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Linear functional clarification (from rudin)

  1. Aug 28, 2009 #1
    In Rudin's Functional Analysis (in theorem 3.4), he says:

    "every nonconstant linear functional on X is an open mapping". X is topological vector space.

    This seems like a strengthening of the open mapping theorem, which requires X to be an F-Space, and that the linear functional to be continuous.

    Indeed, it seems like a discontinuous linear functional from R to R as described in Gelbaum for instance isn't open.

    What am I missing?
  2. jcsd
  3. Aug 28, 2009 #2
    There is no discontinuous linear functional R to R. Linear is in the sense of real scalars.
  4. Aug 28, 2009 #3
    Are you sure? "Counterexamples in Analysis" by Gelbaum p.33 gives an example.
  5. Aug 28, 2009 #4
    p. 33 ... compare it carefully to what you said before.
  6. Aug 28, 2009 #5
    This isn't homework. Can you be more direct?

    As far as I see, the function defined by Gelbaum is a discontinuous linear functional R to R.
  7. Aug 28, 2009 #6
    I think the problem is that Gelbaum's construction is using a different definition of linear. It seems Gelbaum's "linear" doesn't also imply f(a*x) = a*f(x) for scalar a. So that resolves some of my confusion. Does this sound accurate?

    That still leaves my original question, though:

    Is "every nonconstant linear functional on X is an open mapping"? Even assuming that the functional is continuous, do we really not need the stronger condition that X is an F-space?
  8. Aug 29, 2009 #7
    We do not need [itex]F[/itex]-space and we do not need continuous. A linear functional [itex]f[/itex] on a TVS [itex]X[/itex] can be discontinuous. But the point of the "hints" I gave before is that when you restrict [itex]f[/itex] to a line (say a line [itex]L = \{tu : t \in \mathbb{R}\}[/itex] through the origin and the point [itex]u[/itex]), that restriction is continuous, in fact it is of the form [itex]f(tu) = tc, t \in \mathbb{R}[/itex], for some constant [itex]c[/itex]. And (more to the point) it maps open intervals in [itex]L[/itex] to open intervals in [itex]\mathbb{R}[/itex].

    We do have to assume [itex]f[/itex] is not the zero linear functional!

    We claim [itex]f[/itex] is an open map. Let [itex]U \subset X[/itex] be an open set. We must show the image [itex]f(U)[/itex] is open in [itex]\mathbb{R}[/itex]. Let [itex]u \in U[/itex] . We claim that [itex]f(U)[/itex] contains an interval centered at [itex]f(u)[/itex]. This is because of the restriction property mentioned above. OK?

    (Strictly speaking, we still have to adjust something when [itex]f(u) = 0[/itex].)
  9. Aug 29, 2009 #8
    Your post reminds me of the example of a linear functional from R to R when R is viewed as a vector space over the rationals.

    It is easy to construct a discontinuous one from a Q basis.

    Problem: Show that such the graph such a discontinuous linear functional is dense in the plane.
  10. Aug 30, 2009 #9
    Thanks, g_edgar. That clears things up.

    wofsy: that's the same construction that Gelbaum is citing. Am I correct that that construction doesn't have the property that: f(a*x) = a*f(x) for scalar a?
  11. Aug 30, 2009 #10
    if a is a rational number, then f(a*x) = a*f(x).

    The problem has more parts to it. Show that if f is measurable then it is continuous. If f is continuous then it is just multiplication by a scalar.
    What does that tell you about the measurability of a basis for the reals over the rationals?
    Last edited: Aug 30, 2009
  12. Oct 28, 2010 #11
    I don't see the "adjust something" of g_edgar very clearly (for the case f(u)=0).

    None of that is required to complete Rudin's argument, though. In his case, what one can do is simply let gamma=sup Gamma(A). All one needs to prove is that gamma is not in the image of A. Assume that it is in the image of A; that is, gamma=Gamma(a) for some a in A. Then, as A is open, one can find a balanced neighbourhood U of 0 with a+U in A. If u in U, then gamma >= Gamma(a+u)=gamma+Gamma(u), so we conclude that Gamma(u)<=0. But, as U is balanced, this argument should also work for -u, and that gives us a contradiction. The conclusion is that gamma is not in the image of A, and so Gamma(a)<gamma for all a in A.
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