Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  Jonas Meyer Sep 4 '13 at 6:24

1 Answer 1

up vote 1 down vote accepted

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 $g(x)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.