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Proof that a TVS is seminormed iff it is locally convex

  1. Jan 13, 2015 #1


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    There is a guided exercise in the course of functional analysis that I am following where we have to prove that a topological vectors space is seminormed if and only if it is locally convex. There is one step in te proof that I can't figure out.

    Let [itex]X[/itex] be a topological vector space and [itex]\mathcal{U}[/itex] a convex neighbourhood of [itex]0[/itex].
    [tex]\mathcal{V} = \bigcap_{\lambda \in S_1} \lambda \mathcal{U} [/tex].
    Where [itex]S_1[/itex] are the complex numbers of modulus one.

    Use the compactness of [itex]S_1[/itex] to prove that [itex]\mathcal{V}[/itex] is a balanced convex neighbourhood of zero.

    Proving that [itex]\mathcal{V}[/itex] is balanced an convex is straightforward and doesn't require the compactness of [itex]S_1[/itex]. I haven't been able to prove that [itex]\mathcal{V}[/itex] is still a neighbourhood of zero though.


  2. jcsd
  3. Jan 13, 2015 #2
    Try to use the continuity of multiplication by scalars.
  4. Jan 14, 2015 #3


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    OK, I think I've got it.

    Denote by [itex]m : \mathbb{C} \times X \to X : (z, x) \to zx[/itex] the scalar multiplication map. Since it is continuous and [itex]\mathcal{U}[/itex] is a neighbourhoud of zero, the inverse [itex]m^{-1}(\mathcal{U})[/itex] is a neighbourhood of any pair [itex](z, x)[/itex] such that [itex]zx = 0[/itex]. In particular, [itex]m^{-1}(\mathcal{U})[/itex] is a neighbourhood of [itex](\lambda, 0)[/itex] for any [itex]\lambda \in S^1[/itex]. So we can find for every such [itex]\lambda[/itex] an open neighbourhood [itex]\mathcal{O}_{\lambda} \subset \mathbb{C}[/itex] of [itex]\lambda[/itex] and an open neighbourhood [itex]\mathcal{W}_{\lambda} \subset X[/itex] of zero such that [itex]m(\mathcal{O}_{\lambda} \times \mathcal{W}_{\lambda}) \subset \mathcal{U}[/itex]. The [itex]\{ \mathcal{O}_{\lambda}\}[/itex] are an open cover of the compact set [itex]S^1[/itex] so we can take a finite subcover [itex]\{\mathcal{O}_i \}_{i=1,\dots,k}[/itex] where to each [itex]\mathcal{O}_i[/itex] there still corresponds an open neighbourhood [itex]\mathcal{W}_i[/itex] of zero such that [itex]m(\mathcal{O}_i \times \mathcal{W}_i) \subset \mathcal{U}[/itex]. But [itex]\mathcal{W} = \bigcap_{i = 1, \dots, k} \mathcal{W}_i[/itex] is also an open neighbourhood of zero, and for all [itex]i=1,\dots,k[/itex] we have [itex]m(\mathcal{O}_i \times \mathcal{W}) \subset \mathcal{U}[/itex]. Since [itex]\{\mathcal{O}_i \}_{i=1,\dots,k}[/itex] covers [itex]S^1[/itex] we have that [itex]\lambda \mathcal{W} \subset \mathcal{U}[/itex] for any [itex]\lambda \in S^1[/itex]. Therefore, [itex]\mathcal{W} \subset \lambda \mathcal{U}[/itex] for any [itex]\lambda \in S^1[/itex] whence [itex]\mathcal{W} \subset \mathcal{V}[/itex]. Since [itex]\mathcal{W}[/itex] is an open neighbourhood of zero, we conclude that [itex]\mathcal{V}[/itex] is a neighbourhood of zero.

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