DISCLAIMER: I posted the same question a week ago on Mathematics Stack Exchange.

We know by an abstract argument that there exist Banach spaces $E$, $F$, $G$ and maps $E \to F \hookrightarrow G$ such that $E \to F$ is non-nuclear, $F \hookrightarrow G$ is an isometry (metric injection), and the composition $E \to F \hookrightarrow G$ is nuclear. (In other words, the operator ideal of nuclear operators is not injective.) The typical line of reasoning can be found in the corresponding post on MSE, or in [DF93, §9.7]. However, these proofs are non-constructive, which leads me to the following question:

Question. Can we write down explicit examples of Banach spaces $E$, $F$, $G$ and maps $E \to F \hookrightarrow G$ such that $E \to F$ is non-nuclear, $F \hookrightarrow G$ is an isometry, and the composition $E \to F \hookrightarrow G$ is nuclear?

In the (non-constructive) example given in my MSE post, all spaces have the approximation property, but for the moment I do not care about this requirement.


[DF93]: A. Defant, K. Floret, Tensor Norms and Operator Ideals (1993), Mathematics Studies 176, North-Holland.

  • 1
    $\begingroup$ Using the notation/construction of what you wrote on MSE, I think this should be relatively routine once one has explicit witnesses to the fact that the "inclusion" $\ell^2 \tilde{\otimes} \ell^2 \to \ell^2 \tilde{\otimes} L_1$ does not have closed range; you take a sequence of finite-rank tensors in the LHS whose norms become much smaller on the RHS, and then rescale and stack them together. $\endgroup$
    – Yemon Choi
    Jul 10, 2020 at 16:15
  • $\begingroup$ By "relatively routine" I don't mean that this is a trivial amount of work, I just mean that once one has the finite building blocks, the steps one takes to produce the desired counterexample don't require any new fancy tricks $\endgroup$
    – Yemon Choi
    Jul 10, 2020 at 16:17
  • $\begingroup$ @YemonChoi ah yes, that is how these things work. But for this I would have to get a grip on the projective norm. I'll think about it some more at a later time; maybe this approach is not as hard as it seems. $\endgroup$ Jul 14, 2020 at 10:05

1 Answer 1


Take any sequence $a_n$ of scalars that is square summable but not summable. That is the "hard" (in the technical sense) part of the argument. The rest is "soft". Let $T$ be the diagonal operator on $\ell_2$ with diagonal $a_n$. So $T$ is $2$-summing (Hilbert-Schmidt) but not nuclear (trace class). Let $S: \ell_2 \to L_1$ be the isomorphism you mention in your MSE post that maps the $n$-th unit vector basis of $\ell_2$ to the $n$-th Rademacher function. $S^*$ is an operator from $L_\infty$ to $\ell_2$, so is $2$-summing. $T^*$ is $2$-summing obviously, so $T^*S^*$ is nuclear, so $ST$ is nuclear because every space in sight has even the metric approximation property.

  • $\begingroup$ If you want $S$ to be an isometry from $\ell_2$ into $L_1$, map the unit vector basis of $\ell_2$ injectively onto independent Gaussian random variables that are normalized in $L_1$. $\endgroup$ Jul 10, 2020 at 17:07
  • $\begingroup$ Very nice! You make it look so easy. 😉 Seeing as you're not a member of MSE, I'll write up a slightly expanded answer there, aiming at the MSE audience. (Although I guess everyone who understands my question can also understand your answer.) $\endgroup$ Jul 14, 2020 at 10:09
  • $\begingroup$ As a side note, I noticed that $S : \ell_2 \to L_1$ itself is not absolutely $p$-summing for any $p$, since it is not completely continuous. (We have $r_n \rightharpoonup 0$, but $\lVert Sr_n \rVert_{L_1} = 1 \not\to 0$.) $\endgroup$ Jul 14, 2020 at 10:20

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