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I am trying to prove that the closure of $S=\{x\in X : ||x||=1\}$ in weak topology is the closure of $B_1(0)$ . I have a doubt about what i am doing is correct and for that i need to know whether the intersection of kernels of finite linear functionals is non-trivial . I think here comes the main point about $X$ being infinite dimensional .

And it looks like the closure seems to be the whole space , the neighbourhood of a point $x_0$ contains the lines passing through $x_0$, so i am finding it a bit tricky to get an idea about the closure .

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It's not clear what you mean. What is $B_1(0)$? –  Cantor Dec 2 '12 at 13:50
    
@Cantor : It means sphere of radius $1$ with the center as $0$. –  Theorem Dec 2 '12 at 13:53
    
@Theorem: In that case, $B_1(0)$ is the same thing as $S$ and the claim is trivial. –  Chris Eagle Dec 2 '12 at 14:27

2 Answers 2

up vote 1 down vote accepted

You are right in most of what you say, and you seem to be missing a very tiny, though important, basic theorem from linear algebra:

Definition: Let $\,V:=V_{\Bbb F}\,$ be a vector space . A subspace $\,H\,$ is called a hyperplane if it is maximal subspace of $\,V\,$, i.e.: $\,H\,$ is a hyperplace iff for every

$$x\in V\setminus H\;\;,\;\;V=Span\{H,x\}=:\langle\,H\,,\,x\,\rangle$$

If $\,\dim_{\Bbb F}V=n<\infty\,$ , the above condition is equivalent with $\,\dim_{\Bbb F}H=n-1\,$

Theorem:

(1)${}\,\,\,\,\,{}$For any $\,f\in V^*:=Hom_{\Bbb F}(V,\Bbb F)\,$ , either $\,f\,$ is trivial or else $\,f\,$ is onto.

(2)${}{}\,\,\,\,{}$ A subspace $\,H\leq V\,$ is a hyperplane iff $\,H=\ker f\,$ , for some $\,0\neq f\in V^*\,$.

Thus, as you can see, the intersection of two hyperplanes indeed is a non-trivial subspace of the space (with $\,\dim V= n=2\,$ being the only non-trivial exception), and the condition of infinte dimension is not important in this case.

Proof of (2) (Highlights):

Suppose $\,H\leq V\,$ is a hyperplane and let $\,\{x_i\}_{i\in I}\,$ be a basis of $\,H\,$. Take now $\,v\notin Span\{x_i\}_{i\in I}\,$ , then $\,B:=\{x_i\,,\,v\}_{i\in I}\,$ is a basis of $\, V\,$ (why?), and define $\,f:V\to\Bbb F\,$ by extending linearly the function

$$f:B\to\Bbb F\;\;\;,\;\;f(y):=\begin{cases}0&\text{if}\,\,\, y=x_i\,\,,\,\,\text{for some}\,\,i \in I\\{}\\1&\text{if}\,\,y=v\end{cases}$$

Check now that $\,H=\ker f$

If $\,H=\ker f\,$ , for some $\,0\neq f\in V^*\,$ , and if $\,f(x)=t\neq 0\,\,,\,\,x\in V\,$ , then $\,V=\langle\,H,x\,\rangle\,$ , because: let $\,v\in V\,$, then: if $\,f(v)=0\,$ then $\,v\in\ker f\,$, otherwise

$$f(v)=k\neq 0\Longrightarrow \,\exists! z\in\Bbb F\,\,s.t.\,\,k=zt\Longrightarrow f(v)=k=zt=zf(x)=f(zx)\in\langle x\rangle$$

Try to finish and round up the argument now.

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What? The intersection of two different hyperplanes in a two-dimensional space is trivial. –  Chris Eagle Dec 2 '12 at 14:37
    
You are right, and that's the unique non-trivial case. Thanks. –  DonAntonio Dec 2 '12 at 14:40
    
@DonAntonio : 1) is clear , but can you tell me shortly why 2) is true. brief explanation is enough , i will fill in the gaps. –  Theorem Dec 2 '12 at 15:32
    
@Theorem, I'll add an edit to my answer with some gihlights of (2)'s proof. –  DonAntonio Dec 2 '12 at 15:50
    
@DonAntonio : This is surprising to me ! Specially if $H$ is a kernel of $f$ then its hyperplane . –  Theorem Dec 2 '12 at 17:00

The closure is not the whole space. The fact that neighbourhoods contain lines is irrelevant, since the closure contains no neighbourhood.

Now let me show below that the intersection of the kernels of a finite set of functionals has to be non-trivial if $X$ is infinite-dimensional.

If you have functionals $f_1,\ldots,f_n$, consider the bounded linear map $\tilde f:X\to\mathbb C^n$ given by $\tilde f(x)=(f_1(x),\ldots,f_n(x))$. Then $\ker\tilde f=\bigcap_k\ker f_k$. So, if $\bigcap_k\ker f_k=\{0\}$, we would have $\tilde f$ linear and injective; it would then be an injection into a finite-dimensional space, which would imply $X$ is finite-dimensional. In conclusion, $\bigcap_k\ker f_k\ne\{0\}$

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I am missing something here , why would it mean that if $\bar f$ is injective that the space should be finite dimensional . –  Theorem Dec 2 '12 at 19:58
    
Because $\tilde f$ is an isomorphism from $X$ onto its image. The image is a subspace of $\mathbb C^n$, so it's finite-dimensional. Thus, $X$ is isomorphic to a finite-dimensional space, so finite-dimensional. –  Martin Argerami Dec 2 '12 at 20:09

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