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Let $S = (f_{ij})_{ij}$ be a $n \times n$ real symmetric matrix, with functions $f_{ij} \in L^1(\mathbb{R}^d,\mathbb{R})$ in it. We define $\left(\int u S \right)_{ij} = \int u S_{ij}$ as the elementwise matrix of integrals. What is a necessary and equivalent condition to \begin{align*} \min \sigma \left( \int u S \right)\leq 0 \text{ for any $u \in L^{\infty}(\mathbb{R}^d,\mathbb{R})$} \end{align*} in terms of $S$ only (without talking about $u$'s) ? Actually it implies \begin{align*} \min \sigma(S(x)) \leq 0 \leq \max \sigma(S(x)) \text{ a.e on } \mathbb{R}^d \end{align*} by taking sequences $u_n$ converging to $\delta_x$, but the converse is not true thanks to Frederico's counterexample. This is a question raised in a quantum mechanics problem.

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  • $\begingroup$ If the converse includes $\max \sigma(S(x)) \geq 0$, I don't think that holds, e.g., for $S = -I$. $\endgroup$ – Federico Poloni Oct 5 at 12:54
  • $\begingroup$ The matrix $S = - I$ is not possible since it does not respect the condition $\min \sigma \left( \int u S \right)\leq 0$ for all $u$ $\endgroup$ – Jom Oct 5 at 13:09
  • $\begingroup$ I thought $u$ was a non-negative weight vector; so do you confirm that it can have arbitrary signs? The counterexample in my answer should still work anyway, if I am not mistaken. $\endgroup$ – Federico Poloni Oct 5 at 13:11
  • $\begingroup$ $u$ can have any sign, so in your case the condition is $\int u \geq 0$, which is not valid ;) but yes your counterexample works $\endgroup$ – Jom Oct 5 at 13:17
  • $\begingroup$ Thus I'm trying to find an equivalent relation for $\min \sigma \left( \int u S \right)\leq 0$, just in terms of $S$ and without needing $u$'s, if one has an idea... $\endgroup$ – Jom Oct 5 at 13:30
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[This is an answer to the first version of the question, which asked if $\min \sigma(S(x)) \leq 0 \leq \max \sigma(S(x)) \text{ a.e on } \mathbb{R}^d$ implies $\min \sigma \left( \int u S \right)\leq 0 \text{ for any $u \in L^{\infty}(\mathbb{R}^d,\mathbb{R})$}$]

I think the answer is no. A counterexample is simpler to construct if you first suppose that the domain is $[0,1]$ rather than $\mathbb{R}^d$; then you can take $S(x) \in \mathbb{R}^{n\times n}$ diagonal, with $$ S(x)_{ii} = \begin{cases} -1 & x\in [\frac{i-1}n, \frac{i}n),\\ 1 & \text{otherwise}. \end{cases} $$ Then, integrating with weight $u(x) \equiv 1$ gives $\int_{[0,1]} S(x)_{ii} = \frac{n-2}n \geq 0$, but each $S(x)$ has an eigenvalue $-1$.

Now one can make a change of variable to transform the domain into $\mathbb{R}$, introducing a weight $u(x)$ in the process. Similarly, one can extend the function to $[0,1]^d$ by making it constant on the last $d-1$ dimensions, and then transform the domain into $\mathbb{R}^d$ with a change of variable.

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