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Thread: Extreme Value Theorem

  1. #1
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    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.
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  2. #2
    Senior Member Tinyboss's Avatar
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    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 ...
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  3. #3
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    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.
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  4. #4
    Senior Member Tinyboss's Avatar
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    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?
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  5. #5
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    Since $\displaystyle f$ is continuous then the $\displaystyle lim_{x\rightarrow x_{max}}f(x)=f(x_{max}})$
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