Let $(a_k)_{k\geq 1}$ a bounded sequence, $(e_n)_{n\geq 1}$ an orthonormal basis of an Hilbert space $H$. Define $x_n:=\frac{1}{n}\sum_{k=1}^na_ke_k$
Then $x_n\rightarrow0$
I proved that $|<x_n,y>|\rightarrow 0$ hence $x_n$ converges weak to zero. Now my idea was to prove the convergence of x_n in the norm to could conclude that the convergence also holds strongly to 0. But here I became trouble because I can only bound this.. Does it is the good way to do it?
For $x_n=\frac{1}{n}\sum_{k=1}^na_ke_k$ we have,
$$||x_n||^2=\langle\frac{1}{n}\sum_{k=1}^na_ke_k,\frac{1}{n}\sum_{k=1}^na_ke_k\rangle$$
$$=\frac{1}{n^2}\langle\sum_{k=1}^na_ke_k,\sum_{k=1}^na_ke_k\rangle$$
$$=\frac{1}{n^2}\sum_{k=1}^n\sum_{i=1}^n\langle a_ke_k,a_ie_i\rangle$$
$$=\frac{1}{n^2}\sum_{k=1}^n\sum_{i=1}^na_k\overline{a_i}\langle e_k,e_i\rangle$$ $$=\frac{1}{n^2}\sum_{k=1}^na_k\overline{a_k}=\frac{1}{n^2}\sum_{k=1}^n|a_k|^2$$
$$\leq\frac{1}{n^2}\sum_{k=1}^nM^2=\frac{1}{n^2}nM^2\rightarrow0$$
where $M$ is the bound of $(a_k)_{k\in\mathbb N}\subset\mathbb K$.