# Counterexample of non invertible operator

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?

• "Just" remove the wannabe inverse from a Banach space? – 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.

• 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. – Kavi Rama Murthy Dec 27 '18 at 0:00
• -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. – Martin Argerami Dec 27 '18 at 2:51
• @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 $c_{00}$ – sinbadh Dec 27 '18 at 4:59
• @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. – mechanodroid Dec 27 '18 at 10:41