# Commutation between integrating and taking the minimal eigenvalue

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.

• If the converse includes $\max \sigma(S(x)) \geq 0$, I don't think that holds, e.g., for $S = -I$. – Federico Poloni Oct 5 at 12:54
• 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 $u$ – Jom Oct 5 at 13:09
• 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. – Federico Poloni Oct 5 at 13:11
• $u$ can have any sign, so in your case the condition is $\int u \geq 0$, which is not valid ;) but yes your counterexample works – Jom Oct 5 at 13:17
• 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... – Jom Oct 5 at 13:30

[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.