Sequence of Natural Powers of Right Shift Operator in 2-Sequence Space does not Converge in Strong Operator Topology

Theorem

Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $2$-sequence normed vector space.

Let $\map {CL} {\ell^2} := \map {CL} {\ell^2, \ell^2}$ be the continuous linear transformation space.

Let $R \in \map {CL} {\ell^2}$ be the right shift operator over $\ell^2$.

Let $\sequence {R^n}_{n \mathop \in \N}$ be a sequence.

Let $\mathbf 0 \in \map {CL} {\ell^2}$ be the zero mapping.

Then $\sequence {R^n}_{n \mathop \in \N}$ does not converge to $\mathbf 0$ in the strong operator topology.

Let $\mathbf e_1 = \tuple {1, 0, \ldots} \in \ell^2$

Then $R^n \mathbf e_1 = \tuple {\underbrace {\ldots, 0}_{n \text{ terms} }, 1, 0, \ldots}$

So:

$\forall n \in \N : \norm {R^n \mathbf e_1}_2 = 1$

Therefore:

$\ds \lim_{n \mathop \to \infty} \norm {R^n \mathbf e_1}_2 = 1$

Hence, $\sequence {R^n}_{n \mathop \in \N}$ does not converge to $\mathbf 0$ in the strong operator topology.

$\blacksquare$

2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. Strong and weak operator topologies on $\map {CL} {X, Y}$