0
$\begingroup$

I am currently working on a problem related to argmax functions in the context of operations research. I am trying to figure out if the weak limit of a sequence of argmax functions is again a argmax function.

Assume the following conditions:

  • $I = [0,1]$ and $A$ is a compact subset of $\mathbf{R}^n.$
  • $U_i:A \times \mathbf{R} \to \mathbf{R}$ is a continuous function for all $i \in I$.
  • $\{\sigma_n \}$ is a sequence of $\mathbf{R},$ and $g_n: I \to A$ is a Lebesgue measurable function such that $U_i(g_n(i), \sigma_n) = \mathrm{max}_{a \in A}U_i(a,\sigma_n)$ a.e. $i.$

Here, If $\sigma_n \to \sigma \in \mathbf{R}$ and $g_n \to g\in L^1([0,1])$ in the weak topology of $L^1([0,1]),$ then does it holds that $U_i(g(i),\sigma) = \mathrm{max}_{a \in A}U_i(a,\sigma)$ a.e. $i$?

If $g_n$ strongly convege to $g$, then the conclusion follows from the continuity of $U_i$ because a subsequences pointwise converge to $g$. I am wondering if weak convergence is sufficient.

I would greatly appreciate insights or suggestions.

Thank you.

$\endgroup$
2
  • $\begingroup$ $A$ is a compact subset of what? And what does $\arg\max$ mean when the maximum is multiply achieved? Should not you write simply $U_i(g(i),\sigma) = \max_{a \in A} U_i(a,\sigma)$ and the same for $\sigma_n$ ? $\endgroup$ Commented Nov 24, 2023 at 12:02
  • $\begingroup$ Sorry for many mistakes. $A$ is a compact subset of $\mathbf{R}^n.$ For $\mathrm{argmax},$ you are right; $U_i(g(i), \sigma) = \mathrm{max}_{a\in A} U_i(a, \sigma)$ and the same for $\sigma_n.$ I will edit may post. $\endgroup$
    – Saito
    Commented Nov 24, 2023 at 12:18

2 Answers 2

2
$\begingroup$

The answer is negative : let $A = [-1,1]$, $U_i(a,s) = a^2$ for every $i \in [0,1]$ and $(a,s) \in A \times \mathbb{R}$, and $(\sigma_n)_n \to \sigma$ be any convergent sequence of real numbers.

For every $i \in [0,1]$ and $s \in \mathbb{R}$, the maximum of the function $a \mapsto U_i(a,s)$ is achieved only at $\pm 1$. Thus, one may choose Rademacher functions for the $g_n$, namely $g_n(s) = (-1)^{\lfloor 2^n x \rfloor}$ for all $s \in \mathbb{R}$.

The sequence $(g_n)_n$ thus defined converges weakly to the null function. Yet, $g(0,\sigma) < \max_{a \in A}g(a,\sigma)$.

$\endgroup$
1
$\begingroup$

It seems you want the theory of $\Gamma$-convergence (usually stated in terms of minimisers rather than maximisers).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.