Let $T: \ell^2 (\mathbb R) \to \ell^2 (\mathbb R)$ be the left shift operator $(x_1,x_2,x_3, \dots) \mapsto (x_2,x_3,x_4,\dots)$. Let $T^n$ denote a left shift by $n$ positions.
What is $\lim_{n \to \infty} T^n$? Is it $0$?
Edit: I want to have $T^n \to 0$. Let $B(\ell^2(\mathbb R))$ denote the space of bounded linear operators on $\ell^2$ and let us endow it with the strong operator topology or whatever is convenient to achieve the goal.
Best Answer
Note that on $B(\ell^2\def\R{\mathbb R})$ we have different topologies in which we can talk about convergence of $(T^n)$. As $\{T^n \mid n \in \mathbb N\}$ is a norm-bounded set, many of these topologies coincide, for an overview of the topologies see for example this wikipedia article.
I will discuss three topologies here (the IMHO most important ones for norm bounded sets): The norm topology, the strong operator topology, the weak operator topology:
In the norm topology, $(T^n)$ does not converge, as David wrote in his comment, to see this, observe, that - for example $$ (T^{n+1} - T^n)(e_{n+2}) = e_1 - e_2 $$ And hence $\|T^{n+1} - T^n\| \ge \sqrt 2$, so $(T^n)$ is not norm-Cauchy.
In the strong operator topology, we have $T^n \to 0$. To see this, let $x \in \ell^2$. Then $$ \|T^nx\|^2 =\sum_{k=n+1}^\infty|x_k|^2 \to 0 $$ so $T^n x \to 0$ for each $x$, that is $T^n \to 0$ strongly operator.