- #1

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Is

[tex]\underset{i}{\min}\underset{n}{\max}=\underset{n}{\max}\underset{i}{\min}[/tex]?

Thanks

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- Thread starter EngWiPy
- Start date

- #1

- 1,367

- 61

Is

[tex]\underset{i}{\min}\underset{n}{\max}=\underset{n}{\max}\underset{i}{\min}[/tex]?

Thanks

- #2

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- 3,297

[tex]x_{i,n}=\left\{\begin{array}{l} 0 ~\text{if}~i+n~\text{even}\\

1 ~\text{if}~i+n~\text{odd}\end{array}\right.[/tex]

- #3

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[tex]x_{i,n}=\left\{\begin{array}{l} 0 ~\text{if}~i+n~\text{even}\\

1 ~\text{if}~i+n~\text{odd}\end{array}\right.[/tex]

Yeah, right.

- #4

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- 0

Given a real valued function ##P(x,y): X \times Y \rightarrow \mathbb{R}## one has

$$\inf_{x\in X} \sup_{y\in Y} P(x,y) \geq \sup_{y\in Y} \inf_{x\in X} P(x,y)$$

If you aren't familiar with sup (supremum - least upper bound) and inf (infimum - greatest lower bound) you can think of sup as max and inf as min.

I learned this from Robert McCann, it's in the following article. Skip to 2.2 Game Theory for an intuitive 'proof'.

http://www.math.toronto.edu/mccann/papers/FiveLectures.pdf

- #5

- 1,367

- 61

Given a real valued function ##P(x,y): X \times Y \rightarrow \mathbb{R}## one has

$$\inf_{x\in X} \sup_{y\in Y} P(x,y) \geq \sup_{y\in Y} \inf_{x\in X} P(x,y)$$

If you aren't familiar with sup (supremum - least upper bound) and inf (infimum - greatest lower bound) you can think of sup as max and inf as min.

I learned this from Robert McCann, it's in the following article. Skip to 2.2 Game Theory for an intuitive 'proof'.

http://www.math.toronto.edu/mccann/papers/FiveLectures.pdf

Is this inequality true in general?

- #6

- 95

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Yes. To see this pick [itex]x'\in X[/itex] and [itex]y'\in Y[/itex]. Clearly [itex]\sup_Y P(x',y)\geq \inf_X P(x,y')[/itex], and since this is true for all [itex]y'\in Y[/itex] we have [itex]\sup_Y P(x',y)\geq \sup_Y \inf_X P(x,y)[/itex], by definition of the supremum. Similarly, since this is true for all [itex]x'\in X[/itex] we have [itex]\inf_X\sup_Y P(x,y)\geq \sup_Y \inf_X P(x,y)[/itex] by definition of the infimum.Is this inequality true in general?

Last edited:

- #7

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Yes. To see this pick [itex]x'\in X[/itex] and [itex]y'\in Y[/itex]. Clearly [itex]\sup_Y P(x',y)\geq \inf_X P(x,y')[/itex], and since this is true for all [itex]y'\in Y[/itex] we have [itex]\sup_Y P(x',y)\geq \sup_Y \inf_X P(x,y)[/itex], by definition of the supremum. Similarly, since this is true for all [itex]x'\in X[/itex] we have [itex]\inf_Y\sup_Y P(x,y)\geq \sup_Y \inf_X P(x,y)[/itex] by definition of the infimum.

OK, what abou max max or min min, are they interchangeable?

- #8

- 95

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Yes. Consider [itex]\sup_X\sup_Y P(x,y)[/itex]. By definition we have [itex]\sup_X\sup_Y P(x,y)\geq P(x',y')[/itex] for all [itex]x'\in X,y'\in Y[/itex], so [itex]\sup_X\sup_Y P(x,y)\geq \sup_{X\times Y} P(x,y)[/itex] by definition of [itex]\sup[/itex]. Conversely, if [itex]\sup_X\sup_Y P(x,y)\not\leq \sup_{X\times Y} P(x,y)[/itex] then there is [itex]x'\in X[/itex] with [itex]\sup_Y P(x',y)\not\leq \sup_{X\times Y} P(x,y)[/itex], and thus there is [itex]y'\in Y[/itex] with [itex]\sup_Y P(x',y')\not\leq \sup_{X\times Y} P(x,y)[/itex], which would be a contradiction. So [itex]\sup_X\sup_Y P(x,y)= \sup_{X\times Y} P(x,y)[/itex] and a symmetry argument gives [itex]\sup_X\sup_Y P(x,y)= \sup_{X\times Y} P(x,y)=\sup_Y\sup_X P(x,y)[/itex].OK, what abou max max or min min, are they interchangeable?

Edit: We require that the suprema and infima are defined, I forgot to mention that in my original response.

- #9

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- 61

Yes. Consider [itex]\sup_X\sup_Y P(x,y)[/itex]. By definition we have [itex]\sup_X\sup_Y P(x,y)\geq P(x',y')[/itex] for all [itex]x'\in X,y'\in Y[/itex], so [itex]\sup_X\sup_Y P(x,y)\geq \sup_{X\times Y} P(x,y)[/itex] by definition of [itex]\sup[/itex]. Conversely, if [itex]\sup_X\sup_Y P(x,y)\not\leq \sup_{X\times Y} P(x,y)[/itex] then there is [itex]x'\in X[/itex] with [itex]\sup_Y P(x',y)\not\leq \sup_{X\times Y} P(x,y)[/itex], and thus there is [itex]y'\in Y[/itex] with [itex]\sup_Y P(x',y')\not\leq \sup_{X\times Y} P(x,y)[/itex], which would be a contradiction. So [itex]\sup_X\sup_Y P(x,y)= \sup_{X\times Y} P(x,y)[/itex] and a symmetry argument gives [itex]\sup_X\sup_Y P(x,y)= \sup_{X\times Y} P(x,y)=\sup_Y\sup_X P(x,y)[/itex].

Edit: We require that the suprema and infima are defined, I forgot to mention that in my original response.

Yes, of course they are defined.

Thanks

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