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Simple proof of uniform continuity

  1. Mar 28, 2012 #1
    If the function f:D→ℝ is uniformly continuous and a is any number, show that the function a*f:D→ℝ also is uniformly continuous.

    Ok, so I am just learning my proofs so be patient with me, i'm very new at it.

    take a>0, ε>0 and x,y in D. We know |x-y|<δ whenever |f(x)-f(y)|<ε.
    If we take a*f:D→ℝ, we have |a*f(x)-a*f(y)|<ε → |a*[f(x)-f(y)]|<ε→
    a*|f(x)-f(y)|<ε→ |f(x)-f(y)|<ε/a. Therefore, if we use ε/a, the result is proven.

    This just seems a little too easy to me, plus I've only done a few of these on my own. any suggestions/advice are greatly appreciated. Also, do I need to do this separately for a<0?
  2. jcsd
  3. Mar 28, 2012 #2


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    You have this written backwards. As written, you're saying that if you pick values of x and y such that |f(x)-f(y)| is small, then |x-y| must be small as well. But this isn't true -for example if f is a constant function - and isn't what you want with uniform continuity. The actual statement is |f(x)-f(y)|<ε whenever |x-y|<δ (note that you need to state at some point that given epsilon, you can find delta such that this is true)

    Your next string of logic looks good until this point:
    Use ε/a for what? To prove uniform continuity, you need: for any ε that I give you, you must produce δ such that if |x-y|<δ, then |a*f(x)-a*f(y)|<ε. Unless you're suggesting δ = ε/a you need to expand upon this to have a complete proof with all the details
    The a<0 case is handled by factoring the a out of the absolute value signs and getting an |a|, not just an a. Also the case a=0 isn't covered by the proof (but that's a pretty easy case)
  4. Mar 28, 2012 #3


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    Assuming that by a you mean a number and a*f is just multiplication of the function f by a, yes, it really is that easy!
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