The weak limit of a sequence of argmax functions

Question

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.

Answer 1

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, $g(0,\sigma) < \max_{a \in A}g(a,\sigma)$.

Answer 2

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