Simplify $f\Big(\text{argmax}_{y\in \mathcal{Y}} \int_{\mathcal{X}} f(y,x) g(x) dx, x\Big)$

Question

Take:

• a function $f:\mathcal{Y}\times \mathcal{X} \rightarrow \mathbb{R}$, where $\mathcal{Y}$ is finite

• a function $g: \mathcal{X} \rightarrow \mathbb{R}$.

Fix $x\in \mathcal{X}$ and consider $$ (1)\quad f\Big(\text{argmax}_{y\in \mathcal{Y}} \int_{\mathcal{X}} f(y,t) g(t) dt,\text{ } x\Big) $$

Question: Is there a way to simplify (1) by bringing the maximisation problem and/or integral in front of $f(\dots)$?

Comments: Consider $$ f\Big(\text{argmax}_{y\in \mathcal{Y}} f(y,x) , \text{ } x\Big) $$ This is equal to $$ \max_{y\in \mathcal{Y}} f(y,x) $$ I wonder if something similar applies to (1).

Answer 1

Perhaps the analog is: $$ \int_{\mathcal X}f\Big(\operatorname{argmax}_{y\in \mathcal{Y}} \int_{\mathcal{X}} f(y,t) g(t) dt,\; x\Big)\;dx = \max_{y\in \mathcal{Y}} \int_{\mathcal{X}} f(y,t) g(t) dt $$