Thread: Extreme Value Theorem

Here is a proof from my notes that I could use some help with.

Let $\displaystyle I$ and $\displaystyle J$ be closed intervals. Suppose $\displaystyle f:I$ x $\displaystyle J\rightarrow R$ is continuous. Show there are$\displaystyle (x_{min},y_{min}) $and $\displaystyle (x_{max},y_{max})$ in $\displaystyle I$ x $\displaystyle J$, such that $\displaystyle f(x_{min},y_{min})\leq f(x,y)\leq f(x_{max},y_{max})$, for all $\displaystyle (x,y)\in I$ x $\displaystyle J$.

The hint that we are given is to divide the rectangle $\displaystyle I$ x $\displaystyle J$into four subrectangles. Then, the supremum of $\displaystyle f$ over at least one of these subrectangles must be equal to the supremum of $\displaystyle f$ over $\displaystyle I$ x $\displaystyle J$...

I see that they are just applying the Extreme Value Theorem to rectangles but I'm not sure how to word the proof.

The short answer is that the product of compact sets is compact, and a continuous function with compact domain has such points.

But I think you're probably building up to that theorem. I think the idea here is to continue the process: take the rectangle where the supremum of f occurs, and divide it again, and so on. You'll then be able to define a Cauchy sequence in IxJ, which converges since ... and the value of f at the limit point is ...

Ok, so I just divide the rectangles into smaller and smaller rectangles until I have a rectangle that contains only one point,$\displaystyle x_{max}$. So, $\displaystyle x_{max}$ is the limit point. But then I'm not really sure what else to say. I'm guessing that we have to verify that this point $\displaystyle x_{max}$ is actually the supremum of the entire rectangle.

Well, you don't ever reach a rectangle with only one point. But there is a well-defined limit to the process. Now what can you say, for f continuous, about lim f(x) as x approaches some point?

Since $\displaystyle f$ is continuous then the $\displaystyle lim_{x\rightarrow x_{max}}f(x)=f(x_{max}})$