# Uniform continuity and boundedness

1. Dec 9, 2007

### mgiddy911

In my analysis class we were posed the following question:
Give an example of a uniformly continuous function f: (0,1) ---> R'
such that f' exists on (0,1) and is unbounded.

we came up with the example that f(x) = x*sin(1/x) if you interpret the question to mean f' is unbounded, not f itself.

Our teacher then asked us to ponder whether it is possible for the condition to be reversed, ie. what if the unbounded is refering to f(x) not f'(x). Is it possible for f(x) to be uniformly continuous but unbounded on the open interval.

The only thing I have thought would be if it had an infinite discontinuity at the end point. Like if the function went to positive infinity at x=1. Is it possible for what ever curve would do that to still satisfy the necessary conditions to be uniformly continuous?

2. Dec 9, 2007

### ircdan

any uniformly continuous function on a bounded set is bounded

3. Dec 10, 2007

### SiddharthM

the answer is no. ircdan's statement is correct. Note that there do exist unbounded functions (on unbounded domains) that are uniformly continuous, the fixed map from R to R is a straightforward example.

A useful theorem about uniform continuity: if a function,f, is UC on a dense subset D of R then there exists a continuous extension of f into R.

Since (0,1) is dense in the closed unit interval there exists a continuous extension to [0,1]. This extension is continuous on a compact interval and the max/min theorem says that this function is bounded. Clearly these bounds also work for f on (0,1).

That a function that is UC on a bounded domain is bounded can also be proved directly and is not so difficult.

4. Dec 10, 2007

### mgiddy911

Does an open interval such as (0,1) count as a bounded set?
I am familiar with the boundedness theorem, a continuous function on a compact interval is bounded. But this is not a function on a compact interval so I didn't think that theorem applied

5. Dec 10, 2007

### mgiddy911

Ok, thank you. I think I understand now. As I said earlier I am familiar with boundedness theorem, and or the max/min theorems. I was not sure how they applied to the open intervals.

6. Dec 10, 2007

### SiddharthM

if f is UC on (0,1) there exists a function g that is continuous on [0,1] and for all x in (0,1), f(x)=g(x).

this is the meaning of a continuous extension.

7. Dec 10, 2007

### mgiddy911

Thanks again to everyone that helped out.