Results 1 to 5 of 5

Math Help - Uniform Convergence & Boundedness

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    404

    Uniform Convergence & Boundedness

    "Let f_k be functions defined on R^n converging uniformly to a function f. IF each f_k is bounded, say by A_k, THEN f is bounded."

    f_k converges to f uniformly =>|| f_k - f||∞ ->0 as k->∞
    Also, we know| f_k(x)|≤ A_k for all k, for all x.
    But why does this imply that f is bounded? I don't see how to prove it.
    Also, why do we need uniform convergence? (why is pointwise convergence not enough?)

    Any help is appreciated!

    [also under discussion in math links forum]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kingwinner View Post
    "Let f_k be functions defined on R^n converging uniformly to a function f. IF each f_k is bounded, say by A_k, THEN f is bounded."

    f_k converges to f uniformly =>|| f_k - f||∞ ->0 as k->∞
    Also, we know| f_k(x)|≤ A_k for all k, for all x.
    But why does this imply that f is bounded? I don't see how to prove it.
    Also, why do we need uniform convergence? (why is pointwise convergence not enough?)

    Any help is appreciated!

    [also under discussion in math links forum]
    Hint:
    Spoiler:
    \|f\|_{\infty}-\|f_n\|_{\infty}\leqslant \|f-f_n\|_{\infty}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    Quote Originally Posted by Drexel28 View Post
    Hint:
    Spoiler:
    \|f\|_{\infty}-\|f_n\|_{\infty}\leqslant \|f-f_n\|_{\infty}
    OK, so if we choose n large enough, we will have ||f||∞ ≤ ε + A_k. To show f is bounded, I think we can just fix some ε.

    Why is pointwise convergence not enough?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kingwinner View Post
    OK, so if we choose n large enough, we will have ||f||∞ ≤ ε + A_k. To show f is bounded, I think we can just fix some ε.

    Why is pointwise convergence not enough?
    Well, you need to be careful. A_k depends on f_k, right?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    Quote Originally Posted by Drexel28 View Post
    Well, you need to be careful. A_k depends on f_k, right?
    Why does it matter? What should we do instead? Can we take the maximum over all the A_k and use that as a bound?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 31st 2010, 08:09 PM
  2. Uniform boundedness
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: December 17th 2009, 05:46 PM
  3. Uniform boundedness Theorem
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: July 1st 2009, 02:44 PM
  4. Boundedness and Convergence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 22nd 2008, 08:00 PM
  5. Uniform Continuous and Uniform Convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 28th 2007, 03:51 PM

Search Tags


/mathhelpforum @mathhelpforum