If $f,g$ are scalar functions then $\sup\{f(x)-g(y)\}=\sup\{f(x)\}-\inf\{g(y)\}$.

Question

Statement

Let be $A\subseteq\Bbb{R}^n$ and let be $f,g:A\rightarrow\Bbb{R}$ functions. So $\sup\{f(x)-g(y)\}=\sup\{f(x)\}-\inf\{g(y)\}$.

Clearly $f(x)\le\sup\{f(x)\}$ and $-g(y)\le-\inf\{f(y)\}$ and so summing $f(x)-g(y)\le\sup\{f(x)\}-\inf\{g(y)\}$ and so $\sup\{f(x)-g(y)\}\le\sup\{f(x)\}-\inf\{g(y)\}$. Unfortunately I can't prove the other inequality so I ask to do it and if the statement is generally false I ask to show a counterexample and so I ask if it is true when $f=g$. So could someone help me, please?

Answer 1

More generally, let $A, B$ be non-empty sets (not necessarily subsets of $\mathbb{R}^n,$ for any $n$), $f \colon A \to \mathbb{R}$ a function that is bounded above, and $g \colon B \to \mathbb{R}$ a function that is bounded below. Then: \begin{multline*} \sup_{x \in A,\ y \in B}(f(x) - g(y)) = \sup\{ f(x) - g(y) : x \in A, \ y \in B \} \\ = \sup(f(A) - g(B)) = \sup(f(A)) - \inf(g(B)) = \sup_{x \in A}f(x) - \inf_{y \in B}g(y), \end{multline*} because if $E$ is a non-empty subset of $\mathbb{R}$ that is bounded above, and $F$ is a non-empty subset of $\mathbb{R}$ that is bounded below, then the set $E - F = \{ u - v : u \in E, \ v \in F \}$ is non-empty and bounded above, and: $$ \sup(E - F) = \sup E - \inf F. $$ Proof. Let $a = \sup E$ and $b = \inf F.$ Then $u - v \leqslant a - b$ for all $u \in E$ and $v \in F.$ On the other hand, for all $\varepsilon > 0,$ there exist $u \in E$ and $v \in F$ such that $u > a - \frac\varepsilon2$ and $v < b + \frac\varepsilon2,$ whence $u - v > a - b - \varepsilon.$ Therefore $E - F$ is bounded above, with $\sup(E - F) = a - b.$ $\ \square$

Answer 2

Note that

$$\begin{align}f(x) \leqslant \sup_{x,y} (f(x) - g(y)) + g(y) &\implies \sup_x f(x) \leqslant \sup_{x,y} (f(x) - g(y)) + g(y) \\&\implies -g(y) \leqslant \sup_{x,y} (f(x) - g(y)) - \sup_x f(x) \\ &\implies -\inf_y g(y) = \sup_y (-g(y)) \leqslant \sup_{x,y} (f(x) - g(y)) - \sup_x f(x)\end{align} $$

Rearranging the final inequality, we get

$$\sup_x f(x)-\inf_y g(y) \leqslant \sup_{x,y} (f(x) - g(y)) $$

Answer 3

Fix $\varepsilon > 0$, and call $A= \sup_x f(x)$ and $B= \inf_y g(y)$.

There exis $x,y$ such that $$f(x) \ge A - \varepsilon \qquad ; \qquad g(y) \le B + \varepsilon$$ Then $$f(x) - g(y) \ge A-B- 2 \varepsilon $$ By arbitrarity of $\varepsilon$, this shows that $$\sup \{ f(x) - g(y) \} \ge A-B$$