Can $\text{ arg}$ be thought of as operator? Forgive me if the question is to vague.
The argument, denoted by $\text{arg}$, is a commonly used notation.
I am specifically interested in the following use of $\text{arg}$:
\begin{align}
a=\text{ arg} \min_x f(x) \tag{*}
\end{align}
My question is: can $\text{arg}$ in context of $\text{(*)}$ be thought as an operator?
What properties would this operator have?
 A: No, $\text{arg}$ is not an operator. "arg min" and "arg max" are operators, at least in an informal sense, and they apply to functions. To avoid the confusion you're puzzling over, they're often written as single tokens: $\operatorname{argmin}$ and $\operatorname{argmax}$ respectively. Each is well-defined only if the function achieves a min (respectively max), and then, at only a single argument. Your $\text{arg}$ "operator" would apply to numbers $\min_x f(x)$ and $\max_x f(x)$, and to yield the value meant by $\operatorname{argmin}_x f(x)$, this "operator" would have to know $f$ in order to apply $f^{-1}$ to the min or max.
A: If you think it might be an operator, the first thing you should do is define what you think it should be as a mapping.  In other words, define a domain and a co-domain, then define the action of the map on elements of the domain.  Otherwise this is very ambiguous.  For instance, in your example, you want a mapping that takes a function, finds its minimum value, and returns the argument of the function at which the function attains its minimum.  Well, the first issue here is that not all functions have a unique minimum value, and even if they do, that value might not be attained.  So, your operator is not well-defined for a large class of functions.  
You could restrict the domain of your mapping so that it is well defined, for instance let $D$ be the set of continuous, strictly convex functions on $[0,1]$.  Then, you can define the mapping $\text{argmin}:D\rightarrow[0,1]$ in the obvious way, and it'll be well-defined and you can proceed to study its properties.  The major problem here is that I can't think of a good vector space for which $\text{argmin}$ is well-defined, and that's kind of what you would need in order to start thinking of $\text{argmin}$ as an operator in the usual sense.
