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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

    micromass

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    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.
     
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