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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.

Tomasz Kania
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John Steinbeck
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1 Answers1

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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?

Tomasz Kania
  • 16,996