1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Evan William Chandra Apr 17 at 9:13
  • 1
    $\begingroup$ $Ph$ 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 $H_0$ $\endgroup$ – Kavi Rama Murthy Apr 17 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 $\endgroup$ – Evan William Chandra Apr 17 at 9:38
  • 1
    $\begingroup$ No, $(I-P)(H)=h-Ph$ and only the second term is in $H_0$. $\endgroup$ – Kavi Rama Murthy Apr 17 at 9:40
  • 1
    $\begingroup$ in 1) both sides are $0$. The range of $I-P$ is the orthogonal complement of $H_0$. $\endgroup$ – Kavi Rama Murthy Apr 17 at 10:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.