# Concrete example of non-nuclear operator $E \to F$ and isometry $F \hookrightarrow G$ so that the composition $E \to F \hookrightarrow G$ is nuclear

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.

References.

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

• 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. Jul 10, 2020 at 16:15
• 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 Jul 10, 2020 at 16:17
• @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. Jul 14, 2020 at 10:05

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.
• 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 $L_1$. Jul 10, 2020 at 17:07
• 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 $\lVert Sr_n \rVert_{L_1} = 1 \not\to 0$.) Jul 14, 2020 at 10:20