Uniformly continuous function

1. Dec 5, 2009

A-ManESL

Suppose f:[0,$$\infty]\rightarrow \mathbb{R}$$ is a continuous function such that $$\lim_{x\rightarrow \infty} f(x)=1$$. I want to show that f is uniformly continuous. Thanks.

2. Dec 5, 2009

Hurkyl

Staff Emeritus
Every continuous function on a compact set is uniformly continuous.

Did you mean $[0, +\infty)$?

3. Dec 5, 2009

A-ManESL

Yes I meant $$f:[0,\infty)\rightarrow \mathbb{R}$$. Sorry for the mistake. Obviously $$[0,\infty)$$ is not compact and so the above stated result doesn't apply.

4. Dec 5, 2009

Hurkyl

Staff Emeritus
P.S. I was being sneaky -- my question is also a hint.

5. Dec 5, 2009

A-ManESL

Can you please me more explicit? All I can think of is that for large x, f(x) will be close to 1.

6. Dec 5, 2009

Hurkyl

Staff Emeritus
Well, I do have to give you a chance to see for yourself how to fit the ideas you've learned together. Otherwise, you won't learn!

If f really were a function on $[0, +\infty]$, the problem would be easy....

Or, if you don't want to think my way, then we can think your way. What is the definition of uniform continuity? How might we be able to apply this fact you've observed?

p.s. do you understand the proof that any continuous function on a compact set is uniformly continuous? If not, then you really don't have much chance of getting this proof to work.... It would definitely be worth your time to review that proof.

7. Dec 6, 2009

Take_it_Easy

That's very easy to show, once you can use the theorem "f continuous on a compact => f uniformly continuous", even though I had to stress some tecnichal detail.

Take eps > 0.

You have to find a delta > 0 such that for every couple of elements x1, x2 in [0, +infty) such that | x1 - x2 | < delta you have |f(x1) - f(x2)| < eps.

From the definition of limit you have that an M positive real exist such that for every x > M
you are sure |f(x) - 1| < eps/2.

Now on [0,M+1] f is uniformly continuous, so you have a delta1 >0 such that for every x1,x2 in [0,M+1] we have |f(x1) - f(x2)| < eps.

Take delta = min(delta1, 1).

We have trivially that, for every x1,x2 in [0, M+1] with | x1 - x2 | < delta we get
|f(x1) - f(x2)| < eps and delta is in (0,1].

Ok. Now take x1, x2 in [0, + infty) with | x1 - x2 | < delta. If the greatest of the 2 xs lays in [0,M+1] you have, of course, |f(x1) - f(x2)| < eps .
If the greatest xs lays in (M+1, +infty), since | x1 - x2 | < delta <= 1, we have the lowest x stays in (M, +infty). In conclusion we have x1, x2 > M.

So | f(x1) - f(x2)| = | f(x1) - 1 + 1 - f(x2) | <= |f(x1) -1| + |1 - f(x2)| =
|f(x1) - 1| + |f(x2) -1| < eps/2 + eps/2 = eps.