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Prove f-g is uniformly continuous

  1. Mar 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Let f, g : D→R be uniformly continuous. Prove that f-g: D→R is uniformly continuous aswell



    2. Relevant equations

    none

    3. The attempt at a solution

    Okay, I am posting this question because I want to make sure that my solution is correct and if it isn't I would really be thankful if someone pointed out its flaws.


    My solution:

    For all ε >0 there exists δ >0 s.t.

    abs( x - y ) < δ => abs( f(x) - f(y)) <ε

    then

    abs( f(x) - f(y)) <ε/2

    abs( g(x) - g(y)) <ε/2

    now its f - g so

    abs( f(x) - f(y) - g(x) + g(y)) < ε

    abs( f(x) - f(y)) + abs( - g(x) + g(y)) < ε

    abs( f(x) - f(y)) + abs(-1)*abs( +g(x) -g(y)) < ε

    abs( f(x) - f(y)) + abs( +g(x) -g(y)) < ε

    ε/2 + ε/2 <ε

    ε ≤ ε

    Also can anyone describe to me what uniformly continuous means besides from the actual definition. I have a hard time understanding what uniformly continuous means.

    Thanks.
     
  2. jcsd
  3. Mar 27, 2013 #2

    LCKurtz

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    How do you know the same ##\delta## works for both f and g?
     
  4. Mar 27, 2013 #3
    Can't you just choose what ε and δ is?

    It can have many values as long as its > 0. Right?
     
  5. Mar 27, 2013 #4

    LCKurtz

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    No. ##\epsilon## is a given, presumably small, number. If you have a uniformly continuous function ##f##, that means that given any ##\epsilon > 0## there exists a ##\delta##, depending only on ##\epsilon## and not ##x## or ##y## such that if ##|x-y|<\delta##, then ##|f(x)-f(y)|<\epsilon##. You don't get to choose ##\delta## but you are guaranteed that there is one. A different function ##g## would normally need a different ##\delta## for the same ##\epsilon##.

    There are other issues with your writeup, but first you have to figure out how you can get one ##\delta## that works for both ##f## and ##g##.
     
  6. Mar 27, 2013 #5
    I'm going to take a few guesses.

    1. What if both f and g are the same function then they would have the same δ and ε values.

    2. Because they are both uniformly continuous.

    3. Because I am picking the same ε for each function.

    I have tried to understand the whole ε δ thing many times and although I think I kind of get it I don't think that I have a fundamental understanding of it.

    Also, my professor just picked the same δ like I did for a similar problem. I do know that they don't have to be the same.
     
  7. Mar 27, 2013 #6

    LCKurtz

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    They aren't, so that is irrelevant.
    No.
    No.
    Yes, I don't think you quite have it yet either. Think about my question like this. Suppose you have a ##\delta_1## that makes ##|f(x)-f(y)|< \epsilon## when ##|x-y|<\delta_1## and you also have a ##\delta_2## that makes ##|g(x)-g(y)|< \epsilon## when ##|x-y|<\delta_2##. Can you see how to make a ##\delta## from that that works for both?
     
  8. Mar 28, 2013 #7
    Actually now that I'm looking at my professors notes more I do see that he indeed labeled δ1 and δ2 separately even though they are in the end the same.

    So is there anything else wrong with my write up? I just turned in the homework yesterday and I will see what I get on this problem.
     
  9. Mar 28, 2013 #8

    LCKurtz

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    No, they aren't the same. What you want is to to pick ##\delta_1## for ##f## and ##\delta_2## for ##g## so ##|f(x) - f(y)|< \frac \epsilon 2## when ##|x-y|<\delta_1## and ##|g(x) - g(y)|< \frac \epsilon 2## when ##|x-y|<\delta_2##. Then choose ##\delta = \text{min}\{\delta_1,\delta_2\}## so that when ##|x-y|<\delta## they both work.

    Then in your writeup, you don't want to end up with ##\epsilon\le \epsilon##. After the above paragraph you want to say that if ##|x-y|<\delta## then$$
    |f(x)+g(x) - (f(y)+g(y))| \le ... < \epsilon$$where you fill in the dots with a string of inequalities similar to what you did write.
     
    Last edited: Mar 28, 2013
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