What is the exact return of argmax or argmin operator?

Question

In many journals, I saw $\text{argmin}$ and $\text{argmax}$ used as follows: $$ x^* = \underset{i}{\text{argmin}}~f(x_i).\tag{1}\label{1}\\ $$

I think that Eq. \eqref{1} is incorrect, so it should be changed into \begin{align} i^* &= \underset{i}{\text{argmin}}~f(x_i),\tag{2}\label{2}\\ x^* &= x_{i^*}.\tag{3}\label{3} \end{align}

Is it okay if I use Eq. \eqref{1} instead of Eqs. \eqref{2} and \eqref{3}?

Answer 1

I think $$\arg \min _{x \in X} f(x)$$

and $$\arg \min_{i \in Z^+} f(x_i)$$ should return different object and hence writing $(1)$ is not correct.

Also, actually $$\arg \min_{x \in S} f(x) = \{ x \in S \mid f(y) \ge f(x) \; \forall y \in S\}$$

is a set.

Answer 2

Perhaps an equally succint yet less confusing way to express this is:

$$x^* = \underset{x_i}{\text{argmin}}~f(x_i).$$

But ideally, it would be good to have a set that you're enumerating over rather than just some arbitrary expression:

$$X = \{x_1, \ldots, x_n\},$$

$$x^* = \underset{x \in X}{\text{argmin}}~f(x).$$

This way, you actually have a collection of objects to check rather than specifying an arbitrary expression which can't actually be looped over.