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So my question is straightforward. Let $\mathfrak{X}$ be a (complex, if necessary) Banach space and $K\colon\mathfrak{X}\to\mathfrak{X}$ a nonzero compact operator. Denote by $\mathcal{C}(K)$ the commutant of $K$—i.e. the algebra of operators that commute with $K$, and $\mathcal{C}^2(K):=\mathcal{C}(\mathcal{C}(K))$.

For every $T\in \mathcal{C}^2(K)$ that is not a multiple of the identity, is it the case that $\mathcal{C}^2(T)$ also contains a nonzero compact operator?

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  • $\begingroup$ $C^2(T)$ contains $T$. $\endgroup$
    – markvs
    Jul 8, 2021 at 3:41
  • $\begingroup$ Yes, but $T$ may not be compact (outside of finite dimensional spaces). $\endgroup$
    – Jack L.
    Jul 8, 2021 at 3:46
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    $\begingroup$ It's easy to come up with an example of a compact $K$ on a Hilbert space whose double commutant contains an orthogonal projection of infinite dimensional range and kernel. $\endgroup$ Jul 8, 2021 at 5:39
  • $\begingroup$ @NarutakaOZAWA. That’s absolutely true. (If you would write it out/expand it into an answer, I would accept). $\endgroup$
    – Jack L.
    Jul 8, 2021 at 6:02

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