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I know it's similar to this and I proved this inequality, but I'm stuck at finding examples that this inequality is strict.
thanks.

Edit: range is bounded.

mdmi
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2 Answers2

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The inequality can be strict if both $X$ and $Y$ have at least two elements. For an example pick arbitrary elements $x_1 \in X$ and $y_1 \in Y$ and define $$ f(x, y) = \begin{cases} 0 & \text{ if } x=x_1 \text{ and } y = y_1 \\ 1 & \text{ if } x=x_1 \text{ and } y \ne y_1 \\ 1 & \text{ if } x\ne x_1 \text{ and } y = y_1 \\ 0 & \text{ if } x\ne x_1 \text{ and } y \ne y_1 \, . \end{cases} $$ Then $\inf\{f(x,y) : x \in X\} = 0$ for all $y \in Y$, and $\sup\{f(x,y) : y \in Y\} = 1$ for all $x \in X$, and therefore $$ \sup\{\inf\{f(x,y) : x \in X\}: y \in Y\} = 0 < 1 = \inf\{\sup\{f(x,y) : y \in Y\}: x \in X\} \, . $$

Martin R
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  • Thanks, I'm curious that whether we can generalize some patterns about functions shape that this inequality is strict? I thought about it but I couldn't formalize those thoughts – mdmi Nov 22 '21 at 07:55
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    @mdmi: For some $a < b$ you have to assign function values $\le a$ and $\ge b$ in every “row” and every “column.” –  In the case of real intervals, a simple closed-form example is $f(x, y) = |x-y|$. – Martin R Nov 22 '21 at 09:09
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Any nonzero linear function should demonstrate this. For example:

Set $X=\mathbb{R}$ and $Y=\mathbb{R}$ and $f(x,y)=x+y$. Then, for any fixed $y^*\in Y$, we have $\inf_{x\in X}f(x,y)=\inf_{x\in\mathbb{R}}(x+y^*)\to-\infty$, so $$\sup_{y\in Y}\big(\inf_{x\in X}f(x,y))\big)=\sup_{y\in Y}\{-\infty\}=-\infty.$$ Similarly, for any fixed $x^*\in X$, one can observe $\sup_{y\in Y}f(x,y)\to +\infty$ and hence $$\inf_{x\in X}\big(\sup_{y\in Y}f(x,y)\big)=+\infty,$$ which demonstrates strictness of the linequality.

Zim
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  • thank you for your answer, but can you give an example when range of function is bounded? – mdmi Nov 21 '21 at 19:03
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    would have been helpful if it were originally stated in the question. Look to Martin R's hint. – Zim Nov 21 '21 at 19:14