}. The smooth regular-value
facts used below are given in \cite[Sections~2--3]{Milnor1965}.
We prove the needed relative version for continuous families of bundle
sections here. A section of a vector bundle assigns to each base point a
vector in the fiber above it. A zero is \emph{transverse} if the
derivative of the section in any local trivialization maps onto that
fiber.
\begin{lemma}[Continuation modulo two]\label{path:parity}
Let $Q$ be a smooth $d$-dimensional manifold without boundary, and let
$E\to Q$ be a smooth real vector bundle of rank $d$. Suppose
$s_\theta$, $0\leq\theta\leq1$, is a continuous homotopy of sections and
\[
\{(q,\theta)\in Q\times[0,1]:s_\theta(q)=0\}
\]
is compact. If $s_0$ is smooth and has an odd number of zeros, all
transverse, then $s_1$ has a zero.
\end{lemma}
\begin{proof}
Suppose that $s_1$ has no zero. Choose a relatively compact smooth
domain $\Omega\subset Q$ containing the projection of the entire zero
set. Such a domain is obtained by taking a regular level of a smooth
compactly supported function equal to one near that projection.
Fix a smooth fiber metric on $E$. On the compact set
\[
(\partial\Omega\times[0,1])
\ \cup\ (\overline\Omega\times\{1\}),
\]
the norm of the section has a strictly positive minimum.
Reparametrize the homotopy so that it is identically $s_0$ on a small
initial collar. It can then be uniformly approximated on
$\overline\Omega\times[0,1]$ by a smooth section that agrees with $s_0$
on a smaller initial collar. To construct this approximation, use
finitely many bundle charts, approximate the continuous coefficient
functions by smooth functions, and combine them with a smooth
partition of unity. A cutoff supported inside the original collar
splices this approximation with the smooth section $s_0$. Choose the
approximation error smaller than one quarter of the positive boundary
minimum. The approximating section therefore has no zeros near the
spatial boundary or the final boundary.
We can also make this smooth section transverse to zero while keeping
it unchanged near the initial boundary. Here are the finite-dimensional
details. In an initial collar its zeros are already transverse, since
the derivative in the $Q$ directions is onto at each zero of $s_0$.
The remaining possible zeros lie in a compact subset of the interior.
Choose finitely many compactly supported local frame sections whose
values span every fiber on a neighborhood of this subset, with supports
away from the initial boundary and the forbidden boundaries. Adding
linear combinations of these sections gives a finite-dimensional
family whose derivative in the new parameters is onto wherever
transversality is still needed. Restrict the parameters to a small open
ball in $\mathbb R^m$, where $m$ is the number of frame sections.
Smallness preserves the positive gap on the compact complement of the
spanning neighborhood and the initial collar, so no new zeros appear there.
The universal zero set in the interior is a smooth manifold of
dimension $m+1$. Sard's theorem applied to its projection onto
$\mathbb R^m$ provides arbitrarily small regular parameters. At a zero,
write the universal derivative as $Av+Db$, with $v$ a $(q,\theta)$ variation and
$b$ a parameter variation. Surjectivity of the projection on the kernel
means that $Av=-Db$ is solvable for every $b$. Where perturbations are
needed, $D$ is onto, so this is equivalent to $A$ being onto; elsewhere
the section is already transverse. Thus regular parameters make the
fixed-parameter section transverse.
Smallness in the smooth topology preserves the transversality already
present on the overlap with the initial collar, and smallness in norm
preserves the boundary gap.
The zero set of the resulting section is a compact smooth
one-dimensional manifold. Its boundary consists exactly of the zeros
of $s_0$ on $\Omega\times\{0\}$. A compact one-dimensional manifold
has an even number of boundary points, contradicting the assumed odd
count. No orientation of $Q$ or $E$ is required.
\end{proof}
\begin{proof}[Proof of Proposition~\ref{path:existence}]
Put $\kappa=\|K\|_{\mathrm{op}}$ and $\ell=x_+-x_-$, and use the
continuous representative of $u$. Choose a smooth nonnegative
function $\chi$ on $\Sym_N$, invariant under
orthogonal conjugation, compactly supported, and equal to one whenever
$\|B\|_{\mathrm{op}}\leq\kappa$. For example, take
$\chi(B)=\eta(\operatorname{tr}B^2)$ with $\eta$ equal to one on
$[0,N\kappa^2]$ and zero outside a larger bounded interval. The field
$\widehat M=\chi M$ is bounded and globally Lipschitz.
\smallskip
\noindent\emph{The terminal matching problem.}
Let $\mathcal L$ denote the vector space of real lower triangular
$N\times N$ matrices. For $C\in\mathcal L$ and $0\leq\theta\leq1$,
solve
\begin{equation}\label{path:homotopy-ode}
\varepsilon B'=\theta\widehat M(B)-(Cu)(Cu)^{\mathsf T},
\qquad B(x_-)=K.
\end{equation}
The bounded globally Lipschitz field and continuous forcing give a
unique solution throughout $I$. The integral equation and Gronwall's
inequality give continuous dependence on $(C,\theta)$. Thus
\[
\Phi_\theta(C)=B(x_+)+K
\]
is a continuous homotopy of maps from $\mathcal L$ to
$\Sym_N$.
At a zero of $\Phi_\theta$, integration of
\eqref{path:homotopy-ode} and orthonormality of $u$ give
\begin{equation}\label{path:coefficient-identity}
CC^{\mathsf T}
=2\varepsilon K+\theta\int_I\widehat M(B(x))\,dx.
\end{equation}
Consequently,
\begin{equation}\label{path:homotopy-bound}
\|C\|_{\mathrm{HS}}^2
\leq 2\varepsilon\operatorname{tr}K
+\ell\sup_B|\operatorname{tr}\widehat M(B)|.
\end{equation}
For fixed $\varepsilon$, the full zero set in
$\mathcal L\times[0,1]$ is therefore compact.
\smallskip
\noindent\emph{Rows cannot vanish at a matching solution.}
Let $G$ be the group of diagonal sign matrices. Equation
\eqref{path:equivariance}, invariance of $\chi$, and uniqueness of the
flow show that
\begin{equation}\label{path:terminal-equivariance}
\Phi_\theta(SC)=S\Phi_\theta(C)S
\qquad(S\in G).
\end{equation}
Suppose that row $i$ of $C$ vanishes, and let $S_i$ change only its
sign. Then $S_iC=C$, so uniqueness gives $S_iBS_i=B$ throughout the
flow. Hence $Be_i=b_i e_i$. By \eqref{path:coordinate},
\[
\varepsilon b_i'
=\theta\chi(B(x))\mu(k_i^2-b_i^2),
\qquad b_i(x_-)=k_i.
\]
Regarding $\theta\chi(B(x))/\varepsilon$ as a known continuous
coefficient, local Lipschitz uniqueness for this scalar equation gives
$b_i\equiv k_i$. Since $k_i>0$, the terminal value cannot be $-k_i$.
Thus no zero of $\Phi_\theta$ has a vanishing row.
Let $\mathcal L^\times\subset\mathcal L$ be the open subset on which
every row is nonzero. The compact joint zero set lies inside
$\mathcal L^\times\times[0,1]$, and hence a positive distance away
from every zero-row stratum. We do not require $C$ to be invertible.
\smallskip
\noindent\emph{One zero modulo signs persists.}
The action of $G$ on $\mathcal L^\times$ is free: changing the sign
of a nonzero row cannot fix $C$. Its quotient
\[
Q=\mathcal L^\times/G
\]
is therefore a smooth manifold of dimension $d=N(N+1)/2$. Concretely,
each row $i$ belongs to $\mathbb R^i\setminus\{0\}$ and is identified
with its negative. Sign conjugation on symmetric matrices defines
the rank-$d$ vector bundle
\[
E=(\mathcal L^\times\times\Sym_N)/G
\longrightarrow Q,
\qquad
(C,Z)\sim(SC,SZS).
\]
By \eqref{path:terminal-equivariance}, the maps $\Phi_\theta$ descend
to a continuous homotopy of sections of this bundle. Their joint
zero set remains compact.
At $\theta=0$,
\[
\Phi_0(C)=2K-\varepsilon^{-1}CC^{\mathsf T}.
\]
The lower triangular solutions of $CC^{\mathsf T}=2\varepsilon K$
are exactly
\[
C=\operatorname{diag}
(\pm\sqrt{2\varepsilon k_1},\ldots,
\pm\sqrt{2\varepsilon k_N}).
\]
Indeed, the first row determines $C_{11}$, the other entries of the
first column then vanish, and the same argument applies successively
to the remaining principal blocks. These solutions form one $G$-orbit.
At any one of them, writing $C=\operatorname{diag}(c_i)$,
\[
D\Phi_0(C)[H]
=-\varepsilon^{-1}(HC^{\mathsf T}+CH^{\mathsf T})
\qquad(H\in\mathcal L).
\]
Its diagonal entries are $-2c_iH_{ii}/\varepsilon$, and its entries
with $i>j$ are $-c_jH_{ij}/\varepsilon$. Since every $c_i\ne0$, this
linear map is an isomorphism onto the symmetric matrices. The
initial quotient section has exactly one transverse zero.
Lemma~\ref{path:parity} now gives a zero of $\Phi_1$.
\smallskip
\noindent\emph{The endpoints remove the cutoff.}
Take a path obtained from this zero. For a fixed real unit vector
$e$, put $h(x)=e^{\mathsf T}B(x)e$. Whenever $|h(x)|>\kappa$,
\eqref{path:comparison} implies
\begin{align*}
\varepsilon h'(x)
&\leq\chi(B(x))
\mu\bigl(e^{\mathsf T}K^2e-h(x)^2\bigr)
-|e^{\mathsf T}Cu(x)|^2
\leq0.
\end{align*}
Here $e^{\mathsf T}K^2e\leq\kappa^2<h(x)^2$; the last inequality
uses $\chi\geq0$ and remains true where $\chi=0$.
Both endpoint values of $h$ lie in $[-\kappa,\kappa]$.
A component of $\{h>\kappa\}$ cannot start from the initial side:
on that component $h$ is nonincreasing, whereas its left endpoint
has value $\kappa$. Similarly, a component of $\{h<-\kappa\}$
cannot return to the final endpoint: its right endpoint would have
value $-\kappa$, again contradicting nonincrease. Thus
$|h(x)|\leq\kappa$ throughout $I$.
Since $e$ was arbitrary, this proves
\eqref{path:operator-bound}. Therefore $\chi(B(x))=1$ everywhere,
and the path solves the uncut equation \eqref{path:equation}.
Finally, its integrated equation gives, for $0<\varepsilon\leq1$,
\begin{equation}\label{path:uniform-coefficients}
\|C\|_{\mathrm{HS}}^2
\leq 2\operatorname{tr}K
+\ell\sup_{\|B\|_{\mathrm{op}}\leq\kappa}
|\operatorname{tr}M(B)|.
\end{equation}
This is the asserted bound on $C$.
\end{proof}
For the regularized field $M_\delta$, the bound
\eqref{path:uniform-coefficients} may depend on $\delta$.
With $\delta$ fixed, it implies that all pairs $(B(x),Cu(x))$ produced
for $0<\varepsilon\leq1$ lie in one compact subset of
$\Sym_N\times\mathbb R^N$.
This is the compactness needed in the action limit; it requires no
bound on $B'$ uniform in $\varepsilon$.
==== END Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/connecting-path.tex ====
==== BEGIN Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/introduction.tex ====
\section{Introduction}\label{sec:intro}
The Lieb--Thirring inequality bounds the negative eigenvalues of a
Schr\"odinger operator by an integral of its potential. Write
$t_+=\max\{t,0\}$ and $t_-=\max\{-t,0\}$. On the line, let
$0\le W\in L^{\gamma+1/2}(\R)$ and let $H_{-W}$ be the self-adjoint operator
associated with the quadratic form
\[
h_{-W}[v]=\int_\R|v'(x)|^2\,dx-\int_\R W(x)|v(x)|^2\,dx,
\qquad v\in H^1(\R).
\]
For $\gamma>1/2$, this form is closed and bounded below, and its negative
spectrum is discrete. We give the needed form and compactness arguments in
Section~\ref{sec:spectral}. Write
\[
\Tr(H_{-W})_-^\gamma
=\sum_{\lambda_j(H_{-W})<0}|\lambda_j(H_{-W})|^\gamma,
\]
counting multiplicities; before the inequality is proved this sum is allowed
to be infinite. The optimal constant is
\[
L_{\gamma,1}=
\sup_{\substack{0\le W\in L^{\gamma+1/2}(\R)\\W\not\equiv0}}
\frac{\Tr(H_{-W})_-^\gamma}{\displaystyle\int_\R W^{\gamma+1/2}}.
\]
Two natural comparisons are the semiclassical constant
\begin{equation}\label{eq:semiclassical}
L_{\gamma,1}^{\cl}
=\frac1{2\pi}\int_\R(1-\xi^2)_+^\gamma\,d\xi
=\frac{\Gamma(\gamma+1)}{2\sqrt\pi\,\Gamma(\gamma+3/2)}
\end{equation}
and the optimal constant when only the lowest eigenvalue is retained,
denoted by $L_{\gamma,1}^{(1)}$. The one-dimensional Lieb--Thirring
conjecture predicts the one-bound-state constant for
$1/2<\gamma<3/2$ and the semiclassical constant for $\gamma\ge3/2$.
\begin{theorem}\label{thm:main}
For every $1/2<\gamma<3/2$ and every nonnegative
$W\in L^{\gamma+1/2}(\R)$,
\begin{equation}\label{eq:main}
\Tr(H_{-W})_-^\gamma
\le 2\left(\frac{\gamma-1/2}{\gamma+1/2}\right)^{\gamma-1/2}
L_{\gamma,1}^{\cl}\int_\R W^{\gamma+1/2}.
\end{equation}
The constant is optimal and equals $L_{\gamma,1}^{(1)}$.
More precisely, if $r=(\gamma-1/2)^{-1}$, equality is attained by
$W(x)=(r+1)\sech^2(rx)$.
\end{theorem}
The estimate includes rough, unbounded potentials and infinitely many
negative eigenvalues. Corollary~\ref{spectral:signed} gives the corresponding
inequality for signed potentials on their natural dense weighted form
domain. The theorem concerns scalar potentials on the line; the matrices
used below encode finite orthonormal families. Its sharpness assertion
exhibits an equality potential without classifying all optimizers.
\subsection{History and relation to prior work}
Lieb and Thirring introduced these spectral inequalities in their work on
the kinetic energy of fermions and the stability of matter
\cite{LiebThirring1975,LiebThirring1976}. At exponent one, optimizing over
the potential makes the spectral inequality equivalent to a lower bound
on the kinetic energy of an orthonormal family in terms of its density.
On the line, for a finite orthonormal family $(u_j)\subset H^1(\R)$, this
density is $\rho=\sum_j|u_j|^2$, and the corresponding kinetic bound
controls $\int\rho^3$ by $\sum_j\int|u_j'|^2$.
The one-state variational problem goes back to Keller \cite{Keller1961}.
The conjecture asks whether allowing arbitrarily many bound states can
improve on its constant in the lower exponent range. We verify the
one-state equality potential and its normalization directly.
The scalar upper endpoint predates the general Lieb--Thirring inequalities.
Inverse-scattering trace identities for the Korteweg--de Vries equation
give the sharp value $L_{3/2,1}=3/16$
\cite{ZakharovFaddeev1971,GardnerGreeneKruskalMiura1974};
Lieb and Thirring explicitly credit Gardner, Greene, Kruskal and Miura
in \cite[Section~4(B)]{LiebThirring1976}.
Aizenman and Lieb's monotonicity argument extends the semiclassical
equality to all larger exponents \cite{AizenmanLieb1978}.
At the other endpoint, Weidl first proved finiteness of $L_{1/2,1}$
\cite{Weidl1996}, and Hundertmark, Lieb and Thomas obtained its sharp
value $1/2$ \cite[Theorem~1]{HundertmarkLiebThomas1998}.
Laptev and Weidl established the operator-valued upper-endpoint estimate
and used dimension lifting to obtain the sharp semiclassical constants
for every dimension and $\gamma\ge3/2$
\cite[Theorems~2.1, 2.2 and~3.1]{LaptevWeidl2000}.
Quantitative estimates and orthonormal-function methods narrowed the
remaining gap. Hundertmark, Laptev and Weidl proved
$L_{\gamma,1}\le2L_{\gamma,1}^{\cl}$ for $1/2\le\gamma<3/2$
\cite[Theorem~4.1]{HLW2000}.
At exponent one, Dolbeault, Laptev and Loss extended the method of
Eden and Foias to matrix-valued potentials
\cite{EdenFoias1991,DLL2008}. Their all-dimensional semiclassical ratio
was $\pi/\sqrt3$; subsequent bounds were $1.456$ by Frank, Hundertmark,
Jex and Nam \cite[Theorem~1]{FHJN2021} and the reported $1.44655$
of Corso and Ried \cite[Corollary~1.7]{CorsoRied2025}.
These bounds preceded the sharp exponent-one theorem described below.
In the full one-dimensional strict interval, Levitt's numerical
investigation observed convergence to the one-state profile or splitting
into increasingly separated copies \cite[Section~4.1]{Levitt2014}.
Such computations support the conjectured value but do not bound the
contribution of every orthonormal family. That uniform bound is the
obstacle addressed by our action inequality.
Read and Schulz proved the case $\gamma=1$ in
\cite[Theorem~1]{ReadSchulz2026}, obtaining the sharp constant
$4/(3\sqrt3\pi)$. Their proof extends Benguria and Loss's scalar
cumulative-mass argument \cite[Section~2]{BenguriaLoss2004} to
orthonormal families through a matrix correction to the kinetic energy.
Their operator-valued extension also gives the semiclassical ratio
$2/\sqrt3$ in every dimension at exponent one
\cite[Theorem~2 and Section~2]{ReadSchulz2026}.
Our proof retains the individual eigenvalue scales in a finite-interval
action inequality. Its main analytic step is a rank-one comparison for
an energy-dependent regularized matrix field. A continuation argument
chooses a lower triangular coefficient matrix and a path joining two
prescribed matrix endpoints. Together these yield the action inequality
for every $1/2<\gamma<3/2$. Theorem~\ref{thm:main} therefore resolves
the remaining open-interval cases of the one-dimensional Lieb--Thirring
conjecture positively; the case $\gamma=1$ is already covered by
Read--Schulz.
\subsection{Proof strategy}
Put $\sigma=\gamma-1/2\in(0,1)$. The central result is a variational
inequality on a bounded interval, independent of any potential.
Fix $I=(x_-,x_+)$. Given $N$ orthonormal real functions
$u_i\in H^1_0(I;\R)$ and positive numbers
$k_i$, let $u=(u_i)_{i=1}^N$ and $K=\diag(k_i)$. We prove that the action
\[
\mathcal E_K(a)=\int_I\bigl(|a'|^2+a^{\transpose}K^2a
-|a|^{2+2/\sigma}\bigr)\,dx
\]
has supremum at least
$\sum_i\int_{-k_i}^{k_i}(k_i^2-s^2)^\sigma\,ds$
among functions $a=Cu$ with $C$ lower triangular.
Section~\ref{sec:action} states this precisely in
Theorem~\ref{thm:action} and explains the role of triangularity.
The difficulty is to obtain this lower bound for an arbitrary orthonormal
family while retaining its individual scales $k_i$. The argument replaces
a direct optimization over $C$ by a path with prescribed matrix endpoints.
Section~\ref{sec:field} constructs a regularized map $B\mapsto M_\delta(B)$
from real symmetric $N\times N$ matrices to themselves, and a scalar
primitive whose change from $K$ to $-K$ gives the required
endpoint integral. Section~\ref{sec:rankone} proves the analytic comparison
that controls the nonlinear action term when $M_\delta(B)=aa^{\transpose}$.
Its integrated tangent map is an average of unitary
conjugations; a trace inequality removes this average from the comparison.
Section~\ref{sec:path} chooses $C$ and a path from $K$ to $-K$ by
continuation modulo row signs. Compactness and the exclusion of zero rows
make the continuation possible even when $C$ loses rank.
Section~\ref{sec:action-limit} combines the path with the primitive:
failure of the rank-one constraint contributes a negative square to the
action identity. Compactness allows the penalty parameter to tend to zero
at fixed regularization. Only then is the regularization removed.
Section~\ref{sec:spectral} returns to the potential. Take the $u_i$ to be Dirichlet eigenfunctions with
eigenvalues $-k_i^2$, ordered so that $k_1\ge\cdots\ge k_N$.
Lower triangularity lets the eigenfunction equations bound the quadratic
part of the action by $\int W|a|^2$. A scalar maximization bounds the whole
action by a multiple of $\int W^{1+\sigma}$. Finally, cutoff trial spaces
pass from bounded intervals to the line, and finite partial sums exhaust
all negative eigenvalues. The sharpness calculation uses the explicit
potential in Theorem~\ref{thm:main}.
==== END Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/introduction.tex ====
==== BEGIN Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/matrix-field.tex ====
\section{A regularized matrix field}\label{sec:field}
We construct the field and primitive used in Section~\ref{sec:action}.
Throughout this section $K=\diag(k_1,\ldots,k_N)>0$ and $\delta>0$ are fixed.
The scalar case explains the choice of field. For $N=1$, write $K=(k)$.
If $-k\le b\le k$, putting $m=(k^2-b^2)^\sigma$ gives
\[
m(k^2-b^2)=m^{1+1/\sigma}.
\]
Thus this field gives equality in the scalar version of the rank-one
comparison, and a primitive with derivative $-m$ has the required change
$\int_{-k}^k(k^2-b^2)^\sigma\,db$ from $k$ to $-k$.
We seek matrix versions of the comparison and endpoint identity,
with a regularization removed at the end.
The identity
\[
k^2-2sb+s^2=(s-b)^2+k^2-b^2
\]
suggests integrating a scalar function of the quadratic on the left:
its affine dependence on $b$ will let us construct a trace primitive
even when the corresponding matrices $K$ and $B$ do not commute.
Set $\beta=\sigma-\tfrac12\in(-\tfrac12,\tfrac12)$ and define
$f:\R\to\R$ by
\begin{equation}\label{eq:f}
f(0)=0,\qquad f'(y)=(\max\{y,0\}+\delta)^{\beta-1},
\qquad
c=\frac{\sigma}{\displaystyle\int_\R(1+s^2)^{\beta-1}\,ds}.
\end{equation}
The exponent and normalization are chosen so that, for $z\ge0$,
scaling the integration variable gives
\[
c\int_\R f'(s^2+z)\,ds
=\sigma(z+\delta)^{\beta-1/2}
=\sigma(z+\delta)^{\sigma-1}.
\]
Integration in $z$ therefore produces the regularized power
$(z+\delta)^\sigma-\delta^\sigma$. On the scalar interval $[-k,k]$,
we will recover this expression with $z=k^2-b^2$.
Keeping $f'$ constant for negative arguments extends the construction
beyond that interval, as needed for the continuation argument.
The resulting $f$ is $C^1$, strictly increasing, and concave on $\R$.
The integral defining $c$ is finite because $\beta<1/2$; the definition
also includes $\beta=0$ without a limiting convention.
For $B\in\Sym_N$, put
\begin{equation}\label{eq:X-M}
\begin{split}
X_s(B)&=K^2-2sB+s^2I_N,\\
M_R(B)&=c\int_{-R}^R\bigl(f(X_s(B))-f(s^2)I_N\bigr)\,ds,\\
M(B)&=\lim_{R\to\infty}M_R(B).
\end{split}
\end{equation}
Here and below scalar functions of symmetric or Hermitian matrices are
defined by the spectral theorem. The symmetric cutoff in
\eqref{eq:X-M} is part of the definition.
\subsection{Convergence and a primitive}
We first record an elementary estimate that keeps the regularity argument
in the Hilbert--Schmidt norm.
\begin{lemma}\label{lem:HS-calculus}
If a real function $h$ is $L$-Lipschitz on an interval containing the spectra
of two Hermitian matrices $A$ and $B$, then
$\|h(A)-h(B)\|_{\HS}\le L\|A-B\|_{\HS}$.
\end{lemma}
\begin{proof}
Choose orthonormal eigenbases $v_i$ and $w_j$ of $A$ and $B$, with respective
eigenvalues $\lambda_i$ and $\nu_j$. In these two bases the entries of
$h(A)-h(B)$ and $A-B$ are, respectively,
\[
(h(\lambda_i)-h(\nu_j))\langle v_i,w_j\rangle,
\qquad (\lambda_i-\nu_j)\langle v_i,w_j\rangle.
\]
Square, sum over $i,j$, and apply the scalar Lipschitz bound.
\end{proof}
\begin{lemma}\label{lem:field-regularity}
The limit in \eqref{eq:X-M} is locally uniform. The resulting map $M$ is
locally Lipschitz. Moreover,
\begin{equation}\label{eq:J}
J(B)=-\int_0^1\tr\bigl(BM(tB)\bigr)\,dt
\end{equation}
is $C^1$ and satisfies \eqref{eq:framework-gradient}.
\end{lemma}
\begin{proof}
Fix a bounded set of $B$'s. For sufficiently large $|s|$, the spectrum of
$H_s(B)=K^2-2sB$ lies in $[-A|s|,A|s|]$, with $A$ independent of $B$.
On this interval set
\[
h_s(t)=f(s^2+t)-f(s^2)-f'(s^2)t.
\]
All arguments of $f$ are positive. Since
$f''(y)=(\beta-1)(y+\delta)^{\beta-2}$ for $y>0$, the mean value theorem gives
\[
\operatorname{Lip}(h_s)=O(|s|^{2\beta-3}),\qquad
\sup_{|t|\le A|s|}|h_s(t)|=O(|s|^{2\beta-2}).
\]
It follows that
\begin{equation}\label{eq:tail}
f(X_s(B))-f(s^2)I_N=-2s f'(s^2)B+R_s(B),
\end{equation}
where $R_s(B)=f'(s^2)K^2+h_s(H_s(B))$ has norm
$O(|s|^{2\beta-2})$. By Lemma~\ref{lem:HS-calculus}, its Lipschitz constant
as a function of $B$ has the same bound: the factor $2|s|$ from $H_s$ multiplies
$O(|s|^{2\beta-3})$. The first term on the right of \eqref{eq:tail} is odd
in $s$ and cancels at every symmetric cutoff; the remainder is integrable.
On bounded $s$-intervals, $f$ is Lipschitz and the same lemma applies directly.
This proves both assertions about $M$.
For completeness we justify the primitive without differentiating $M$.
Let $F$ be any $C^2$ primitive of $f$. The trace differentiation formula is
\begin{equation}\label{eq:trace-derivative}
d\tr F(A)[H]=\tr(f(A)H).
\end{equation}
For polynomials it follows by cyclicity of trace. Approximation of $F$ and
$F'$ on a compact spectral interval by a polynomial and its derivative
proves the general formula. For $s\ne0$, the map
$B\mapsto f(X_s(B))-f(s^2)I_N$ is therefore a gradient: a potential is
\[
-\frac{1}{2s}\tr F(X_s(B))-f(s^2)\tr B.
\]
At $s=0$ the field is constant and is again a gradient. Consequently each
$M_R$ has zero integral around every piecewise $C^1$ closed curve, by
integration first along the curve and then in $s$. Its negative has the
primitive
$J_R(B)=-\int_0^1\tr(BM_R(tB))\,dt$.
Local uniform convergence gives $J_R\to J$ and $M_R\to M$ on compact sets.
Passing to the limit in the line-segment identity for $J_R$ yields
\[
J(B+H)-J(B)=-\int_0^1\tr(M(B+tH)H)\,dt.
\]
Continuity of $M$ proves that $J$ is $C^1$ with the stated derivative.
\end{proof}
\subsection{Scalar comparison and boundary values}
Define
\begin{equation}\label{eq:mu}
\mu(z)=c\int_\R\bigl(f(s^2+z)-f(s^2)\bigr)\,ds,
\qquad z\in\R.
\end{equation}
This integral is absolutely convergent: on bounded $z$-sets its integrand
has tail $O(|s|^{2\beta-2})$.
\begin{lemma}\label{lem:mu}
The function $\mu$ is locally Lipschitz and strictly increasing, with
$\mu(0)=0$. For $z\ge0$,
\begin{equation}\label{eq:mu-positive}
\mu(z)=(z+\delta)^\sigma-\delta^\sigma,
\qquad \mu'(z)=\sigma(z+\delta)^{\sigma-1}.
\end{equation}
For every fixed $b,z\in\R$,
\begin{equation}\label{eq:shift}
c\lim_{R\to\infty}\int_{-R}^R
\bigl(f((s-b)^2+z)-f(s^2)\bigr)\,ds=\mu(z).
\end{equation}
\end{lemma}
\begin{proof}
The tail estimate above, also for differences in $z$, proves local
Lipschitz continuity. Strict increase follows from strict increase of $f$
at every $s$. Differentiation under the integral for $z\ge0$ gives
\[
\mu'(z)=c\int_\R(s^2+z+\delta)^{\beta-1}\,ds
=\sigma(z+\delta)^{\beta-1/2}
=\sigma(z+\delta)^{\sigma-1}.
\]
Integration from zero proves \eqref{eq:mu-positive}.
For the shift, write $g(s)=f(s^2+z)$. It is even, and
$g'(s)=2s f'(s^2+z)=O(|s|^{2\beta-1})\to0$ as $|s|\to\infty$.
Suppose first that $b\ge0$. The change in the integral of $g$ under the
shift is
\[
\int_R^{R+b}g(t)\,dt-\int_{R-b}^{R}g(t)\,dt.
\]
Its absolute value is at most
$b^2\sup_{|t-R|\le b}|g'(t)|$, which tends to zero. Evenness handles
$b<0$ as well. Subtracting $f(s^2)$ now proves \eqref{eq:shift}.
\end{proof}
\begin{lemma}[Directional comparison]\label{lem:jensen}
For every $B\in\Sym_N$ and every real unit vector $e$,
\begin{equation}\label{eq:jensen}
e^{\transpose}M(B)e\le
\mu\bigl(e^{\transpose}K^2e-(e^{\transpose}Be)^2\bigr).
\end{equation}
If a standard basis vector $e_i$ is an eigenvector of $B$ with eigenvalue
$b_i$, then
\begin{equation}\label{eq:common-eigenvector}
M(B)e_i=\mu(k_i^2-b_i^2)e_i.
\end{equation}
For the primitive in \eqref{eq:J},
\begin{equation}\label{eq:boundary-J}
J(-K)-J(K)=\sum_{i=1}^N\int_{-k_i}^{k_i}\mu(k_i^2-s^2)\,ds.
\end{equation}
\end{lemma}
\begin{proof}
Scalar concavity and the spectral decomposition of $X_s(B)$ give
\[
e^{\transpose}f(X_s(B))e\le f(e^{\transpose}X_s(B)e).
\]
Set $b=e^{\transpose}Be$ and $z=e^{\transpose}K^2e-b^2$. The scalar argument
on the right is $(s-b)^2+z$. Integrate at a symmetric cutoff and use
Lemma~\ref{lem:mu} to obtain \eqref{eq:jensen}. This uses scalar concavity
only; no operator concavity on the whole real line is asserted.
For \eqref{eq:common-eigenvector}, $e_i$ is also an eigenvector of $X_s(B)$,
with eigenvalue $(s-b_i)^2+k_i^2-b_i^2$. Functional calculus and
\eqref{eq:shift} give the identity. Finally, restrict $dJ=-\tr(M\,dB)$ to
diagonal matrices and integrate along the diagonal segment from $K$ to
$-K$. Each coordinate contributes
\[
-\int_{k_i}^{-k_i}\mu(k_i^2-s^2)\,ds
=\int_{-k_i}^{k_i}\mu(k_i^2-s^2)\,ds,
\]
as claimed.
\end{proof}
In particular $M(K)=M(-K)=0$. We shall also use the following symmetry.
If $S$ is a diagonal matrix with entries $\pm1$, then $SKS=K$ and
\begin{equation}\label{eq:sign-equivariance}
M(SBS)=SM(B)S.
\end{equation}
This follows already at each cutoff. The matrix $K$ stays fixed throughout;
we do not require equivariance under conjugations that move $K$.
==== END Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/matrix-field.tex ====
==== BEGIN Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/rank-one.tex ====
\section{The rank-one comparison}\label{sec:rankone}
The field $M$ was chosen so that its rank-one values control the nonlinear
term in the action. We now prove that property. The parameters $K,\sigma$
and $\delta$ remain fixed as in Section~\ref{sec:field}.
\begin{proposition}\label{prop:rankone}
For every $B\in\Sym_N$ and every $a\in\R^N$,
\begin{equation}\label{eq:rankone}
M(B)=aa^{\transpose}\quad\Longrightarrow\quad
a^{\transpose}(K^2-B^2)a\ge |a|^{2q}.
\end{equation}
\end{proposition}
The proof compares $X_s(B)$ with a matrix that is a function of $B$ alone.
Its tangent map, integrated in $s$, is an average of unitary conjugations.
Trace monotonicity then removes this average from the needed inequality.
\subsection{A tangent inequality on positive matrices}
Divided differences describe the derivative of matrix functional calculus,
as in the work of Daleckii and Krein \cite{DaleckiiKrein1965}; see
\cite[Theorem~2.3.1]{Hiai2010} for a modern finite-dimensional formulation.
We prove the required derivative formula and tangent inequality directly
from the resolvent representation below.
\begin{lemma}\label{lem:resolvent-tangent}
For $y\ge0$ the function $f$ of \eqref{eq:f} has the representation
\begin{equation}\label{eq:resolvent-f}
f(y)=c_\beta\int_0^\infty v^\beta
\left(\frac1{\delta+v}-\frac1{y+\delta+v}\right)\,dv,
\qquad
c_\beta^{-1}=\int_0^\infty\frac{t^\beta}{(1+t)^2}\,dt.
\end{equation}
If $X,Y\ge0$ are Hermitian matrices, then
\begin{equation}\label{eq:tangent}
f(X)\le f(Y)+T_Y(X-Y),
\end{equation}
where, in an eigenbasis of $Y$ with eigenvalues $y_i$,
\begin{equation}\label{eq:divdiff}
(T_Y Z)_{ij}=f[y_i,y_j]Z_{ij},\qquad
f[x,y]=\int_0^1 f'(\tau x+(1-\tau)y)\,d\tau.
\end{equation}
For $x\ne y$, this is $(f(x)-f(y))/(x-y)$; for $x=y$ it is $f'(x)$.
\end{lemma}
\begin{proof}
The integrand in \eqref{eq:resolvent-f} is $O(v^\beta)$ near zero and
$O(v^{\beta-2})$ at infinity. The range $-1<\beta<1$ suffices for convergence.
Differentiation gives
\[
c_\beta\int_0^\infty\frac{v^\beta}{(y+\delta+v)^2}\,dv
=(y+\delta)^{\beta-1}=f'(y),
\]
and both sides of \eqref{eq:resolvent-f} vanish at zero.
Put $A=Y+(\delta+v)I$ and $Z=X-Y$. Since $A$ and $A+Z$ are positive
definite, the resolvent identity gives
\[
-(A+Z)^{-1}
=-A^{-1}+A^{-1}ZA^{-1}
-A^{-1}Z(A+Z)^{-1}ZA^{-1}
\le -A^{-1}+A^{-1}ZA^{-1}.
\]
Integrating proves \eqref{eq:tangent}. The coefficient multiplying $Z_{ij}$
in its linear term is
\[
c_\beta\int_0^\infty
\frac{v^\beta}{(y_i+\delta+v)(y_j+\delta+v)}\,dv=f[y_i,y_j].
\]
This also covers repeated eigenvalues. All these integrals converge
absolutely, since $\delta>0$.
\end{proof}
\subsection{The integrated tangent map}
Suppose that $M(B)=aa^{\transpose}$ and $m=|a|^2>0$. By
Lemma~\ref{lem:jensen}, for every unit $e$,
\[
0\le e^{\transpose}M(B)e
\le\mu\bigl(e^{\transpose}K^2e-(e^{\transpose}Be)^2\bigr).
\]
Strict increase of $\mu$ implies
$e^{\transpose}K^2e\ge(e^{\transpose}Be)^2$. Consequently
\begin{equation}\label{eq:X-positive}
e^{\transpose}X_s(B)e
=(s-e^{\transpose}Be)^2
+e^{\transpose}K^2e-(e^{\transpose}Be)^2\ge0
\end{equation}
for every real $s$: the matrices $X_s(B)$ are positive semidefinite.
To apply the tangent inequality, we compare $X_s(B)$ with
$(sI-B)^2+dI$, which is diagonal in an eigenbasis of $B$.
Choose $d>0$ so that its integrated scalar value is $\mu(d)=m$:
\begin{equation}\label{eq:D-d-Y}
D=K^2-B^2,\qquad
d=(m+\delta^\sigma)^{1/\sigma}-\delta,\qquad
Y_s=(sI-B)^2+dI.
\end{equation}
This choice matches the nonzero eigenvalue of $M(B)$.
When we pair the integrated tangent inequality with $M(B)$,
the identity $M(B)^2=mM(B)$ will cancel the contribution $mI$
against the left side.
We have $X_s(B)=Y_s+(D-dI)$; we do not assert that $D\ge0$.
Apply Lemma~\ref{lem:resolvent-tangent} to $X_s(B)\ge0$ and $Y_s>0$,
and denote the tangent map at $Y_s$ by $T_s$.
Subtract $f(s^2)I$ and integrate symmetrically.
In an eigenbasis of $B$, the shift identity \eqref{eq:shift} gives
$c\int_\R(f(Y_s)-f(s^2)I)\,ds=mI$, with the integral understood
at symmetric cutoffs. Thus
\begin{equation}\label{eq:M-tangent}
M(B)\le mI+\mu'(d)\mathcal P(D-dI),
\qquad \mathcal P=\frac{c}{\mu'(d)}\int_\R T_s\,ds.
\end{equation}
The integral of the tangent maps converges absolutely: for large $|s|$,
their matrix coefficients are $O(|s|^{2\beta-2})$.
\begin{lemma}\label{lem:average}
In an eigenbasis of $B$ with eigenvalues $z_i$, the map $\mathcal P$ is
\begin{equation}\label{eq:P-kernel}
(\mathcal P Z)_{ij}=P_{ij}Z_{ij},\qquad
P_{ij}=\int_0^1
\left(1+\frac{\tau(1-\tau)(z_i-z_j)^2}{d+\delta}\right)^{\sigma-1}\,d\tau.
\end{equation}
There is an even probability measure $\nu$ on $\R$ such that
\begin{equation}\label{eq:P-average}
\mathcal P(Z)=\int_\R e^{itB}Ze^{-itB}\,d\nu(t).
\end{equation}
In particular $\mathcal P(I)=I$, and $\mathcal P$ is self-adjoint for the
trace pairing on Hermitian matrices.
\end{lemma}
\begin{proof}
The eigenvalues of $Y_s$ are $y_i=(s-z_i)^2+d$. Formula
\eqref{eq:divdiff} yields
\[
f[y_i,y_j]=\int_0^1
\bigl((s-\tau z_i-(1-\tau)z_j)^2+d+\delta
+\tau(1-\tau)(z_i-z_j)^2\bigr)^{\beta-1}\,d\tau.
\]
These scalar integrands are nonnegative. Tonelli's theorem and a shift in
$s$ evaluate their integral by the same calculation as $\mu'(d)$ in
Lemma~\ref{lem:mu}, giving \eqref{eq:P-kernel}.
For $h\ge0$ and $x\in\R$, the gamma integral gives
\begin{equation}\label{eq:gamma-mixture}
(1+hx^2)^{\sigma-1}
=\frac1{\Gamma(1-\sigma)}
\int_0^\infty v^{-\sigma}e^{-v}e^{-vhx^2}\,dv.
\end{equation}
The density $v^{-\sigma}e^{-v}/\Gamma(1-\sigma)$ has total mass one.
For fixed $v,h$, $e^{-vhx^2}$ is the characteristic function of a centered
Gaussian with variance $2vh$; when $h=0$, use the point mass at zero.
Choose $\tau$ uniformly in $[0,1]$ and put $h=\tau(1-\tau)/(d+\delta)$.
The resulting mixture of these symmetric Gaussian measures is an even
probability measure $\nu$, independent of $i,j$. Its characteristic
function at $z_i-z_j$ is precisely $P_{ij}$, which proves
\eqref{eq:P-average}. The diagonal coefficients are one. Finally, changing
$t$ to $-t$ in \eqref{eq:P-average} and using cyclicity of trace proves
self-adjointness.
\end{proof}
The integrated tangent inequality \eqref{eq:M-tangent} controls
$\mathcal P(D)$, while the desired conclusion involves $D$ itself. The
next step compares them in precisely the trace pairing with $M(B)$.
\subsection{Trace monotonicity and contraction}
\begin{lemma}\label{lem:trace-average}
With $B,D$ and $\mathcal P$ as above,
\begin{equation}\label{eq:trace-average}
\tr(M(B)D)\ge\tr(M(B)\mathcal P(D)).
\end{equation}
\end{lemma}
\begin{proof}
We use an elementary instance of the generalized Klein trace inequality
\cite[Proposition~3, p.~289]{Petz1994}: for any increasing real function $g$ and
Hermitian matrices $A_1,A_2$,
\begin{equation}\label{eq:trace-monotonicity}
\tr\bigl((g(A_1)-g(A_2))(A_1-A_2)\bigr)\ge0.
\end{equation}
Indeed, in eigenbases of the two matrices the trace equals
\[
\sum_{i,j}(g(\lambda_i)-g(\nu_j))(\lambda_i-\nu_j)
|\langle v_i,w_j\rangle|^2\ge0.
\]
Apply this to $g=f$, $A_1=UX_s(B)U^*$ and $A_2=X_s(B)$, with
$U=e^{itB}$. Since $U$ commutes with $(sI-B)^2$, expansion of the trace gives
\begin{equation}\label{eq:trace-fixed-s}
0\le\tr\bigl(f(X_s(B))(2D-UDU^*-U^*DU)\bigr).
\end{equation}
The coefficient $2D-UDU^*-U^*DU$ has trace zero. We may therefore subtract
$f(s^2)I$ inside the trace, integrate over $[-R,R]$, and pass to the
symmetric-cutoff limit defining $M$. The result is
\[
0\le\tr\bigl(M(B)(2D-UDU^*-U^*DU)\bigr).
\]
This passage is also uniform in $t$: the coefficient has norm at most
$4\|D\|_{\HS}$, whereas $M_R(B)\to M(B)$ in matrix norm.
Average with respect to the even measure $\nu$ from Lemma~\ref{lem:average}.
Both conjugation terms average to $\mathcal P(D)$, proving the claim.
\end{proof}
\begin{proof}[Proof of Proposition~\ref{prop:rankone}]
If $a=0$, both sides of \eqref{eq:rankone} vanish. Otherwise the preceding
construction applies. Multiply the matrix inequality
\eqref{eq:M-tangent} by the positive semidefinite matrix $M=aa^{\transpose}$
and take its trace. Since $M^2=mM$ and $\mathcal P(I)=I$, we obtain
\[
m\tr M\le m\tr M+\mu'(d)
\bigl(\tr(M\mathcal P(D))-d\tr M\bigr).
\]
Here $\mu'(d)>0$. Lemma~\ref{lem:trace-average} therefore gives
\[
a^{\transpose}Da=\tr(MD)
\ge\tr(M\mathcal P(D))\ge d\tr M=dm.
\]
For nonnegative $x,y$ and $0<\sigma<1$, $(x+y)^\sigma\le x^\sigma+y^\sigma$.
Apply this with $x=d$, $y=\delta$ to get
$m=(d+\delta)^\sigma-\delta^\sigma\le d^\sigma$.
Thus $dm\ge m^{1+1/\sigma}=|a|^{2q}$, completing the proof.
\end{proof}
==== END Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/rank-one.tex ====
==== BEGIN Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/spectral.tex ====
\section{From the action inequality to spectral moments}
\label{sec:spectral}
We apply Theorem~\ref{thm:action} first to ordered Dirichlet
eigenfunctions. We then pass to the line using finite-dimensional trial
spaces. This last step treats all negative eigenvalues, without assuming
in advance that their moment is finite.
Throughout this section, $0<\sigma<1$, $p=1+\sigma$,
$q=1+1/\sigma$, and $\gamma=\sigma+1/2$. Write
\begin{equation}\label{spectral:constant}
A_\sigma=\frac{\sigma^\sigma}{(1+\sigma)^{1+\sigma}},
\qquad
\ell_\sigma=\frac{A_\sigma}{\mathrm B(1/2,1+\sigma)},
\end{equation}
where $\mathrm B$ denotes the beta function.
\subsection{Quadratic forms and negative spectrum}
For $0\leq W\in L^p(\R)$, define
\[
h_{-W}[v]=\int_\R |v'|^2-\int_\R W|v|^2,
\qquad v\in H^1(\R).
\]
We record the facts needed to define its operator and to exhaust its
negative eigenvalues. We use the association of a densely defined closed
semibounded form with its self-adjoint operator, and the spectral
description of the operator and its form domain, in
\cite[Theorem~2.13, Theorems~3.7--3.8 and (3.50)--(3.51)]{Teschl2009}.
All theorem numbers here refer to the 2009 first edition.
\newpage
\begin{lemma}\label{spectral:forms}
Let $p>1$ and $0\leq W\in L^p(\R)$. The form $h_{-W}$ is closed
and bounded below on $H^1(\R)$. For its associated self-adjoint operator
$H_{-W}$, every spectral projection onto $(-\infty,-\alpha]$, with
$\alpha>0$, has finite rank. Thus its negative spectrum consists of
eigenvalues of finite multiplicity, with possible accumulation only at
zero. On a bounded interval $I$, the corresponding Dirichlet form on
$H^1_0(I)$ is closed and bounded below and has compact resolvent.
\end{lemma}
\begin{proof}
Put $p'=p/(p-1)$. The one-dimensional Sobolev inequality and
interpolation give
\[
\|v\|_{2p'}^2
\leq \|v\|_\infty^{2/p}\|v\|_2^{2-2/p}
\leq 2^{1/p}\|v'\|_2^{1/p}\|v\|_2^{2-1/p}.
\]
Consequently H\"older's and Young's inequalities imply that, for every
$\eta>0$, there is $b_{\eta,W}<\infty$ such that
\begin{equation}\label{spectral:form-bound}
\int_\R W|v|^2
\leq \eta\|v'\|_2^2+b_{\eta,W}\|v\|_2^2.
\end{equation}
The KLMN Theorem \cite[Theorem~6.24]{Teschl2009} now gives closure and semiboundedness, with
form domain exactly $H^1(\R)$.
Multiplication by $\sqrt W$ is compact from $H^1(\R)$ to $L^2(\R)$.
Indeed, choose nonnegative bounded, compactly supported $W_n$ converging
to $W$ in $L^p$. Multiplication by $\sqrt{W_n}$ is compact by local
Rellich compactness, and
\[
\| (\sqrt W-\sqrt{W_n})v\|_2^2
\leq \|W-W_n\|_p\|v\|_{2p'}^2
\leq C_p\|W-W_n\|_p\|v\|_{H^1}^2.
\]
Thus these multiplication operators converge in operator norm.
If the spectral subspace of $H_{-W}$ for $(-\infty,-\alpha]$ were
infinite-dimensional, it would contain an orthonormal sequence $v_n$.
Semiboundedness places this entire spectral subspace in the form domain.
The lower bound obtained from \eqref{spectral:form-bound} with
$\eta=1/2$, together with $h_{-W}[v_n]\leq-\alpha$, bounds this
sequence in $H^1$. After passing to a subsequence, it converges weakly
to zero in $H^1$, because it is orthonormal in $L^2$. Compactness then
gives $\sqrt Wv_n\to0$ in $L^2$, contradicting
\[
-\alpha\geq h_{-W}[v_n]\geq-\|\sqrt Wv_n\|_2^2.
\]
This proves the negative-spectrum assertion. On a bounded interval,
the same form estimate follows by zero extension. The form norm is
equivalent to the $H^1_0$ norm, whose embedding into $L^2(I)$ is compact;
hence the Dirichlet operator has compact resolvent.
\end{proof}
\subsection{The Dirichlet estimate}
The order of the eigenfunctions is what makes the triangular
coefficient restriction in Theorem~\ref{thm:action} useful.
\begin{proposition}\label{spectral:dirichlet}
Let $I$ be a bounded interval and $0\leq W\in L^p(I)$. If
$-k_1^2,\ldots,-k_N^2$ are the first $N$ negative Dirichlet eigenvalues
of $-d^2/dx^2-W$, ordered so that $k_1\geq\cdots\geq k_N>0$, then
\begin{equation}\label{spectral:dirichlet-bound}
\sum_{j=1}^N k_j^{2\gamma}
\leq \ell_\sigma\int_I W^p.
\end{equation}
\end{proposition}
\begin{proof}
Choose real orthonormal eigenfunctions $u_1,\ldots,u_N$; these exist
because the Dirichlet form is real. Put $u=(u_1,\ldots,u_N)^{\mathsf T}$
and $K=\operatorname{diag}(k_1,\ldots,k_N)$. For every real lower
triangular matrix $C$, write $a=Cu$. Its $i$th component lies in the
span of $u_1,\ldots,u_i$, and the eigenfunction equations in form sense
give
\[
\int_I \bigl(|a_i'|^2-W|a_i|^2+k_i^2|a_i|^2\bigr)
=\sum_{j\leq i} C_{ij}^2(k_i^2-k_j^2)\leq0.
\]
Summing over $i$ and putting $m=|a|^2$, we obtain
\[
\mathcal E_K(a)\leq\int_I(Wm-m^q)
\leq A_\sigma\int_I W^{1+\sigma}.
\]
The last inequality follows by maximizing $Wm-m^q$ over $m\geq0$;
the maximizer is $(W/q)^\sigma$.
Theorem~\ref{thm:action} therefore yields
\[
\sum_{j=1}^N\int_{-k_j}^{k_j}(k_j^2-s^2)^\sigma\,ds
\leq A_\sigma\int_I W^{1+\sigma}.
\]
Substituting $s=k_jt$ in each integral gives
$\mathrm B(1/2,1+\sigma)k_j^{2\sigma+1}$, proving
\eqref{spectral:dirichlet-bound}.
\end{proof}
The beta--gamma identity identifies the constant without any limiting
argument in $\sigma$:
\begin{align}\label{spectral:constant-identification}
\ell_\sigma
&=\left(\frac{\sigma}{1+\sigma}\right)^\sigma
\frac{\Gamma(\sigma+3/2)}{\sqrt\pi\,\Gamma(\sigma+2)}
\notag\\
&=2\left(\frac{\sigma}{1+\sigma}\right)^\sigma
L_{\gamma,1}^{\mathrm{cl}}.
\end{align}
\subsection{Exhaustion of the line}
Let $0\leq W\in L^p(\R)$. If there are no negative eigenvalues, the
trace inequality is immediate. Otherwise fix any finite number $N$
of them, counted with multiplicity and ordered increasingly:
\[
\lambda_1\leq\cdots\leq\lambda_N<0.
\]
Let $\phi_1,\ldots,\phi_N\in H^1(\R)$ be corresponding orthonormal
eigenfunctions. Choose smooth cutoffs $\chi_R$ equal to one on
$[-R,R]$, supported in $I_R=(-2R,2R)$, and satisfying
$0\leq\chi_R\leq1$ and $\|\chi_R'\|_\infty\leq c/R$. Then
\[
v_{j,R}=\chi_R\phi_j\longrightarrow\phi_j
\quad\hbox{in }H^1(\R).
\]
H\"older's inequality and the embedding $H^1\subset L^{2p'}$ show
that the potential form is continuous under this convergence.
For clarity, let $G_R$ and $Q_R$ be the Gram and form matrices of the
$N$ functions $v_{j,R}$. Then
\[
G_R\longrightarrow I_N,
\qquad
Q_R\longrightarrow\Lambda
:=\operatorname{diag}(\lambda_1,\ldots,\lambda_N)
\]
in matrix norm. In particular the functions are linearly independent
for large $R$. If $g_R=\|G_R-I_N\|_{\mathrm{op}}<1$ and
$e_R=\|Q_R-\Lambda\|_{\mathrm{op}}$, their Rayleigh quotients differ
uniformly from the limiting coefficient-space quotients by at most
\[
\eta_R=\frac{e_R+\|\Lambda\|_{\mathrm{op}}g_R}{1-g_R}
\longrightarrow0.
\]
Applying the form-domain min--max principle
\cite[Theorem~4.10 and the following form-domain extension]{Teschl2009}
to the span of
$v_{1,R},\ldots,v_{j,R}$ gives, for every $j\leq N$,
\[
\lambda_j(I_R)\leq\lambda_j+\eta_R,
\]
where $\lambda_j(I_R)$ is the $j$th Dirichlet eigenvalue on $I_R$.
All these levels are negative for sufficiently large $R$.
Proposition~\ref{spectral:dirichlet} now implies
\[
\sum_{j=1}^N(-\lambda_j-\eta_R)^\gamma
\leq \sum_{j=1}^N|\lambda_j(I_R)|^\gamma
\leq \ell_\sigma\int_{I_R}W^p
\leq \ell_\sigma\int_\R W^p.
\]
Letting $R\to\infty$ proves the estimate for every finite partial
sum. Taking their supremum gives
\begin{equation}\label{spectral:line-bound}
\Tr(H_{-W})_-^\gamma
\leq \ell_\sigma\int_\R W^p.
\end{equation}
Thus the full moment is finite even when there are infinitely many
negative eigenvalues. No approximation of $W$ by smoother potentials
is needed.
\subsection{Signed potentials}
To include unbounded positive parts, we specify the form domain in the
signed-potential statement.
\begin{corollary}\label{spectral:signed}
Let $V$ be real measurable with $V_-\in L^p(\R)$, and suppose
\[
\mathcal D_V
=\left\{v\in H^1(\R):\int_\R V_+|v|^2<\infty\right\}
\]
is dense in $L^2(\R)$. This holds in particular if
$V_+\in L^1_{\mathrm{loc}}(\R)$. The form
\[
h_V[v]=\int_\R|v'|^2+\int_\R V_+|v|^2-\int_\R V_-|v|^2,
\qquad v\in\mathcal D_V,
\]
is closed and bounded below. Its associated operator $H_V$ has
discrete negative spectrum and satisfies
\[
\Tr(H_V)_-^\gamma\leq\ell_\sigma\int_\R V_-^p.
\]
\end{corollary}
\begin{proof}
The positive form $\int|v'|^2+\int V_+|v|^2$ is closed on
$\mathcal D_V$, by completeness of $H^1$ and of the weighted $L^2$
space. Estimate~\eqref{spectral:form-bound}, with $W=V_-$, proves
closure and semiboundedness after subtraction of the negative part.
If $V_+\in L^1_{\mathrm{loc}}$, then
$C_c^\infty(\R)\subset\mathcal D_V$, proving the asserted density.
The compactness argument in Lemma~\ref{spectral:forms} still applies.
On a negative spectral subspace bounded away from zero, the same relative
form bound controls the $H^1$ norm of unit vectors, since the positive
potential term can be discarded. Compactness of multiplication by
$\sqrt{V_-}$ then excludes an infinite orthonormal sequence in that
subspace: the integral against $V_-$ would tend to zero along a subsequence,
whereas its negative form energy stays bounded away from zero. Thus every
such spectral projection has finite rank.
For $v\in\mathcal D_V\subset H^1(\R)$, one has
$h_V[v]\geq h_{-V_-}[v]$. The min--max principle, with this inclusion
of form domains, bounds each negative eigenvalue of $H_V$ below by
the corresponding negative eigenvalue of $H_{-V_-}$; in particular,
if $H_V$ has $j$ negative eigenvalues, so does $H_{-V_-}$. Summation and
\eqref{spectral:line-bound} prove the result.
\end{proof}
\subsection{Sharpness}
Put $r=1/\sigma>1$ and consider
\[
W(x)=(r+1)\sech^2(rx),
\qquad
\psi(x)=\sech^{1/r}(rx).
\]
Direct differentiation gives
\[
\frac{\psi'}{\psi}=-\tanh(rx),
\qquad
\frac{\psi''}{\psi}=1-(r+1)\sech^2(rx).
\]
The function $\psi$ and its first two derivatives decay exponentially.
Thus $\psi\in H^2(\R)$, the operator domain for this bounded $W$,
and $H_{-W}\psi=-\psi$.
Substitution $t=rx$, followed by $z=\tanh t$, gives
\[
\int_\R W^p
=\frac{(r+1)^p}{r}\,\mathrm B(1/2,p),
\qquad
\left(\int_\R W^p\right)^{-1}=\ell_\sigma.
\]
The eigenvalue $-1$ gives $\Tr(H_{-W})_-^\gamma\geq1$; the upper
bound \eqref{spectral:line-bound} gives the reverse inequality.
This completes the proof of Theorem~\ref{thm:main}.
Hence this potential attains equality, and the constant in
\eqref{spectral:constant-identification} is optimal. This establishes
$L_{\gamma,1}=L_{\gamma,1}^{(1)}$ throughout $1/2<\gamma<3/2$.
==== END Sharp-One-Dimensional-Lieb-Thirring-Constants//sections/spectral.tex ====
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