Uniform Continuity and Supremum

[renjean]

thanks!

 

[Dick]

Homework Statement

Homework Equations

Give an example of a function f that is uniformly continuous on [-1,1] such that
sup{ [f(x)-f(y) / [x-y] } = infinity

The Attempt at a Solution

I have tried to come up with functions for hours but I am just not getting it. Any help would be appreciated.

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

 

[LCKurtz]

Hint: That's a difference quotient.

 

[micromass]

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

 

[renjean]

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

 

[micromass]

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

 

[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.

 

[micromass]

I thought that any continuous function on a closed and bounded interval is also uniformly continuous.

Correct.

And in the case of x*sin(1/x) the derivative appears to go to infinity at x=0.

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)

 

[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].

 

[micromass]

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].

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.

 

[chairbear]

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

 

[micromass]

That should do it.

 

[chairbear]

thanks

 

[Dick]

Thank you for your help. I was wondering if you could help me to get started on another question I have.

I have to prove that for a function f with f'(0)=0, there's a sequence xn that converges to 0 for all n such that f'(xn) converges to 0. and xn can't = 0 for any n.

I'm not sure exactly how to get started on this, because I'm not sure if it's supposed to be a rigorous proof, or if I'm just supposed to come up with a sequence that satisfies the conditions for some f.

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

 

[chairbear]

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

 

[Dick]

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

Then maybe use the mean value theorem?

 

[I like Serena]

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

 

[micromass]

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

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