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| Chemboy |
 Posted: Thu Jul 03, 2008 10:33 am Post subject: extreme value theorem |
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I came across the following in some notes I'm looking at online...
"If f(x,y) is continuous in some closed, bounded set D in R2 then there are points in D, (x1,y1) and (x2,y2) so that f(x1,y1) is the absolute maximum and f(x2,y2) is the absolute minimum of the function in D."
I'm thinking that this doesn't always hold true. Am I correct? I think it wouldn't hold true for a flat plane (for example, z=1 or something), and it seems there could be a region with an absolute minimum or maximum, but not the other. For example, something with a 'dip' or a 'hill,' that slopes smoothly, and where the value at the edge of the region is the same on the entire edge. The examples seem pretty obvious, I'd just like to hear it from someone other than myself that there are exceptions to the rule...
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| serpicojr |
| Posted: Thu Jul 03, 2008 11:26 am Post subject: |
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It's a rule for reason: it's true. Your suggestion of a counterexample doesn't work because a plane, although closed, is unbounded.
There's a wonderful theorem called the Heine-Borel Theorem that says, in part, if you have a sequence xn in a closed, bounded subset of Rm, then there exists a subsequence xnk (where n1 < n2 < ... is an increasing sequence of integers) such that xnk converges. Since the set is closed, the limit point must be in the set.
Now if you had a continuous, unbounded function f on a closed, bounded set, you'd have to be able to find a sequence of points xn so that, say, |f(xn)| approaches infinity. But we know there is a subsequence xnk of xn that converges to a point x in the set. But then, since f and the absolute value are continuous, we must have |f(xnk)| approaches |f(x)|. This is a contradiction to the fact that |f(xn)| and hence |f(xnk)| approach infinity. |
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| Chemboy |
| Posted: Thu Jul 03, 2008 12:15 pm Post subject: |
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I don't know that I understand that completely, but I think I understand what you're saying fairly well... Could you kind of say that a function can always take a value as you increase x, while considering the values on the set you're looking at, they approach a limit...? I'm not sure that's quite right, but it's my attempt of explaining it in terms I can understand better...
Could you explain just how a plane is closed but unbounded, and how it (or any function) could be bounded?
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| serpicojr |
| Posted: Thu Jul 03, 2008 2:50 pm Post subject: |
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| Chemboy wrote: |
| Could you kind of say that a function can always take a value as you increase x, while considering the values on the set you're looking at, they approach a limit...? |
I'm not sure I understand your question, but continuity of a function f(x) from a subset D of Rn to, say, R is equivalent to this statement:
For every sequence xn of D that converges to an element x of D, we have that f(xn) converges to f(x).
That's what I was using in my previous post.
| Quote: |
| Could you explain just how a plane is closed but unbounded, and how it (or any function) could be bounded? |
Any plane is unbounded because it contains points arbitrarily far away from the origin. A bounded set in Rn is one you can enclose in some ball. |
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