# Homework Help: Uniform Continuity 2

1. Oct 31, 2005

### Icebreaker

"Let $$f:[a,b]\rightarrow [a,b]$$ be defined such that $$|f(x)-f(y)|\leq a|x-y|$$ where 0<a<1. Prove that f is uniformly continuous and (other stuff)."

Let e>0 and let d=e/a. Whenever $$0<|x-y|<d, |f(x)-f(y)|\leq a|x-y|<ad=e$$. f is therefore by definition uniformly continuous.

Did I do this right? It seems too good to be true. It doesn't seem right because I did not use the fact that 0<a<1.

Last edited by a moderator: Oct 31, 2005
2. Oct 31, 2005

3. Nov 1, 2005

### HallsofIvy

Ordinary continuity says "For any x0, for any $$\epsilon$$>0, there exist $$\delta$$> 0 such that if $$|x-x_0|< \delta$$, then $$|f(x)-f(x_0)|< \epsilon$$".
Uniform continuity says "For any $$\epsilon$$>0, there exist $$\delta$$> 0 such that, for any x0, if $$|x-a_0|< \delta$$, then $$|f(x)-f(x_0)|< \epsilon$$".

The difference is that, for uniform continuity, the same $$\delta$$ works for all x0.
If $$|f(x)-f(y)|< a|x-y|$$, then in particular, for any x0, $$|f(x)-f(x_0)|< a|x-x_0|$$, then given any $$\epsilon$$, we can choose $$\delta= \frac{\epsilon}{a}$$ for all x0.

That's essentially what you are saying since you seem to be using a slightly different (equivalent) definition of uniform continuity.

4. Nov 1, 2005

### Icebreaker

I just find it odd that it gave me a property (0<a<1) which I did not need. Thanks for your help.

5. Nov 1, 2005

### HallsofIvy

What was the "(other stuff)"? 0< a< 1 may be need for that. It means that f is a "contraction" function and, among other things, has a unique "fixed point".

6. Nov 1, 2005

### Icebreaker

The other thing was to show that there exists an e in [a,b] such that f(e) = e.

7. Nov 1, 2005

### EnumaElish

Isn't there a theorem you can use?

8. Nov 1, 2005

### Icebreaker

Assuming f(a) is not a and f(b) is not b, I've basically proved that a continuous function defined on [a,b] must intersect the identity function g(x)=x at least once. I haven't been able to prove uniqueness. And yes, there are lots of theorems I can use.