# The weak limit of a sequence of argmax functions

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.

• $A$ 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 $\sigma_n$ ? Commented Nov 24, 2023 at 12:02
• 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. Commented Nov 24, 2023 at 12:18

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, $$g(0,\sigma) < \max_{a \in A}g(a,\sigma)$$.
It seems you want the theory of $$\Gamma$$-convergence (usually stated in terms of minimisers rather than maximisers).