Let $E$ be a Banach space, $\mathcal B(E)$ the Banach space of linear bounded operators and $\mathcal I$ the set of all invertible linear bounded operators from $E$ to $E$. We know that $\mathcal I$ is an open set, and if $E$ is finite dimensional then $\mathcal I$ is dense in $\mathcal B(E)$. It's not true that $\mathcal I$ is dense if we can find $T\in\mathcal B(E)$ injective, non surjective with $T(E)$ closed in $E$, since such an operator cannot be approximated in the norm on $\mathcal B(E)$ by elements of $\mathcal I$ (in particular $E$ has to be infinite dimensional).

So the question is (maybe a little vague): is there a nice characterization of $\overline{\mathcal I}^{\mathcal B(E)}$ when $E$ is infinite dimensional? Is the case of Hilbert space simpler?

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    $\begingroup$ The Hilbert case seems subtle enough. See this paper for a characterization. In the case of a separable Hilbert space an operator $T$ belongs to the closure of $\mathcal{I}$ if and only if $\dim{\ker{T}} = \dim{\ker{T^\ast}}$ or the range of $T$ is not closed. $\endgroup$ – t.b. Dec 16 '11 at 18:30
  • $\begingroup$ @t.b. Thanks for the paper. Indeed, the characterization is not simple when the Hilbert space is not separable, so I don't know if we can hope something of similar of Banach spaces. $\endgroup$ – Davide Giraudo Dec 16 '11 at 18:58
  • $\begingroup$ The closure of $\mathcal I$ is $B(E)$ if $E$ is hereditarily indecomposable. $\endgroup$ – Jonas Meyer Dec 16 '11 at 19:55
  • $\begingroup$ @JonasMeyer Is it a standard result? Where can we found a proof? $\endgroup$ – Davide Giraudo Dec 16 '11 at 20:06
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    $\begingroup$ @DavideGiraudo: Operators on such a space are scalar plus strictly singular, and in particular have countable spectrum. Therefore there are aribtrarily small elements of the resolvent set, so each $T\in B(E)$ is the limit of a sequence of invertible operators $T+\lambda_n I$ with $\lambda_n\to 0$. I only know about this from doing a little research to answer another question. $\endgroup$ – Jonas Meyer Dec 16 '11 at 20:09

In 1, we can find a characterization in the case of Hilbert spaces. For $T$ a bounded operator, let $T=U|T|$ be the polar decomposition of $T$, and $E(\cdot)$ be the spectral measure of $|T|$. Define $$\operatorname{ess\, nul}(T):=\inf\{\dim E[0,\varepsilon]H,\varepsilon>0\}.$$ Then $T$ is in the closure of invertible operators for the norm if and only if $\operatorname{ess\, nul}(T)=\operatorname{ess\, nul}(T^*)$.

1 Bouldin, Richard Closure of invertible operators on a Hilbert space. Proc. Amer. Math. Soc. 108 (1990), no. 3, 721–726.

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There cannot be a nice characterisation, I believe, as for some Banach spaces the closure is everything. See my answer here.

A partial answer to your question would be that the set of Fredholm operators with index 0 is always in the closure of invertible operators.

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