Results 1 to 3 of 3

Thread: finding a sequence of continuous bounded functions

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    5

    finding a sequence of continuous bounded functions

    I want to find a sequence (f_k)of continuous bounded functions on [0,1]

    converging pointwise to a continuous limit, but such that

    the sup-norm of f_k tends to infinity when k tends to infinity...

    is there any?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member jakncoke's Avatar
    Joined
    May 2010
    Posts
    387
    Thanks
    80
    Nope, i don't think there are any. Since if $\displaystyle f_k \rightarrow f $ pointwise then
    $\displaystyle \lVert f_k \lVert_{sup} \rightarrow \lVert f \lVert_{sup} $ which would mean f is not bounded on [0, 1] (since $\displaystyle \lVert f_k \lVert_{sup} \rightarrow \infty $ ) thus it must not continuous on [0, 1].
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by losin90 View Post
    I want to find a sequence (f_k)of continuous bounded functions on [0,1]

    converging pointwise to a continuous limit, but such that

    the sup-norm of f_k tends to infinity when k tends to infinity...

    is there any?
    What you need to do is to ensure that the function $\displaystyle f_k$ has a narrow spike somewhere. For example, define $\displaystyle f_k(x) = k^2x$ in the interval $\displaystyle 0\leqslant x<1/k$, $\displaystyle f_k(x) = \tfrac2k - k^2x$ in the interval $\displaystyle 1/k\leqslant x<2/k$, and $\displaystyle f_k(x) = 0$ in the remainder of the unit interval. Then $\displaystyle f_k(x)\to0$ as $\displaystyle k\to\infty$, for each (fixed) point x in the unit interval.

    Quote Originally Posted by jakncoke View Post
    Nope, i don't think there are any. Since if $\displaystyle f_k \rightarrow f $ pointwise then
    $\displaystyle \lVert f_k \lVert_{sup} \rightarrow \lVert f \lVert_{sup} $ which would mean f is not bounded on [0, 1] (since $\displaystyle \lVert f_k \lVert_{sup} \rightarrow \infty $ ) thus it must not continuous on [0, 1].
    That is not true, because pointwise convergence does not imply convergence in the sup norm.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Examples of bounded/continuous sets and functions
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Oct 20th 2011, 07:30 AM
  2. Replies: 1
    Last Post: Nov 13th 2010, 12:44 AM
  3. continuous bounded functions
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Mar 22nd 2010, 06:53 PM
  4. [SOLVED] is space of bounded continuous functions separable?
    Posted in the Advanced Math Topics Forum
    Replies: 9
    Last Post: Jan 18th 2009, 08:16 PM
  5. Bounded, Continuous and Closed Functions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 27th 2007, 06:05 PM

Search Tags


/mathhelpforum @mathhelpforum