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When is

$$ \min_{x\in X,y\in Y}f(x,y)=\min_{x\in X}(\min_{y\in Y}f(x,y))? $$

Moaleroa
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SuperM
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  • Are these $\min$'s or $\inf$'s? (not certain it will change the answer, but if $\min$'s (i.e., the minimum has to be actually reached for some value $(x,y)$, as is the case in particular when $X$ and $Y$ are finite), I would venture they are the same no matter what $f$ is. – Clement C. Aug 25 '14 at 11:54
  • @ClementC. I guess this is ment for $f(X,Y)$ to be a compact set (closed and bounded). So using $\min$ should be ok. – Darth Geek Aug 25 '14 at 12:00
  • $f$ is then assumed to be continuous? – Clement C. Aug 25 '14 at 12:03
  • @ClementC. do you have any counterexample in mind such that the relation doesnt hold? given that min always exists? – Seyhmus Güngören Aug 25 '14 at 12:06
  • Not really — trying to think if there could be such cases, actually. But I am not clear on whether the OP's question is on the assumptions on $f$, or on $X,Y$, or on both. – Clement C. Aug 25 '14 at 12:07
  • Hi everyone. I have noty made any particular assumptions on $X$, $Y$ or $f$. However, all I need is an existence result of the form: if $X$ is...... and $f$ is ..... then the relation holds. – SuperM Aug 25 '14 at 12:11
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    @SuperM Maybe you could change your $\min$ into $\inf$, so that the answers stays concentrate on the separability of $f$ on $X \times Y$. You can then always state that these $\min$ exists and get a similar result. – Surb Aug 25 '14 at 12:23
  • To support my last comment, see e.g. http://math.stackexchange.com/questions/18605/max-and-min-versus-sup-and-inf – Surb Aug 25 '14 at 12:27

2 Answers2

2

Let $f:X \times Y \to \mathbb{R}$, and let us prove that $$\inf_{(x,y)\in X \times Y}f(x,y) = \inf_{x \in X}\left( \inf_{y \in Y}f(x,y)\right).$$ Therefore by definition of the infimum we have a sequence $(x_n) \subset X$ such that $$\lim_{n \to \infty} \left(\inf_{y \in Y} f(x_n,y)\right) =\inf_{x\in X}\left( \inf_{y \in Y} f(x,y)\right),$$ furthermore for every $n \in \mathbb{N}$, there is some sequence $(y_m^n) \subset Y$ such that $$\lim_{m \to \infty} f(x_n,y^n_m) = \inf_{y \in Y} f(x_n,y).$$ Now, note that $$f(x_n,y_m^n) \geq \inf_{(x,y)\in X \times Y}f(x,y)$$ for every $m,n \in \mathbb{N}$ and thus $$ \inf_{x \in X} \left(\inf_{y\in Y}f(x,y)\right) =\lim_{n \to \infty}\left(\lim_{m \to \infty}f(x_n,y^n_m)\right) \geq \lim_{n \to \infty} \left( \lim_{m \to \infty} \inf_{(x,y)\in X\times Y}f(x,y) \right) = \inf_{(x,y)\in X \times Y}f(x,y).$$ Still using the definition of the infimum, we have the existence of a sequence $(x_k,y_k) \subset X \times Y$ such that $$ \inf_{(x,y)\in X \times Y}f(x,y) =\lim_{k \to \infty}f(x_k,y_k),$$ but note that for every $k \in \mathbb{N}$ we have $$f(x_k,y_k) \geq \inf_{y \in Y}f(x_k,y) \geq \inf_{x \in X} \left( \inf_{y \in Y}f(x,y)\right).$$ It finally follows that $$\inf_{(x,y)\in X \times Y}f(x,y) = \lim_{k \to \infty}f(x_k,y_k) \geq \lim_{k \to \infty}\inf_{x \in X}\left( \inf_{y \in Y}f(x,y)\right)=\inf_{x \in X}\left( \inf_{y \in Y}f(x,y)\right),$$ which proves the first statement. Now, if for every $z \in X$, $\min_{(x,y)\in X \times Y}f(x,y), \min_{y \in Y}f(z,y)$ and $\min_{x\in X} \left( \min_{y\in Y} f(x,y)\right)$ exist, $$\min_{(x,y)\in X \times Y}f(x,y)= \inf_{(x,y)\in X \times Y}f(x,y), \\ \min_{y \in Y}f(z,y)=\inf_{y \in Y}f(z,y),\\ \min_{x\in X} \left( \min_{y\in Y} f(x,y)\right)=\inf_{x\in X} \left( \inf_{y\in Y} f(x,y)\right)$$ and your result follows directly.

Surb
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  • Thanks @Surb....I see the result holds all the time for inf, but when is it true for min? – SuperM Aug 27 '14 at 12:03
  • @SuperM It is always true for min provided that they exists. Basically, each time you can prove that one of these minimum exists, you may turn the corresponding $\inf$ in a $\min$. I hope it is more clear to you... Anyway you should maybe have a look at the link I posted in the OP problem to get a better understanding of the difference between min and inf. – Surb Aug 27 '14 at 12:52
  • Ok...I see...Thanks a lot! – SuperM Aug 27 '14 at 15:54
  • @SuperM Thank you for this very interesting question :). – Surb Aug 27 '14 at 16:07
1

This equation is always true.

You proof this by contradiction:

1) It is clear that $$\min_{x\in X,y\in Y}f(x,y)>\min_{x\in X}(\min_{y\in Y}f(x,y))$$ cannot hold true.

2) Let us assume $$\min_{x\in X,y\in Y}f(x,y)<\min_{x\in X}(\min_{y\in Y}f(x,y))$$ and let $x^\star $ and $y^\star $ minimizer.

Now, suppose you search over all possible $x$ for the smallest value of $\min_{y\in Y}f(x,y))$ and suppose that for each $x$ you can find the optimal $y$.

Then, choosing $x^\star$ will give you $y^\star$ and you arrive at $\min_{x\in X,y\in Y}f(x,y)$

As 1) and 2) cannot be true, you can conclude that equality holds always.

The Pheromone Kid
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