Results 1 to 4 of 4

Math Help - Bounded linear operator

  1. #1
    Junior Member
    Joined
    Apr 2011
    Posts
    35

    Bounded linear operator

    Question
    Show that the range \mathcal{R}(T) of a bounded linear operator T: X \rightarrow Y is not necessarily closed.
    Hint: Use the linear bounded operator T: l^{\infty} \rightarrow l^{\infty} defined by (\eta_{j}) = T x, \eta_{j} = \xi_{j}/j, x = (\xi_{j}).

    Attempt to solution

    My idea was to find an element y \in l^{\infty} that does not belong to the range and then try to build a convergent sequence in \mathcal{R}(T) that has limit  y . The element y = (1, 1, \ldots) satisfy the criteria because T^{-1}y = \{ x\}, with x = (\xi_{j}), \xi_{j} = j, but, clearly, x \not\in l^{\infty}, therefore, y \not\in \mathcal{R}(T). The problem arise when I try to build the sequence, because (T x_{m}) with x_{m} \in l^{\infty} cannot converge to  y . Briefly, my problem is that I canīt find a limit point of \mathcal{R}(T) that doesnīt belong to \mathcal{R}(T).
    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 kinkong View Post
    Question
    Show that the range \mathcal{R}(T) of a bounded linear operator T: X \rightarrow Y is not necessarily closed.
    Hint: Use the linear bounded operator T: l^{\infty} \rightarrow l^{\infty} defined by (\eta_{j}) = T x, \eta_{j} = \xi_{j}/j, x = (\xi_{j}).

    Attempt to solution

    My idea was to find an element y \in l^{\infty} that does not belong to the range and then try to build a convergent sequence in \mathcal{R}(T) that has limit  y . The element y = (1, 1, \ldots) satisfy the criteria because T^{-1}y = \{ x\}, with x = (\xi_{j}), \xi_{j} = j, but, clearly, x \not\in l^{\infty}, therefore, y \not\in \mathcal{R}(T). The problem arise when I try to build the sequence, because (T x_{m}) with x_{m} \in l^{\infty} cannot converge to  y . Briefly, my problem is that I canīt find a limit point of \mathcal{R}(T) that doesnīt belong to \mathcal{R}(T).
    That's the right idea. How about something like defining the m^{\text{th}} element of your sequence has 1 in the first m positions and zero afterwards-- i.e. (1,\cdots,\underbrace{1}_{m^{\text{th}}},0,\cdots,  0,\cdots). This is clearly in \text{im}(T) since it is equal to T(1,\cdots,m,0,\cdots) and it converges to (1,\cdots,1,\cdots)
    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
    7
    Quote Originally Posted by Drexel28 View Post
    How about something like defining the m^{\text{th}} element of your sequence has 1 in the first m positions and zero afterwards-- i.e. (1,\cdots,\underbrace{1}_{m^{\text{th}}},0,\cdots,  0,\cdots). This is clearly in \text{im}(T) since it is equal to T(1,\cdots,m,0,\cdots) and it converges to (1,\cdots,1,\cdots)
    The trouble there is that (1,\ldots,1,0,\ldots) does not converge to (1,\ldots,1,\ldots) in the space l^\infty (because the norm there is the sup norm). To avoid this difficulty, you could modify the construction slightly by taking y = \bigl(1,\tfrac1{\sqrt2},\tfrac1{\sqrt3},\ldots \bigr). Then (1,\ldots,\tfrac1{\sqrt m},0,\ldots,0,\ldots) is in \text{im}(T) since it is equal to T(1,\ldots,\sqrt m,0,\ldots), and it does converge to (1,\ldots,\tfrac1{\sqrt m},\ldots) in the l^\infty norm.
    Last edited by Opalg; June 2nd 2011 at 12:54 AM. Reason: corrected typos
    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 Opalg View Post
    The trouble there is that (1,\ldots,1,0,\cdots) does not converge to (1,\ldots,1,\ldots) in the space l^\infty (because the norm there is the sup norm). To avoid this difficulty, you could modify the construction slightly by taking y = \bigl(1,\tfrac1{\sqrt2},\tfrac1{\sqrt3},\ldots \bigr). Then (1,\cdots,\tfrac1{\sqrt m},0,\cdots,0,\cdots) is in \text{im}(T) since it is equal to T(1,\ldots,\sqrt m,0,\ldots)' and it does converge to (1,\ldots,\tfrac1{\sqrt m},\ldots) in the l^\infty norm.
    You're right, that was a stupid mistake.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear and bounded operator
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: June 20th 2011, 12:56 AM
  2. Bounded linear operator
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: June 2nd 2011, 10:03 PM
  3. Replies: 2
    Last Post: December 5th 2009, 01:30 AM
  4. Replies: 2
    Last Post: January 1st 2009, 02:54 PM
  5. Norm of a Bounded Linear Operator
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 1st 2009, 06:39 AM

Search Tags


/mathhelpforum @mathhelpforum