6
$\begingroup$

We know the following classical statement: "Let $X$ a Banach space and $T:X\to X$ a bounded operator such that $\|T\|<1$. Then $I-T$ is invertible".

When we review the proof, it is easy to note how completeness of $X$ is required. But, do you know some example of a non Banach space $X$ such that there is a bounded operador $T$ with $\|T\|<1$ and $I-T$ is not invertible?

$\endgroup$
  • $\begingroup$ "Just" remove the wannabe inverse from a Banach space? $\endgroup$ – Hagen von Eitzen Dec 26 '18 at 19:42
9
$\begingroup$

Consider the non-Banach space $c_{00}$ of all finitely-supported sequences equipped with the supremum norm. Let $S : c_{00} \to c_{00}$ be the unilateral shift. Consider $T = \frac12 S$. Then $\|T\| = \frac12 < 1$ but the inverse $(I-T)^{-1}$ is supposed to be given by $$(I-T)^{-1} = \sum_{n=0}^\infty T^n = \sum_{n=0}^\infty \frac1{2^n}S^n$$

However, this sum doesn't converge in the operator norm. Indeed, it doesn't even converge pointwise: for the canonical vector $e_1 \in c_{00}$ we have $S^ne_1 = e_{n+1}$ so it would be $$(I-T)^{-1}e_1 = \sum_{n=0}^\infty \frac1{2^n}S^ne_1 = \sum_{n=0}^\infty \frac1{2^n}e_{n+1} = \left(1, \frac12, \frac14, \ldots\right)\notin c_{00}$$

Therefore $(I-T)^{-1}$ doesn't exist so $I-T$ isn't invertible.


For a more formal proof that $I-T$ is not invertible, we'll show that $e_1$ is not in the image of $I-T$.

Indeed, assume that $\exists x = (x_n)_n \in c_{00}$ such that $(I-T)x = e_1$. Then

$$(1,0,0, \ldots) = e_1 = (I-T)x = \left(x_1, x_2 - \frac12 x_1, x_3-\frac12x_2, \ldots\right)$$

which implies $x_1 = 1$ and $x_{n+1} = \frac12 x_n, \forall n \in \mathbb{N}$. Induction gives $x_n = \frac1{2^{n-1}},\forall n \in \mathbb{N}$ or $$x = \left(1, \frac12, \frac14, \ldots\right)\notin c_{00}$$ which is a contradiction.

$\endgroup$
  • 1
    $\begingroup$ This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $\sum T^{n}$. May be true, but certainly not obvious to me. $\endgroup$ – Kavi Rama Murthy Dec 27 '18 at 0:00
  • $\begingroup$ -1: the example works, but the argument is sketchy. I have never seen a proof of "if $\sum_nT^n$ doesn't converge, then $I-T$ is not invertible". If such a proof exists, I would like to see it. $\endgroup$ – Martin Argerami Dec 27 '18 at 2:51
  • 1
    $\begingroup$ @KaviRamaMurthy example is valid since this $T$ has a natural extension to all $\ell^\infty$, which is complete. Therefore, by the original statement about Banach spaces, this $T$ is invertible and its inverse is $\sum T^n$. So $T$ is indeed invertible in all $\ell^\infty$, but it does not in $c_{00}$ $\endgroup$ – sinbadh Dec 27 '18 at 4:59
  • 1
    $\begingroup$ @MartinArgerami You are right, I have added an explicit proof that $I-T$ is not surjective. I wanted to demonstrate where the original argument fails if the space is not complete. $\endgroup$ – mechanodroid Dec 27 '18 at 10:41

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.