# About uniform continuity and derivative

1. Apr 5, 2013

### A Dhingra

hello
(pardon me if this is a lame question, but i got to still ask)

If a function is uniformly continuous (on a given interval) then is it required for the derivative of the function to be continuous?
I was thinking as per the definition of Uniform continuity, f(x) should be as close to f(y) as possible, for x around y, meaning change in f should not be sudden at some point within the interval; it should rise or fall in a uniform manner, suggesting the slope of the function at different points should change gradually and be real always (i.e. never infinite), that implies the derivative should be continuous in the open interval. So if we have a function whose derivative is continuous then can we say it is uniformly continuous on the closed interval?

(I must mention i won't be able to understand very rigorous proofs, so if possible explain this either by counter example, if any, or geometrically. )

thank you..

2. Apr 5, 2013

### Staff: Mentor

Consider f: [-1,1]->R, f(x)=x^2 sin(1/x) for x!=0, f(0)=0

3. Apr 6, 2013

### Stephen Tashi

A uniformly continuous function may not have a derivative at some points. If your mean to assume the derivative exists everywhere, you should state this.

4. Apr 7, 2013

### A Dhingra

So in the interval [-1,1] function f(x) is continuous.
And its uniform continuity can be easily seen on [-1, 1]
The derivative of the function is =(1/x)( 2x^2 sin(1/x)- x cos(1/x))
at x=0, it is -cos(1/x) = not defined could be any value between [-1,1]

So??

5. Apr 7, 2013

### A Dhingra

stephen, the above example states what you said...
Okay if i talk about the converse, i.e. if a function has derivative over the interval then the function is uniformly continuous, is this right??

6. Apr 7, 2013

### Staff: Mentor

The derivative of the function I posted is well-defined at x=0 (it is 0). The approach you tried does not work at x=0.

7. Apr 7, 2013

### Stephen Tashi

To say "the interval" is not specific. Do you mean an open interval? a closed interval?

8. Apr 8, 2013

### Bacle2

To follow up on Stephen, differentiability implies continuity; continuity in a compact subset implies....

9. Apr 8, 2013

### A Dhingra

Should be closed interval.....

"To follow up on Stephen, differentiability implies continuity; continuity in a compact subset implies.... "
Does this imply uniform continuity? if the differentiation of function f(x) is g(x), is also continuous ...

10. Apr 8, 2013

### Stephen Tashi

Yes, a function that is differentiable everywhere on a closed interval is uniformly continuous on that interval.

What are you asking? If f(x) is differentiable at x and g(x) = f'(x) then g(x) itself need not be continuous at x.

11. Apr 8, 2013

### Bacle2

Take, for example, f(x)=|x|; it is not differentiable at 0 . Find its integral ( it is continuous, so the indefinite integral exists ) , and use the FTCalculus:

f(x) = -x^2/2 , if x<0
f(0) =0
f(x) =x^2/2

Now, f is continuous, and, in a compact interval, it is uniformly continuous, and f'(x)=|x|.

To folllow up on mfb's point, the derivative at 0 should be:

Lim_h->0 f(h)/h .

(Sorry, mfb, don't mean to speakfor you, just thought you may be away for a while)

Last edited: Apr 8, 2013
12. Apr 8, 2013

### lavinia

any continuous function on a compact set - such as a closed interval - is uniformly continuous. But such a function need not be differentiable anywhere.

If the function is continuously differentiable the its derivative is uniformly continuous on a compact set.

13. Apr 8, 2013

### Bacle2

Actually, take a function with countably-many points of discontinuity, e.g., the Dirichlet function. Then its Riemann integral exists, since functions only need tobe a.e. continuous in order to be Riemann integrable. Now,consider the integral of the Dirichlet function over a compact set . It is uniformly continuous, and its derivative ( by the FTCalc) has countably-infinite many points of discontinuity.

EDIT: Please see my corrections below, in #16.

Last edited: Apr 8, 2013
14. Apr 8, 2013

### micromass

Staff Emeritus
Isn't the integral of the Dirichlet function simply the constant zero function?

Anyway, there can never exist a function whose derivative is the Dirichlet function. The reason is that derivatives always need to satisfy the intermediate value property. This is Darboux's theorem: http://en.wikipedia.org/wiki/Darboux's_theorem_(analysis) . However, the Dirichlet function does not satisfy the intermediate value property.

15. Apr 8, 2013

### lavinia

take a sample path from a continuous Brownian motion on a closed interval. It is uniformly continuous
and is nowhere differentiable.

16. Apr 8, 2013

### Bacle2

My bad; I meant the Thomae function; the Dirichlet function is nowhere-continuous , so it cannot be Riemann integrable. Of course, you want to exclude cases like a function being a.e. 0 , like the Cantor function; the condition is absolute continuity,which allows you to recover the function from integration

EDIT: I keep mixing functions in my head; please see my post after this one.

Last edited: Apr 8, 2013
17. Apr 8, 2013

### micromass

Staff Emeritus
Yes, I was talking about the Thomae function as well (for some reason, I always called that the Dirichlet function, but now I see that it is something different). The same remarks hold.

18. Apr 8, 2013

### Bacle2

EDIT: Well, yes, it is a.e. 0 ; just take copies/iterates of |x| in a closed interval. So it is covered (modulo my mistake) in my previous post,where I excluded functions that are a.e. 0, but mistakenly mentioned Thomae. So I think the Darboux Thm is overkill here; I think it is obvious issue that being a.e. 0 means the integral is 0, so that you cannot recover the original function, and absolute continuity.

Last edited: Apr 8, 2013
19. Apr 9, 2013

### A Dhingra

Does every one agree to this?

If yes....I think i got the answer.
(don't mind but i don't want to get into reimann integral and those functions again, i had them last semester )

20. Apr 9, 2013

### Staff: Mentor

That can be proven, we don't need votes ;).

Note that "If the function is continuously differentiable" is a requirement you did not have in the first post.