# Uniform Convergence?

1. Apr 10, 2007

### pivoxa15

If f is uniformly continous then does that mean f^-1 its inverse is also?

Last edited: Apr 10, 2007
2. Apr 10, 2007

### StatusX

Do you mean uniformly continuous? Then no, even if f is invertible. For example, take f=x^3, which is uniformly continuous, but whose inverse, x^1/3, has infinite slope at x=0, and so isn't uniformly continuous.

3. Apr 10, 2007

### Dick

You mean uniformly continuous, right? And the answer is no. Consider tan(x) and arctan(x).

4. Apr 10, 2007

### pivoxa15

Yes. Uniformly continous. f=x^3 is not uniformly continous, however.

5. Apr 10, 2007

### StatusX

Ok, fine, take it on the interval [-1,1], or redefine it outside a region like this so the slope doesn't blow up. You should notice that the problem comes from the origin, so changing it's behavior far away from this point won't help anything.

6. Apr 11, 2007

### pivoxa15

Why is the problem at the origin? It seems to me that the bound on y is increading at the end of the domains. So harder to keep it fixed when the domain is infinite. So confining the domain is an option as you stated.

7. Apr 11, 2007

### pivoxa15

Is h(x)=sqrt(x) on [0,infinity) uniformly continuous?

8. Apr 11, 2007

### matt grime

The 'problem' is that which makes the inverse not uniformly continuous.

Anyway, why did you assume that the domain had to be all of R? Clearly it doesn't need to be - it can be anything - and with the domain [0,1] for x^3, you have a counter example. If someone posts a bit of help, maybe think about what they wrote for a while before telling them its wrong.

9. Apr 11, 2007

### StatusX

Sorry, x^1/3 on a compact set is uniformly continuous by heine-cantor. I must have been thinking of Lipschitz continuity. Just use dick's example.

10. Apr 11, 2007

### Dick

x^(1/3) on [0,1] IS uniformly continuous. It just doesn't have a bounded derivative.

11. Apr 11, 2007

### pivoxa15

So is this the reasoning:

Uniform continuity means choose any fixed epsilon>0 such that there exists a minimum delta in the domain. In this case the minimum delta=d(0,x) for every fixed epsilon. d(x,y) denotes the distance between x and y in the domain using the standard metric.

However if for the same function on (0,1] then there isn't a min delta for every fixed epsilon because for any epsilon, one can keep going towards the origin so that delta can always be made smaller.

x^3 would be uniformly continous on [0,1] but not on [0,1).

12. Apr 11, 2007

### Dick

It's still uniformly continuous on [0,1). Where did you find a rule that says delta has to be smaller than the distance to the end of your domain?

13. Apr 11, 2007

### pivoxa15

So x^3 is uniformly continous on [0,1)?
The minimum delta would be d(x,1) but 1 is not in the domain. Is that a problem?

If that is not a problem than how is x^(1/3) on [0,1] uniformly continuous?

Can you rephrase your question? I don't understand it very well.

Last edited: Apr 11, 2007
14. Apr 11, 2007

### Dick

The point is that |x-x0|<delta doesn't mean that all such x need to be in the domain. There is no such rule in the definition of uniform continuity.

15. Apr 12, 2007

### matt grime

That is a very strange way of defining uniform continuity.

f is uniformly continous if for all e, there is a d such that |x-y|<d implies |f(x)-f(y)|<e.

What has 'the minimum delta' got to do with anything?