I am trying to understand the following proof in Trace Ideals and Their Applications by Barry Simon (Proposition 2.1):
Let $\mathcal{J}$ be a two-sided ideal in $\mathcal{L}(\mathcal{H})$ containing an operator $A$ which is not compact. Then $\mathcal{J} = \mathcal{L}(\mathcal{H})$.
I have two main questions about the proof. First of all, the author uses the spectral decompositon of $|A|$ and that $$ |A| = \lim_{a\to 0} |A|P_{[a,\infty]} $$ He then claims that if all the $P_{[a,\infty]}$ were finite dimensional, $|A|$ would be compact. Why is this?
The other question: the author uses that fact that for an infinite dimensional spectral projection $P_{[a,\infty]}$, there exists an isometry $V$ from $\mathscr{H}$ onto $\text{Ran} \, P_{[a,\infty]}$ and does not give a proof. I have no idea where I can find this.
I would love to be directed to other sources that can flesh out the details missing in the above proof.
I'll fill in some details.
Let's prove that $|A|$ is compact given the finite-dimensional assumption on its spectral projections. If the $P_{[a, \infty)}$ were all finite dimensional, then they'd be compact, so the operators $|A|P_{[a, \infty)}$ are also all compact, since $K(H)$, the collection of compact operators, is an ideal. Now we see that $|A|$ can written as a limit of some compact operators. But $K(H)$ is closed in $B(H)$, so $|A|$ is therefore compact.
Something worth mentioning: the statement you're trying to prove is only true if $H$ is separable (i.e. has countable dimension). So we can construct your isometry as follows. Choose $\{f_i\}$ a countable basis for the range of $P$ (assumed infinite dimensional) and $\{e_i\}$ a countable basis for $H$. Then $e_i \mapsto f_i$ is your desired isometry.