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I am not sure if this question makes sense, or if it is trivial, but does there exists an infinite dimensional Banach space (necessarily without the approximation property) such that no compact, non-nuclear operator is the norm limit of finite rank operators?

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  • $\begingroup$ Isn't this equivalent to asking: is there a space X on which every approximable operator is nuclear? (and so you want to assume X infinite-dimensional to avoid trivialities) $\endgroup$
    – Yemon Choi
    May 1, 2016 at 15:11
  • $\begingroup$ I believe it is an open question if there exists a Banach space X for which every operator is "scalar plus nuclear". If such a space existed it would have the property that every compact operator is nuclear, which seems to satisfy your requirements $\endgroup$
    – Yemon Choi
    May 1, 2016 at 15:13
  • $\begingroup$ You are right, that would do it, but I think my question is weaker. In "scalar plus nuclear" there would not be any compact, non-nuclear operators. $\endgroup$
    – Markus
    May 1, 2016 at 15:16
  • $\begingroup$ Is the folloing space appropriate? $\endgroup$ May 2, 2016 at 19:27
  • $\begingroup$ There is a banach space such that the only bounded operators are the compact perturbation of scalar operators. $\endgroup$ May 2, 2016 at 19:30

1 Answer 1

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Pisier constructed a Banach space such that the operator norm is equivalent to the nuclear norm on the finite rank operators. Consequently, no compact non nuclear operator on his space is the norm limit of finite rank operators. However, it is open whether there exists a compact non nuclear operator on his space!

Pisier, Gilles Counterexamples to a conjecture of Grothendieck. Acta Math. 151 (1983), no. 3-4, 181–208.

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  • $\begingroup$ Hah, I had a vague recollection that his 1983 example might answer the question, but I couldn't remember the details off the top of my head. $\endgroup$
    – Yemon Choi
    May 1, 2016 at 15:17
  • $\begingroup$ Is the lack of compact, non-nuclear operators equivalent with the "scalar plus nuclear"? It seems to be a weaker requirement $\endgroup$
    – Markus
    May 1, 2016 at 16:29
  • $\begingroup$ I don't know if Pisier's construction always gives a space on which there operators that are not scalar plus nuclear, but they can contain $\ell_1\oplus \ell_2$ isomorphically, and if follows from known results that on any space that contains $\ell_1\oplus \ell_2$ isomorphically there is an operator that has closed, infinite dimensional and infinite codimensional.range. $\endgroup$ May 2, 2016 at 23:14

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