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Math Help - x_0 element of closure of span of M

  1. #1
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    x_0 element of closure of span of M

    Let M be any subset of a normed space X.
    Show that an x_0 element of X is an element of A=closure(span M)

    if and only if

    f(x_0)=0 for every f element X' such that

    f(x)=0 for all x elements of M (restriction of f to M is 0).
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  2. #2
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    Proof =>
    Let  x_o \in X and  x_o \in span(M), we know that x_o can be written as a linear combination of elements in  M Say  x_o = \sum a_i b_i where  a_i is scalar and b_i \in M.

    Let  f \in X^1 be arbitrary and such that  f(x) = 0 for all x \in M .

    We have  f(x_o)=f(\sum a_i b_i ) = a_i \sum f(b_i) by linearity. Since  b_i \in M,  f(b_i) = 0 . Therefore f(x_o) = 0 .
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  3. #3
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    Proof <=

    Since the statement applies to all linear functionals, choose  f such that  f(x) = 0 \ \forall x \in M and  f(y) <> 0 \ \forall \ y \notin M . This means that  f (z ) = 0 implies z in M.
    Suppose to the contrary that x_o \in X and x_o \notin span(M) with  f(x_o ) = 0 . Then  x_o = \sum a_i b_i + c for some  c \notin M . Since  \sum f (b_i) = 0 , we have  f(x_o) = f(\sum a_i b_i) + f(c)= 0 + f(c) = 0 . This implies  c \in M . Contradiction.

    Note: This is my first answer I've posted. YAY!
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  4. #4
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    Let E=\textrm{span}(M) and suppose x_0\in\bar E. Then there is a sequence x_n\in E for which ||x_n-x_0||\to 0.

    If f\in X^* vanishes on E then f(x_n)=0 for all n, so that |f(x_0)|=|f(x_0)-f(x_n)|=|f(x_0-x_n)|\leq ||f||||x_0-x_n||\to 0, giving f(x_0)=0, and so f must vanish on x_0.

    For the converse, suppose x_0\notin\bar E so that \inf_{x\in E}d(x_0,x)=d(x_0,E)=k>0.

    Let E_1=E+\textrm{span}(x_0). Elements of E_1 are of the form x+\lambda x_0 for x\in E and scalar \lambda, and this expression is unique since x_0\notin E.

    Define a linear functional f on E_1 by f(x+\lambda x_0)=\lambda k. Then f vanishes on E and f(x_0)=k. We will show that ||f||=1.

    Firstly, if \lambda\neq0 then ||x+\lambda x_0||=|\lambda|||x_0+\lambda^{-1}x||\geq|\lambda|k since -\lambda^{-1}x\in E and d(x_0,E)=k.

    Hence |f(x+\lambda x_0)|=|\lambda|k\leq||x+\lambda x_0|| (also true if \lambda=0) and so ||f||\leq 1.

    For any \epsilon>0 with there is an element x\in E for which ||x_0-x||<k+\epsilon. Since f(x_0-x)=f(x_0)=k it follows that

    ||f||\geq\frac{|f(x_0-x)|}{||x_0-x||}>\frac k{k+\epsilon} for all \epsilon>0, so ||f||\geq 1. Therefore ||f||=1.

    By the Hahn-Banach theorem, f can be extended to all of X as a linear functional and the extension has norm 1, so f\in X^*. For this extension, f(x)=0 for all x\in E but f(x_0)=k\neq 0.

    So, if x_0\notin\bar E then there is a linear functional which vanishes on E but not on x_0. So if all linear functionals that vanish on E also vanish on x_0 then x_0\in\bar E.
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