Every Bounded Seq. in a Hilbert Space has a Weakly Con. Subseq.

# Every Bounded Sequence in a Hilbert Space has a Weakly Convergent Subsequence

 Theorem 1: Let $H$ be a Hilbert space. Then every bounded sequence in $H$ has a weakly convergent subsequence.
• Proof: Let $(h_n)$ be a bounded sequence in $H$. Let:
(1)
\begin{align} \quad H_0 = \mathrm{cl} (\mathrm{span} (h_1, h_2, ...)) \end{align}
• Then $H_0$ is separable as the set of all finite linear combinations of points in $(h_n)$ with rational coefficients is a countable and dense subset of $H_0$. For each $n \in \mathbb{N}$ let $f_n : H_0 \to \mathbb{R}$ be defined for all $h \in H_0$ by:
(2)
\begin{align} \quad f_n(h) = \langle h, h_n \rangle \end{align}
(3)
\begin{align} \quad |f_n(h)| = |\langle h, h_n| \leq \| h \| \| h_n \| = \| h_n \| \| h \| \end{align}
• So $(f_n)$ is a bounded sequence of linear functional on the separable space $H_0$. By Helley's Theorem we have that $(f_n)$ has a weak-* convergent subsequence, say $(f_{n_k})$ weak-* converges to $f_0 \in H_0$. By The Riesz Representation Theorem for Hilbert Spaces there exists an $h_0 \in H_0$ such that $f_0(h) = \langle h, h_0 \rangle$ for all $h \in H_0$. So $(f_{n_k})$ weak-* converges to $f_0(h)$, and so for every $h \in H$:
(4)
\begin{align} \quad \lim_{k \to \infty} f_{n_k}(h) &= \langle h, h_0 \rangle \\ \quad \lim_{k \to \infty} \langle h, h_{n_k} \rangle &= \langle h, h_0 \rangle \end{align}
• Let $P$ be the orthogonal projection of $H$ onto $H_0$. Then for each $k \in \mathbb{N}$ we have that:
(5)
\begin{align} \quad \langle (I - P)(h), h_{n_k} \rangle = \langle (I - P(h), h_0 \rangle \end{align}
• Hence, for each $h \in H$ we have that:
(6)
\begin{align} \quad \lim_{k \to \infty} \langle h_{n_k}, h \rangle = \langle h_0, h \rangle \end{align}
• So from the characterization of Weak Convergence in Hilbert Spaces we have that $(h_{n_k})$ weakly converges to $h_0 \in H$. So every bounded sequence in $H$ has a weakly convergent subsequence. $\blacksquare$