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[tex]l^{p} space[/tex]

  1. Nov 22, 2007 #1
    1. The problem statement, all variables and given/known data
    Let 1[tex]\leq[/tex]p[tex]\leq[/tex][tex]\infty[/tex] and suppose ([tex]\alpha_{ij}[/tex] is a matrix such that (Af)(i)=[tex]\sum^{\infty}_{j=1}[/tex][tex]\alpha[/tex][tex]_{ij}[/tex]f(j) defines an element Af of [tex]l^{p}[/tex] for every f in [tex]l^{p}[/tex]. Show that A is a bounded linear functional on [tex]l^{p}[/tex]


    2. Relevant equations
    Isn't this obvious if we apply theorem that says following are equivalent for A:X-->X a linear transformation on normed space?
    (a)A is bounded linear functional
    (b)A is continuous at some point
    (c)There is a positive constant c such that ||Ax||[tex]\leq[/tex]c||x|| for all x in X.


    3. The attempt at a solution
    Isn't the contiunuity of f obvious? So by the theorem, I think A is bounded linear functional on [tex]l^{p}[/tex]. Could you guys correct me if I am wrong? Or if I am right could you just say it is right?

    Thanks
     
    Last edited: Nov 22, 2007
  2. jcsd
  3. Nov 22, 2007 #2

    morphism

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    Er, the continuity of f? f is an element of l^p, not a function. Also, how is A a functional on l^p? To me a functional on X is a mapping into the scalar field, and I believe this is standard terminology; A looks like a mapping from l^p to l^p (namely f [itex]\mapsto[/itex] Af).

    Edit:
    Looking at what you said again, I think you might've meant to say "continuity at f". If so, what f?
     
    Last edited: Nov 22, 2007
  4. Nov 23, 2007 #3
    Maybe then I should say that A is bounded linear transformation?
    But still isn't continuity of A obvious by construction?
     
  5. Nov 23, 2007 #4

    morphism

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    Why is it obvious? Can you post your proof?
     
  6. Nov 24, 2007 #5
    Sorry. It's not obvious. It seems continuous though.
    I know for given epsilon > 0, I need to find delta>0 such that ||f||<delta implies
    ||Af||<epsilon. Hmm.. How can I find such delta? Or use cauchy sequence?
     
  7. Nov 24, 2007 #6

    morphism

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    It's not that it's a difficult problem, but I just wouldn't say that the continuity of A is obvious. Anyway, I would use characterization (c) in the first post. Namely, try to compute ||Ax||_p given an arbitrary sequence x=(x_1, x_2, ...) in l^p. It will help to split this into three cases, depending on whether p=1, 1<p<[itex]\infty[/itex], or p=[itex]\infty[/itex]. Holder's inequality (and Cauchy-Schwarz) will be helpful in the first two cases.

    If you need any more hints, post back.

    Also, a much more interesting (and more difficult!) problem is to prove that the same map, x [itex]\mapsto[/itex] Ax, is continuous if it maps sequence in l^p to sequences in l^p', where 1 < p, p' < [itex]\infty[/itex]. You might want to try to tackle this one if you're looking for a challenge.
     
    Last edited: Nov 24, 2007
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