# Proving an operator from $\ell^\infty \to \ell^\infty$ is invertible

(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.

• 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 $AS$). Jul 26 '17 at 13:20
• 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. Jul 26 '17 at 13:38
• May I ask what is the origin of this problem. It might be worth including this information into the question. Jul 26 '17 at 14:05
• 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. Jul 26 '17 at 14:09
• Unfortunately it is in italian, here is the url if you want dm.unipi.it/~abbondandolo/teaching/archivio/istituzioni-2010-11/… 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 $n$.