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Proving l^{\infty} (sequence space w/infinity norm) is not separable, and dual spaces

  1. Jan 27, 2006 #1

    benorin

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    I would like to prove that [itex] \ell^{\infty}[/itex], namely the Banach space whose elements are sequences of complex numbers that have a fininte infinity-norm (a.k.a. the supremum-norm,) that is for [itex]\alpha = \left\{ \alpha_k \right\}_{k=1}^{\infty},[/tex]

    [tex] \ell^{\infty}=\left\{\alpha : \sup\left\{ \left| \alpha_n \right| : n\in\mathbb{N}\right\}<\infty \right\}, [/tex] normed by [tex]\| \alpha\|_{\infty}=\sup\left\{ \left| \alpha_n \right| : n\in\mathbb{N}\right\}[/tex]

    is not separable, that is that it does not possess a countable dense basis. I do not well understand what it means for a space to be separable :confused: : does it mean that any element (e.g. any sequence, vector, function, ...) of that space can be expressed as either a linear combination of the elements (say, functions) of some countable (basis?) set or a limit thereof?

    Would someone please clearly explain this topic that I might more fully understand it, and, perhaps, the concept of dual spaces: specifically, why is [itex] \ell^{\infty}=(\ell^{1})^*[/itex]? where X* denotes the dual space of X and where [itex] \ell^{1}[/itex] is the Banach space of sequences of complex numbers defined by

    [tex] \ell^{1}=\left\{\alpha : \sum_{k=1}^{\infty} \left| \alpha_k \right| <\infty \right\}, [/tex] normed by [tex]\| \alpha\|_{1}=\sum_{k=1}^{\infty} \left| \alpha_k \right|[/tex]

    And, why, despite this relationship, is [itex] (\ell^{\infty})^*=(\ell^{1})^{**}\neq \ell^{1}[/itex]? I realize that I have asked alot, but I would rather that sufficient information be put forth that I could join-in the discussion.

    Thanks,
    --Ben
     
  2. jcsd
  3. Jan 27, 2006 #2

    matt grime

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    Let me do the second one first, and get it out the way.

    l^infinity is the dual space of l^1 'becuase it is'. you can work out what the linear functionals on l^1 are and they are precisely the space of bounded sequences.

    the reflexive property fails 'because it does', ie you can find a linear functional on l^infinity that is not in l^1, obviously l^1 is contained in the dual space of l^infinity but it is not all of it.

    you should try to prove these things to your self. but do not try to work out what the dual space of l^infinity actually is, because it will give you a headache.
     
  4. Jan 27, 2006 #3

    benorin

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    I'm a little confused...

    Regarding functionals on the dual space of a Banach space, I'm a little confused... :redface:

    So let X be a Banach space, a functional on X is a mapping of vectors in X to complex numbers (or whatever the scalar field is,) correct? Then a functional on X* is what? a vector in X that maps functionals in X* to complex numbers?
     
  5. Jan 27, 2006 #4

    matt grime

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    add in the word linear and yes



    why don't you put X* in the sentence above, since that is its definition?

    Incidentally there is an existential proof that the double dual of l^1 is not l^1.

    Thm: Let X be a banach space, then if X^* is separable so is X.

    Cor. l^infinity dual is not l^1.

    Proof of cor: if l^infinity dual were l^1(which is separable) then applying the theorem l^infinity would be separable, but we know it isn't.
     
    Last edited: Jan 27, 2006
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