The ovals problem of Benguria and Loss

Bend a closed loop of length $2\pi$ and use its squared curvature as the potential of a Schrödinger operator: can the ground-state energy ever drop below 1, the value for a circle? Benguria and Loss conjectured that it cannot, which would give a sharp Lieb–Thirring inequality. The best lower bound has climbed from 1/4 to about 0.96, and the conjecture is still open.

Problem

Let $\gamma$ be a simple closed convex curve in the plane of length $2\pi$, parametrized by arc length $s$, with curvature $\kappa(s)$. Let $\lambda_0(\gamma)$ be the lowest eigenvalue of the periodic operator

$$H_\gamma = -\frac{d^2}{ds^2} + \kappa(s)^2 \quad \text{on } [0, 2\pi),$$

and let $C = \inf_\gamma \lambda_0(\gamma)$. The unit circle has $\kappa \equiv 1$ and $\lambda_0 = 1$, so $C \le 1$. Obtain upper and lower bounds for $C$ that are as strong as possible; the ovals conjecture asserts $C = 1$.

Background

Benguria and Loss (2004) showed that $C$ determines the sharp constant in a one-dimensional Lieb–Thirring inequality for Schrödinger operators with two bound states, and found a two-parameter family of ovals, containing the circle and degenerating to a segment traversed twice, all with $\lambda_0 = 1$. The quantity had appeared slightly earlier in Burchard and Thomas's work on a dynamical Euler elastica, where they proved $C \ge 1/4$, a bound that is sharp for open curves. Bernstein and Mettler (2015) characterized the Benguria–Loss ovals as the absolute minima of two related geometric functionals and explained their symmetries through projective differential geometry.

What is known

• Upper bound: $C \le 1$, attained by the circle and the Benguria–Loss ovals.
• Lower bounds: $1/4$ (Burchard and Thomas); $1/2$ (Benguria and Loss, 2004); $0.6085$ (Linde, 2006); $0.81$ (Linde, 2025); and $\tfrac{\sqrt{\pi}}{2}\left(\tfrac{\Gamma(7/6)}{\Gamma(5/3)}\right)^3 \approx 0.9618$ (Suragan, September 2026 preprint), which holds for closed curves in any dimension.
• Numerics: Georgiev, Gómez-Serrano, Tao and Wagner (2025) searched over spline curves and test functions and recovered the circle, but found nothing below 1.

What would count as progress

• A proof of the conjecture, $C = 1$, or a rigorous lower bound above $0.9618$. An independent check of the September 2026 argument would also be valuable.
• A counterexample: a closed convex curve of length $2\pi$ with $\lambda_0 < 1$. Given the curve, $\lambda_0$ can be bounded above rigorously by a test function in the Rayleigh quotient $\int (\phi'^2 + \kappa^2 \phi^2)\, ds / \int \phi^2\, ds$, so a counterexample comes with a certificate checkable in seconds.

References

1. R. D. Benguria and M. Loss, "Connection between the Lieb–Thirring conjecture for Schrödinger operators and an isoperimetric problem for ovals on the plane", Contemporary Mathematics 362 (2004), 53–61. doi:10.1090/conm/362/06604
2. H. Linde, "A lower bound for the ground state energy of a Schrödinger operator on a loop", Proceedings of the AMS 134 (2006), 3629–3635. doi:10.1090/s0002-9939-06-08483-8
3. J. Bernstein and T. Mettler, "One-dimensional projective structures, convex curves and the ovals of Benguria & Loss", Communications in Mathematical Physics 336 (2015), 933–952. doi:10.1007/s00220-014-2275-7
4. H. Linde, "An improved bound for the ground state of a Schrödinger operator on a loop" (2025). arXiv:2504.20229
5. D. Suragan, "An improved bound for the ovals problem" (2026). arXiv:2609.10775

Source

Imported from AlphaEvolve Repository of Problems, problem 19, curated by Google DeepMind. Open according to the source as of 2026-10-20. The statement above is written in our own words; see the source for the original wording, history and discussion.