Orthogonal Projection in Hilbert Space to prove weak convergence

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 $$$$\tag{1} \langle (I-P)(h),h_{n_{k}}\rangle = \langle (I-P)(h),h_{0}\rangle$$$$ Hence, $$\forall h \in H$$, we have $$$$\tag{2} \lim\limits_{k\to\infty}\langle h_{n_{k}},h\rangle = \langle h_{0},h\rangle$$$$

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.

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).
• 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? – Evan William Chandra Apr 17 at 9:13
• $Ph$ 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 $H_0$ – Kavi Rama Murthy Apr 17 at 9:18
• Last question, $-P(h)$ is in $H_{0}$ and therefore $(I-P)(h)$ is also in $H_{0}$. I just want to confirm this – Evan William Chandra Apr 17 at 9:38
• No, $(I-P)(H)=h-Ph$ and only the second term is in $H_0$. – Kavi Rama Murthy Apr 17 at 9:40
• in 1) both sides are $0$. The range of $I-P$ is the orthogonal complement of $H_0$. – Kavi Rama Murthy Apr 17 at 10:21