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This is my question:

Is $\mathcal{G}(\ell_2)$ is dense in $\mathcal{B}(\ell_2)$?

I am attempting to show that it is not by showing that the right-shift - call it $T:\ell_2 \rightarrow \ell_2$ - cannot be made invertible via a small perturbation. We know that the spectrum of $T$ is $$\sigma(T) = \overline {\mathcal D} = \{z \in \Bbb C \ | \ |z| \le 1 \}.$$ This includes $0$, showing that $T$ itself is not invertible (pretty clear as not surjective, although is injective). Also, it shows us that if $T$ is perturbed by a small multiple of the identity (in fact, any multiple less than or equal to $1$), then it is still non-invertible.

The issue that I'm having is showing that for any $S \in \mathcal{B}(\ell_2)$ with $\|S\|=1$, given $\epsilon > 0$ small enough, $T + \epsilon S$ is not invertible. If we choose $S$ injective, then $T + \epsilon S$ is injective, so we need $S$ not surjective. We know by the open mapping theorem that any surjective (linear and continuous) operator is an open map - can I use this to show obtain a contradiction if $T + \epsilon S$ is invertible?

If I could be given a hint, then I'd be most appreciative. I would, however, only like a small hint, as I would like to learn from this question, so please don't just blast out the answer!
As a note, the first part of the question asked to show that $\mathcal{F}(\ell_2)$ (the set of finite rank operators on $\ell_2$) is dense in $\mathcal{K}(\ell_2)$ (the set of compact operators on $\ell_2$). I can't see how this could be used, but maybe...?
Thank you! :)

As a final note, hopefully this isn't a duplicate question: it seems a bit odd to me that no-one has asked it before, but I couldn't find it. Apologies if it is.

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  • $\begingroup$ The set of all invertible operators is open. You can prove this by showing that if $\|T\|<1$, then $I-T$ is invertible. $\endgroup$ – Vincent Boelens Dec 1 '14 at 20:23
  • $\begingroup$ Yes, that is one of the things that I've covered. Just wasn't sure how to use this! Need to find a ball (open or closed) about the right shift for which no element in it is invertible... $\endgroup$ – Sam T Dec 1 '14 at 21:12
  • $\begingroup$ Sorry, I was confused before, openness of the invertible operators doesn't help of course. $\endgroup$ – Vincent Boelens Dec 1 '14 at 21:23
  • $\begingroup$ It was my first port of call for thought, but didn't help. No worries - think I've got it now though. :) $\endgroup$ – Sam T Dec 1 '14 at 21:34
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I suppose that $0\in\mathbb{N}$. If you follow the other convention, replace $e_0$ with $e_1$ in the following.

If $T+ \epsilon S$ is surjective, then there is an $x\in \ell_2$ with $\lVert x\rVert = 1$ and

$$(T+\epsilon S)(x) = c\cdot e_0$$

for some $c\neq 0$.

Then $\lVert\epsilon Sx\rVert^2 = \lVert c\cdot e_0 - Tx\rVert^2 = \lvert c\rvert^2 + 1$ and hence $\lVert \epsilon S\rVert \geqslant \sqrt{1+\lvert c\rvert^2} > 1$, therefore $\epsilon > 1$.

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  • $\begingroup$ This was the sort of thing that I was thinking. I then know that $(\epsilon S x)_1 = c$ (meaning first component) because $(Tx)_1 = 0$. Surely this is ok? Or does it mean that if it is the case, then $T + \epsilon S$ is not injective, ie non-trivial kernet? $\endgroup$ – Sam T Dec 1 '14 at 21:14
  • $\begingroup$ (Just typing a comment now explaining that I think I've got it, so if you see this before it's arrived, hold fire :)!) $\endgroup$ – Sam T Dec 1 '14 at 21:23
  • $\begingroup$ So that I can fit it in in one go, I've made it an answer (as hopefully it's also correct!). If you could check it, then I'd be most appreciative! :) $\endgroup$ – Sam T Dec 1 '14 at 21:34
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Ok, so I think I've got it (I use $\Bbb N = \{1,2,...\}$): Let $S' \in \mathcal{B}$ be given by $S'y = Sy - \langle e_1,Sy \rangle e_1$. $T + \epsilon S'$ is an isomorphism for small enough $\epsilon \ (>0)$ from $H_0 = \ell_2$ to $H_1 = \{ x \in \ell_2 | \langle e_1,x \rangle = 0 \}$. Now suppose that $e_1 \in \text{range}(T + \epsilon S)$: there exists $x \in B_H$ (ie $x \in \ell_2$ with $\|x\| = 1$) such that $(T + \epsilon S)x = c e_1$, for $c \neq 0$; in fact, $c = \langle e_1, Sx \rangle$. Then $$(T + \epsilon S)x = Tx + \epsilon S' + \langle e_1, Sx \rangle e_1$$ $$\Rightarrow (T + \epsilon S')x = 0$$ by definition of $c$. This is a contradiction since $x \neq 0$ and $T + \epsilon S$ is an isomorphism, so in particular is injective. Hence $e_1 \not\in \text{range}(T + \epsilon S)$, and so $T + \epsilon S$ is not surjective, and hence not invertible.

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  • $\begingroup$ Correct. I've added the argument I had in mind to my answer, it amounts to the same, but is shorter. $\endgroup$ – Daniel Fischer Dec 2 '14 at 10:15
  • $\begingroup$ Ah yes, I see, because it's orthogonal so bring out Pythagoras. Interesting that your shows that we require $\epsilon > 1$, whereas I just showed that it didn't work for $\epsilon$ "small enough". I guess that $1$ is an intuitive bound as the set of eigenvalues is the unit disc. Thank you. :) $\endgroup$ – Sam T Dec 2 '14 at 11:29

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