Results 1 to 3 of 3

Math Help - finding fixed points

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    30

    Cool finding fixed points

    PROBLEM:
    Let a and b be real numbers with a<b. Suppose that f:[a,b]-->R is a continuous function that satisfies f([a,b]) is a set of [a,b].
    Prove that f has a fixed point.
    {The hint is: g=f(x)-x and use IVT}

    ATTEMPT AT PROOF:
    so using IVT if there is a and b then there must be a c so, a<c<b
    which means f(a)<f(c)<f(b) or f(a)>f(c)>f(b). if f([a,b]) is a set of [a,b] then at most they are the same interval. if f is a fixed point then f(x)-x = 0 so, g=0.


    OKay here is where i am stuck ..... if g(a)>0 and g(b)<0 then there is a fixed point and if g(a)<0 and g(b)>0 there is a fixed point but is that all they can ever be? .... is there an instant where g(a)<0 and g(b)<0 and still be existent with a fixed point in this interval.




    WHY MUST THERE BE A FIXED POINT IN THIS INTERVAL?
    Also can someone please give an example of what f:[a,b] --> R actually looks like. Thank you for your time
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,793
    Thanks
    1688
    Awards
    1
    You are given that  f\left( {\left[ {a,b} \right]} \right) \subseteq \left[ {a,b} \right] which implies \left( {\forall x \in \left[ {a,b} \right]} \right)\left[ {a \leqslant f(x) \leqslant b} \right].
    Use the hint g(x)=f(x)-x so g(a)\ge 0 and g(b)\le 0.
    If either of those is equal to 0 you are done. WHY?
    Now finish.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,977
    Thanks
    1643
    Quote Originally Posted by derek walcott View Post
    PROBLEM:
    Let a and b be real numbers with a<b. Suppose that f:[a,b]-->R is a continuous function that satisfies f([a,b]) is a set of [a,b].
    Prove that f has a fixed point.
    {The hint is: g=f(x)-x and use IVT}

    ATTEMPT AT PROOF:
    so using IVT if there is a and b then there must be a c so, a<c<b
    which means f(a)<f(c)<f(b) or f(a)>f(c)>f(b). if f([a,b]) is a set of [a,b] then at most they are the same interval. if f is a fixed point then f(x)-x = 0 so, g=0.


    OKay here is where i am stuck ..... if g(a)>0 and g(b)<0 then there is a fixed point and if g(a)<0 and g(b)>0 there is a fixed point but is that all they can ever be? .... is there an instant where g(a)<0 and g(b)<0 and still be existent with a fixed point in this interval.
    But g(a)< 0 and g(b)< 0 is not possible.
    I think you have forgotten that "f([a,b]) is a set of [a,b]". If f(a)= a, you are done- a is a fixed point. So you can assume that f(a)\ne a. And since f(a) must be in [a,b], f(a)> a. If f(b)= b you are done so you can assume f(b)\ne b and that implies that f(b)< b.
    Defining g(x)= f(x)- x, then, g(a)= f(a)-a> 0 and g(b)= f(b)- b< 0.



    WHY MUST THERE BE A FIXED POINT IN THIS INTERVAL?
    Also can someone please give an example of what f:[a,b] --> R actually looks like. Thank you for your time
    For example, f:[0,1]-> [0,1] defined by f(x)= 1- x is such a function. It's fixed point, of course, is 1/2.

    On the other hand f(x)= x+ 1 is a continuous function from [0,1]--> R which does NOT have a fixed point. The theorem does not hold here because this function does not map an interval [a,b] into itself which is a crucial hypothesis.

    (Blast! Plato got in just ahead of me again! )
    Last edited by HallsofIvy; February 13th 2010 at 08:38 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Fixed Points
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: February 10th 2011, 06:44 PM
  2. Finding fixed points - help please!
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: May 2nd 2010, 03:10 AM
  3. Fixed Points
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 21st 2010, 08:55 AM
  4. fixed points
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 7th 2009, 02:25 PM
  5. fixed points
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 25th 2009, 09:59 AM

Search Tags


/mathhelpforum @mathhelpforum