 Solution

By weakly convergent sequence is bounded, $M={\sup }_n\Vert x_n{\Vert }^2$ is finite. Choose any $n_1$. Once $n_1<\cdots <n_k$ have been chosen, put $s_k={\sum }_{j=1}^k{x}_{n_j}$. Since $s_k$ is fixed, weak convergence to zero gives $⟨s_k,x_n⟩\to 0$. Choose $n_{k+1}>n_k$ with $|⟨s_k,x_{n_{k+1}}⟩|\le 1$. This gives the required one-sided inequality as well.

The useful estimate is for the squared norm. Indeed, $\Vert s_1{\Vert }^2\le M$, and the inner product expansion givesBy mathematical induction, $\Vert s_k{\Vert }^2\le Mk+2(k-1)\le (M+2)k$. HenceThis proves norm convergence of the averages, the weak Banach–Saks theorem in a Hilbert space. The printed inequality omits the square on its left side. Our stronger estimate also proves that printed bound, since $\sqrt{(M+2)k}\le (M+2)k$ for $k\ge 1$, but the printed bound alone only bounds the averages and cannot imply their norm convergence to zero.