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Distribution interpretation

  1. Dec 31, 2008 #1
    Since [itex]C_0^\infty[/itex] (the space of smooth functions with compact support) is reflexive, we should, in theory, be able to identify every distribution (object that lives in the dual space [itex]C_0^{\infty '}[/itex]) with a corresponding actual function in [itex]C_0^\infty[/itex]. Is it all interesting or useful to do so?

    For example, a constant function can be interpreted as a distribution. Then what is the corresponding smooth function with compact support for the constant function?
     
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  3. Dec 31, 2008 #2
    I am reading some online lecture notes that say it is reflexive but do not give proof.

    The reason why the constant function wouldn't necessairily be a counterexample is if the isomorphism between [itex]C_0^\infty[/itex] and [itex]C_0^{\infty '}[/itex] is something other than as identifing a function with its dual. Not very intuitive I suppose, but it's possible.

    EDIT: err, this post was a response to a post by someone else above, but there is some sort of forum bug that has deleted it. (?)
     
  4. Dec 31, 2008 #3

    Hurkyl

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    Are you sure about that? This isn't a formal consequence of duality; there are plenty of other situations where an object X can be isomorphic to X**, but not really have any relationship with X*. I don't know this case well enough to have a good expectation whether or not your claim is true.
     
  5. Dec 31, 2008 #4

    Hurkyl

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    I deleted my post because I was being an idiot; I misread the definitions.
     
  6. Dec 31, 2008 #5
    Err hmm. Is this really true? By Hahn-Banach, we have embeddings such that [itex]X \subset X' \subset X'' \subset ...[/itex], but by reflexivity we have a bijection such that [itex]X \cong X''[/itex]. Doesn't this mean that [itex]X \cong X'[/itex], or am I being stupid here? (always a possibility...)
     
  7. Dec 31, 2008 #6

    Hurkyl

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    The other situations I was referring to were non-vector space situations (e.g. groups or modules).

    The embedding X --> X* is clear. But do we really have an embedding X* --> X**? That is non-obvious to me....
     
  8. Dec 31, 2008 #7
    My experience is mostly with normed vector spaces rather than the more general algebraic setting and algebraic duals. However, if X is a normed vector space, then its continuous dual X' (set of all bounded linear functionals) is also a normed vector space under the induced norm (ie: [itex]||\phi|| = sup_{||x|| = 1} |\phi(x)|[/itex]. Then there is really no difference in going from X to X' versus X' to (X')' = X'', versus X''''''' to X''''''''. In each case you are just taking the continuous dual of a normed vector space and can use the Hahn-Banach to find an embedding.
     
    Last edited: Dec 31, 2008
  9. Dec 31, 2008 #8

    Hurkyl

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    The embedding X --> X* that I see makes crucial use of the fact X is actually an inner product space: I'm thinking of the embedding [itex]i(x)(y) = \langle x, y \rangle[/itex].
     
  10. Dec 31, 2008 #9

    Hurkyl

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    Aha! If we have an embedding X --> X* where X* is an inner product space, then X must also be an inner product space.

    Is there an inner product on [itex](C_0^{\infty})^*[/itex]? I suspect not, therefore there cannot exist an embedding [itex](C_0^{\infty})^* \subseteq (C_0^{\infty})^{**} [/itex]
     
  11. Dec 31, 2008 #10
    Ahh right. [itex]C_0^\infty[/itex] is not even metrizable.

    As a side note, you don't actually need an inner product, just a norm will do. For each element x of X, you would associate the functional [itex]\phi[/itex] in X' such that [itex]\phi(x) = ||x||[/itex] and [itex]|| \phi || = 1[/itex].
     
  12. Dec 31, 2008 #11

    Hurkyl

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    (Red part mine: changed 'the' to 'a', since [itex]\phi[/itex] is not uniquely determined by those properties) Hrm. It's clear that this provides a function X --> X*. But it's nonobvious that this function is either injective or linear. (And it's clearly not norm-preserving; I think you meant [itex]\phi(x) = ||x||^2[/itex] and [itex]|| \phi || = ||x||[/itex]?)
     
  13. Dec 31, 2008 #12
    This is (IMO) the crucial idea behind Hahn-Banach - that you can take a bounded linear functional defined on a subspace (the subspace spanned by vector x) and extend it to the whole space without increasing the norm.

    Starting with the simple functional [itex]\phi_0(y) = ||y||[/itex] defined for all y along the line spanned by x, we have [itex]||\phi_0|| = 1[/itex]. Then we can extend this functional to the whole space and construct a functional [itex]\phi[/itex] that agrees with [itex]\phi_0[/itex] along the line spanned by x, and has the same norm! Thus [itex]\phi(x) = ||x||[/itex] and [itex]||\phi|| = 1[/itex].

    The [itex]\phi[/itex] thus constructed is linear and norm preserving by Hahn-Banach, and you can show it is unique by demonstrating two functionals [itex]\phi[/itex] and [itex]\psi[/itex] that satisfy these properties must have the same null space. The null space is, in some sense, the space "perpendicular" to x, even though technically we don't have an inner product.
     
    Last edited: Dec 31, 2008
  14. Dec 31, 2008 #13

    Hurkyl

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    D'oh, I guess I wasn't clear -- you claim the existence of a function [itex]i : X \to X^*[/itex], which you defined by saying [itex]i(x) = \phi[/itex], where [itex]\phi[/itex] is a functional satisfying [itex]\phi(x) = ||x||[/itex] and [itex]|| \phi || = 1[/itex]. i is clearly not a linear map, because [itex]||i(rx)|| = 1[/itex], but [itex]||r i(x)|| = |r|[/itex], so [itex]i(rx) \neq r i(x)[/itex] whenever [itex]|r| \neq 1[/itex].

    If you make the change I suggested, the above argument fails, but it's still nonobvious that i should be injective or linear.
     
  15. Dec 31, 2008 #14
    You're right, what you said in post 11 is correct. I'm on crack.

    It's ok though because you can show it is 1-1 by using the following fact: if given [itex]\phi[/itex] in X' and a particular x in X, x not in the kernel of [itex]\phi[/itex], then any y in X has a unique expression as [itex]y = u + \alpha x[/itex] for some u in the kernel of [itex]\phi[/itex] and scalar [itex]\alpha[/itex].

    Now suppose that [itex]\phi[/itex] is the functional corresponding to both points x and y, and express [itex]y = u + \alpha x[/itex]. Then [itex]\phi(y) = \alpha\phi(x)[/itex]. If we restrict ourselves to points x and y on the unit ball, then this means [itex]\phi[/itex] can only be the functional for both x and y if [itex]\alpha=1[/itex] (and u=0). ie: x and y are the same point.

    Considering only the points on the unit ball is sufficient, since we require [itex]||\phi||=||x||[/itex] and [itex]||\phi|| = ||y||[/itex]. Thus the functionals corresponding to any 2 points must be distinct and so i is 1-1!
     
    Last edited: Dec 31, 2008
  16. Dec 31, 2008 #15

    Hurkyl

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    Okay, I think I have a counterexample: the spaces [itex]l^3(\mathbb{R})[/itex] and [itex]l^{3/2}(\mathbb{R})[/itex] are dual to each other. Assume we have an inclusion [itex]i : l^3 \to l^{3/2}[/itex]. The construction you give tells us that for a standard basis vector [itex]\hat{e}_n[/itex], we must have [itex]i(\hat{e}_n) = \hat{e}_n[/itex].

    Now, consider the following:
    [tex]|| \hat{e}_1 + \hat{e}_2 ||_3 = \sqrt[3]{2}[/tex]
    [tex]|| \hat{e}_1 + \hat{e}_2 ||_{3/2} = \sqrt[3]{4}[/tex]
    and so i cannot be norm-preserving.

    Even better, consider the sequence [itex]v_n = n^{-2/3}[/itex]. This is in [itex]l^3[/itex] because [itex]\sum_{n = 1}^{+\infty} v_n^3 = \sum n^{-2} = \pi^2 / 6[/itex]. However, [itex]v_n[/itex] diverges in [itex]l^{3/2}[/itex]. But [itex]v_n[/itex] is clearly a limit of finite linear combinations of basis vectors. So i cannot be continuous.

    Conclusion: for the space [itex]X = l^3(\mathbb{R})[/itex], there does not exist an inclusion [itex]X \to X^*[/itex] of Banach spaces.
     
  17. Dec 31, 2008 #16
    Hmm interesting. It may be 1-1 but it's definitely not norm preserving like you say. For some reason I thought it was a well known result that you can embed any NLS into its dual, but apparently that is completely false. I must have been confusing the NLS case with the IPS case. How strange that there is always such an embedding into the double dual but not necessairily the dual.

    Thanks, this has forced me to tidy up some things I thought I understood better than I did.
     
  18. Dec 31, 2008 #17

    Hurkyl

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    Well, the double dual is a very fundamental thing; the point being that you simply have a bivariate 'evaluation' map from V* and V into R. (With the obvious generalization to any sort of 'function object' besides dual spaces) Given any sort of bivariate map from X and Y to Z, you can always 'transpose' it: for any element x of X, 'plugging in x' gives you a map from Y to Z. (And conversely for elements y of Y)
     
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