# Show that a uniformly continuous function on a bounded, open interval is bounded

1. Oct 6, 2011

### krissycokl

1. The problem statement, all variables and given/known data
Suppose that the function $f|(a,b)→ℝ$ is uniformly continuous. Prove that $f|(a,b)→ℝ$ is bounded.

2. Relevant equations
A function $f|D→ℝ$ is uniformly continuous provided that whenever {un} and {vn} are sequences in D such that lim (n→∞) [un-vn] = 0, then lim (n→∞) [f(un) - f(vn)] = 0.

A function $f|D→ℝ$ is bounded if there exists a real number M such that |f(x)| ≤ M for all x in D

Every bounded sequence has a convergent subsequence.

3. The attempt at a solution

$f|(a,b)→ℝ$ is uniformly continuous. Then for all sequences un and vn in (a,b) such that lim (n→∞) [un - vn] we have lim (n→∞) [f(un) - f(vn)] = 0.

Suppose that f is not bounded. Then for all real numbers M, there exists a number x in (a,b) such that |f(x)|> M. Further, for all natural numbers n, there exists an xn in (a,b) such that |f(xn)|> n . Then {xn} is a sequence in the bounded, open interval (a,b). Thus {xn} has a convergent subsequence {xnk}.

Aaand...I'm not even really sure where I was heading with that. Direction would be greatly appreciated. This is a basic real analysis course, we haven't talked about metric spaces, Cauchy-continuity, or any of that stuff. So, whatever proof the text wants should use rather simple concepts. Thanks in advance for your time.

2. Oct 6, 2011

### micromass

Do you think you can extend f to a continuous function on [a,b]? That is, can you define f(a) and f(b) in a continuous way??

3. Oct 6, 2011

### Bacle

Another approach, similar to Micromass'ss idea: you know if f is continuous, it is bounded in [a-1/n,b+1/n] , for some integers n, or just in [a-e,b+e], for e>0. Now , use uniform continuity at a+e, and at b-e, to find a bound in an e-interval about each. As Micromass said; think of what happens with, e.g., f(x)=1/x in (0,1).

Don't mean to upstage you, Micromass, just to give a related version.

Last edited: Oct 6, 2011
4. Oct 6, 2011

### krissycokl

I'm not familiar with this assertion: if f is continuous, it is bounded in [a-1/n,b+1/n]
For instance, f = 1/x is continuous in (0,1) but is unbounded and discontinuous in the closed interval [0,1].

[STRIKE]But, let me try to follow other than that...

For the lower endpoint:
Since f is continuous in the interval (a,b), f is continuous at a+e for some ε>0, ε<(b-a). Thus if lim (n→∞) xn = a + e, then lim (n→∞) f(xn) = f(a+e)

Then it follows that for all ε1>0, there exists a natural number N such that |xn-(a+ε)|<ε1 for all n≥N
and there exists a natural number N2 such that |f(xn-f(a+ε)|<ε1 for all n≥N2.[/STRIKE]

Uhm. I'm stuck again.

Oh, wait! Maybe, rereading what you've said...

Can I simply turn f|(a,b)→ℝ into a piecewise function f|[a,b]→ℝ, and
define a sequence {xn} such that a = lim (n→∞) xn and
define a sequence {yn} such that b = lim (n→∞) yn and I then set
f(a) = lim (n→∞) f(xn) and
f(b) = lim (n→∞) f(yn)

My problem with this last step is visualizing it. For example, I really can't see what value I could assign to f(0) in the case f(x) = 1/x such that the function would be continuous at x = 0. But I will continue despite my doubts...

Then f is continuous at a and b by definition. Then f is continuous on the closed, bounded interval [a,b].

Then, by the Extreme Value Theorem f has both a maximum and minimum value in [a,b] and thus f is bounded in [a,b].

Is that right? If any of it is wrong, I think it might be the piecewise definition part. It just smells too easy. But thanks for the help!

5. Oct 6, 2011

### Bacle

Careful, krissy:

If f is continuous in (a,b) implies f is continuous in [a+1/n,b-1/n], does not imply that

f continuous in (0,1), then f also continuous on [0,1]; read carefully.

6. Oct 6, 2011

### krissycokl

This is what you previously said.

Did you perhaps mean it is bounded in [a+1/n,b-1/n]? The signs on the 1/n terms are different between your first post and your more recent one.

I did not use that logic in my proof, however. Aside from that, then, does my work make sense?

7. Oct 8, 2011

### Bacle

You're right, I made a mistake; it should be [a+1/n, b-1/n], altho the statement is true for _some_, tho not for all n.

The thing that confuses me about your proof is that I do not see where you used the fact that f is u.continuous on (a,b). And, remember that 1/x is not uniformly continuous on (0,1). My idea was to use the fact that for all ε>0 there exists a fixed δ>0 , so that |f(x)-f(y)|<ε
for |x-y|<δ , once you know that f is bounded on a compact interval.