How to write "the parameter maximizing the maximum of the maximum value of two functions continuous in the domain of maximization"

Question

Say you have $f(x),g(x)$ continuous where they need to be and you want to express the following:

Give me the biggest value of $f$ for $x \leq X_f$ , give me the biggest value of $g$ for $x \leq X_g$, get the larger of the two, and return the $x$ that maximized it.

Like say $f=x$ and $g=x^2$ , and $X_f = 2$, $X_g = 1.5$, then the biggest $f$ on that interval is $f(w_f)=2$ and the biggest $g$ on that interval is $g(w_g)=2.25$, so since $2.25>2$ I want to return $w_g=1.5$.

What is the correct way to express this?

Edit: If you are wondering about the motivation, suppose $f$ and $g$ are two utility functions that are continuous for $X \leq \max(X_f,X_g)$. You know that you can only input $x \leq X_f$ into $f$ and $x \leq X_g$ into $g$ and you want to find out which one of $f$ and $g$ is the biggest over those domains and get the $x$ where the max of the max of those two functions occurs so you can plug it into your opponent's utility function.

Answer 1

I think it would be very difficult to express that in a single expression with no preliminary explanation. Here is a way of doing it (that is in my opinion, terrible, compared to just explaining what you mean / introducing auxiliary notation): $$\operatorname{arg\,max}\mathopen{\bigg[}\max\mathopen{\bigg\{}\!\min(f,\infty\cdot\chi_{(-\infty,X_f]}+(-\infty)\cdot\chi_{(X_f,\infty)}),\min(g,\infty\cdot\chi_{(-\infty,X_g]}+(-\infty)\cdot\chi_{(X_g,\infty)})\mathclose{\bigg\}}\mathclose{\bigg]}$$ Given a set $X$ and a subset $S\subset X$, the function $\chi_S$ (known as the indicator function, or characteristic function, of $S$) is defined as $$\chi_S(x)=\begin{cases} 1& \text{if }x\in S,\\ 0 & \text{if }x\notin S. \end{cases}$$ Thus, $$\bigg[\infty\cdot\chi_S+(-\infty)\cdot\chi_{X\setminus S}\bigg](x)=\begin{cases} \infty& \text{if }x\in S,\\ -\infty & \text{if }x\notin S, \end{cases}$$ and thus for any function $f:X\to\mathbb{R}$, $$\min(f,\infty\cdot\chi_S+(-\infty)\cdot\chi_{X\setminus S})(x)=\begin{cases} \min\{f(x),\infty\}=f(x)& \text{if }x\in S,\\ \min\{f(x),-\infty\}=-\infty & \text{if }x\notin S. \end{cases}$$ Given another subset $T\subset X$ and a function $g:X\to\mathbb{R}$, we therefore have that $$\max\bigg\{\!\min(f,\infty\cdot\chi_S+(-\infty)\cdot\chi_{X\setminus S}),\,\min(g,\infty\cdot\chi_T+(-\infty)\cdot\chi_{X\setminus T})\!\bigg\}(x)=\begin{cases} \max\{f(x),g(x)\} & \text{if }x\in S\text{ and }x\in T,\\ \max\{f(x),-\infty\}=f(x) & \text{if }x\in S\text{ and }x\notin T,\\ \max\{-\infty,g(x)\}=g(x)& \text{if }x\notin S\text{ and }x\in T,\\ \max\{-\infty,-\infty\}=-\infty& \text{if }x\notin S\text{ and }x\notin T.\end{cases}$$ If $x\in X$ maximizes the above function (i.e. it is an element of the $\operatorname{arg\,max}$ of it), then regardless of which of the four cases holds for that $x$, it is necessarily either an $x$ that maximizes $f(x)$ on $S$, or an $x$ that maximizes $g(x)$ on $T$.

Answer 2

As I interpret your question, one of the bounds $X_f$ and $X_g$ is redundant because while you vary $x$, you have to observe both inequalities $x \leq X_f$ and $x \leq X_g$ at the same time. Hence you can discard the higher bound.

Find $x_0 \leq \min(X_f, X_g)$ such that $$ \max(f(x_0), g(x_0)) = \max_{x \leq \min(X_f, X_g)}\max(f(x), g(x)). $$

Answer 3

A. If you are satisfied with having returned just one of the maximizing $x^*$'s (and I think you are), then this does the job:

$$x^*=\min\underset{x\le\max\{X_f,X_g\}}{\operatorname{\arg\max}}\max\left\{f\left(\min\{x,X_f\}\right),g\left(\min\{x,X_g\}\right)\right\}.$$

(Note that "$\arg\max$" returns a set. Also note that the first "$\min$" is essential and is not just an arbitrary way to extract one value of $x^*$ out of a set of $x^*$'s.)

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B. It is probably easier to define

$$h(x)=\max\left[\left\{f(y)~\left|~\forall y:x=y\le X_f\right.\right\}\cup\left\{g(y)~\left|~\forall y:x=y\le X_g\right.\right\}\right]$$ for all $x\le\max\left\{X_f,X_g\right\}$ first. Then you can use $$x^*\in\underset{{x\le\max\{X_f,X_g\}}}{\operatorname{\arg\max}}h(x),$$ which gives you all maximizing $x^*$'s (as a bonus).

Or, in one go:

$$x^*\in\underset{{x\le\max\{X_f,X_g\}}}{\operatorname{\arg\max}}\max\left[\left\{f(y)~\left|~\forall y:x=y\le X_f\right.\right\}\cup\left\{g(y)~\left|~\forall y:x=y\le X_g\right.\right\}\right].$$