In $\ell^p$, if an operator commutes with left shift, it is continuous?

Question

Our professor put this one in our exam, taking it out along the way though because it seemed too tricky. Still we wasted nearly an hour on it and can't stop thinking about a solution.

What we have: The left shift $L : \ell^p \to \ell^p$ $$L(x_1,x_2,x_3,\ldots) = (x_2,x_3,\ldots)$$

and another operator $T$. We should prove that if $TL=LT$, then $T$ is continuous.

We had defined subspaces $$ X_k = \{ (x_i) : x_i = 0 \text{ for } i>k \} $$ and seen that these are $T$-invariant and the restrictions $T : X_k \to X_k$ continuous (obvious). The hint was to use closed-graph-theorem to show that $T$ is continuous. Of course we can truncate any sequence to then lie in $X_k$, however I do not see how convergence of the truncated sequences relates to convergence of the images under $T$.

Any help please?

Answer 1

The statement is false, as I discovered here. Since not everybody has access to the paper, let me provide a summary of the argument:

Let $R=\mathbb C[t]$, $L$ the left shift operator, and view $\ell^p$ as an $R$-module by defining $t\cdot x=Lx$. Let $X=\sum \ker L^i \subset \ell^p$ be the subspace of eventually-zero sequences.

Lemma: Given a PID $P$, a $P$-module $M$ is injective if and only if it is divisible, i.e., if for every $p\in P$, $pM=M$.

Lemma: $X$ is an injective $R$-module.

Proof. Since a PID is a UFD, we can check divisibility (and hence injectivity) using irreducible elements. Over $R$, the irreducible elements are linear polynomials. Thus, we need to show that if $x$ is an eventually-null sequence and $\lambda\in \mathbb C$, there exists a $y$ such that $Ly-\lambda y=x$. This is straight forward.

Because $X$ is injective, the inclusion $X\subset \ell^p$ splits, there exists a (non-unique) projection map $P:\ell^p\to X$. Since this is a map of $R$-modules, it commutes with $L$.

Lemma: $P$ is not continuous.

Proof. Suppose that $P$ were continuous. Then $X=\ker (P-\operatorname{Id})$ is closed. However, this is absurd, as every sequences can be approximated by finitely many terms (i.e., $X$ is a dense proper subspace).

Answer 2

my idea is that :

$L$ is continuous $\Longrightarrow$ $LT$ is continuous $\Longrightarrow$ $T$ is continuous

you can use the closed-graph-theorem to the left shift firstly :

$L:l_{p}$$\longrightarrow$$l_{p}$

since the $l_{p}$ is banach space

then its product space $X\times{Y}$ is banach space and the image of $L$ is a colsed subset of $X\times{Y}$ since $L$ is a left shift

therefore, $L$ is continuous

another condition you used is that ：

since, if $LT=TL$

then, $(LT)^{n}=L^{n}T^{n}$

because of, $\Arrowvert$$AB$$\Arrowvert$$\le$$\Arrowvert$$A$$\Arrowvert$$\Arrowvert$$B$$\Arrowvert$

so, $\Arrowvert$$(LT)^{n}$$\Arrowvert$$\le$$\Arrowvert$$L^{n}$$\Arrowvert$$\Arrowvert$$T^{n}$$\Arrowvert$$\Longrightarrow$$\Arrowvert$$(LT)^{n}$$\Arrowvert$$^{1/n}$$\le$$\Arrowvert$$L^{n}$$\Arrowvert$$^{1/n}$$\Arrowvert$$T^{n}$$\Arrowvert$$^{1/n}$

as you describe, $$ X_k = \{ (x_i) : x_i = 0 \text{ for } i>k \} $$

that means your subspace is $c_{00}$ , and in this space there are just finite terms not equal to zero.

therefore, by the property of its subspace $c_{00}$ :

since the restrictions $T : X_k \to X_k$ is invariant and the norms on $c_{00}$ is $\Vert$$\Vert$$_{\infty}$

which imply that,

$\Arrowvert$$(LT)^{n}$$\Arrowvert$$\le$$\Arrowvert$$L^{n}$$\Arrowvert$$\Arrowvert$$T^{n}$$\Arrowvert$$\Longrightarrow$$\Arrowvert$$(LT)^{n}$$\Arrowvert$$^{1/n}$$\le$$\Arrowvert$$L^{n}$$\Arrowvert$$^{1/n}$$\Arrowvert$$T^{n}$$\Arrowvert$$^{1/n}$$\le$$\Arrowvert$$L^{n}$$\Arrowvert$$^{1/n}$$max$$\arrowvert$$x_{i}$$\arrowvert$$^{1/n}$

$=$$\Arrowvert$$L^{n}$$\Arrowvert$$^{1/n}$

holds as $n\longrightarrow$$\infty$

consequently, $LT\in{BL(X)}$$\Longrightarrow$$T\in{BL(X)}$