Results 1 to 7 of 7

Math Help - Geometric proof for theorem about continuous functions

  1. #1
    Junior Member
    Joined
    Mar 2011
    Posts
    34

    Geometric proof for theorem about continuous functions

    Hello

    I'm trying to prove that if f : [a,b] ---> [a,b] is continuos, there exists an x in [a,b] such that f(x) = x

    I've formed a sort of geometric proof - if you take the square [a,b] and draw the diagonal down the middle representing the identity function, then any function defined on the interval [a,b] has to intersect the line at least once.

    I'm having trouble formalizing it though. Help would be greatly appreciated...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Try experimenting with the new function g(x) = f(x) - x, and think about the Intermediate Value Theorem.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2011
    Posts
    34
    Okay, I see that if g intersects with the x axis, then at that point the value of f is x. But I don't understand why the point of intersection has to be x...and why can't the values of f all be greater than x, so that there is no intersection?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,803
    Thanks
    1692
    Awards
    1
    Quote Originally Posted by moses View Post
    I'm trying to prove that if f : [a,b] ---> [a,b] is continuos, there exists an x in [a,b] such that f(x) = x
    Notice that f(a)\ge a~\&~f(b)\le b.
    So if g(x)=f(x)-x then g(b)\le 0\le g(a).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2011
    Posts
    34
    shouldn't it be g(a) <= 0 <= g(b) ?

    Anyway I see now why g gets the value zero, from the Intermediate Value Theorem, but why does this have to happen at x?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,803
    Thanks
    1692
    Awards
    1
    Quote Originally Posted by moses View Post
    shouldn't it be g(a) <= 0 <= g(b) ?
    No indeed. f(a)\ge a implies f(a)-a\ge 0
    And f(b)\le b implies f(b)-b\le 0.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2011
    Posts
    34
    Got it!!! I think...if g(b) <= 0 <= g(a) then from the i.v.t there exists c in [a,b] such that g(c) = 0, and then f(c) - c = 0 and then f(c) = c

    Whoo thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: May 23rd 2011, 02:36 AM
  2. Replies: 15
    Last Post: May 25th 2009, 06:31 PM
  3. Continuous Functions Proof
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 5th 2009, 04:30 PM
  4. Replies: 6
    Last Post: May 4th 2009, 11:29 AM
  5. Proof about uniformly continuous functions and densness of Q
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 9th 2009, 07:26 PM

Search Tags


/mathhelpforum @mathhelpforum