Results 1 to 2 of 2

Math Help - D bounded => f is bounded on D

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    104

    D bounded => f is bounded on D

    I need some help with a proof
    Assume \to \mathbb{R}" alt="f\to \mathbb{R}" /> is uniformly continous on D.
    Show that D being bounded implies f is bounded on D.

    All I can think of is to assume f is unbounded and show that it leads to a contradiction, but not sure how to proceed with the proof.
    any suggestions? tHanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jun 2009
    Posts
    113
    Precompactness is the key. Let f be uniformly continuous on D\subseteq  \mathbb{R}^n. You get d>0 such that the oscilation of fin each ball of radius d is smaller than 1. Every bounded subset in \mathbb{R}^n is precompact, hence you can find x_1,\ldots, x_n\in D such that D\subset\cup_{i=!}^{n}B(x_i,d). Now you can easily get a bound for |f|.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Closure of a totaly bounded set is totally bounded
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: April 8th 2011, 07:42 PM
  2. Replies: 2
    Last Post: September 14th 2010, 06:15 AM
  3. Replies: 3
    Last Post: March 17th 2010, 06:12 PM
  4. Replies: 4
    Last Post: October 12th 2009, 08:43 PM
  5. Replies: 0
    Last Post: July 1st 2008, 02:47 AM

Search Tags


/mathhelpforum @mathhelpforum