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Let $X,Y \ne\varnothing$ and $g:X\times Y \rightarrow \mathbb{R}$. Show that $$\sup_{y \in Y} \inf_{x \in X}g(x,y) \leq \inf_{x \in X} \sup_{y \in Y} g(x,y).$$

First note that $$ g(\overline{x},\overline{y}) \leq \sup_{y \in Y}g(\overline{x},y)$$ for all $\overline{x} \in X$ and $\overline{y} \in Y$. Then take the infimum wrt $X$ on both sides, giving $$ \inf_{x \in X}g(x,\overline{y}) \leq \inf_{x \in X}\sup_{y \in Y}g(x,y) $$ which now holds for all $\overline{y} \in Y$. Thus we can take the supremum over $Y$ on the lhs to give the desired result.

Question

Why taking the supremum over $Y$ on the lhs in the last inequality won't affect the inequality?

Could someone explain please?

I found this solved exercise here How to show $ \sup \inf g(x,y) \leq \inf \sup g(x,y)$?

Yiorgos S. Smyrlis
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user441848
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2 Answers2

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Since the the last inequality you wrote holds for all $\bar{y}\in Y$, we have that the right hand side is an upper bound for the left hand side. The supremum of the left hand side is the least upper bound, so it is less than or equal to all other upper bounds of the left hand side, in particular it is less than or equal to the right hand side. This is the reason we have $\sup_{y\in Y}\inf_{x\in X}g(x,y)\leq \inf_{x\in X}\sup_{y\in Y}g(x,y)$, not necessarily that we just take the sup of both sides, the fact that that works out is somewhat coincidental.

Jürgen Sukumaran
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In the last inequality $$ \inf_{x \in X}g(x,\overline{y}) \leq \inf_{x \in X}\sup_{y \in Y}g(x,y), \quad \text{for all $\,\overline{y}\in Y$}, $$ the RHS does NOT depend on $\overline{y}$, whereas the LHS does. So it is basically an inequality of the form $$ f(\overline{y})\le A \quad \text{for all $\,\overline{y}\in Y$}, $$ and hence $A$ is an upper bound of all the values $f(\overline{y})$, for $\,\overline{y}\in Y$. Thus $A$ is great or equal to the least upper bound of the values $f(\overline{y})$, for $\,\overline{y}\in Y$, and therefore $$ \sup_{\overline{y}\in Y} f(\overline{y})\le A \quad \text{for all $\,\overline{y}\in Y$}. $$ and hence $$ \sup_{\overline{y}\in Y}\inf_{x \in X}g(x,\overline{y}) \leq \inf_{x \in X}\sup_{y \in Y}g(x,y). $$

Yiorgos S. Smyrlis
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