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Uniform continuity

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data

    if f and g are 2 uniformly continuous functions on X --> R

    show that f+g is uniformly continuous on X

    3. The attempt at a solution

    I tried showing that f+g is Lipschitz because all Lipschitz functions are uniformly continuous.

    So i end up with d(x_1,x_2) < delta then d([f(x1)+g(x1)],[f(x2)+g(x2)])<epsilon. how can I show this is lipshcitz
  2. jcsd
  3. Nov 18, 2008 #2
    forget about lipschitz

    you know that if you pick x1 and x2 close to eachother, then fx1 and fx2 are at most a distance close to eachother, namely epsilon. And also gx1 and gx2 are at most a distance epsilon. so their summed distance is 2 *epsilon.

    so now you can define epsilon' as 2epsilon... so for any distance delta between x1 and x2 you now have a distance epsilon' between f+gx1 and f+gx2, that upholds that "for any epsilon' >0 there is a delta >0 such that the distance between x1 and x2 <delta implies that the distance between f+gx1 and f+gx2 is smaller than epsilon' "... it's just the same delta as before.

    or at least thats what I make of it. It's 1 AM here and I can't finish my own problem so I went to look at other peoples' problems... but I don't have a clear mind right now.

    Oh yeah, and when you write this down... you'll want to use nice math symbols and make things a bit more rigorous... but I think that should be no problem.
    Last edited: Nov 18, 2008
  4. Nov 18, 2008 #3
    Let e>0 be given. Since f(x) is uniformly continuous on X, there exists s > 0, such that y in neighborhood of x, N_s (x), implies |f(y)-f(x)| < e/2. Similarly, since g(x) uniformly continuous, y in N_s (x) implies |g(y) - g(x)| < e/2. Since |f(y)+g(y) - f(x)-g(x)| <= |f(y) - f(x)| + |g(y) - g(x)| < e/2 + e/2, we have |(f+g)(y) - (f+g)(x)| < e, for all y in N_s (x). That is (f+g) uniformly continuous, this completes the proof.
  5. Nov 19, 2008 #4
    I don't think we're supposed to write out the entire answer like that
  6. Nov 19, 2008 #5


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    Homework Helper

    VERY DEFINITELY NOT. It's against the forum rules. "Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made." Don't do that again.
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