Is this the correct interpretation of $\sup_{a \in A} \inf_{b \in B} f(a, b)$?

Question

Suppose $f:A \times B \to \mathbb{R}$. I am trying to understand the notation $$\sup_{a \in A} \inf_{b \in B} f(a, b)$$ My interpretation is this: $$\inf_{b \in B} f(a, b) = \{\inf_{b \in B}f(a, b)|a \in A \}$$ That is, $\inf_{b \in B}$ applied to a two or more variable function is a set.

Of course, once we have a set, there can only be one interpretation: $$\sup_{a \in A} \inf_{b \in B} f(a, b)= \sup \{\inf_{b \in B}f(a, b)|a \in A \}.$$ I would understand the meaning if $\inf$ was applied to a one variable function. But when it's applied to a two variable function, I find the situation confusing.

Is the above the correct interpretation?

Thank you very much.

Answer 1

For fixed $a \in A$, the infimum $\inf_{b \in B} f(a,b)$ is a real number, or possibly $-\infty$. This means we may define a function $g: A \to \overline {\mathbb R}$ by $$g(a) = \inf_{b \in B}f(a,b), \quad \text{for all } a \in A.$$ Then $\sup_{a \in A}\inf_{b \in B} f(a,b)$ is the supremum of $g$ over $A$, i.e. $$\sup_{a \in A}\inf_{b \in B} f(a,b)= \sup_{a \in A} g(a).$$ So think of $\inf_{b \in B}f(a,b)$ as a function of $a$, and then take the supremum of this function.

Answer 2

If we generalize the @averagemonkey's answer, the theme is: In a string of sup's and inf's as you go to the right freeze every variable you pass by, then compute the last sup or inf thus eliminating one of the variables, then work backwards as you successively compute a sup or inf at a time and get rid of one more variable. Hope this helps.