# Uniform Continuity and Supremum

1. Dec 14, 2011

### renjean

thanks!!!!

Last edited: Dec 14, 2011
2. Dec 14, 2011

### Dick

What kinds of functions have you tried and why do you think they aren't working?

3. Dec 14, 2011

### LCKurtz

Hint: That's a difference quotient.

4. Dec 14, 2011

### micromass

Staff Emeritus
Further hint: find a function whose derivative is infinite/doesn't exist...

5. Dec 14, 2011

### renjean

Does x*sin(1/x) work since its derivative is undefined at x=0 which is in [-1,1]?

6. Dec 14, 2011

### micromass

Staff Emeritus
Well,

1) is it uniform continuous??
2) Do the difference quotients go to infinity??

Just saying that the derivative is undefined isn't really enough...

PS There is a much easier example

7. Dec 14, 2011

### chairbear

I thought that any continuous function on a closed and bounded interval is also uniformly continuous. And in the case of x*sin(1/x) the derivative appears to go to infinity at x=0.

8. Dec 14, 2011

### micromass

Staff Emeritus
Correct.

Not really. Rather, the derivative does not exist (since it oscillates too much). Nevertheless the supremum you mention does indeed go to infinity. (you might want to give a further proof if it is not clear)

9. Dec 14, 2011

### chairbear

I feel silly for making things more complicated than necessary. What was the easier example you had in mind? I was thinking square root x would work if the interval was [0,1].

10. Dec 14, 2011

### micromass

Staff Emeritus
Don't feel silly. Your example is very elegant.

The square root is indeed the one I had in mind. You just need to modify it a bit.

11. Dec 14, 2011

### chairbear

So it would just be sqrt(x+1) for the [-1,1] interval as another solution?

12. Dec 14, 2011

### micromass

Staff Emeritus
That should do it.

13. Dec 14, 2011

### chairbear

thanks

Last edited: Dec 14, 2011
14. Dec 14, 2011

### Dick

It's not even true. There are functions that differentiable at x=0, that aren't even differentiable anywhere else.

15. Dec 14, 2011

### chairbear

Sorry, there's a condition also that f: R-->R and must be differentiable on R

16. Dec 14, 2011

### Dick

Then maybe use the mean value theorem?

17. Dec 15, 2011

### I like Serena

Just out of curiosity, renjean and chairbear, what is the reason you deleted your questions?

18. Dec 15, 2011

### micromass

Staff Emeritus
It's because they are cheating. Please report these kind of things.

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