A necessary and sufficient condition for weak convergence in a Hilbert space

Question

The problem is:

Suppose $H$ is Hilbert and $\{e_n\}_{n = 1}^\infty$ is its orthonormal basis. Prove $x_n \rightharpoonup x_0$ if and only if:

1. $||x_n||$ is uniformly bounded (i.e. $\exists M >0$ s.t. $||x_n|| \le M$ holds for arbitrary $n \in \mathbb{N}_+$).
2. $\langle x_n, e_k \rangle \xrightarrow{n \to \infty} \langle x_0, e_k\rangle, \, k = 1,2,\dots$.

The necessary condition is easy to follow, but I get stuck in the proof of the sufficient condition.

My attempt:

(Necessary; Easy)

Since $x_n \rightharpoonup x_0$, we know $\sup_\limits{n} ||x_n|| < \infty$. Thus, $||x_n||$ is uniformly bounded. For any $k \in \mathbb{N}_+$, the function $\langle e_k, \cdot\rangle \in H^*$. Thus, we have $\langle e_k, x_n\rangle \xrightarrow{n \to \infty} \langle e_k, x_0\rangle$. This implies $\lim_\limits{n \to \infty}\langle x_n, e_k \rangle = \lim_\limits{n \to \infty} \overline{\langle e_k , x_n\rangle} = \overline{\lim_\limits{n \to \infty} \langle e_k , x_n\rangle} = \overline{\langle e_k, x_0\rangle} =\langle x_0, e_k\rangle$. Since $\{e_n\}_{n = 1}^\infty$ is an orthonormal basis of $H$, we know $x_n = \sum_\limits{ k= 1}^\infty \langle e_k, x_n\rangle e_k$ and $x_0 = \sum_\limits{ k= 1}^\infty \langle e_k, x_0 \rangle e_k$.

(Sufficient)

Conversely, suppose $||x_n||$ is uniformly bounded and $\langle x_n, e_k \rangle \xrightarrow{n \to \infty} \langle x_0, e_k\rangle, \, k = 1,2,\dots$. $\forall f \in H^*$, we have $|f(x_n) - f(x_0)| = |f(x_n - x_0)| = \left| f\left( \sum_\limits{k = 1}^\infty \langle e_k, x_n - x_0\rangle e_k\right) \right| \le ||f|| \cdot \left\| \sum_\limits{k = 1}^\infty \langle e_k, x_n - x_0\rangle e_k \right\|$. By Minkowski inequality, we have $\left\| \sum_\limits{k = 1}^\infty \langle e_k, x_n - x_0\rangle e_k \right\| \le \sum_\limits{k = 1}^\infty |\langle e_k, x_n - x_0 \rangle| \cdot||e_k|| = \sum_\limits{k = 1}^\infty |\langle e_k, x_n - x_0 \rangle| $

But how to carry on?

Weak convergence is defined by:

If $x_n \rightharpoonup x_0$, then $\forall f \in H^*$ (the space of all continuous and linear functional), $f(x_n) \to f(x_0)$.

Answer 1

By the Riesz-Fischer theorem, every $f\in H^*$ has the form $$f=\sum_{k=1}^{\infty}a_ke_k$$ where $\sum_ka_k^2<\infty$. Let $N$ be a natural number. Write $$\langle x_n-x_0,f\rangle=\sum_{k=1}^Na_k\langle x_n-x_0,e_k\rangle+\sum_{k>N}a_k\langle x_n-x_0,e_k\rangle$$ Using the Cauchy-Scwhartz inequality and the fact that the $e_k$'s are an orthononrmal basis, the second summand on the r.h.s can estimated as follows:

$$\begin{aligned} \left|\sum_{k>N}a_k\langle x_n-x_0,e_k\rangle\right|\leq \sum_{k>N}|a_k||\langle x_n-x_0,e_k\rangle|&\leq\left(\sum_{k>N}a_k^2\right)^{1/2}\left(\sum_{k>N}|\langle x_n-x_0,e_k\rangle|^2\right)^{1/2}\\ &\leq \left(\sum_{k>N}a_k^2\right)^{1/2}\|x_n-x_0\| \end{aligned}$$ Since $\sum_ka_k^2<\infty$ we can choose $N$ so that the last expression is as small as required, because we know that the $x_n$'s are uniformly bounded in norm. Fixing such $N$, the finite part of the sum, namely, $\sum_{k=1}^Na_k\langle x_n-x_0,e_k\rangle$, is easily seen to tend to zero as $n\to\infty$, because $\langle x_n-x_0,e_k\rangle$ tends to zero for each $k$, and there are only a finite number of $k$'s in the finite sum term.

Answer 2

$\langle (x_n-x_0), y \rangle \to 0$ if $ y \in \operatorname{span}(e_k)_k$. Let $y \in H$ be arbittary and $\epsilon >0$. Then there exists $y_0 \in \operatorname{span}(e_k)_k$ such that $\|y-y_0\|<\epsilon$. Now $|\langle (x_n-x_0), y \rangle| \leq \langle (x_n-x_0), y_0 \rangle|+(M+\|x_0\|)\epsilon$ and the first term tends to $0$. Hence $x_n \to x_0$ weakly.