Let $X$ and $Y$ be nonempty sets, and let $f:X\times Y \rightarrow \mathbb{R}$ be a bounded function.

a) Prove that $$\sup_{y\in Y}\left(\inf_{x\in X}f(x,y)\right)\leq \inf_{x\in X}\left(\sup_{y\in Y}f(x,y)\right)$$

b) Give an example (with proof) where the inequality is strict.

I am stumped. First of all, I am having a hard time understanding the notation. The way I understand it, in the most basic terms, we have that the least upperbound of the greatest lower bound of the function is less than or equal to the greatest lower bond of the least upper bound of the function. But does this even make sense? sup and inf are unique, so wouldn't $$\sup_{y\in Y}\left(\inf_{x\in X}f(x,y)\right) = \inf_{x\in X}f(x,y)?? $$ and likewise for the inf of the sup...am I missing something? Plus I'm not sure what implications the subscripts $x\in X$ and $y\in Y$ carry.

Please help me get an understanding of what the notation means and let me know if my reasoning is off.

  • $\begingroup$ This question was previously posted. math.stackexchange.com/q/483607 After it was closed, OP edited to its present form, then deleted and reposted here. $\endgroup$ Sep 4, 2013 at 6:24

2 Answers 2


Let $f(0,y)=0$ for all $y\ne 0$, and let $f(1,y)=0$ for all $y\ne 1$. Let $f(x,y)=1$ in all other cases. This includes $x=0,y=0$ and $x=1$, $y=1$.

Look at the left-hand side. For any fixed $y$, we have $\inf_{x\in X} f(x,y)=0$. For if $y=0$ we can take $x=1$, and if $y\ne 0$ we can take $x=0$.

Taking the sup of all these $0$'s leaves us at $0$.

Now look at the right-hand side. For any $x$, we have $\sup_{y\in Y} f(x,y)=1$, since for any $x$ there is an $y$ such that $f(x,y)=1$. Taking the inf over all $x$ leaves us at $1$.

Thus the inequality can be strict.

We used the names $0$ and $1$ for certain elements of $X$ and $Y$. They could equally well have been called $x_0,x_1,y_0,y_1$.

Remark: We consider the meaning of an expression such as such as $\sup_{y\in Y}f(x,y)$. The function $f$ is bounded. Fix $x$, and look at $f(x,y)$ as $y$ varies over $Y$. The supremum over all $y$ of $f(x,y)$ is sort of the greatest possible value of $f(x,y)$ for that fixed value of $x$. Not really greatest, it is least upper bound, but for visualization we can think of it as the greatest. So $\sup_{y\in Y} f(x,y)$ is a function of $x$, say $g(x)$. Then, in the expression on the right, we sort of take the smallest possible value of $g(x)$.

  • 4
    $\begingroup$ How does one prove part a of the question? $\endgroup$
    – TheNotMe
    Oct 13, 2017 at 21:23
  • $\begingroup$ @TheNotMe Both of these people agree on a point :$\sup_{y\in Y} \left(\inf_{x \in X} f(x,y)\right) <\inf_{x \in X} \left( \sup_{y \in Y} f(x,y) \right). $ but this answer states that equality isn't always working and the other proved this case: less than OR equal. so the final conclusion : the only case that always happens is LESS THAN ;) $\endgroup$
    – Arian
    Oct 11 at 5:59

The proof for part a):

By the definition of $\sup$, $$\sup_{y\in Y} \left(\inf_{x \in X} f(x,y)\right) - \epsilon < \inf_{x \in X} f(x,y^\ast)$$ for some $y^\ast \in Y$. Also, $$ \inf_{x \in X} f(x,y^\ast) \leq \inf_{x \in X} \left( \sup_{y \in Y} f(x,y) \right), $$ because $f(x,y^\ast) \leq \sup_{y \in Y} f(x,y)$ for every $x$. Combine the two above we have $$\sup_{y\in Y} \left(\inf_{x \in X} f(x,y)\right) - \epsilon < \inf_{x \in X} \left( \sup_{y \in Y} f(x,y) \right).$$ This holds for any $\epsilon>0$. Therefore, $$\sup_{y\in Y} \left(\inf_{x \in X} f(x,y)\right) \leq \inf_{x \in X} \left( \sup_{y \in Y} f(x,y) \right). $$ $$\tag*{$\blacksquare$}$$


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