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