Strong operator convergence of unilateral shift. Giving an infinite dimensional separable Hilbert space $X$, with an o.n. basis $(e_n)$, we have the standard left and right shift operators $L$ and $R$. I am trying to study the convergence (in the strong operator topology) of $R^nL^n$ and $L^nR^n$. For the former I suspect that it converges to 0 ($L^n$ deletes the first $n$ entries and the $R^n$ puts 0's in their place). For the latter $L^nR^n$ leaves nothing changed so i assume the sequence converges to the identity map. However I do not know how to formally prove this? For starters $R^n$ does not converge in the strong topology (while $L^n does) so I cannot use convergence of compositions.

Since $LR=I$, you immediately get that $L^nR^n=I$ for all $n$, no limits required. Formally, this is shown by induction: if $L^nR^n=I$, then $L^{n+1}R^{n+1}=L^nLRR^n=L^nR^n=I$.
The other way around, the composition $R^nL^n$ equals the orthogonal projection onto $\overline{\operatorname{span}}\{e_n,e_{n+1},\ldots\}$. So indeed it goes to zero sot.
