1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Show that f Uniform Differentiable implies f' Uniform Continuous

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    A function [itex]f:(a,b)\to R[/itex] is said to be uniformly differentiable iff [itex]f[/itex] is differentiable on [itex](a,b)[/itex] and for each [itex]\epsilon > 0[/itex], there is a [itex]\delta > 0[/itex] such that [itex]0 < |x - y| < \delta[/itex] and [itex]x,y \in (a,b)[/itex] imply that [itex]\left|\frac{f(x) - f(y)}{x - y}-f'(x)\right| < \epsilon[/itex].

    Prove that if f is uniformly differentiable on [itex](a,b)[/itex], then [itex]f'[/itex] is continuous on [itex](a,b)[/itex].


    3. The attempt at a solution

    This is my first time being presented with the definition of uniform differentiability. I suppose that I am looking to show that the definition of uniform differentiability implies [itex]|f'(y) - f'(x)|< \epsilon[/itex]... However, I'm having a hard time doing that. Any help would be appreciated.
     
  2. jcsd
  3. Nov 1, 2011 #2
    Here's one thing to think about: What is it about uniform differentiability that would give you the continuity in f'? What does it mean to be continuous in f, but discontinuous in f'? A good example would be a function like [itex]f(x) = x^2\sin(1/x), x \neq 0, f(x) = 0, x = 0[/itex]. This function is continuous everywhere but does not have a continuous derivative. Thus it is not uniformly differentiable. Why is that?
     
  4. Nov 1, 2011 #3
    All that I have been given at this point is that if f is differentiabile at a point, then f is continuous at that point. So all I could say that it would make f uniformly continuous...

    So, I guess if f is uniformly differentiable, then it has a derivative everywhere, and that should make f' be continuous everywhere.

    Your explanation makes sense, but I guess I should be specific. I'm having trouble understanding the meaning of: [itex]\left|\frac{f(x) - f(y)}{x-y} - f'(x)\right| < \epsilon[/itex]
     
  5. Nov 1, 2011 #4
    I'm a bit confused...are we trying to show that f' is continuous or uniformly continous?
     
  6. Nov 1, 2011 #5
    Just continuous I believe... I may have misstated the problem in the title. Apologies.
     
  7. Nov 1, 2011 #6
    I don't think I've ever heard of uniformly differentiable before... but f(x)-f(y)/(x-y) is the slope of the line connecting f(x) and f(y). The inequality is saying that that slope gets arbitrarily close to the derivative of f at x. You might want to ask: how is that different from the definition of differentiable?

    Here's a hint for the proof:
    |f(y) - f(x)| = |f(y) - [f(y)-f(x)/(y-x)] + [f(x)-f(y)/(x-y)] - f(x)|
     
  8. Nov 1, 2011 #7
    In the first slope you put (y-x) and in the next one you put (x-y)... Is that right?
     
  9. Nov 1, 2011 #8
    Yes, but I'm actually adding and subtracting the same thing because I multiplied by -1 on the top and bottom.
     
  10. Nov 1, 2011 #9
    Ok then, I see what you're doing. I did not notice that the f(x) and the f(y) were also switched.

    That helped me finish it, thank you very much.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Show that f Uniform Differentiable implies f' Uniform Continuous
Loading...