(Just to fix some notation, $\ell^\infty = \ell^\infty(\mathbb{N\cup \{0\}},\mathbb{C})$ in what follows)

Let $S:\ell^\infty\to\ell^\infty$ be the left shift operator and let $A:\ell^\infty\to\ell^\infty$ be the the multiplication by the element $a\in \ell^\infty$. In other words we have that $S(x)_n = x_{n+1}$ and $A(x)_n=a_n x_n$.

Suppose that $a\in \ell^\infty$ satisfies $$(1)\quad \quad |\Pi_{k=0}^{n-1}a_{j+k}|\leq c\alpha^n \ \ \forall j\geq 0$$ where $c>0, \alpha \in (0,1)$. My problem is the following:

Prove that $I-AS$ is invertible ($I$ is the identity operator).

Unfortunately $\|AS\|_{\mathcal{\ell^\infty}}= \|a\|_{\ell^\infty}$ that can be greater than $1$ so we cannot exploit the Neumann series to define the inverse.

I managed to prove, though, that $I-AS$ is injective, indeed if $y = AS y$ then we would have (by $(1)$) that $$y_n = a_n y_{n+1} = a_n a_{n+1}\dots a_{n+N} y_{n+N+1}\leq ||y|||_{\infty}\Pi_{k=0}^{N}a_{n+k}|\leq ||y|||_{\infty} c\alpha^{N+1} $$ for any $N>0$ thus taking the limit $y_n = 0 $ for any $n$.

Now this would be sufficient to conclude if for example we manage to prove that $AS$ is a compact operator. But this is not the case since the image of the unit ball through $S$ is the ball itself and $A$ is not compact (for example consider a sequence $a$ such that $a_{2n} = 0$ $a_{2n+1}=1$ so that condition $(1)$ is satisfied).

The only way I see to solve this problem is therefore to prove that $I-AS$ is surjective or maybe that is Fredholm of index $0$. And here I got stuck since I tried to write the inverse obtaining $$ (I-AS)^{-1}(y)_n = y_n + \sum_{k\geq 0} a_{n+k} y_{n+k+1}$$ but I don't see a way to exploit relation $(1)$ to prove the series is convergent.

  • $\begingroup$ If we could show $\|AS\|<1$, then we could use Neumann series. But since $c$ is arbitrary, I guess we cannot get some estimate on the norm of $A$ (or $AS$). $\endgroup$ Jul 26 '17 at 13:20
  • $\begingroup$ In the last paragraph you are asking how to show that $\sum_{k\geq 0} a_{n+k} y_{n+k+1}$ is convergent. I guess this follows simply from the fact that $A$ is function from $\ell_\infty$ to $\ell_\infty$. But you still need some kind of uniform bound (the same for each $n$), to get that the result belongs to $\ell_\infty$ again. $\endgroup$ Jul 26 '17 at 13:38
  • $\begingroup$ May I ask what is the origin of this problem. It might be worth including this information into the question. $\endgroup$ Jul 26 '17 at 14:05
  • 1
    $\begingroup$ It is a point of an exercise that was part of a test for an exam about Functional Analysis at my University some years ago. The other point of the exercise was to prove that $I-AS$ is invertible but assuming that $||a||_{\ell^\infty}<1$ and this can be solved with Neumann series. Unfortunately it doesn't help much. There was also a third point but it doesn't help either. $\endgroup$ Jul 26 '17 at 14:09
  • 1
    $\begingroup$ Unfortunately it is in italian, here is the url if you want dm.unipi.it/~abbondandolo/teaching/archivio/istituzioni-2010-11/… $\endgroup$ Jul 26 '17 at 14:13

It seems that there is a problem in your computation of the inverse. Indeed, let $\left(y_n\right)_{n\geqslant 0}$ be an element of $\ell^\infty$. Then we would like to prove that there exists some $x\in\ell^\infty $ such that $x-ASx=y$, which is equivalent to $$x_n=a_nx_{n+1}+y_n \mbox{ for each }n\geqslant 0. $$
One can prove by induction that for any $N\geqslant 1$, $$\tag{* } x_n=y_n+\sum_{i=1}^Ny_{n+i} \prod_{l=0}^{i-1}a_{n+ l} +x_{n+N+1}\prod_{l=0}^Na_{n+l}. $$ Using the assumption (1) in the opening post and the fact that $\left\lvert y_{n+i}\right\rvert \leqslant \left\lVert y\right\rVert_\infty$ for each $n$ and $i$, one get the convergence of $\sum_{i=1}^{+\infty} y_{n+i} \prod_{l=0}^{i-1}a_{n+ l}$. Since $\left\lvert x_{n+i}\right\rvert$ is bounded independently of $n$ and $i$, an application of (1) gives that $x_{n+N+1}\prod_{l=0}^Na_{n+l}$ goes to $0$ as $N$ goes to infinity. Therefore, letting $N$ going to infinity in (1), one obtains that $$x_n=y_n+\sum_{i=1}^{+\infty} y_{n+i} \prod_{l=0}^{i-1}a_{n+ l}.$$ Using again (1), one gets that $x_n$ is bounded independently of $n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.