One-bound-state optimality in one dimension for $1/2 < \gamma < 3/2$
In plain words
In one dimension the conjecture says the worst-case potential well, the one with the most binding for its size, has only a single bound state when the exponent $\gamma$ is between $1/2$ and $3/2$. Prove it.
Precise statement
For $-\mathrm{d}^2/\mathrm{d}x^2 + V$ on $L^2(R)$, with $L_{\gamma,1}$ the sharp constant in $\operatorname{sum}_j \left|E_j\right|^\gamma \le L_{\gamma,1} \operatorname{int} V_-^{(\gamma + 1/2)} dx$, prove $L_{\gamma,1} = L^1_{\gamma,1}$ for $1/2 < \gamma < 3/2$, where $L^1_{\gamma,1}$ is the sharp constant over potentials with exactly one negative eigenvalue. Known endpoints: equality at $\gamma = 1/2$ (Hundertmark, Lieb, Thomas 1998) and $L_{\gamma,1} = L^{\mathrm{cl}}_{\gamma,1}$ for $\gamma \ge 3/2$.
What would settle it
A proof that multi-bound-state potentials never beat the single-bound-state value for $1/2 < \gamma < 3/2$, or a numerical-plus-rigorous counterexample.
Status in the literature
The 2021 Frank-Gontier-Lewin disproof applies in $d = 1$ only for $\gamma > 3/2$, so the conjecture remains consistent with current results; at $\gamma = 1$ the best bound is $L_{1,1}/L^{\mathrm{cl}}_{1,1} \le 1.456$ against the conjectured $2/\sqrt{3}$, about 1.155 (Frank, Hundertmark, Jex, Nam 2021).