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!

Bounded Function

  1. Oct 18, 2005 #1
    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. jcsd
  3. Oct 18, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    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]?
  4. Oct 19, 2005 #3
    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.
  5. Oct 19, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    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].
  6. Oct 19, 2005 #5
    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.
  7. Oct 19, 2005 #6
    The interval must be closed.
  8. Oct 19, 2005 #7
    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
  9. Oct 19, 2005 #8
    True. However, I have found a proof using Bolzano-Weierstrass theorem. Thanks for the inputs.
  10. Oct 19, 2005 #9


    User Avatar
    Science Advisor
    Homework Helper

    When you have a chance can you post a sketch of that proof?
  11. Oct 19, 2005 #10
    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).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Bounded Function
  1. Bounded functions (Replies: 0)

  2. Bounded Function (Replies: 2)