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Prove or disprove about Dual space

  1. Nov 13, 2012 #1
    1. X : Banach space

    Z : closed subspace of X

    Prove or disprove that X* ⊆ Z*

    where Z* and X* are dual space of Z and X, respectively.

    2. X : normed space and f : X → R : linear functional.

    Assume that ∃a∈X and r∈(0,1] such that f(B(a,r))=R(Real numbers)

    where B(a,r) is open ball. Prove that f is discontinuous

    Proof. 2) I think it must suppose f is continuous and find a contradiction with f(B(a,r))=R

    But I can't find some contradiction.

    1) We think it doesn't satisfy X* ⊆ Z*. But we will show X* ⊆ Z* and find some contradiction. Let f ∈ X*. Thus f : X → R is bounded linear functional. We must to show that f : Z → R is bounded linear functional or find some contradiction.
     
  2. jcsd
  3. Nov 13, 2012 #2
    You have that [itex] Z \subseteq X [/itex] right? Linear functionals must be linear maps into [itex] \mathbb R [/itex] for all elements on which they act. A way of thinking about this is that the bigger a space is, the more restrictions are imposed on the dual functions, hence we would suspect that the dual spaces reverse inclusion!

    Now check this. As you started, let [itex] f \in X^* [/itex] so that [itex] f^*: X \to \mathbb R [/itex] is linear and bounded. You want to know how it acts on a closed subspace of X. So first of all, could the function fail to be linear? Secondly, could the function fail to be bounded?

    For the second question, think about the definition of continuity. The function f is continuous at a if for all [itex] \epsilon >0 [/itex] there exists [itex] \delta >0 [/itex] such that [itex] f(B(a,\delta)) \subseteq B(f(a),\epsilon). [/itex] Could you function be continuous at 'a'? Why or why not?
     
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