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?

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

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.

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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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