Two part question about $\sup_y\inf_xf(x,y)\leqslant\inf_x\sup_yf(x,y)$

Question

Background Info:

Let $X,Y\neq\emptyset$ and let $f:X\times Y\to\mathbb{R}$ have a bounded range in $\mathbb{R}$ . Also, let

$f_{1}(x)=\sup\{f(x,y):\: y\in Y\}$ and $f_{2}(y)=\sup\{f(x,y):\: x\in X\}$

Establish the Principle of Iterated Suprema:

\begin{align*} \sup\{f(x,y):\: x\in X,y\in Y\} &=\sup\{f_{1}(x):\: x\in X\}\\ &=\sup\{f_{2}(y):\: y\in Y\} \end{align*}

I proved the Principle of Iterated Suprema but now I am stuck on the proceeding question

The Question let $f$ and $f_{1}$ be as in the preceding exercise and let

$$g_{2}(y)=\inf\{f(x,y):\: x\in X\}$$

Prove that $$\sup\{g_{2}(y):y\in Y\}\leqslant\inf\{f_{1}(x):\: x\in X\}$$

Show that strict inequality can hold. We sometimes express this inequality as $$\underset{y\quad x}{\sup\inf}f(x,y)\leqslant\underset{x\quad y}{\inf\sup}f(x,y)$$

I don't really know where to start. A hint would be greatly appreciated! Thanks

Answer 1

Hints:

• $g(y) \leq \sup\limits_y\ g(y)$
• $h(x) \leq k(x) \Rightarrow \inf\limits_x\ h(x) \leq \inf\limits_x\ k(x)$
• $l(y) \leq M \Rightarrow \sup\limits_y\ l(y) \leq M$

Answer 2

Note that $$ g_2(y) = \inf\{f(x,y) : x\in X\} \leq f(x,y) \quad\forall x \in X, y\in Y $$ and similarly $$ f(x,y) \leq f_1(x) \quad\forall x\in X, y\in Y $$ Thus, for all $x,y$, $$ g_2(y)\leq f_1(x) $$ Taking sup over $y\in Y$, we get $$ \sup \{g_2(y) : y\in Y\} \leq f_1(x) $$ This is true for all $x\in X$, so $$ \sup \{g_2(y) : y\in Y\} \leq \inf \{ f_1(x) : x\in X\} $$

Now for the counterexample, let $X = \mathbb{N}$ and $Y = \mathbb{N}$. Then $f(x,y)$ can be represented by an infinite matrix; where the $x$ variable represents the rows, and $y$ represents the columns.

Let's say you want to construct such a matrix whose row infimum is zero for any row, and whose column supremum is 1 for any column. Just take $f$ to be the matrix $$ f(0,j) = (1,0,1,0,1,\ldots ) $$ $$ f(1,j) = (0,1,0,1,0,\ldots ) $$ and $f(2n,j) = f(0,j)$, and $f(2n+1,j) = f(1,j)$ for all $n$,

Answer 3

So, the claim is that \begin{align*}\sup\{g_{2}(y):y\in Y\} &\leqslant \inf\{f_{1}(x):\: x\in X\}\\ \sup\{\inf\{f(x,y):\: x\in X\}:y\in Y\} &\leqslant \inf\{\sup\{f(x,y):\: y\in Y\}:\: x\in X\}\end{align*}

Since we know that $\text{Ran}g_{2}$ is a bounded subset of $\mathbb{R}$ , we know that $\sup\{g_{2}(y):y\in Y\})$ exists, likewise since $\text{Ran}f_{1}$ is a bounded subset of $\mathbb{R}$ , we know that $\inf\{f_{1}(x):\: x\in X\}$ exists. Let us examine an arbitrary $x'\in X$ and an arbitrary $y'\in Y$ we have

\begin{align*} g_{2}(y') \quad &? \quad f_{1}(x')\\ \inf\{f(x,y'):\: x\in X\} \quad &? \quad \sup\{f(x',y):\: y\in Y\} \end{align*}

Let $f(x_{0},y')=\inf\{f(x,y'):\: x\in X\}$ and $f(x',y_{0})=\sup\{f(x',y):\: y\in Y\}$ , then we have that $(\forall x)(f(x_{0},y')\leqslant f(x,y'))$ and $(\forall y)(f(x',y)\leqslant f(x',y_{0}))$ . In words, we have that is doesn’t matter what $y'$ you choose $(f(x',y')\leqslant f(x',y_{0})$ , and for not matter what $x'$ you choose $(f(x_{0},y')\leqslant f(x',y')$ . That is that $(f(x_{0},y')\leqslant f(x',y_{0})$ , i.e. $g_{2}(y')\leqslant f_{1}(x')$ . Since our selection of $x'$ and $y'$ was arbitrary we have that $(\forall x,y)(g_{2}(y)\leqslant f_{1}(x))$ . So, it follows that $\sup\{g_{2}(y):y\in Y\}\leqslant\inf\{f_{1}(x):\: x\in X\}$ .