# Is there a name for the “projection” of a function under argmax?

Let $A$ and $B$ be closed sets (subsets of $\mathbb{R}^m$ and $\mathbb{R}^n$, say), and let $f : A \times B \rightarrow \mathbb{R}$ be a continuous function.

Consider the function $g : A \rightarrow B$ defined by

$g(x) = \underset{y \in B}{\operatorname{arg}\max} f(x,y)$

assuming some tie-breaking strategy for $f(x,y_1) = f(x,y_2)$. Clearly, $g$ may have discontinuities (but perhaps only countably many?).

If $A$ and $B$ are intervals of $\mathbb{R}$, $g$ corresponds to looking down the $y$-axis at the 3D graph of $f$, and marking the points on the "skyline".

Does this correspond to some known operation that has a established name?

My motivation is merely that I'm thinking about using this for a fun side project, so I'd like to know if it has a name and any known interesting properties beyond piecewise continuity(?). I've tagged this recreational accordingly; if it's an inappropriate question for MathOverflow, I apologize.

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I'm baffled by the notation: what is "arg" here? Generally that is used for the argument of a complex number, but that can't be the case here. –  Robin Chapman Jul 1 '10 at 6:18
arg here means the argument (i.e the y) that achieves the max value. so g(x) is the value of y that maximizes f(x,y) –  Suresh Venkat Jul 1 '10 at 6:28
I think it's the "argument that maximizes"; that is, fix a value of x and ask which y leads to the largest value of f(x,y). –  Michael Burge Jul 1 '10 at 6:31
I believe this is standard notation in certain contexts (en.wikipedia.org/wiki/Arg_max , planetmath.org/encyclopedia/ArgMin.html). Thanks to Suresh and Michael Burge for restating my definition very clearly in terms that do not use arg max. –  Rahul Jul 1 '10 at 7:39
A lot will depend on your tie-breaking rule. If f is constant, you will have no problem finding nowhere continuous and even nonmeasurable functions g. –  Michael Greinecker Jul 1 '10 at 9:11

In the special case where the correspondence is single valued (for example if $f(x,\cdot)$ is always strictly concave), you actually get a continuous function, however.
It might be useful to know that the value function, the function $v:X\to\mathbb{R}$ given by $v(x)=f(x,y^* )$ with $y^*$ being in the argmax of $f(x,y)$, is always continuous.