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I need your help with this problem that I founded it in a lecture notes. Then, the problem says:

Let $ X $ be a normed space. Show that if a sequence $ (x_n) _ {n \in \mathbb {N}} $ in $ X $ converges weakly to $ x $ then $ x\in Y $, where $ Y $ is the closure of the vector space generated by $\{x_n : n \in \mathbb{N} \} $.

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what have you done? –  Holdsworth88 Aug 28 '12 at 17:59
    
from what I have noticed, people are more likely to help those who have a high acceptance rate and put effort into their questions. Perhaps showing a bit of effort and accepting answers to you previous questions will help. –  Holdsworth88 Aug 28 '12 at 18:55
    
For convex sets, the norm closure coincides with the weak closure. –  copper.hat Aug 28 '12 at 20:16
    
@Holdsworth88 I should review all topics of my courses undergraduate for a exam that I have that pass, is the final, so I don't have much time, and I'm here, because with your help I could get it. For make this review quickly. Excuse me. –  d555 Aug 29 '12 at 4:31
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3 Answers 3

Suppose that $x\notin Y$.

Then, since $\{x\}$ is a compact convex set and $Y$ is a closed linear subspace (and therefore a closed convex set also), there exists a linear functional $h\in X^*$ such that $\langle h,x\rangle >0$ and $\langle h, x_i\rangle =0$ for every $i$.

It is a straightforward aplication of the Mazur-Eidelheit separation theorem, can also be proved using Hahn-Banach theorem.

Now observe that $\langle h,x_i\rangle =0<\langle h,x\rangle$ so we cannot have the convergence $\langle h,x_i\rangle \rightarrow\langle h,x\rangle$, which is a contradiction since $x_i\rightarrow x$ in the weak sense. Q.E.D.

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For a convex set in a normed space, the closure and the weak closure coincide.

Let $S = \mathbb{sp}\{x_n\}$. Since $S$ is convex, we have $Y = \overline{S} = {\overline{S}}^w$ (the latter denoting the weak closure).

Since $x_n \to x$ weakly, we have $x \in {\overline{S}}^w$, hence $x \in \overline{S}$.

To see why $x$ must be in ${\overline{S}}^w$, suppose $x \notin {\overline{S}}^w$. Then there is a weak open set containing $x$ that does not intersect $S$. The open set must contain a weak neighborhood of the form $\{y \, |\, |\phi_i(x-y)| < \epsilon, \, \, i=1,...,n \}$. However, since $\phi_i(x_n) \to \phi(x)$, we quickly obtain a contradiction.

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Hint: Proof by contradiction, using Hahn-Banach to produce a linear functional.

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