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I am trying to solve the following problem but I don't really getting it where to start from , which way to think . Any help would be appreciated.

If $(X,\|.\|)$ is a seperable Banach space,

a) Why is it true that the unit sphere ie $S=\{x\in X ;\|x\|=1\}$ is seperable in Relative topology ?

b) Is $(x_n)_{n\in \mathbb N } \subset S$ dense sequence and also $T:l^1(\mathbb N) \to X $ defined by $T((a_n)_{n\in\mathbb N})=\sum_{n\in \mathbb N }a_n x_n$ , why is $T$ bounded and surjective ?

Is is true that $X$ is topologically isomorphic to $\ell^1(\mathbb {N})/\mathrm{Ker}(T)$

Thanks

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a) is a bit surprising. Is this really true? –  Rudy the Reindeer Nov 20 '12 at 8:35
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(a) is not true in general. It's true exactly iff $X$ is separable. Did you omit this condition? –  martini Nov 20 '12 at 8:37
    
@MattN. This has to be true, I heard this statement made by Prof , but i was not satisfied because he didn't explain why its true . –  Theorem Nov 20 '12 at 8:39
    
I don't understand your question b). Where is defined the operator $T$ ? $\ell^1$ ? $\ell^2$ ? $\ell^{\infty}$ ? –  Ahriman Nov 20 '12 at 8:45
    
@Ahriman : Sorry , i had not written down this question so i couldn't exactly reproduce it , i hope now it makes sense , Thank you . –  Theorem Nov 20 '12 at 8:50

1 Answer 1

up vote 0 down vote accepted

a) A metric space is separable iff it admits a countable base of open sets. From this, it easily follows that every subset of a separable metric space is still a separable metric space.

In your particular setting, there is a simpler proof : if $\pi$ is the radial projection onto the unit sphere, and if $D$ is countable dense subset of $X$, then $\pi(D)$ is a countable dense subset of the unit sphere.

b) Boundedness of $T$ follows immediately from the definition : how can you bound $\left| \sum_n a_n x_n \right|$ in terms of $\|a\|_1 = \sum_n |a_n|$ ? For the surjectivity, notice that each $x_n$ belongs to the range of $T$. Hence the range of $T$ contains the closure of the linear span of $\{x_n\}$ which is the whole space. The fact that $X$ is topologically isomorphic to $\ell^1 / \ker(T)$ then follows from the open mapping theorem.

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In the surjectivity part of b) you are assuming that the range of $T$ is closed -- which isn't clear a priori. You need to be a bit more careful: from the usual back-and-forth method in the proof of the open mapping theorem you can deduce: if $T$ maps the unit ball of $Y$ densely into the unit ball of $X$ then $T\colon Y \to X$ is onto (this is sometimes called the Banach-Schauder theorem). –  commenter Nov 20 '12 at 10:33
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On second reading, maybe you are not assuming that the range is closed... However, I think that a justification for "Hence the range of $T$ contains the closure of the linear span" should be given. –  commenter Nov 20 '12 at 10:46
    
@Ahriman : how can i say that $\pi(D)$ is dense subset of unit sphere ? –  Theorem Nov 23 '12 at 7:04

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