How to write "the parameter maximizing the maximum of the maximum value of two functions continuous in the domain of maximization" 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 $2.25&gt;2$ 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.
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
$$\chi_S(x)=\begin{cases}
1&amp; \text{if }x\in S,\\
0 &amp; \text{if }x\notin S.
\end{cases}$$
Thus,
$$\bigg[\infty\cdot\chi_S+(-\infty)\cdot\chi_{X\setminus S}\bigg](x)=\begin{cases}
\infty&amp; \text{if }x\in S,\\
-\infty &amp; \text{if }x\notin S,
\end{cases}$$
and thus for any function $f:X\to\mathbb{R}$,
$$\min(f,\infty\cdot\chi_S+(-\infty)\cdot\chi_{X\setminus S})(x)=\begin{cases}
\min\{f(x),\infty\}=f(x)&amp; \text{if }x\in S,\\
\min\{f(x),-\infty\}=-\infty &amp; \text{if }x\notin S.
\end{cases}$$
Given another subset $T\subset X$ and a function $g:X\to\mathbb{R}$, we therefore have that
$$\max\bigg\{\!\min(f,\infty\cdot\chi_S+(-\infty)\cdot\chi_{X\setminus S}),\,\min(g,\infty\cdot\chi_T+(-\infty)\cdot\chi_{X\setminus T})\!\bigg\}(x)=\begin{cases}
\max\{f(x),g(x)\} &amp; \text{if }x\in S\text{ and }x\in T,\\
\max\{f(x),-\infty\}=f(x) &amp; \text{if }x\in S\text{ and }x\notin T,\\
\max\{-\infty,g(x)\}=g(x)&amp; \text{if }x\notin S\text{ and }x\in T,\\
\max\{-\infty,-\infty\}=-\infty&amp; \text{if }x\notin S\text{ and }x\notin T.\end{cases}$$
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 $T$.
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
$$
\max(f(x_0), g(x_0)) = \max_{x \leq \min(X_f, X_g)}\max(f(x), g(x)).
$$