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Thread: 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 $\displaystyle x_o \in X $ and $\displaystyle x_o \in span(M)$, we know that $\displaystyle x_o$ can be written as a linear combination of elements in $\displaystyle M $ Say $\displaystyle x_o = \sum a_i b_i $ where $\displaystyle a_i $ is scalar and $\displaystyle b_i \in M$.

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

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

    Since the statement applies to all linear functionals, choose $\displaystyle f $ such that $\displaystyle f(x) = 0 \ \forall x \in M $ and $\displaystyle f(y) <> 0 \ \forall \ y \notin M $. This means that $\displaystyle f (z ) = 0 $ implies z in M.
    Suppose to the contrary that $\displaystyle x_o \in X $ and $\displaystyle x_o \notin span(M) $ with $\displaystyle f(x_o ) = 0 $. Then $\displaystyle x_o = \sum a_i b_i + c$ for some $\displaystyle c \notin M $. Since $\displaystyle \sum f (b_i) = 0 $, we have $\displaystyle f(x_o) = f(\sum a_i b_i) + f(c)= 0 + f(c) = 0 $. This implies $\displaystyle 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 $\displaystyle E=\textrm{span}(M)$ and suppose $\displaystyle x_0\in\bar E$. Then there is a sequence $\displaystyle x_n\in E$ for which $\displaystyle ||x_n-x_0||\to 0$.

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

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

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

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

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

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

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

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

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

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