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Showing composition of functions are uniformly continuous

  1. Sep 27, 2012 #1
    Showing the sum of functions are uniformly continuous

    1. The problem statement, all variables and given/known data
    Suppose f and g are uniformly continuous on an interval I. Prove f + g are uniformly continuous on I.

    2. Relevant equations



    3. The attempt at a solution

    Let ε >0

    By definition, since f and g are uniformly continous on I, there exists a [itex]\delta[/itex]_1 such that |f(y)-f(x)| < ε for all x,y in I that satisfy |x-y| < [itex]\delta[/itex]_1

    Similarly, for g, there exists a [itex]\delta[/itex]_2 such that |g(y)-g(x)| < ε for all |y-x| < [itex]\delta[/itex]_2

    Then, for all x,y in I, |f(y)+g(y)-(f(x)+g(x))| ≤ |f(y)-f(x)|+|g(y)-g(x)| by the triangle inequality. This implies |f(y)+g(y)-(f(x)+g(x))| < 2ε

    Choose [itex]\delta[/itex]=min{[itex]\delta[/itex]_1,[itex]\delta[/itex]_2} * 1/2

    Then |f(y)+g(y)-(f(x)+g(x))| < 2ε for all x,y in I that satisfy |y-x| < [itex]\delta[/itex]

    ∴f+g is uniformly continuous on I.

    Is this correct?
     
    Last edited: Sep 27, 2012
  2. jcsd
  3. Sep 27, 2012 #2

    LCKurtz

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    Yes, but a couple of minor changes would make it slightly nicer. For your first two inequalites use ##\frac \epsilon 2## so it will only add to ##\epsilon## at the end. And choosing ##\delta##, there is no need to multiply by 1/2. Finally, where you have:

    |f(y)+g(y)-(f(x)+g(x))| ≤ |f(y)-f(x)|+|g(y)-g(x)| by the triangle inequality. This implies |f(y)+g(y)-(f(x)+g(x))| < 2ε

    you could instead just write

    |f(y)+g(y)-(f(x)+g(x))| ≤ |f(y)-f(x)|+|g(y)-g(x)| < ε/2+ε/2 =ε.

    Nice work. And, by the way, you have shown the sum of two u.c. functions is u.c. If you really meant the composition, you did the wrong problem :uhh:
     
    Last edited: Sep 27, 2012
  4. Sep 27, 2012 #3
    Oops! I did mean the sum. I am not sure why I wrote that... Thank you for the suggestions. I will rewrite it to make it look nicer!
     
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