Show that the sequence $x_N$ converges weakly and compute the weak limit

Solution 1:

Observe that $\left\langle x_N, e_j \right\rangle = \frac{1}{\sqrt{N}} \to 0$, as $N \to \infty$. Thus, by linearity, for any $y = a_1 e_1 + \ldots + a_r e_r$, we have $\left\langle x_N, e_j \right\rangle \to 0$. Let $y \in H$, with $y = \sum_{n=1}^\infty a_n e_n$ and $\epsilon > 0$. Get $R$ so large that $\| y - \sum_{k=1}^R a_n e_n\|_H < \epsilon$. Then,

$$|\left\langle x_N, y \right\rangle| \leq \left| \left\langle x_N, \sum_{k=1}^R a_n e_n \right\rangle \right| + \left| \left\langle x_N, \sum_{k=R+1}^\infty a_n e_n \right\rangle \right| \leq \underbrace{\left| \left\langle x_N, \sum_{k=1}^R a_n e_n \right\rangle \right|}_{\to 0 } + \epsilon$$

Where the last inequality is Cauchy-Schwartz with $\|x_N\| = 1$. We may send $N \to \infty$ and conclude to $x_N$ converges to zero weakly.

Solution 2:

$\{x_n\}$ converges weakly to 0.

It suffices to show that $$ \langle x_n,y\rangle\to 0, $$ for all $y\in H$.

Let $y\in H$, then $y=\sum_{n=1}^\infty a_ne_n$, where $a_n=\langle y,e_n\rangle$, and $\|y\|^2=\sum_{n=1}^\infty a_n^2$. Hence $$ \langle x_n,y\rangle=\left\langle \frac{1}{n}\sum_{k=1}^ne_k,\sum_{n=1}^\infty a_ne_n\right\rangle=\frac{1}{n}\sum_{k=1}^n a_k. $$ Clearly, since the sequence $\{a^n\}$ is summable, then $a^2\to 0$, and hence $a_n\to 0$ as well. So does the sequence $$ b_n=\frac{1}{n}\sum_{k=1}^n a_k. $$ See for example Stolz–Cesàro theorem