1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Real Analysis: Properties of Continuity

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Suppose f is continuous on [0,2]and thatn f(0) = f(2). Prove that there exists x,y in [0,2] such that |y-x| = 1 and f(x) = f(y)

    2. Relevant equations

    3. The attempt at a solution

    I got the following 1 line proof.

    Suppose g(x) = f(x + 2) - f(x) on I = [0,2]

    this proofs that |x - y| = 1 for x = 1, y = 2

    and f(x) = f(y)

  2. jcsd
  3. Nov 9, 2008 #2
    As you have defined g, unless you know more about f, then g is only defined at 0 and no other points in [0,2]. I mean g(1)=f(1+2)-f(1)=f(3)-f(1), but what is f(3)?

    And how does what you have written show that f(1)=f(2) certainly there are continuous functions on [0,2] with f(0)=f(2) but that do not satisfy f(1)=f(2).

    What do you know about the function h(x)=f(x+1)-f(x)?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Real Analysis: Properties of Continuity