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Continuous functional s.t. $f(x_0)\ne 0$

  1. Aug 31, 2014 #1
    I read that in any locally convex topological space [itex]X[/itex], not necessarily a Hausdorff space but with linear operations continuous, for any ##x_0\ne 0## we can define a continuous linear functional [itex]f:X\to K[/itex] such that [itex]f(x_0)\ne 0[/itex].

    I cannot find a proof of that anywhere and cannot prove it myself. Please correct me if I am wrong, but I think that if [itex]A[/itex] is closed in [itex]X[/itex] and [itex]x_0\in A[/itex] there exist a continuous linear functional rigorously separating [itex]x_0[/itex] and [itex]A[/itex], but I am not sure whether we can use that...

    I have also tried with Minkowski functional, but with no result...
    [itex]\infty[/itex] thanks for any help!
     
    Last edited by a moderator: Aug 31, 2014
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  3. Aug 31, 2014 #2

    micromass

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    This follows from the Hahn-Banach theorems. One of the versions of this theorems state that

    A proof can be found in Kadison & Ringrose "Fundamentals of the Theory of Operator Algebras, Vol I", Theorem 1.2.14 page 22.

    Now you only need to define a suitable operator ##\rho_0## on ##\textrm{span}\{x_0\}## and apply the previous theorem.
     
  4. Aug 31, 2014 #3
    ...and we can choose [itex]\rho_0:\alpha x_0\mapsto \alpha[/itex].
    However, I think it may be necessary that ##V## be Hausdorff, since theorem 1.2.6, on which 1.2.14 is based, needs the hypothesis. ##\infty## thanks!
     
  5. Aug 31, 2014 #4

    micromass

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    Yes, you need Hausdorff. Indeed, assume we have for each ##x_0\neq 0## a continuous functional ##f:V\rightarrow K## such that ##f(x_0)\neq 0##.

    Then take ##x,y\in V## distinct points. We can find a continuous linear functional ##f:V\rightarrow K## such that ##f(x-y)\neq 0##. Thus we have ##f(x)\neq f(y)##, and thus we can find open sets ##U## and ##V## in ##K## such that ##f(x)\in U## and ##f(x)\in V## and ##U\cap V = \emptyset##. Then ##x\in f^{-1}(U)## and ##y\in f^{-1}(V)## and ##f^{-1}(U)## and ##f^{-1}(V)## are disjoint open sets in ##V##. As such ##V## is Hausdorff.
     
  6. Aug 31, 2014 #5
    Functional analysis is so fascinating...
    Thank you so much!!!
     
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