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Functional analysis - question about separable dual spaces

  1. Oct 1, 2012 #1
    Suppose X is a normed space and X*, the space of all continuous linear functionals on X, is separable. My professor claims in our lecture notes that we KNOW that X* contains functionals of arbitrarily large norm. Can someone explain how we know this, please?
     
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  3. Oct 1, 2012 #2

    micromass

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    Isn't that pretty obvious?? If T is a nonzero linear functional, then we can look at nT (for n a positive integer). This has norm [itex]n\|T\|[/itex] and can be made large.
     
  4. Oct 3, 2012 #3
    It certainly is obvious, if we take for granted that there exist nonzero linear functionals! But is that true for an arbitrary normed space?
     
  5. Oct 3, 2012 #4

    micromass

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    Of course not, since the space could be {0}, in which case there are no nonzero functionals. Other normed spaces do have nonzero linear functionals. This follows from the Hahn-Banach theorem.
     
  6. Oct 8, 2012 #5
    But is this really the only kind of space where we have no nonzero functionals? For example, if [itex]f\in X^*[/itex], we know that the inverse image under [itex]f[/itex] of every open set in [itex]\mathbb C[/itex] is open, so in particular we must have

    [tex]
    f^{-1} \left( \left\{ |z| < 1 \right\} \right) = \left\{ x\in X : |f(x)| < 1 \right\}.
    [/tex]

    open. But this set is also convex, so our [itex]X[/itex] must contain at least some open convex sets. Doesn't this fail for, say, the nonzero Banach space [itex]\ell^p[/itex] for [itex]0<p<1[/itex]?
     
  7. Oct 8, 2012 #6

    micromass

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    Indeed. You have now discovered a more general criterion for the existence of nonzero functionals: there must be "enough" convex sets. Enough means here that at any point x in X, there must be a neighborhood basis consisting out of convex sets. A topological vector space satisfying this criterion is called locally convex. By (a more general form of) the Hahn-Banach theorem, we can show that any locally convex topological space has nonzero functionals.

    Sure, this fails. But [itex]\ell^p[/itex] for [itex]0<p<1[/itex] are not Banach spaces!! There is no way to put a norm on [itex]\ell^p[/itex]. The usual norm

    [tex]\|(x_n)_n\|_p=\sqrt[p]{\sum |x_k|^p}[/tex]

    doesn't satisfy the triangle inequality. The best we can do is put a metric on the space by

    [tex]d_p((x_n)_n,(y_n)_n)=\sum |x_k-y_k|^p[/tex]

    Under this metric, the space is indeed complete. But this metric does not come from a norm, and is no Banach space (nor locally convex topological vector space).
     
  8. Oct 8, 2012 #7
    Interesting! I can see why the "standard" p-norm doesn't work, but how does one go about proving that we can't put a norm (of any sort) on this space?
     
  9. Oct 8, 2012 #8

    micromass

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    Well, it seems that you already know that [itex]\ell^p[/itex] (for 0<p<1) has no nonzero functionals. That would already show it, since any normed space admits nonzero continuous functionals.
     
  10. Oct 9, 2012 #9
    I'm not quite sure how to get this result from the (complex) Hahn-Banach theorem...can you help me with this? Maybe I just have the "wrong" formulation of the theorem; the one I'm looking at talks about defining a real-valued function [itex]p(x)[/itex] such that [itex]p(\alpha x + \beta y) \leq |\alpha|p(x) + |\beta|p(y)[/itex] on some subspace of [itex]X[/itex]. It's certainly not hard to think of such a function (try [itex]p(x) = \|x\|[/itex]) or such a subspace (just try [itex]\{0\}[/itex])...but you then need to have a (nonzero) functional [itex]\lambda[/itex] that satisfies [itex]|\lambda(x)| \leq p(x)[/itex] to move any further, and I can't in general think of one!
     
  11. Oct 9, 2012 #10

    micromass

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    OK. So take [itex]p(x)=\|x\|[/itex].

    Now, take [itex]x_0[/itex] an arbitrary nonzero vector. Then [itex]span(x_0)[/itex] is a closed subspace of our normed space (closed because finite-dimensional). Define a functional

    [tex]f:span(x_0)\rightarrow \mathbb{R}:tx_0\rightarrow t\|x_0\|[/tex]

    now apply Hahn-Banach.
     
  12. Oct 9, 2012 #11
    Thanks a lot for your help, micromass!

    So, just to recap: It is not enough to say that one has "functionals of arbitrarily large norm" in an arbitrary normed space; we must specify that the normed space is nonzero. But, once we have done that, we automatically have nonzero functionals, since every normed space is locally convex.
     
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