---
abstract: 'To determine the sharp constants for the one dimensional Lieb–Thirring inequalities with exponent $\gamma \in (1/2,3/2)$ is still an open problem. According to a conjecture by Lieb and Thirring the sharp constant for these exponents should be attained by potentials having only one bound state. Here we exhibit a connection between the Lieb–Thirring conjecture for $\gamma=1$ and an isporimetric inequality for ovals in the plane.'
address:
- 'Department of Physics, P. Universidad Católica de Chile, Casilla 306, Santiago 22, Chile'
- 'School of Mathematics, Georgia Institute of Technology'
author:
- 'Rafael D. Benguria'
- Michael Loss
title: 'Connection between the Lieb–Thirring conjecture for Schrödinger operators and an isoperimetric problem for ovals on the plane'
---

[^1]

[^2]

Introduction
============

The Lieb–Thirring inequalities are one of the main tools in the proof of the stability of matter [@LiTh75] (see also the review article [@Li76] or [@Li00]). Let $H=-\Delta + V$ be the Schrödinger operator acting on $L^2({\mathbb{R}}^n)$, $n \ge 1$ and denote by $e_1 \le e_2 \le \dots < 0$ the negative eigenvalues of $H$. The Lieb–Thirring inequalities are given by $$\sum_{j \ge 1} {\vert e_j \vert}^{\gamma} \le L_{\gamma,n} \int_{{\mathbb{R}}^n} 
V_{-}(x)^{\gamma+n/2} \, dx,
\label{eq:1.1}$$ where $V_{-}(x) \equiv \max(-V(x),0)$ is the negative part of the potential. The above inequalities hold for $\gamma \ge 1/2$ when $n=1$, for $\gamma >0$ when $n=2$, and for $\gamma \ge 0$ for $n \ge 3$. The case $\gamma=1/2$, $n=1$ was established by T. Weidl [@We96]. The case $\gamma=0$, $n \ge 3$ was established independently by M. Cwikel, E.H. Lieb and G.V. Rosenbljum. One can show in general that $L_{\gamma,n} \ge L_{\gamma,n}^c$, where $$L_{\gamma,n}^c = 2^{-n} \pi^{-n/2} \frac{\Gamma(\gamma+1)}
{\Gamma(\gamma+1+n/2)}
\label{eq:1.2}$$ are the semiclassical constants. Define $R_{\gamma,n} \equiv L_{\gamma,n}/ L_{\gamma,n}^c \ge 1$. Aizenman and Lieb proved that $R_{\gamma,n}$ decreases as $\gamma$ increases [@AiLi78]. In [@LiTh76] it is proven that $L_{3/2,1} = L_{3/2,1}^c$ and thus, $L_{\gamma,1} = L_{\gamma,1}^c$, for all $\gamma \ge 3/2$. For $n>1$, Laptev and Weidl [@LaWe00a] proved $L_{3/2,n} = L_{3/2,n}^c$ hence, $L_{\gamma,n} = L_{\gamma,n}^c$, for all $\gamma \ge 3/2$. The sharp constant for $\gamma=1/2$ and $n=1$, $L_{1/2,1}=1/2$ was proved in [@HuLiTh98]. For best constants up to date see [@HuLaWe00].

For $n=1$, the sharp constants $L_{\gamma,1}$ are not known for values of $\gamma$ in the interval $(1/2,3/2)$. However, in 1976 Lieb and Thirring [@LiTh76] conjectured that the sharp constants are attained for potentials that have only one bound state, and therefore $$L_{\gamma,1} \equiv L_{\gamma , 1}^1 = \frac{1}{\sqrt{\pi}} 
\frac{1}{\gamma-1/2} 
\frac{\Gamma(\gamma+1)}{\Gamma(\gamma+1/2)}
\left(\frac{\gamma-1/2}{\gamma+1/2} \right)^{\gamma+1/2}.
\label{eq:1.3}$$

In this manuscript we establish a connection between the Lieb–Thirring conjecture for $\gamma=1$ and $n=1$ and an isoperimetric inequality for closed curves in the plane which are smooth, have positive curvature and length $2 \pi$. The rest of the article is organized as follows. In Section 2, we provide a new and direct method for maximizing the lowest eigenvalue of one dimensional Schrödinger operators. In Section 3 we establish the aforementioned connection with a problem for closed curves in the plane. We should emphasize that the isoperimetric problem that we allude to is also still open.

Maximizing the first eigenvalue
===============================

The problem of maximizing the lowest eigenvalue of the one-dimensional Schrödinger operator on the line subject to a constraint on integrals of powers of the potential was first considered by Joseph Keller in 1961 [@Ke61]. See also [@LiTh76; @AsHa87; @BeVe92; @Ve02].

Consider the Schrödinger operator, $$H=-\frac{d^2}{dx^2} + V
\label{eq:2.1}$$ defined on $L^2({\mathbb{R}})$, and let $-\lambda_1$ be the lowest eigenvalue. Then, $$\lambda_1^{\gamma} \le L_{\gamma , 1}^1 \int_{-\infty}^{\infty} 
V_{-}(x)^{\gamma+1/2} \, dx,
\label{eq:2.2}$$ for all $\gamma>1/2$, where the sharp constants $L_{\gamma,1}^1$ are given by, $$L_{\gamma,1}^1 = \frac{1}{\sqrt{\pi}} \frac{1}{\gamma-1/2} 
\frac{\Gamma(\gamma+1)}{\Gamma(\gamma+1/2)}
\left(\frac{\gamma-1/2}{\gamma+1/2} \right)^{\gamma+1/2}.
\label{eq:2.3}$$ Keller’s proof uses the Direct Calculus of Variations. When the exponent $\gamma=1$ there is a very simple argument to compute the best constant. We give the full argument in the sequel, because it is important in our later derivation of the connection between the Lieb–Thirring conjecture and an isoperimetric inequality for ovals in ${\mathbb{R}}^2$.

Let $u_1$ and $-\lambda_1$ be the normalized ground state and the lowest eigenvalue of the Schrödinger operator $H=-d^2/dx^2 - V$ on $L^2({\mathbb{R}})$, where $V \ge 0$. Thus, $$-u_1''-V \, u_1 = - \lambda_1 u_1, \qquad \mbox{in ${\mathbb{R}}$}.
\label{eq:2.4}$$ Multiplying (\[eq:2.4\]) by $u_1$ and integrating in ${\mathbb{R}}$, we get $$\lambda_1 = \int_{{\mathbb{R}}} V \, u_1^2 \, dx - \int_{{\mathbb{R}}} (u_1')^2 \, dx.
\label{eq:2.5}$$ Since, $$V \, u_1^2 \le K \, V^{3/2} + \frac{4}{27 \, K^2} u_1^6,$$ for all $K >0$, from (\[eq:2.5\]) we get, $$\lambda_1 \le  K \int_{{\mathbb{R}}} V^{3/2} \, dx + \frac{4}{27 \, K^2} 
\int_{{\mathbb{R}}} u_1^6 \, dx - \int_{{\mathbb{R}}} (u_1')^2 \, dx. 
\label{eq:2.6}$$ However, if $\int_{{\mathbb{R}}} u_1^2 \, dx=1$, $$\int_{{\mathbb{R}}} (u_1')^2 \, dx \ge \frac{\pi^2}{4} \int_{{\mathbb{R}}} u_1^6 \, dx,
\label{eq:2.7}$$ so, choosing $K=4/(3 \sqrt{3} \pi)$, we finally get $$\lambda_1 \le \frac{4}{3 \sqrt{3} \pi} \int_{{\mathbb{R}}} V^{3/2} \, dx = L_{1,1}^1 
\int_{{\mathbb{R}}} V^{3/2} \, dx
\label{eq:2.8}$$ which is Keller’s result for $\gamma=1$. For completeness we give an elementary proof of (\[eq:2.7\]). First make the change of variables $x \to s$ given by $$s = \int_{-\infty}^{x} u_1^2 \, dy.
\label{eq:2.9}$$ Here, $s: 0 \to 1$, and $ds/dx=u_1^2$. With this change of variables we have, $$\int_{-\infty}^{\infty} u_1^6 \, dx = \int_0^1 u_1^4 \, ds, 
\label{eq:2.10}$$ and $$\int_{-\infty}^{\infty} (u_1')^2 \, dx = \int_0^1 (\dot u_1)^2 u_1^2 \, ds, 
\label{eq:2.11}$$ where $\dot u_1 \equiv d u_1/ds$. Since $u_1$ goes to zero at $x=\pm \infty$ we have $u_1(s=0)=u_1(s=1)=0$. Finally, if we call $w \equiv u_1^2$, $\int_{-\infty}^{\infty} u_1^6 = \int_0^1 w^2 \, ds$ and $\int_{-\infty}^{\infty} (u_1')^2 \, dx 
=(1/4) \int_0^1 {\dot w}^2 \, ds$. In terms of $w(s)$, (\[eq:2.7\]) is given by $$\int_0^1 {\dot w}^2 \, ds \ge \pi^2 \int_0^1 w^2 \, ds,
\label{eq:2.12}$$ which follows from the fact that the first Dirichlet eigenvalue of the interval $(0,1)$ is $\pi^2$. One can obtain the cases with $\gamma \neq 1$ in (\[eq:2.2\]) in a similar way (see the Appendix).

Maximizing the sum of the first two eigenvalues and the connection with a geometric problem in ${\mathbb{R}}^2$.
================================================================================================================

Consider the Schrödinger operator $$H=-\frac{d^2}{dx^2} - V,$$ on $L^2({\mathbb{R}})$ with $V \ge 0$ such that $\int V^{3/2} \, dx < \infty$. Assume $H$ has at least two negative eigenvalues, and denote by $-\lambda_1$ and $-\lambda_2$ the lowest two eigenvalues and $u_1$, $u_2$ the corresponding normalized eigenfunctions. As before, we have $$\lambda_1 = \int_{{\mathbb{R}}} V \, u_1^2 \, dx - \int_{{\mathbb{R}}} (u_1')^2 \, dx,
\label{eq:3.1}$$ and $$\lambda_2 = \int_{{\mathbb{R}}} V \, u_2^2 \, dx - \int_{{\mathbb{R}}} (u_2')^2 \, dx.
\label{eq:3.2}$$ Adding these two equations and using the pointwise bound, $$V (u_1^2+u_2^2)  \le K \, V^{3/2} + \frac{4}{27 \, K^2} (u_1^2+u_2^2)^3,$$ we get $$\lambda_1+\lambda_2 \le  K \int_{{\mathbb{R}}} V^{3/2} \, dx + 
\frac{4}{27 \, K^2} \int_{{\mathbb{R}}} 
(u_1^2+u_2^2)^3 \, dx
- \int_{{\mathbb{R}}} \left((u_1')^2+(u_2')^2 \right)\, dx. 
\label{eq:3.3}$$ In order to prove the Lieb–Thirring conjecture for $\gamma=1$ in the special case of potentials having only two eigenvalues, it would be enough to prove $$\int_{{\mathbb{R}}} \left((u_1')^2+(u_2')^2 \right)\, dx \ge \frac{\pi^2}{4}
\int_{{\mathbb{R}}} (u_1^2+u_2^2)^3 \, dx,
\label{eq:3.4}$$ for any pair of functions $u_1$, $u_2$ such that $\int_{{\mathbb{R}}} u_1^2 \, dx= 
\int_{{\mathbb{R}}} u_2^2 \, dx=1$, and $\int_{{\mathbb{R}}} u_1 \, u_2 \, dx=0$ (i.e., for any pair of mutually orthogonal, normalized functions). For then, it would follow from (\[eq:3.3\]) and (\[eq:3.4\]) that $$\lambda_1+\lambda_2 \le L_{1,1}^1 \int_{{\mathbb{R}}} V^{3/2} \, dx.
\label{eq:3.5}$$ To prove (\[eq:3.4\]) is still an open problem. Here we will show that (\[eq:3.4\]) is equivalent to an (open) isoperimetric inequality for ovals on the plane. To establish this connection, we perform a change of variables similar to the one used in the previous section to prove Keller’s result on the lowest eigenvalue. First we change the independent variable $$x \to s \equiv \pi \int_{-\infty}^{x} \left(u_1^2+u_2^2 \right) \, dy.
\label{eq:3.6}$$ Since $u_1$ and $u_2$ are both normalized, it follows that $s$ runs from $0$ to $2 \pi$. From (\[eq:3.6\]) we have $$\frac{ds}{dx} = \pi \left(u_1^2+u_2^2 \right).$$ Moreover, set $$u_1=\rho \cos \theta, \qquad \mbox{and} \qquad u_2=\rho \sin \theta,
\label{eq:3.7}$$ so that $$u_1^2+u_2^2= \rho^2, \qquad \mbox{and} \qquad 
{u_1'}^2+{u_2'}^2= {\rho'}^2 + \rho^2 {\theta'}^2.
\label{eq:3.8}$$ With this change of variables we can write $$\int_{{\mathbb{R}}}\left({u_1'}^2+{u_2'}^2 \right) \, dx =
\pi \int_{0}^{2 \pi}
\left(\rho^2 {\dot \rho}^2 + \rho^4 {\dot \theta}^2 \right) \, ds,
\label{eq:3.9}$$ and $$\int_{{\mathbb{R}}}\left(u_1^2+u_2^2 \right)^3 \, dx =
\frac{1}{\pi} \int_{0}^{2 \pi} \rho^4 \, ds. 
\label{eq:3.10}$$ Furthermore, set $$R=\rho^2,$$ and $$\varphi= 2\theta.$$ In these new variables, the desired inequality (\[eq:3.4\]) is equivalent to $$\frac
{\int_0^{2\pi} \left({\dot R}^2+R^2{\dot \varphi}^2 \right) \, ds}
{\int_0^{2\pi} R^2\, ds}
\ge 1,
\label{eq:3.11}$$ subject to the fact that $u_1$ and $u_2$ are orthonormal, fact that we ought to express in terms of the new variables. In the new variables, $$0=\int_{{\mathbb{R}}} u_1 \, u_2 \, dx = \frac{1}{2 \pi}\int_0^{2 \pi} \sin 
\varphi (s) \, ds.$$ Concerning the other side constraints (i.e., the fact that $u_1$ and $u_2$ are normalized), given the definition of $s$ and the fact that $s$ runs from $0$ to $2\pi$, it is enough to consider the combination $$0=\int_{{\mathbb{R}}} (u_1^2-u_2^2) \, dx = \frac{1}{\pi}\int_0^{2 \pi} 
\cos \varphi (s) \, ds.$$ Thus, in the new variables, the fact that $u_1$ and $u_2$ are orthonormal imply $$\int_0^{2 \pi} \sin \varphi (s) \, ds= \int_0^{2 \pi} \cos \varphi (s) \, ds=0.
\label{eq:3.12}$$ These latter conditions can be given a simple geometrical interpretation. If one considers a closed curve in ${\mathbb{R}}^2$ and denote by $\cos \varphi (s)$ and $\sin \varphi (s)$ the components of the unit tangent, with respect to a fixed frame, as a function of arc–length, (\[eq:3.12\]) just says that the curve in question is closed. Moreover, the curvature of the curve is given by $$\kappa(s)=\frac{d \varphi}{ds}.
\label{eq:3.13}$$ Let’s denote by $C$ a closed curve in the plane, of length $2 \pi$, with positive curvature, and let $$H(C) \equiv -\frac{d^2}{ds^2} + \kappa^2
\label{eq:3.14}$$ acting on $L^2(C)$ with periodic boundary conditions. Then, (\[eq:3.4\]), and for that matter (\[eq:3.11\]), is equivalent to saying that the lowest eigenvalue of $H(C)$, $\lambda_1(C)$ say, is larger or equal to $1$, for any closed curve on the plane of length $2 \pi$. It is a simple fact to see that if $C$ is a circle of length $2 \pi$, the lowest eigenvalue of $H(C)$ is precisely $1$. Unfortunately we are far from proving the desired bound for general curves. It is relatively simple to show that the lowest eigenvalue of the Hamiltonian $H(C)$ is bounded below by $1/2$. To see this one first notes that the corresponding eigenfunction can be chosen to be positive. The quadratic form $$(f, H(C) f) = \int_0^{2 \pi} |f'(s)|^2 ds + 
\int _0^{2 \pi} \kappa^2(s) f(s)^2 ds$$ can be written as $$\int_0^{2 \pi} | \frac{d}{ds} (e^{i \varphi(s)} f(s))|^2 ds \ ,$$ which we have to minimize over non negative functions $f$ satisfying $\int_0^{2 \pi} f(s)^2 ds =1$. Expanding the function $e^{i\varphi(s)} f(s)$ into a Fourier series $$e^{i\varphi(s)} f(s) = \sum_{n=-\infty}^\infty c_n 
\frac{ e^{ins}}{\sqrt{2 \pi}} \ ,$$ we find that since $f(s) \geq 0$ $$|c_0|^2 \leq \frac{1}{2 \pi} (\int_0^{2 \pi} f(s) ds)^2 \ .$$ Moreover, since the functions $1/\sqrt{2 \pi}$ and $e^{i \varphi(s)}/\sqrt{2 \pi}$ are orthogonal in the innerproduct of $L^2([0,2 \pi])$ we find that $$|c_0|^2 + \frac{1}{2 \pi} (\int_0^{2 \pi}f(s) ds)^2 \leq 
\int_0^{2\pi} f(s)^2 ds =1 \ .$$ Thus, $$|c_0|^2 \leq \min\{\frac{1}{2 \pi} (\int_0^{2 \pi} f(s) ds)^2, 
1-\frac{1}{2 \pi} (\int_0^{2 \pi} f(s) ds)^2\} \leq 1/2 \ .$$ Since $\sum_n |c_n|^2 =1$ we learn that $$\sum_{n \not= 0} |c_n|^2 \geq \frac{1}{2} \ .$$ Clearly, $$(f, H(C) f) = \sum_{n=-\infty}^{\infty} n^2 |c_n|^2
\geq \sum_{n \not= 0}  |c_n|^2 \geq \frac{1}{2}  \ ,$$ hence $\lambda_1 (C) \ge 1/2$.

[*Remarks:*]{}

i\) A word of warning should be made at this point. In principle, the function $R$ defined from the eigenfunctions $u_1$ and $u_2$, via $\rho$ through equation (\[eq:3.8\]) above, must vanish at $s=0$ and $s=2 \pi$. For the [*curve problem*]{}, however, we drop this boundary condition. Thus, a priori the conjecture for the [*curve problem*]{} is stronger than the Lieb–Thirring conjecture for the two bound states, although we believe it amounts to the same.

ii\) The best bound to date on $L_{1,1}$ is the bound of Eden and Foias [@EdFo91] who proved, $$L_{1,1} \le \frac{2}{9} \sqrt{3} \approx 0.3849 \dots
\label{eq:ed-fo-bound}$$ Our bound $\lambda_1(C) \ge 1/2$ yields the bound $$L_{1,1} \le \frac{4}{9 \pi} \sqrt{6} \approx 0.3465 \dots,$$ which although better than (\[eq:ed-fo-bound\]), only applies to Schrödinger operators with two bound states. Just for comparison, the conjectured sharp value for $L_{1,1}$ is $4\sqrt{3}/(9 \pi) \approx 0,2450$.

iii\) In recent years several authors have obtained isoperimetric inequalities for the lowest eigenvalues of a variant of $H(C)$, and we give a short summary of the main results in the sequel. Consider the Schrödinger operator $$H_g(C) \equiv -\frac{d^2}{ds^2} + g \kappa^2
\label{eq:3.15}$$ defined on $L^2(C)$ with periodic boundary conditions. As before, $C$ denotes a closed curve in ${\mathbb{R}}^2$ with positive curvature $\kappa$, and length $2 \pi$. Here, $s$ denotes arclength. If $g<0$, the lowest eigenvalue of $H_g(C)$, say $\lambda_1(g,C)$ is uniquely maximized when $C$ is a circle [@DuEx95]. When $g=-1$, the second eigenvalue, $\lambda_2(-1,C)$ is uniquely maximized when $C$ is a circle [@HaLo98]. If $0,g \le 1/4$, $\lambda_1(g,C)$ is uniquely minimized when $C$ is a circle [@ExHaLo99]. It is an open problem to determine the curve $C$ that minimizes $\lambda_1(g,C)$ in the cases, $1/4 < g \le 1$, and $g<0, g \neq -1$. If $g > 1$ the circle is not a minimizer for $\lambda_1(g,C)$ (see, e.g., [@ExHaLo99; @Ha02] for more details on the subject).

To conclude this section we give an alternative interpretation of the minimization principle (\[eq:3.11\]) subject to the side constraints (\[eq:3.12\]). Interpret now $s$ as time (instead of arclength) and, given $R(s)$ and $\varphi(s)$ as before, define $$x(s) = R(s) \cos \varphi (s),$$ and $$y(s) = R(s) \sin \varphi (s).$$ Then, the minimization problem (\[eq:3.11\]), (\[eq:3.12\]) is equivalent to the following, $$\frac{\int_0^{2 \pi} \left( {\dot x}^2 + {\dot y}^2 \right) \, ds}
{\int_0^{2 \pi} \left( x^2 + y^2 \right) \, ds} \ge 1,
\label{eq:3.16}$$ where $x(s)$ and $y(s)$ are periodic, of period $2 \pi$ and satisfy the side constraints, $$\int_0^{2 \pi} \frac{x(s)}{\sqrt{x(s)^2+y(s)^2}} \, ds=
\int_0^{2 \pi} \frac{y(s)}{\sqrt{x(s)^2+y(s)^2}} \, ds= 0.
\label{eq:3.17}$$ Notice that (\[eq:3.16\]) certainly holds if one replaces the side constraints (\[eq:3.17\]) by $\int_0^{2 \pi} x(s) \, ds =0$ and $\int_0^{2 \pi} y(s) \, ds =0$, for then both functions $x(s)$ and $y(s)$ would be orthogonal to the constants and one would have $\int_0^{2 \pi} {\dot x}^2 \, ds \ge \int_0^{2 \pi} x(s)^2 \, ds$ and $\int_0^{2 \pi} {\dot y}^2 \, ds \ge \int_0^{2 \pi} y(s)^2 \, ds$, independently.

Appendix
========

To obtain inequality (\[eq:2.2\]) for $\gamma \neq 1$ we start from equation (\[eq:2.5\]) as before. Using Hölder’s inequality we get $$\lambda_1 \le 
\left(
\int_{-\infty}^{\infty} V^{\gamma+(1/2)} \, dx
\right)^{2/2\gamma+1}
\left(\int_{-\infty}^{\infty}u_1^{2 (2\gamma+1)/(2 \gamma-1)} \, dx
\right)^{(2\gamma-1)/(2\gamma+1)} 
- \int_{-\infty}^{\infty} (u_1')^2 \, dx.
\label{eq:a0}$$ We claim that if $\int_{-\infty}^{\infty} u_1^2 \, dx=1$, $$\int_{-\infty}^{\infty} {u_1'}^2 \, dx \ge c(\gamma) 
\left( \int_{-\infty}^{\infty} 
u_1^{2(2\gamma+1)/(2\gamma-1)} \,  dx \right)^{2 \gamma-1},
\label{eq:a1}$$ where $$c(\gamma)= 
\left[\sqrt{\frac{\pi}{2}} \frac{\gamma^{\gamma} \Gamma(\gamma+1/2)}
{\Gamma(\gamma+1)(\gamma-1/2)^{\gamma-1/2}}
\right]^2.$$ Using the claim and denoting $$A \equiv \left( \int_{-\infty}^{\infty} V^{\gamma + (1/2)}\, dx \right)^{2/2\gamma+1},$$ and $$Y \equiv  \left( \int_{-\infty}^{\infty} 
u_1^{2(2\gamma+1)/(2\gamma-1)} \, dx \right)^{2 \gamma-1},$$ we get $$\lambda_1 \le A Y^{1/(2\gamma+1)} - c(\gamma) Y.  
\label{eq:a2}$$ Maximizing the left side of (\[eq:a2\]) over $Y$ (for $\gamma> 1/2$), we get $$\lambda_1 \le \tilde c(\gamma) \left( \int_{-\infty}^{\infty} 
V^{\gamma + (1/2)}\, dx \right)^{1/\gamma},
\label{eq:a3}$$ where $$\tilde c (\gamma) = \frac{2 \gamma}{c(\gamma)^{1/(2 \gamma)}
(2\gamma+1)^{(2\gamma+1)/(2 \gamma)}}.$$ Hence, $$\lambda_1^{\gamma} \le L_{\gamma,1}^1 \int_{-\infty}^{\infty} 
V^{\gamma + (1/2)}\, dx.
\label{eq:a4}$$ To conclude we need only to prove the claim (\[eq:a1\]) whenever $\int_{-\infty}^{\infty}
u_1^2 \, dx=1$. Introducing the same change of variables as in Section 2, i.e., $$x \to s = \int_{-\infty}^{x} u_1^2 \, dy,$$ and $$w\equiv u_1^2,$$ the claim reduces to proving $$\frac{1}{4} \int_0^1 {\dot w}^2 \, ds \ge c(\gamma) \left(
\int_0^1 w^{2/2\gamma-1} \, ds \right)^{2 \gamma -1},
\label{eq:a5}$$ which follows from Sobolev’s inequality in one dimension.

Acknowledgements
================

It is a pleasure to thank the organizers of the Pan-American Advanced Studies Institute (PASI) on Partial Differential Equations, Inverse Problems and Non-Linear Analysis for their kind invitation to present these results. We thank Mark S. Ashbaugh, Evans Harrell and Elliott Lieb for useful discussions.

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[^1]: This work was supported in part by Fondecyt (Chile), Projects \# 102–0844 and \# 702–0844

[^2]: Work partially supported by NSF Grant DMS 03–00349
