Is $\underset{i\in\mathcal{I}}{\text{arg max}\,\,} f_i(x)$ a notation abuse? Let $\mathcal{I} = \{1,2,\ldots,n\}$ be the index set of all my functions (so I have $n$ many functions). For any $x \in \mathbb{R}$, find $i$ that solves:
$$
\underset{i\in\mathcal{I}}{\text{arg max}\,\,} f_i(x)
$$
So, $i$ is not the argument of function $f_i$. Instead, $i$ is the index of function $f_i$. E.g. we have functions $f_1, f_2, \ldots, f_n$, and our goal is to find the index of the function that achieves maximum value when given input $x$.
But my problem is that $i$ is really not an "argument". Or is it? I think "arg max" implis that I'm returning the argument of a function that maximizes it.
But I'm clearly not returning the argument. In fact, the argument is constant at $x$. Instead, I'm returning the index of a function that is maximum for the input $x$.
Therefore, Question 1: should I do it this way, instead?
$$
\underset{i\in\mathcal{I}}{\text{function index max}\,\,} f_i(x)
$$
 A: $\operatorname{argmax}_i f_i(x)$ is no more an abuse of notation than $\operatorname{argmax}_x g(x)$. (I'm omitting the domain of the variable for simplicity; this should cause no confusion)
The point is that neither $f_i(x)$ nor $g(x)$ are functions, or even formulas expressing functions: they are formulas for numbers. For example, $f_i(x)$ is a formula expressing a number that depends on the variables $f$, $i$, and $x$.
(for simplicity of language, I'm assuming all functions here are number-valued; this is by no means essential)
There isn't any fundamental difference between the presence of $i$ in $f_i(x)$ and the presence of $x$ in $g(x)$; in both cases, the subscript on $\operatorname{argmax}$ means to abstract its given formula as a function in the indicated variable.
For example, $\operatorname{argmax}_x g(x)$ means $\operatorname{argmax}(g)$, where the where the latter use of $\operatorname{argmax}$ is meant as an operator on functions.
Similarly, if we define a family of functions $h_x$ by $h_x(y) = f_y(x)$, then $\operatorname{argmax}_i f_i(x)$ means the same thing as $\operatorname{argmax}(h_x)$
