Right shift powers converge in wot not sot

Put $b_k=F(e_k)$ and test $F$ on $v={\sum }_{k<N}{\bar{b}}_k{e}_k$. By F3, $B_N=F(v)={\sum }_{k<N}\mathrm{\mid }{b}_k{\mathrm{\mid }}^2\le \mathrm{\parallel }F\mathrm{\parallel }\sqrt{B_N}$, giving $B_N\le \mathrm{\parallel }F{\mathrm{\parallel }}^2$, also when $B_N=0$. Thus $b\in {\mathrm{\ell }}^2$. Truncations converge in the square-sum norm, so $F(x)={\lim }_N{\sum }_{k<N}{x}_k{b}_k={\sum }_k{x}_k{b}_k$, with absolute convergence: F2 bounds every finite sum ${\sum }_{k<N}\mathrm{\mid }{x}_k{b}_k\mathrm{\mid }$ by $\mathrm{\parallel }x{\mathrm{\parallel }}_2\mathrm{\parallel }b{\mathrm{\parallel }}_2$, and the nonnegative partial sums converge to their finite supremum. The same finite inequality applied to tails gives the infinite tail bound used below.

Consequently $F(R^{n}x)={\sum }_{k\ge 0}{x}_k{b}_{k+n}$ and $\mathrm{\mid }F(R^{n}x)\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_2({\sum }_{j\ge n}\mathrm{\mid }{b}_j{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}\to 0$. This proves WOT convergence. But $\mathrm{\parallel }{R}^{n}x{\mathrm{\parallel }}_2^2={\sum }_k\mathrm{\mid }{x}_k{\mathrm{\mid }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, so at $x=e_0$ the norms remain one. Thus SOT convergence to zero fails. $\mathrm{\square }$