Differentiable and uniformly continuous?

In summary, if f is differentiable and its derivative is bounded on the open interval (a,b), then f is uniformly continuous on (a,b). This can be shown using the mean value theorem or by manipulating the definition of uniform continuity.
  • #1
icantadd
114
0
differentiable and uniformly continuous??

Homework Statement


Suppose f:(a,b) -> R is differentiable and | f'(x) | <= M for all x in (a,b). Prove f is uniformly continuous on (a,b).


Homework Equations


The definition of uniform continuity is:
for any e there is a d s.t. | x- Y | < d then | f(x) -f(y | < e.




The Attempt at a Solution


Intuitively, if f is differentiable it is continuous. If its derivative is bounded it cannot change fast enough to break continuity. The interval is bounded, and the function must be bounded on the open interval. It seems that there is not way that the function cannot be uniformly continuous. But how do I say that? Or am I on the wrong track altogether.
 
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  • #2


icantadd said:

Homework Statement


Suppose f:(a,b) -> R is differentiable and | f'(x) | <= M for all x in (a,b). Prove f is uniformly continuous on (a,b).

Homework Equations


The definition of uniform continuity is:
for any e there is a d s.t. | x- Y | < d then | f(x) -f(y | < e.

The Attempt at a Solution


Intuitively, if f is differentiable it is continuous. If its derivative is bounded it cannot change fast enough to break continuity. The interval is bounded, and the function must be bounded on the open interval. It seems that there is not way that the function cannot be uniformly continuous. But how do I say that? Or am I on the wrong track altogether.
Watch this:
[tex]
|f(x) -f(y)| = \frac{|f(x) -f(y)|}{|x -y|} \cdot |x-y| < M \cdot \delta
[/tex]

Can you verify these steps?
 
  • #3


dirk_mec1 said:
Watch this:
[tex]
|f(x) -f(y)| = \frac{|f(x) -f(y)|}{|x -y|} \cdot |x-y| < M \cdot \delta
[/tex]

Can you verify these steps?

So long | x-y | not equal 0.

Yeah, I got the same thing. Actually a little bit differently, I used the mean value theorem, because you don't know that the function is bounded, only its derivative. The mean value theorem gets you to the point where you know that f '(c) = [f(x) - f(y)] / (x-y) is <= M (because f '(x) <= M for all x). Then you just multiply both sides by | x - y | and get the same end result. As long as | x - y| < e/M |f(x) - f(y) | < e.

Thank you for your help!
 

1. What is the difference between differentiable and uniformly continuous functions?

Differentiable functions are those that have a defined derivative at every point in their domain, while uniformly continuous functions have a limit on the variation of their values as the input changes.

2. How do I know if a function is differentiable or uniformly continuous?

A function is differentiable if it has a defined derivative at every point in its domain. It is uniformly continuous if the variation of its values is limited as the input changes.

3. Are all differentiable functions also uniformly continuous?

No, not all differentiable functions are uniformly continuous. However, all uniformly continuous functions are differentiable.

4. What is the importance of differentiable and uniformly continuous functions in mathematics and science?

Differentiable and uniformly continuous functions are important in understanding the behavior of functions and their derivatives. They are used in various fields such as physics, engineering, and economics to model and analyze real-world phenomena.

5. Can a function be differentiable but not uniformly continuous?

Yes, a function can be differentiable but not uniformly continuous. This occurs when the function has a defined derivative at every point in its domain, but its values vary drastically as the input changes.

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