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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?

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Any thoughts or ideas? At this moment, I even consider a counterexample possible ;) –  Dario Jan 27 at 19:53
    
Mmh I can't provide a solution but googling the title I found lots of articles about operators which are translation invariant. It's a shame I can't understand very well some of the techniques and build a tailored proof for this exercise. And I'm very interested in an answer to this question :) –  Riccardo Mar 1 at 20:36

2 Answers 2

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).

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One can probably construct an $I-P$ directly by taking a basis for $\ell^p/X$ (the "tails" of sequences) and explicitly choosing coset representatives using the axiom of choice. However, this algebraic approach is fairly clean. –  Aaron Mar 25 at 23:11
    
No wonder I cannot rigorously prove it...! –  Bombyx mori Apr 15 at 21:26

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)}$

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