*DISCLAIMER: I [posted](https://math.stackexchange.com/q/3743628) 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](https://math.stackexchange.com/q/3743628), 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](https://math.stackexchange.com/q/3743628), 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.



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
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.


