Results 1 to 5 of 5

Math Help - Limit

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    133

    Limit

    I have a function defined as:

    \begin{displaymath}f(x) = \left\{\begin{array}{lr}\frac{1}{k} & : x \in [0,k]\\0 & : x \notin [0,k]\end{array}\right.\end{displaymath}

    I am asked to show that f_{k} \to 0 in L^{\infty}(\mathbb{R}) as k \to \infty

    To be honest, I don't understand the question completely. When I make a skecth of the first few functions it's clear that f_{k} \to 0. It's the "in L^{\infty}(\mathbb{R})" which is confusing me.

    Could someone clarify?

    Thanks a bunch.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by surjective View Post
    I have a function defined as:

    \begin{displaymath}f(x) = \left\{\begin{array}{lr}\frac{1}{k} & : x \in [0,k]\\0 & : x \notin [0,k]\end{array}\right.\end{displaymath}

    I am asked to show that f_{k} \to 0 in L^{\infty}(\mathbb{R}) as k \to \infty

    To be honest, I don't understand the question completely. When I make a skecth of the first few functions it's clear that f_{k} \to 0. It's the "in L^{\infty}(\mathbb{R})" which is confusing me.

    Could someone clarify?

    Thanks a bunch.

    Please define L^\infty(\mathbb{R})...didn't you mean l^\infty(\mathbb{R}) ?

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    Posts
    133
    Hey,

    No, the exercise says L^{\infty}(\mathbb{R})} and is defined as:

    L^{\infty}(\mathbb{R})}= \left\lbrace  f: \mathbb{R} \to \mathbb{C} | \text{f is bounded} \right\rbrace
    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 surjective View Post
    Hey,

    No, the exercise says L^{\infty}(\mathbb{R})} and is defined as:

    L^{\infty}(\mathbb{R})}= \left\lbrace  f: \mathbb{R} \to \mathbb{C} | \text{f is bounded} \right\rbrace
    Presumably what they mean is that f_k\to 0 in whatever metric is placed on L^{(\infty)}\left(\mathbb{R}\right) ( think \mathcal{B}\left(\mathbb{R},\mathbb{C}\right) is a more common notation--at least it is the notation used by such authors as Simmons). Since L^\infty\left(\mathbb{R}\right) is evidently a vector space(algebra) one can guess that the metric is induced by a norm/inner product. Did they give you one?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by surjective View Post
    Hey,

    No, the exercise says L^{\infty}(\mathbb{R})} and is defined as:

    L^{\infty}(\mathbb{R})}= \left\lbrace  f: \mathbb{R} \to \mathbb{C} | \text{f is bounded} \right\rbrace

    Ok...never saw this notation, and I presume the norm or distance function here is ||f||:=\sup\limits_{x\in\mathbb{R}} |f(x)| ?

    If this is so, we have ||f_k||=\sup\limits_{x\in\mathbb{R}}|f_k(x)|=\frac  {1}{k}\xrightarrow [k\to\infty]{}0...piece of cake, or not.

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 8th 2010, 11:29 AM
  2. Replies: 1
    Last Post: February 5th 2010, 03:33 AM
  3. Replies: 16
    Last Post: November 15th 2009, 04:18 PM
  4. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 05:05 PM
  5. Replies: 15
    Last Post: November 4th 2007, 07:21 PM

Search Tags


/mathhelpforum @mathhelpforum