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Prove that f is uniformly differentiable

  1. Jan 7, 2012 #1
    1. The problem statement, all variables and given/known data


    Suppose f ' is continuous on [a, b] and ε > 0. Prove that there exists ∂ > 0 such that | [f(t)-f(x)]/[t-x] - f '(x) | < ε whenever 0 < |t - x| < ∂, a ≤ x ≤ b, a ≤ t ≤ b.

    2. Relevant equations

    Definitions of continuity and differentiability

    3. The attempt at a solution


    Fix x in [a, b] and ε > 0. Since f ' is continuous at x, there exists a ∂ > 0 such that |f '(x) - f '(t) | < ε whenever 0 < |x - t| < ∂, a ≤ t ≤ b.

    Now I'm not sure how this gets me | [f(t)-f(x)]/[t-x] - f '(x) | < ε. I've tried a lot of different things. Can you give me an itsy bitsy hint?
     
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  3. Jan 7, 2012 #2

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    What is the definition of f'(x)?
     
  4. Jan 7, 2012 #3
    limt-->x [f(t)-f(x)]/(t-x), provided the limit exists, and I don't know that it does exist in this instance.
     
  5. Jan 7, 2012 #4

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    It is given that f' is continuous on [a,b].
    That implies the limit exists.

    Now what is the definition of the corresponding limit?
     
  6. Jan 7, 2012 #5
    f '(x) = limt-->x [f(t)-f(x)]/[t-x],

    so |f '(t) - f '(x)| = | limp-->x [f(p)-f(x)]/[p-x] - limq-->t [f(q)-f(t)]/[q-t] |. In other words, if we choose ε > 0 then for some ∂ > 0 we've simultaneously have

    | [f(p)-f(x)]/[p-x] - f '(x) | < ε
    and
    | [f(q)-f(t)]/[q-t] - f '(t) | < ε

    whenever
    0 < |p - x|< ∂
    and
    0 < |q - t| < ∂.
     
  7. Jan 7, 2012 #6
    So you're saying that f ' exists on [a, b], which is what we're trying to prove in the first place ..... (?)
     
  8. Jan 7, 2012 #7

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    Your problem statement does not ask you to proof that f' exists.
    It asks to proof that some expression with epsilon turns out true.

    As for the existence of f'.
    What is the definition of continuity (for f')?


    That is not the definition of the corresponding limit.

    Suppose ##L = \lim\limits_{t \to x} ~ g(t)##.
    What is the definition of this limit?
     
  9. Jan 7, 2012 #8
    I actually didn't write down the entire problem. In parentheses it then says "This could be expressed by saying that f is uniformly differentiable on [a, b]." You're telling me that f is uniformly differentiable on [a, b] since f ' is continuous (and thus existant) on [a, b]. Right?



    L is the number for which |L - g(t)| can be made as small we'd like with the appropriate ∂ > 0 and t within the range 0 < |t - x| < ∂.
     
  10. Jan 8, 2012 #9

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    I'm only telling you that f is differentiable on [a,b] since f' is continuous (and thus existant) on [a, b].
    "Uniformly" is as yet not in your problem statement.



    Yes. From this definition the epsilon expression in your problem statement follows immediately.
    That is, pending on the meaning of the word "whenever" - I'm not sure what that means.
    Did you perhaps intend "for every t and x"?

    So this does not yet mean "uniformly differentiable".
    Do you have a definition handy for that?
     
  11. Jan 8, 2012 #10
    I'm almost there. Here's what I know:

    Given x in [a, b] and ε > 0:

    f ' is defined on [a, b], so f is continuous on [a, b] (Theorem 5.10), so the Mean Value Theorem can be applied on any subinterval of [a, b]

    f ' is continuous on [a, b], so there is a ∂1 > 0 such that | f '(x) - f '(t) | < ∂1 whenever 0 < |t - x| < ∂1.

    f '(x), f '(t) of course exist, so there exist ∂2, ∂3 > 0 such that

    | [f(s) - f(t)]/[s - t] - f ' (t) | < ε​
    and
    | [f(y) - f(x)]/[y - x] - f ' x) | < ε​

    whenever
    0 < |s - t| < ∂2
    and
    0 < |y - x| < ∂3.​




    Somehow I can cut and paste the information (possibly changing ε to ε/2 or something else during the process, possibly invoking the Mean Value Theorem) and obtain

    | [f (t) - f(x)] / [t-x] - f '(x) | < ε.​

    I'm just not sure how to do it because it's ton of information, some of it possibly unnecessary. But I'm close, aren't I?
     
  12. Jan 10, 2012 #11
    Figured out that problem today (Or did I????). Please critique.


    screen-capture-6-4.png




    Also, I hope I don't get drunk and forget to turn in my homework, as was often the case last quarter.
     
  13. Jan 10, 2012 #12

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    Well, I'm still not seeing the definition of uniform differentiability...

    In particular it needs to hold for any t and x in [a,b] with 0<|t-x|<δ, but I'm not seeing anything where you addressed that.
     
  14. Jan 10, 2012 #13
    I started with an arbitrary x in [a, b], then picked a t with 0<|t-x|<∂. That, or vice versa (pick t in [a, b] and then an x satisfying 0<|t-x|<∂), should be the correct way.
     
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