Notices
Results 1 to 4 of 4

Thread: extreme value theorem

  1. #1 extreme value theorem 
    Moderator Moderator AlexP's Avatar
    Join Date
    Jul 2006
    Location
    NY
    Posts
    1,838
    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 R<sup>2</sup> then there are points in D, (x<sub>1</sub>,y<sub>1</sub>) and (x<sub>2</sub>,y<sub>2</sub>) so that f(x<sub>1</sub>,y<sub>1</sub>) is the absolute maximum and f(x<sub>2</sub>,y<sub>2</sub>) 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...


    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    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 x<sub>n</sub> in a closed, bounded subset of R<sup>m</sup>, then there exists a subsequence x<sub>n<sub>k</sub></sub> (where n<sub>1</sub> < n<sub>2</sub> < ... is an increasing sequence of integers) such that x<sub>n<sub>k</sub></sub> 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 x<sub>n</sub> so that, say, |f(x<sub>n</sub>)| approaches infinity. But we know there is a subsequence x<sub>n<sub>k</sub></sub> of x<sub>n</sub> that converges to a point x in the set. But then, since f and the absolute value are continuous, we must have |f(x<sub>n<sub>k</sub></sub>)| approaches |f(x)|. This is a contradiction to the fact that |f(x<sub>n</sub>)| and hence |f(x<sub>n<sub>k</sub></sub>)| approach infinity.


    Reply With Quote  
     

  4. #3  
    Moderator Moderator AlexP's Avatar
    Join Date
    Jul 2006
    Location
    NY
    Posts
    1,838
    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?
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
    Reply With Quote  
     

  5. #4  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Chemboy
    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 R<sup>n</sup> to, say, R is equivalent to this statement:

    For every sequence x<sub>n</sub> of D that converges to an element x of D, we have that f(x<sub>n</sub>) converges to f(x).

    That's what I was using in my previous post.

    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 R<sup>n</sup> is one you can enclose in some ball.
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •