1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Show that a uniformly continuous function on a bounded, open interval is bounded

  1. Oct 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose that the function [itex]f|(a,b)→ℝ[/itex] is uniformly continuous. Prove that [itex]f|(a,b)→ℝ[/itex] is bounded.

    2. Relevant equations
    A function [itex]f|D→ℝ[/itex] 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 [itex]f|D→ℝ[/itex] 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

    [itex]f|(a,b)→ℝ[/itex] 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. jcsd
  3. Oct 6, 2011 #2
    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??
  4. Oct 6, 2011 #3
    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
  5. Oct 6, 2011 #4
    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!
  6. Oct 6, 2011 #5
    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.
  7. Oct 6, 2011 #6
    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?
  8. Oct 8, 2011 #7
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook