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Let $H$ be a hilbert space and $(h_{n})_{n\in\mathbb{N}}$ be a bounded sequence in $H$.

Define $H_{0}:= \text{cl}(\text{span}(h_{1},h_{2},...))$. Then, $H_{0}$ is a separable space since the set of all finite linear combinations of points in $(h_{n})_{n \in\mathbb{N}}$ with rational coefficients is countable and dense subset of $H_{0}$. Define $\forall n \in\mathbb{N}$, $f_{n} : H_{0}\to\mathbb{R}$ as $f_{n}(h) = \langle h,h_{n}\rangle$ for any $h\in H_{0}$. ($\langle \, \cdot \, \, \cdot \,\rangle$ is inner product in $H$)

By Riesz Representation theorem and Helley's Theorem we can obtain $f_{0} \in H_{0}$ such that there exists a subsequence $(f_{n_{k}})_{k\mathbb{N}}$ such that it weak-*converges to $f_{0}$. Moreover, $\exists!h_{0},\forall h \in H_{0}, f_{0}(h) = \langle h,h_{0}\rangle$

Hence, $\forall h \in H_{0}$ we have $\langle h,h_{n_{k}}\rangle\to\langle h,h_{0}\rangle$ as $k\to\infty$.

Let $P$ be an orthogonal projection of $H$ onto $H_{0}$. Then, $\forall k\in\mathbb{N}$, we have \begin{equation}\tag{1} \langle (I-P)(h),h_{n_{k}}\rangle = \langle (I-P)(h),h_{0}\rangle \end{equation} Hence, $\forall h \in H$, we have \begin{equation}\tag{2} \lim\limits_{k\to\infty}\langle h_{n_{k}},h\rangle = \langle h_{0},h\rangle \end{equation}

My questions are :
1. Why do we need "rational" coefficients? I dont see the point to consider only rational coefficients of the span here
2. How do we obtain (1) and why (1) implies (2)?

Any help is much appreciated since I am trying to prove that any bounded sequence in Hilbert space has a weakly convergent subsequence.

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You take rational coefficients to make the set countable. Even if you take single (non-zero)vector and take all linear combinations, i.e. all scalar multiples of it the set is not countable because the real line is not countable.

1) is just definition of weak* convergence. Saying that $f_{n_k}$ converges to $f_0$ in weak* topology is equivalent to $ \langle y, h_{n_k} \rangle \to \langle y, h_0 \rangle$ for every $y$ in $H$. So you get 1) by taking $y=(I-P)(h)$.

1) implies 2) follows from the fact that $h=P(h)+(I-P)(h)$. Since $P(h) \in H_0$ we already know that $ \langle P(h), h_{n_k} \rangle \to \langle P(h), h_0 \rangle$. Now add this to 1) to get 2).

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  • $\begingroup$ Thank you but I am not familiar with projection operator. $h$ is in $H$ and $P(h)$ is in $H_{0}$. It does not make sense since that $h = P(h) + (I-P)(h)$ since $h$ and $P(h)$ are not in the same space. How to make sense of this? $\endgroup$ Commented Apr 17, 2019 at 9:13
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    $\begingroup$ $Ph$ and $(I-P)(h)$ are both in $H$. ($H_0$ is a subspace of $H$ and you have to think of $P$ as a map from $H$ into $H$ with range contained in the subspace $H_0$ $\endgroup$ Commented Apr 17, 2019 at 9:18
  • $\begingroup$ Last question, $-P(h)$ is in $H_{0}$ and therefore $(I-P)(h)$ is also in $H_{0}$. I just want to confirm this $\endgroup$ Commented Apr 17, 2019 at 9:38
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    $\begingroup$ No, $(I-P)(H)=h-Ph$ and only the second term is in $H_0$. $\endgroup$ Commented Apr 17, 2019 at 9:40
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    $\begingroup$ in 1) both sides are $0$. The range of $I-P$ is the orthogonal complement of $H_0$. $\endgroup$ Commented Apr 17, 2019 at 10:21

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