Results 1 to 9 of 9

Math Help - analysis - continuous functions

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    114

    analysis - continuous functions

    Just going through past papers as my anaysis exam is on Weds, but I'm quite stuck/unsure about 2 questions I've come across...


    If g:[0,1] ---> [0,1] is a continuous function, show that there exists x in [0,1] such that g(x) = x.

    Let F(x) = g(x) - x be defined on [0,1].
    F is continuous on [0,1] since it is the difference of 2 continuous functions.
    If F(0) = 0 or if F(1) =0 then we have our fixed point.
    Otherwise, F(0) = g(0) - 0 > 0.
    F(1) = g(1) - 1 < 0.

    Since g(x) in [0,1] for all x in [0,1], by IVT there exists c in [0,1] such that F(c) = 0, hence g(c) = c.

    Is this an OK answer, and would it be worthy of 8 marks?


    Let f: R ---> R be a continuous function, such that f(x) in the rationals for all x. Show that f is a constant function.

    This one I'm really stuck on and although it seems alot easier I have no idea what to say for it...can anyone please help?

    Thanks in advance!
    Last edited by hunkydory19; April 26th 2008 at 06:34 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi
    Quote Originally Posted by hunkydory19 View Post
    Since g(x) in [0,1] for all x in [0,1], by IVT there exists c in [0,1] such that F(c) = 0, hence g(c) = c.
    I don't understand why you write "Since g(x) in [0,1] for all x in [0,1]". Saying " F(0)F(1)<0 and F is continuous on [0,1] hence by IVT..." would be enough.

    Let f: R ---> R be a continuous function, such that f(x) in the rationals for all x. Show that f is a constant function.

    This one I'm really stuck on and although it seems alot easier I have no idea what to say for it...can anyone please help?
    One possibility : assume that f is not a constant function i.e. there exists x_0 \in \mathbb{R} such that f(x_0)\neq f(1) and then try to show that it is nonsense using the fact that \mathbb{Q} is a dense subset of \mathbb{R}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by hunkydory19 View Post
    Just going through past papers as my anaysis exam is on Weds, but I'm quite stuck/unsure about 2 questions I've come across...


    If g:[0,1] ---> [0.1] is a continuous function, show that there exists x in [0,1] such that g(x) = x.

    Let F(x) = g(x) - x be defined on [0,1].
    F is continuous on [0,1] since it is the difference of 2 continuous functions.
    If F(0) = 0 or if F(1) =0 then we have our fixed point.
    Otherwise, F(0) = g(0) - 0 > 0.
    F(1) = g(1) - 1 < 0.

    Since g(x) in [0,1] for all x in [0,1], by IVT there exists c in [0,1] such that F(c) = 0, hence g(c) = c.

    Is this an OK answer, and would it be worthy of 8 marks?
    [snip]
    I'd be inclined to set it out in the following way:

    Let f(x) = g(x) - x be defined on [0,1].

    f is continuous on [0,1] since it is the difference of two continuous functions.

    0 \leq g(x) \leq 1 \Rightarrow -x \leq g(x) - x \leq 1 - x \Rightarrow -x \leq f(x) \leq 1 - x.

    Therefore 0 \leq f(0) \leq 1 and -1 \leq f(1) \leq 0.

    Therefore f(0) \geq 0 and f(1) \leq 0.

    Therefore, by the IVT there exists c in [0, 1] such that f(c) = 0 \Rightarrow g(c) - c = 0 \Rightarrow g(c) = c.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    I'd be inclined to set it out in the following way:

    Let f(x) = g(x) - x be defined on [0,1].

    f is continuous on [0,1] since it is the difference of two continuous functions.

    0 \leq g(x) \leq 1 \Rightarrow -x \leq g(x) - x \leq 1 - x \Rightarrow -x \leq f(x) \leq 1 - x.

    Therefore 0 \leq f(0) \leq 1 and -1 \leq f(1) \leq 0.

    Therefore f(0) \geq 0 and f(1) \leq 0.

    Therefore, by the IVT there exists c in [0, 1] such that f(c) = 0 \Rightarrow g(c) - c = 0 \Rightarrow g(c) = c.
    Now you should prove that any continuous bounded function over the real numbers f(x) must cross the function g(x) = x.
    Last edited by mr fantastic; April 26th 2008 at 02:10 PM. Reason: Forgot to include "over the real numbers"
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2008
    Posts
    114
    Cheers mr fantastic, that does indeed seem like a better answer.

    By any chance do you have an hints/blatent help, for the second question...as i tried the approach flying squirrel suggested but couldnt get anywhere...

    Thanks again!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Quote Originally Posted by hunkydory19 View Post
    as i tried the approach flying squirrel suggested but couldnt get anywhere...
    Assume f is not constant on \mathbb{R} : there exists x_0\in \mathbb{R} such that f(x_0)\neq f(1).
    The idea is to apply the definition of continuity at x_0 : \forall \varepsilon >0, \, \exists\, \delta >0 \, | \, | x-x_0 | < \delta \Rightarrow | f(x)-f(x_0) | <\varepsilon
    As \mathbb{Q} is a dense subset of \mathbb{R}, given \delta>0, there exists a rational number q\,|\, |x_0-q|<\delta hence by choosing \varepsilon=\frac{| f(1)-f(x_0)|}{2} we get that : | f(q)-f(x_0) | = | f(1)-f(x_0) | < \frac{| f(1)-f(x_0)|}{2} which is a contradiction. Hence f is constant on \mathbb{R}.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by hunkydory19 View Post
    Let f: R ---> R be a continuous function, such that f(x) in the rationals for all x. Show that f is a constant function.
    Using the IVT this proof is almost trivial.
    If f is not constant then there two values f\left( {x_1 } \right) \ne f\left( {x_2 } \right).
    Between any two numbers there is an irrational number.
    So some number between x_1 \,\& \,x_2 is mapped onto it by IVT.
    So that is a contradiction.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jan 2008
    Posts
    114
    Thanks Plato that is an amazing proof, nice and short would this be worthy of 7 marks in an exam though?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by hunkydory19 View Post
    Thanks Plato that is an amazing proof, nice and short would this be worthy of 7 marks in an exam though?
    Forget about marks and focus on:
    1. The maths.
    2. The presentation of the maths (eg. the way a proof is constructed).

    The marks will take care of themselves ......
    Last edited by mr fantastic; April 26th 2008 at 02:28 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Analysis - Continuous Functions
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 7th 2010, 09:26 AM
  2. Real Analysis - Combinations of Continuous Functions #2
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 22nd 2008, 12:14 AM
  3. Real Analysis Continuous Functions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 7th 2008, 07:16 AM
  4. Real Analysis continuous functions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 29th 2007, 12:46 PM
  5. Analysis Continuous Functions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 23rd 2007, 05:19 AM

Search Tags


/mathhelpforum @mathhelpforum