The operator $\min$ gets used both in combination with functions and sets, i.e. $\min f(x)$ is equal to the smallest value the function $f$ attains, and $\min V$, where $V$ some set, is equal to the element of $V$ with the smallest value among all elements.
Does the same hold for $\arg \min$? Obviously, $\arg \min f(x)$ is equal to the $x$-value(s) which correspond(s) with the minimum of the function $f$. More importantly, consider the following:
$$\textbf{x}_{\textrm{LN}} = \textrm{arg min}_{\textbf{x}}\{ \|\textbf{x}\|^2_{2} \ | \ \textbf{x} \in X \}$$
where $X$ some set. In my case,
$$X=\{\textbf{x}\in\mathbb{R}^{n} \ | \ A\textbf{x}=\textbf{b}\}$$
Does by the above notation $\textbf{x}_{\textrm{LN}}$ equate to the vector $\textbf{x}$ with the smallest (squared) Euclidean norm which lies in the set X? That is what I would like it to be. If not, what notation would be correct?
Thanks in advance for your help,
Anil