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 Post subject: Dense subspacePosted: Sat Jan 16, 2016 12:04 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Let $\displaystyle{\left(X,||\cdot||\right)}$ be an $\displaystyle{\mathbb{R}}$ - normed space and $\displaystyle{Y}$ a

subspace of $\displaystyle{\left(X,+,\cdot\right)}$ such that, if $\displaystyle{f\in X^{\star}=\mathbb{B}(X,\mathbb{R})}$ with $\displaystyle{f|_{Y}=\mathbb{O}}$ , then

$\displaystyle{f=\mathbb{O}}$ . Prove that $\displaystyle{Y}$ is dense on $\displaystyle{\left(X,||\cdot||\right)}$ .

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 Post subject: Re: Dense subspacePosted: Sat Jan 16, 2016 12:05 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Suppose that $\displaystyle{Y}$ is not dense on $\displaystyle{\left(X,||\cdot||\right)}$ . Then,

the set $\displaystyle{\overline{Y}\neq X}$ is a closed subspace of $\displaystyle{\left(X,||\cdot\right)}$ and

according to $\displaystyle{\rm{Hahn-Banach}}$ - theorem, there exists $\displaystyle{f\in X^{\star}}$ such that

$\displaystyle{f|_{\overline{Y}}=\mathbb{O}}$ and $\displaystyle{||f||=1}$ . But now,

$\displaystyle{y\in Y\implies y\in \overline{Y}\implies f(y)=0\implies f_{Y}=\mathbb{O}}$ and according to the

hypothesis, $\displaystyle{f=\mathbb{O}}$, a contradiction, since $\displaystyle{||f||=1}$ .

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