Is it possible to express this property in a compact way?

Question

If $f_{a}: \mathbb{R} \to \mathbb{R}$ for all $a \in \mathbb{R}$, and if there is some $x \in \mathbb{R}$ such that $f_{a}(x) \leq f_{a}(y)$ for all $y \in \mathbb{R}$ and all $a \in \mathbb{R}$, I am after a compact way to express this property of $x$, preferably in terms of argmax or argmin.

Taking argmin$_{y, a}f_{a}(y)$ certainly gives an undesired result (for the unwanted additional argument).

Answer 1

I would write something like:

There exists some $x\in \Bbb{R}$ that reaches the minimum of $f_a$ uniformly in $a$, that is, $f_a(x) \leqslant f_a(y)$ for all $y \in \Bbb{R}$ and all $a \in \Bbb{R}$.

Answer 2

The difficulty is not only does $f_a$ have an $\operatorname{argmin}$, but that $\operatorname{argmin}$ is $a$-independent. You could write $\{x\}=\{\operatorname{argmin}_{y\in \Bbb R}f_a(y)|a\in\Bbb R\}$.