Good evening,


I have a question concerning non-invertible operators. 


*Let $H$ be a Hilbert space and $T$ a non-invertible bounded operator on $H.$ Is it true that $T$ is the limit of some sequence of invertible bounded operators on $H$?*


We can see that is true when $H$ is finitely dimensional. So the interesting case is infinitely dimensional. Does anyone know any information about this? 


Thanks in advance,


Duc Anh



is a non-invertible operator a boundary point of the group of invertible operators?
This is true if $0$ belongs to the boundary of the spectrum of $T$ (more or less from the definition of the spectrum) but fails in general. The unilateral left shift $S: \ell^2({\mathbb N})\to\ell^2({\mathbb N})$ satisfies $S^\ast S=I\neq SS^\ast$.


If $(T\_n)$ were a sequence of invertible operators with $T\_n\to S$ in norm, then $S^\*T\_n\to I$ in norm, which means that for all sufficiently large $n$ the operator $S^\*T\_n$ is invertible, which means $S^\*$ is invertible, a contradiction.


