Proving Uniform Continuity for Functions with a<1

[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&lt;|x-y|&lt;d, |f(x)-f(y)|\leq a|x-y|&lt;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.

 

[EnumaElish]

Seems right to me. See also.

 

[HallsofIvy]

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

The difference is that, for uniform continuity, the same \delta works for all x₀.
If |f(x)-f(y)|&lt; a|x-y|, then in particular, for any x₀, |f(x)-f(x_0)|&lt; 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.

 

[Icebreaker]

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

 

[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".

 

[Icebreaker]

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

 

[EnumaElish]

Isn't there a theorem you can use?

 

[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.