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Phys-68 · physics · mathematics · Physics

Every extremizer of the sharp matrix Lieb–Thirring bound

Status OpenLLM status Open unverified

Statement

Definitions and mathematical statement

Statement recorded in the supplied source.
Fix $m\geq1$ and $1/2<\gamma<3/2$. Let $W:\mathbb R\to M_m(\mathbb C)$ be measurable, Hermitian, and nonnegative, with $\int_{\mathbb R}\operatorname{tr}(W^{\gamma+1/2})\,dx<\infty$. Define $H=-d^2/dx^2-W$ on $L^2(\mathbb R;\mathbb C^m)$, and put \[C_\gamma=\left(\frac{\gamma-1/2}{\gamma+1/2}\right)^{\gamma-1/2}\frac{\Gamma(\gamma+1)}{\sqrt\pi\,\Gamma(\gamma+3/2)}.\] Does equality in $\operatorname{Tr}(H_-^\gamma)\leq C_\gamma\int_{\mathbb R}\operatorname{tr}(W^{\gamma+1/2})\,dx$ hold exactly when a constant unitary change of basis makes $W(x)$ a diagonal collection of zero channels and channels \[(r+1)a_j^2\operatorname{sech}^2(r a_j(x-x_j)),\qquad r=(\gamma-1/2)^{-1},\quad a_j>0,\quad x_j\in\mathbb R?\] Each nonzero scalar channel is required to have exactly one negative eigenvalue, equal to $-a_j^2$.

Short English statement

Fix $m\geq1$ and $1/2<\gamma<3/2$. Let $W:\mathbb R\to M_m(\mathbb C)$ be measurable, Hermitian, and nonnegative, with $\int_{\mathbb R}\operatorname{tr}(W^{\gamma+1/2})\,dx<\infty$. Define $H=-d^2/dx^2-W$ on $L^2(\mathbb R;\mathbb C^m)$, and put \[C_\gamma=\left(\frac{\gamma-1/2}{\gamma+1/2}\right)^{\gamma-1/2}\frac{\Gamma(\gamma+1)}{\sqrt\pi\,\Gamma(\gamma+3/2)}.\] Does equality in $\operatorname{Tr}(H_-^\gamma)\leq C_\gamma\int_{\mathbb R}\operatorname{tr}(W^{\gamma+1/2})\,dx$ hold exactly when a constant unitary change of basis makes $W(x)$ a diagonal collection of zero channels and channels \[(r+1)a_j^2\operatorname{sech}^2(r a_j(x-x_j)),\qquad r=(\gamma-1/2)^{-1},\quad a_j>0,\quad x_j\in\mathbb R?\] Each nonzero scalar channel is required to have exactly one negative eigenvalue, equal to $-a_j^2$.

Sources

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Phys-68-A1 · Literature report · LLM-generated · unknown ·

OpenAI claim summary for Phys-68

All equality cases of that matrix inequality are claimed to be channels in a constant unitary basis, each either zero or a scalar single-state soliton. The question retains these same parameters and conventions. Within these assumptions, this is a claimed affirmative answer to the stated question.

Every extremizer of the sharp matrix Lieb–Thirring bound

OpenAI's claim

OpenAI presents a proof in the selected manuscript. All equality cases of that matrix inequality are claimed to be channels in a constant unitary basis, each either zero or a scalar single-state soliton.

Mathematical statement of the claimed result

Result 1.
Fix a finite matrix size $m\ge1$, $1/2<\gamma<3/2$, and $p=\gamma+1/2$. For measurable Hermitian $W(x)\ge0$ with $\int_{\mathbb R}\operatorname{tr}(W^p)<\infty$, form $H_W=-d^2/dx^2-W$ by its $H^1$ quadratic form. The source claims \[ \sum_{\lambda_j(H_W)<0}|\lambda_j|^\gamma \le C_\gamma\int_{\mathbb R}\operatorname{tr}(W^p),\quad C_\gamma=\left(\frac{\gamma-1/2}{\gamma+1/2}\right)^{\gamma-1/2} \frac{\Gamma(\gamma+1)}{\sqrt\pi\Gamma(\gamma+3/2)}. \] Writing $r=(\gamma-1/2)^{-1}$, equality occurs exactly for a constant unitary $U$, $0\le k\le m$, $a_j>0$, $x_j\in\mathbb R$, and \[ W(x)=U\operatorname{diag}(w_1(x),\ldots,w_k(x),0,\ldots,0)U^*,\quad w_j(x)=(r+1)a_j^2\operatorname{sech}^2(ra_j(x-x_j)) \] almost everywhere. Each nonzero channel has just the negative eigenvalue $-a_j^2$, and $k=0$ means the zero potential.

Outline of the argument

The action estimate is refined to keep each convexity and trace defect. Vanishing defects force pointwise matrix relations and fixed invariant channels, after which the scalar differential equations identify the independent soliton scales and centers.

Formal verification

For family 262, the documented result is the sharp scalar one-dimensional inequality in the intermediate exponent range, including all negative eigenvalues. The matrix and full equality-classification companions are not selected. No Comparator is assigned to the selected manuscript here; the family page selects other papers or supporting statements. This site has not executed the Lean code. The scope descriptions do not certify the remaining manuscript arguments.

Status caveat

OpenAI attributes these manuscripts to work by an internal model. Its release cautions that arguments without formalization can contain mistakes. This catalogue has not independently checked the proofs; the labels report the mathematical claims attributed to OpenAI.

Sources

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