Let $(x_n)_{n≥1}$ be a sequence of points in a Hilbert space $H$ such that $C\equiv \lim \inf _{n\to \infty} \|x_n\| < \infty $. Prove that there exists a weakly convergent subsequence $x_{nj} \rightharpoonup x$, for some point $x \in H$ satisfying $\|x\| \le C$.
$\textbf{My attempt}$
It suffices to show the inequality $\|x\| \le C$ and hence given that every bdd sequence admits a weakly convergent subsequence $\exists \ x_{nj} \rightharpoonup x$.
Let $(e_k)$ be the orthonormal basis for the space $H$, we have that
$$\|x\|^2 = (x,x) = \sum_{n=1}^\infty x_n^2 = \sum_{n=1}^\infty |(x_n,e_n)e_n|^2 \le \|x_n\|^2$$
where I used Bessel's inequality in the last term so taking $\lim \inf$;
$$\|x\| \le \lim \inf _{n\to \infty}\|x_n\| = C$$
Best Answer
It suffices to show that every bounded sequence has a weakly convergent subsequence (why: take $(x_{n_k})\subset(x_n)$ such that $\|x_{n_k}\|\to\liminf\|x_n\|=C$. If we show that $(x_{n_k})$ -which is bounded- has a weakly convergent subsequence, we are done.) Moreover we can assume that the bound is 1, since we can scale by $1/M$ if the bound is $M$ and get ourselves in this situation.
So the problem is equivalent to this: Let $H$ be a Hilbert space and $(x_n)\subset H$ a sequence with $\|x_n\|\leq1$. Then $(x_n)$ has a weakly covnergent subsequence.
To prove this, we consider the linear functionals $\langle-,x_n\rangle:H \to\mathbb{C}$. These are all bounded with norm $\leq1$. By the Banach-Alaoglu theorem, the closed unit ball of $H^*$ is weak-* compact, thus $(\langle-,x_n\rangle)_{n=1}^\infty$ has a weak-* convergent subsequence, i.e. there exists $(x_{n_k})\subset (x_n)$ and $\varphi\in H^*$ with $\|\varphi\|\leq 1$ such that for all $x\in H$ it is $$\langle x,x_{n_k}\rangle\to\varphi(x)$$ But the Riesz representation theorem says that there exists a unique $y\in H$ such that $\varphi=\langle-,y\rangle$, therefore we have found $(x_{n_k})\subset(x_n)$ and $y\in H$ such that for all $x\in H$ it is $\langle x,x_{n_k}\rangle\to\langle x,y\rangle$, which is what we wanted.