Extreme Value Theorem

**zebra2147** · Nov 6th 2010, 04:21 PM

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

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

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

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

**Tinyboss** · Nov 6th 2010, 06:15 PM

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 ...

**zebra2147** · Nov 7th 2010, 10:43 AM

Ok, so I just divide the rectangles into smaller and smaller rectangles until I have a rectangle that contains only one point, $x_{max}$ . So, $x_{max}$ 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 $x_{max}$ is actually the supremum of the entire rectangle.

**Tinyboss** · Nov 8th 2010, 05:19 AM

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?

**zebra2147** · Nov 8th 2010, 05:34 AM

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

Thread: Extreme Value Theorem

LinkBack

Thread Tools

Display