Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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. – t.b. Dec 16 '11 at 18:30
@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. – Davide Giraudo Dec 16 '11 at 18:58
The closure of $\mathcal I$ is $B(E)$ if $E$ is hereditarily indecomposable. – Jonas Meyer Dec 16 '11 at 19:55
@JonasMeyer Is it a standard result? Where can we found a proof? – Davide Giraudo Dec 16 '11 at 20:06
@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. – 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.