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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.

tom b.
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    What is the norm of $T^n$? (What is $T^n(e_{n+1})$?) – David Mitra Jul 30 '14 at 14:26
  • In what topology? Strongly (and hence weakly) operator, we have $T^n \to 0$. In norm, as @DavidMitra said, $T^n$ does not converge. – martini Jul 30 '14 at 14:33
  • @martini I was thinking in the norm topology. But what I'm really trying to do is to come up with an example with $T^n \to 0$ so I'll use the strong operator topology. – tom b. Jul 30 '14 at 14:43
  • @martini I'm not familiar with SOT but I'm reading about it now. – tom b. Jul 30 '14 at 14:46

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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.

  • In the weak operator topology, we have $T^n \to 0$, as this topology is weaker than the strong operator topology, so strong convergence implies weak convergence.
martini
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  • I'm sorry but here it is stated that unilateral shift does not converge to $0$ in SOT. Is Wikipedia wrong? – tom b. Jul 30 '14 at 14:48
  • @tomb. They seem to talk about the right shift $S\colon (x_1, x_2, \ldots) \to (0, x_1, x_2, \ldots)$ there. For this operator $S^n \to 0$ (weakly operator), but $S^n x \not\to 0$ for any $x \ne 0$, so $S^n \not\to 0$ (strongly operator). This example also shows (as $S = T^$) that $T \mapsto T^$ is not strongly operator continuous. – martini Jul 30 '14 at 14:51