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Came across this little practice exercise, and I couldn't properly convince myself of this relation:

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

My thinking was starting with $\inf_{x \in X}g(x,\overline{y}) \leq\sup_{y \in Y} g(\overline{x},y)$, but does that even hold for all $\overline{x} \in X$ and $\overline{y} \in Y$?

BallzofFury
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1 Answers1

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Well silly me, I think I've found a decent explanation. 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.

BallzofFury
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    Let me strongly recommend to replace each formula similar to $$ g(x,y) \leq \sup_{y \in Y}g(x,y)$$ by its correct formulation $$ g(x,y) \leq \sup_{z \in Y}g(x,z).$$ – Did Jan 23 '15 at 14:03
  • Good point, I've incorporated that now. – BallzofFury Jan 23 '15 at 15:23
  • Could you explain why the inequality $\le$ does not change when you take the $\sup$ on the lhs (last line) ? – user441848 Jun 02 '18 at 06:02