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Let $A(t)$ be a symmetric $n\times n$ matrix that continuously depend on $t\in [0,1]$. Let $\lambda_1(t)$ stand for the smallest eigenvalue for $A(t)$.

Question. Does there exist a Lebesgue measurable vector $v_1(t)$ on $(0,1)$ that is not zero for almost every t and satisfies $A(t)v_1(t)=\lambda_1(t)v_1(t)$ almost everywhere on $(0,1)$ such that $$ v_1 \in L^{\infty}((0,1);\mathbb R^n)?$$

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    $\begingroup$ I think the Courant minimax principle will probably help? $\endgroup$
    – Daniel Shapero
    Commented Oct 10, 2023 at 4:07
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    $\begingroup$ Yes, $\lambda_1$ is continuous (see for example Kato's book), and then it's clear, though probably quite tedious to prove formally, that $v_1$ can be chosen as a measurable function. Being in $L^{\infty}$ is no extra requirement since we can normalize. $\endgroup$
    – Christian Remling
    Commented Oct 10, 2023 at 14:11

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It should not be hard to prove - e.g. by some minmax-characterization - that $\lambda_1$ is measurable. (As already remarked in the comments by Christian Remling, $\lambda_1$ is actually continuous, but measurability of $\lambda_1$ is sufficient for the following argument.)

Then the existence of a measurable selection $v_1$ of the zeroes (e.g. on the unit sphere) of the superposition operator generated by the Caratheodory function $f(t,v)=A(t)v-\lambda_1(t)v$ is a trivial special case of a Filippov type implicit function theorem. See, for instance, Theorem 3' in https://core.ac.uk/download/pdf/81993694.pdf.

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