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Show that f Uniform Differentiable implies f' Uniform Continuous

  • Thread starter Aryth
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  • #1
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Homework Statement


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].


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.
 

Answers and Replies

  • #2
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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?
 
  • #3
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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?
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]
 
  • #4
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I'm a bit confused...are we trying to show that f' is continuous or uniformly continous?
 
  • #5
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I'm a bit confused...are we trying to show that f' is continuous or uniformly continous?
Just continuous I believe... I may have misstated the problem in the title. Apologies.
 
  • #6
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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)|
 
  • #7
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In the first slope you put (y-x) and in the next one you put (x-y)... Is that right?
 
  • #8
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Yes, but I'm actually adding and subtracting the same thing because I multiplied by -1 on the top and bottom.
 
  • #9
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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.
 

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