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PostPosted: Thu Jul 14, 2016 10:48 am 
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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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PostPosted: Thu Jul 14, 2016 10:48 am 
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Joined: Mon Nov 09, 2015 1:52 pm
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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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