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How can I prove the following statement?

Let X be a normed vector space, $x_n$ a bounded sequence in X, x $\in$ X and X* the dual space of X.

To show: $x_n$ weakly converges to $x$ as $n\rightarrow \infty$ if and only if there exists a subset $B \subseteq X^*$ with $X^*$ is the closure of $L(B)$ (the linear hull of $B$), such that $y^*(x_n) \rightarrow y^*(x)$ for every $y^* \in B$.

(where $x_n$ converges weakly to $x$ iff for every $x^*: x^*(x_n)\rightarrow x^*(x)$).

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    $\begingroup$ @Michael Well, $x^*\in X^*$, and $X^*$ is the dual space of $X$. Need I say more? $\endgroup$ – Harald Hanche-Olsen May 24 '17 at 21:40
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    $\begingroup$ You cannot prove it without an additional assumption. Weakly convergent sequences are bounded. $\endgroup$ – Daniel Fischer May 24 '17 at 21:49
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    $\begingroup$ My edit was to change "$x^*(x_n)\to x^*(x)$ for every $x^*\in B^*$" to "$y^*(x_n)\to y^*(x)$ for every $y^*\in B^*$" for clarity only. $\endgroup$ – DanielWainfleet May 25 '17 at 3:36
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    $\begingroup$ So if you assume there is a $B \subseteq X^*$ that satisfies $X^* = cl(L(B))$, you want to show that for all $y \in X$ we have $<x_n,y>\rightarrow<x,y>$. What if you define $f(x) = <x,y>$ and, since this is a linear function in $X^*$, there must be a function $g \in L(B)$ that is "close" to $f$, and $g$ has a specific form as a linear combination of funciotns in $B$. I suspect that the definition of "closure" and "distance between functions" in this function space will eventually be used with bounded $\{x_n\}$ sequence (I am not sure what those exact definitions are, likely $||f||/||x||$). $\endgroup$ – Michael May 25 '17 at 14:46
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    $\begingroup$ Okay, now I managed to do it and finally needed boundedness of $x_n$. Thanks a lot for all the help! $\endgroup$ – Infinite_28 May 25 '17 at 19:01
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This is wrong. Let $X = l^2$ which is a separable Hilbert Space. Define $B:= \{e_n : n\in N\}$ and take the sequence $\{ke_k\}_{k=1}^{\infty}$. Clearly for any $e_n \in B$ $$\left \langle e_n , k e_k \right \rangle \rightarrow 0 ,~ as ~ k \rightarrow \infty $$ And $cl (L(B) ) = l^2 =X^*.$

But $\{ke_k\}_{k=1}^{\infty}$ is not weak convergent, since it is unbounded.

P.S. $e_n = (0,0,0...0,1,0,...0,0,0)$ the nth coordinate is $1$.

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  • $\begingroup$ I think this is a good example in line with the Daniel Fisher comment above. $\endgroup$ – Michael May 25 '17 at 14:37
  • $\begingroup$ You're welcome. $\endgroup$ – Red shoes May 25 '17 at 19:41
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$\Rightarrow$ : Set $B=X^*$

$\Leftarrow$ : Let $B$ be a subset of $X^*$ with the properties that $X^* = l(L(B))$ and $ \forall y^*\in B :y^*(x_n)−>y^∗(x)$

T.p.: $z^∗(x_n) \rightarrow z^∗(x) \forall z^∗\in X^∗$. Let $z^∗ \in X^∗$be arbitrarily chosen. Because $cl(L(B))=X^∗: \exists a^∗$, which is a linear combination of elements in B, i.e.: $a^∗=\sum \limits_{i=1}^m \lambda_i * a_i^* ,a_i^* \in B \forall i.\\$ Let $\epsilon > 0$ be arbitrarily chosen. Then, using the operator norm we get:

$\lVert z^∗(x_n)−z^∗(x) \rVert <=\lVert z^∗(x_n)−a^∗(x_n) \rVert +\lVert a^∗(x_n)−a^∗(x) \rVert+ \lVert a^∗(x)−z^∗(x) \rVert <= \\ \lVert z^∗−a^∗\rVert * \sup_n \lVert x_n \rVert +\lVert a^∗(x_n)−a^∗(x) \rVert +\lVert a^∗−z^∗\rVert ∗ \lVert x \rVert < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$.

The first and third term become arbitrarily small ($<\frac{\epsilon}{3}$) because $cl(L(B))=X^*$ and $x_n$ is bounded, the second one because it is given that holds $\forall a_i^* \in B$ and for $\epsilon > 0$ arbitrary: $\lVert a^∗(x_n)−a^∗(x) \rVert = \lVert a^∗(x_n - x) \rVert = \lVert \sum \limits_{i=1}^m \lambda_i * a_i^* (x_n-x) \rVert <= \sum \limits_{i=1}^m |\lambda_i| \lVert a_i^*(x_n - x) \rVert <= m*max_i|\lambda _i|*\lVert a_i^*(x_n - x) \rVert < \frac{\epsilon}{3}$, when $\lVert a_i^*(x_n - x) \rVert < \frac{\epsilon}{3*m*max_i|\lambda_i|} $.

Thanks a lot for helping me!

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    $\begingroup$ Your proof is generally good, some stylistic and clarifying comments are: (i) In latex $\implies$, $\Leftarrow$, $\leq$, $\geq$ are [b]ack[s]lash-implies, bs-Leftarrow, bs-leq, bs-geq. (ii) A more understandable first-sentence might be "Let $B$ be a subset of $X^*$ with the properties that...", (ii) WHen you define $a^*$ you never specify its proximity to $z^*$, (iii) As there are two things changing ($n\rightarrow\infty$ and proximity between $a^*$ and $z^*$) you might say "Let $\epsilon>0$ and let $a^*$ be a linear combination of elements of $B$ such taht $||a^*-z^*||\leq \epsilon$." $\endgroup$ – Michael May 30 '17 at 1:32
  • $\begingroup$ I also wonder if it can be assumed that $||x||<\infty$? $\endgroup$ – Michael May 30 '17 at 1:33
  • $\begingroup$ Thanks for your hints! @Michael: How could a norm of a single point not be $< \infty$? $\endgroup$ – Infinite_28 May 30 '17 at 20:40
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    $\begingroup$ I guess you could have an infinite dimensional space of real-valued sequences $\{a_n\}_{n=1}^{\infty}$ and consider the norm $\sqrt{\sum_{n=1}^{\infty} a_n^2}$. So the sequence $(1, 0, 0, 0, 0, ...)$ has finite norm, but the sequence $(1,1,1,1...)$ does not. $\endgroup$ – Michael May 31 '17 at 0:03
  • $\begingroup$ Okay, I see. But in our lecture we defined a norm such that it is a function from a vector space to the positive real numbers. So a norm, as we defined it, can never be infinity. Now I am really a little bit confused.. $\endgroup$ – Infinite_28 Jun 1 '17 at 17:45

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