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 Post subject: The dual space of $X$ giving information about $X$Posted: Thu Jul 14, 2016 10:48 am
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Joined: Tue Nov 10, 2015 8:25 pm
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Let $\displaystyle X$ be a normed space and let $\displaystyle X^{*} = B(X,\mathbb{R})$ be its dual space. Show that

1. For all $\displaystyle x,y \in X$ such that $\displaystyle x \neq y$ there exists $\displaystyle f \in X^{*}$ such that $\displaystyle f(x) \neq f(y)$. $\\$
2. If $\displaystyle Y$ is a proper closed subspace of $\displaystyle X$ and if $\displaystyle x_{0} \in X \smallsetminus Y$, then there exists $\displaystyle f \in X^{*}$ with $\displaystyle ||f||=1$ such that $\displaystyle f(y)=0,\forall y \in Y$ and $\displaystyle f(x_{0}) = d(x_{0},Y)$. $\\$
3. If $\displaystyle X^{*}$ is seperable, then so is $\displaystyle X$. Is the converse true?

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 Posted: Thu Jul 14, 2016 10:48 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Hi Nickos.

1. Let $\displaystyle{x\,,y\in X}$ such that $\displaystyle{x\neq y}$. Let $\displaystyle{u=x-y\neq \overline{0}}$.

Consider the subspace $\displaystyle{Y=<u>=\left\{k\,u: k\in\mathbb{R}\right\}}$ of $\displaystyle{\left(X,||\cdot||\right)}$ .

Obviously, $\displaystyle{\dim_{\mathbb{R}}Y=1}$ .

We define $\displaystyle{g:Y\longrightarrow \mathbb{R}}$ by $\displaystyle{g(k\,u)=k\,||u||}$ .

Let $\displaystyle{y_1\,,y_2\in Y}$ and $\displaystyle{a\in\mathbb{R}}$. So, $\displaystyle{y_1=k_1\,u\,\,,y_2=k_2\,u}$ for

some $\displaystyle{k_1\,,k_2\in\mathbb{R}}$ and then

$\displaystyle{g(y_1+y_2)=g\,((k_1+k_2)\,u)=\left(k_1+k_2\right)\,||u||=k_1\,||u||+k_2\,||u||=g(y_1)+g(y_2)}$

$\displaystyle{g(a\,y_1)=g\,(a\,(k_1\,u))=g\,((a\,k_1)\,u)=(a\,k_1)\,||u||=a\,(k_1\,||u||)=a\,g(y_1)}$

which means that $\displaystyle{g}$ is a linear map. Also, if $\displaystyle{y=k\,u\in Y\,\,,k\in\mathbb{R}}$, then :

$\displaystyle{\left|g(y)\right|=\left|g\,(k\,u)\right|=\left|k\,||u||\right|=\left|k\right|\,||u||=||k\,u||=||y||}$ .

According to $\displaystyle{\rm{Hahn-Banach}}$ - theorem, there is $\displaystyle{f:X\longrightarrow \mathbb{R}}$

with $\displaystyle{f\in X^{\star}}$, $\displaystyle{\left|f(x)\right|\leq ||x||\,,\forall\,x\in X}$ and

$\displaystyle{f(y)=g(y)\,,\forall\,y\in Y}$ .

$\displaystyle{f(u)=g(u)=g\,(1\cdot u)=1\cdot ||u||=||u||\neq 0}$ so :

$\displaystyle{f(x-y)\neq 0\iff f(x)-f(y)\neq 0\iff f(x)\neq f(y)}$ .

2.We observe that $\displaystyle{x_0\neq \overline{0}}$ cause $\displaystyle{x_0\notin Y}$ . So, we define $\displaystyle{Z=<Y\cup\left\{x_0\right\}>}$ ,

$\displaystyle{g:Z\longrightarrow \mathbb{R}\,,g(y+k\,x_0)=k\,d\,(x_0,Y)}$

and we continue as in 1.

3.The set $\displaystyle{S=\left\{f\in X^{\star}: ||f||=1\right\}}$ is a seperable subset of $\displaystyle{\left(X^{\star},||\cdot||\right)}$ cause

$\displaystyle{\left(X^{\star},||\cdot||\right)}$ is seperable (left as an exercise) .

There exists dense subset $\displaystyle{D=\left\{f_{n} :n\in\mathbb{N}\right\}}$ of $\displaystyle{S}$ ,

which means that $\displaystyle{\left(\forall\,s\in S\right)\,\left(\forall\,\epsilon>0\right)\,\left(\exists\,d\in D\right): ||d-s||<\epsilon}$ .

Exercise: Prove that : $\displaystyle{\left(\forall\,n\in\mathbb{N}\right)\,\left(\exists\,x_{n}\in\left\{x: ||x||\leq 1\right\}\right) : f_{n}(x_n)>\dfrac{1}{2}}$ .

Setting $\displaystyle{Y=<\left\{x_{n}: n\in\mathbb{N}\right\}>}$ , we have that $\displaystyle{Y\simeq \mathbb{N}}$ and $\displaystyle{\overline{Y}=X}$ (why ?)

, which means that $\displaystyle{\left(X,||\cdot||\right)}$ is seperable.

The converse is not true cause $\displaystyle{\mathbb{l_{1}}^{\star}=\mathbb{l}_{\infty}}$ and

$\displaystyle{\mathbb{l}_{\infty}}$ is not seperable.

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