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Let $E$ be a normed vector space (Banach space, if you like).

Is $GL(E)$, the set of invertible and continuous endomorphism of $E$, dense in $L(E)$, the set of continuous endomorphism of $E$?

I specify that I know the answer if $dim(E)<\infty$, with classical arguments about the spectrum of matrices, and, I know that $GL(E)$ is open in $L(E)$, even if $dim(E)=\infty$ (if $E$ is a Banach space), using the formula $(I-u)^{-1}=\sum_{n\in\mathbb{N}}u^n$ for $u$ small enough.

So the remaining question I would like to ask is about the density of $GL(E)$ in $L(E)$, and in the case it is not, about its closure.

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  • $\begingroup$ Searching a bit results in: math.stackexchange.com/questions/92079/… $\endgroup$ – Mariano Suárez-Álvarez Jan 10 '16 at 22:21
  • $\begingroup$ Thanks, couldn't find it, though I assure you I tried... (sorry, a bit new to this, I guess I don't master the research process yet) $\endgroup$ – John Steinbeck Jan 11 '16 at 23:32
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    $\begingroup$ Use google (and add «site:math.stackexchange.com» to your search to restrict results to ones in this site) $\endgroup$ – Mariano Suárez-Álvarez Jan 11 '16 at 23:38
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For all classical infinite-dimensional Banach spaces invertible operators are not dense but there are instances when they are.

In a Banach algebra, invertible elements are dense if and only if left-invertible elements are dense and if this is so, left-invertible elements are already invertible. However, in the case of classical Banach spaces you always have non-invertible, left invertible elements (for example, isomorphisms onto subspaces of codimension 1).

This is explained in detail in Section 4.2 of my article with Sz. Draga, When is multiplication in a Banach algebra open?

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