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Math Help - Annihilators of Banach Spaces

  1. #1
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    Annihilators of Banach Spaces

    Hi All,

    Could you please help with the following?

    Let X be a normed vector space with Banach dual space X'. Let  f_{1}, \ ... \ f_{n} be in X' such that  Ker(f) \supset \bigcap_{i} Ker (f_{i}) . Show that

     T: (f_{1}(x),...,f_{n}(x))\rightarrow f(x)

    is a well-defined linear map on a subspace of  \mathbb{F}^{n}. Extending T to all of  \mathbb{F}^{n}, show that there are  \alpha_{1},...,\alpha_{n} \in \mathbb{F} such that  f=\sum_{j=1}^{n} \alpha_{j}f_{j}

    Conclude that, for any finite-dimensional subspace M of X',  ( M_{o} ) ^{o} =M

    Now, by definition, for  M \subset X, N \subset X'

    M^{o}= \{f \in X': \ f(x)=0 \ \forall x \in M\}
     N_{o}=\{ x \in X; \ f(x)=0 \ \forall f \in N \}

    Now, I can show that T is well-defined and linear on the subspace  (ranf_{1}) x ... x (ranf_{n})=:K .
    By the Hahn-Banach theorem, I can extend this to a function  G \in \( \mathbb{F}^{n} \)'  such that  G|_{K}=T and  \|G\|_{ ( \mathbb{F}^{n} )'} = \|T\|_{K'} . But where do I go from here?
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  2. #2
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    The key thing that you need to notice is that every linear functional T on \mathbb{F}^{n} is of the form Ty = \alpha_1y_1 + \alpha_2y_2 + \ldots + \alpha_ny_n\;\,(y = (y_1,y_2,\ldots,y_n)\in\mathbb{F}^{n}), for some scalars \alpha_1,\, \alpha_2,\,\ldots, \alpha_n. All you have to do then is to take (y_1,y_2,\ldots,y_n) = (f_1(x),f_2(x),\ldots,f_n(x)).
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  3. #3
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    Thanks a lot for this!
    Could you please check I've got the last part, ( ( M_{o} ) ^{o} =M ), right?

    So, want to show  M \subset ( M_{o} ) ^{o}\ , \ so \ take \ f \in M \leq X' \ \Rightarrow  f \in X' Want to show  f \in ( M_{o} ) ^{o} ,

    i.e. that (by defn of   ( M_{o} ) ^{o} )  f \in X' \ and \ f(x)=0 \ \forall x \in M_{o}
    i.e. (by defn of  M_{o} )  f(x)=0 \ \forall x \in X : (f(x)=0 \forall f \in M )
    which is true, as we started with  f \in M

    Similarly for the other direction. What I don't get - where did we use the  Ker(f) \supset \bigcap_{i} Ker (f_{i}) bit? And where in this last part did we use any of the previous bits of the question?

    Thanks in advance!
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  4. #4
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    Quote Originally Posted by Mimi89 View Post
    What I don't get - where did we use the  Ker(f) \supset \bigcap_{i} Ker (f_{i}) bit?
    The place where you needed  \ker(f) \supset \bigcap_{i} \ker (f_{i}) was right at the beginning of the question, in order to show that T is well defined. If the definition  T: (f_{1}(x),...,f_{n}(x))\rightarrow f(x) is to make sense, it must be true that f(x) depends only on f_{1}(x),...,f_{n}(x). So you need to be sure that if (f_{1}(x),...,f_{n}(x)) = (f_{1}(y),...,f_{n}(y)) then f(x) = f(y). But that is equivalent to saying that if f_i(x-y) = 0 for i=1,2,...,n, then f(x-y)=0. That in turn is equivalent to the condition  \ker(f) \supset \bigcap_{i} \ker (f_{i}) .
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  5. #5
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    Quote Originally Posted by Mimi89 View Post
    Conclude that, for any finite-dimensional subspace M of X',  ( M_{o} ) ^{o} =M .
    For the last part of the question, you are given a finite-dimensional subspace M\subset X'. Let f_1,\ldots,f_n be a basis for M. Then

    M_o = \{ x \in X:  f(x)=0\  \forall f \in M \} = \{x\in X:f_i(x)=0\ (1\leqslant i\leqslant n)\} = \bigcap_{i} \ker (f_{i}),

    and (M_o)^o = \{f\in X':f(M_o)=0\} = \{f\in X':\ker(f)\supset \bigcap_{i} \ker (f_{i})\}.

    But it follows from the earlier part of the question that that last condition implies that f is a linear combination of f_1,\ldots,f_n, which says that f\in M. Therefore (M_o)^o\subset M. You have shown how to prove the reverse inclusion M\subset (M_o)^o, and so (M_o)^o = M.
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