Continuous function on intervals

1. Dec 11, 2008

rapple

1. The problem statement, all variables and given/known data
f:(0,1]->R be a continuous function. Is it possible that f does not have an absolute min or max. Give counter examples

2. Relevant equations

3. The attempt at a solution
Since f is partially bounded, if I break the interval down into smaller sub intervals, each will have an abs min and max. So I will have at least one of the extremes.

2. Dec 11, 2008

Dick

The problem says 'give counterexamples'. Do that. They aren't hard to find.

3. Dec 11, 2008

VeeEight

"if I break the interval down into smaller sub intervals, each will have an abs min and max."

If you break the interval into a union smaller intervals, you will always have one interval of the form (0, a) where a is some real in (0,1]. What is the difference between [0,1] and (0,1]? Does a continuous function on [0,1] attain a min/max?

Think of some functions on (0,1] that may not have a min or max (you might want to think of f(x) as x approaches 0)

4. Dec 12, 2008

rapple

How about f(x) = 1 if x=1 and f(x) = 1/(x-1) if x Not= 1. Then as x->0 , f(x) is increasing and as x-> 1, f(x) is decreasing. Since it is 1 at one, it is not the minimum.

5. Dec 12, 2008

Dick

That f is also not continuous on (0,1]. Keep it simple. How about f(x)=1/x?

6. Dec 12, 2008

rapple

Then f has an abs min at x=1. even though it is continuous

7. Dec 12, 2008

Dick

Oh, I thought you wanted no max or no min. You want neither max NOR min. How about a function that oscillates with increasing increasing frequency and magnitude as x->0? Can you give an example of one of those?

8. Dec 12, 2008

rapple

How about sin(1/x)/x. But I don't think it is continuous because of the sharp peaks. Is it a form of sin function?

9. Dec 12, 2008

Dick

That's perfect. It's continuous on (0,1]. The only place it has a problem is at x=0. But x=0 is not in (0,1].

10. Dec 12, 2008

rapple

Thank you. You seem to have a way of making it happen!