# Bounded Function

1. Oct 18, 2005

### Icebreaker

If f is defined on [a,b] and for every x in [a,b] there is a d_x such that if is bounded on [x-d_x, x+d_x]. Prove that f is bounded on [a,b].

This question seems very odd. If every point, and indeed the neighbourhood of every point is bounded, then of course the function itself must be bounded. Of course I doubt this can be passed as a proof, so any suggestions would be helpful.

Last edited by a moderator: Oct 19, 2005
2. Oct 18, 2005

### EnumaElish

The statement "[x-d_x, x+d_x] is bounded" is a tautology. Is it meant to say f is bounded on [x-d_x, x+d_x]?

3. Oct 19, 2005

### Icebreaker

Yes, I believe so. I've made the modification but on the question sheet it was written exactly as I had copied it. My professor has a habit of handing out handwritten assignments.

4. Oct 19, 2005

### EnumaElish

Suppose that f was not bounded on [a,b]. Then there must exist a point c in [a,b] such that f(c) > M for all real M.

Does that sound like the right beginning?

Proceed to show a contradiction with boundedness on [x-d_x,x+d_x] for all x in [a,b].

5. Oct 19, 2005

### Muzza

Consider 1/x on (0, 1) then.

If we assume that all d_x are strictly greater than 0, then I believe you can use the compactness of [a, b] to solve this problem.

6. Oct 19, 2005

### Icebreaker

The interval must be closed.

7. Oct 19, 2005

### Muzza

You missed my point. I was objecting to your implication that the statement was rather trivial (or obvious) just because the function was bounded on every neighbourhood in its domain.

Last edited: Oct 19, 2005
8. Oct 19, 2005

### Icebreaker

True. However, I have found a proof using Bolzano-Weierstrass theorem. Thanks for the inputs.

9. Oct 19, 2005

### EnumaElish

When you have a chance can you post a sketch of that proof?

10. Oct 19, 2005

### Icebreaker

This will be sketchy, but the proof itself is not mine:

Assume the contrary, that f is unbounded. By Weierstrass, we can find a sequence x_n that converges to some number c in I = [a,b] and f(x_n)>n for every n. For this c, let d_c be as in the hypothesis. Then, for n large enough, x_n is in (c-d_c, c+d_c). This contradicts our hypothesis that f is bounded on (c-d_c, c+d_c).