Question. Are there concrete examples of Banach spaces $E$ and $G$, a closed subspace $F \subseteq G$, and a non-nuclear operator $E \to F$ for which the composition $E \to F \hookrightarrow G$ is nuclear?
Background: it is well-known that such operators exist, because the ideal of nuclear operators is not injective (see [DF93; §9.7]). The typical (non-constructive) example runs as follows:
Example. (Compare [DF93; §9.8].) Let $\{r_n\}_{n=1}^\infty \subseteq L^2[0,1]$ be the Rademacher functions $r_n(x) = (-1)^{\lfloor 2^n x\rfloor}$, and let $R \subseteq L^2[0,1]$ be their closed linear span. Since $\{r_n\}_{n=1}^\infty$ is an orthonormal system in the Hilbert space $L^2[0,1]$, we may identify $R$ with $\ell^2$.
It follows from the Khintchine inequality that all $L^p$ norms ($1 \leq p < \infty$) are equivalent on $R$, so in particular $R$ is a complete (and therefore closed) subspace of $L^1[0,1]$. However, by the Dunford–Pettis theorem, $L^1[0,1]$ has no reflexive, infinite-dimensional, complemented subspaces, so in particular $\ell^2 \cong R \subseteq L^1[0,1]$ is not complemented.
Let $T : \ell^2 \cong R \hookrightarrow L^1[0,1]$ denote the inclusion, and consider the following commutative diagram:
(Everything is well-behaved because all spaces have the approximation property.) Evidently $\mathfrak{K}(\ell^2,\ell^2)$ is closed in $\mathfrak{K}(\ell^2,L^1[0,1])$, so it follows that $\mathfrak{N}(\ell^2,L^1[0,1]) \cap \mathfrak{K}(\ell^2,\ell^2)$ is closed in $\mathfrak{N}(\ell^2,L^1[0,1])$. We claim that this subspace is larger than the image of $\mathfrak{N}(\ell^2,\ell^2)$. To that end, note that \begin{align*} ((\ell^2)' \mathbin{\tilde\otimes_\pi} \ell^2)' &\cong \mathscr{B}\textit{i}\ell((\ell^2)' \times \ell^2) \cong \mathfrak{L}(\ell^2 , (\ell^2)'') \cong \mathfrak{L}(\ell^2 , \ell^2);\\[1ex] ((\ell^2)' \mathbin{\tilde\otimes_\pi} L^1[0,1])' &\cong \mathscr{B}\textit{i}\ell((\ell^2)' \times L^1[0,1]) \cong \mathfrak{L}(L^1[0,1] , (\ell^2)'') \cong \mathfrak{L}(L^1[0,1] , \ell^2). \end{align*} Under these isomorphisms, the adjoint of $\text{id} \mathbin{\tilde\otimes_\pi} T$ becomes $$ (\text{id} \mathbin{\tilde\otimes_\pi} T)' : \mathfrak{L}(L^1[0,1], \ell^2) \to \mathfrak{L}(\ell^2 , \ell^2), \ S \mapsto ST, $$ which has dense range (because $\text{id} \mathbin{\tilde\otimes_\pi} T$ is injective) but is not surjective (we have $\text{id} \notin \text{ran}((\text{id} \mathbin{\tilde\otimes_\pi} T)')$ because $\ell^2$ is not complemented in $L^1[0,1]$). Hence, by duality, $\text{id} \mathbin{\tilde\otimes_\pi} T$ does not have closed range, so in particular $$ \mathfrak{N}(\ell^2,\ell^2) \subsetneq \mathfrak{N}(\ell^2,L^1[0,1]) \cap \mathfrak{K}(\ell^2,\ell^2). $$ In other words, there is a non-nuclear compact operator $\ell^2 \to \ell^2$ for which $\ell^2 \to \ell^2 \hookrightarrow L^1[0,1]$ is nuclear.
This proof is non-constructive; I am asking for explicit examples.
References.
[DF93]: A. Defant, K. Floret, Tensor Norms and Operator Ideals (1993), Mathematics Studies 176, North-Holland.
The following answer is due to Bill Johnson on MathOverflow (link).
Hence a concrete example is given by any operator $S : \ell^2 \to \ell^2$ which is Hilbert–Schmidt but not trace-class (for instance, a diagonal operator whose diagonal is square summable but not absolutely summable); then $S$ is not nuclear but the composition $\ell^2 \stackrel{S}{\longrightarrow} \ell^2 \hookrightarrow L^1[0,1]$ is nuclear.
References.
[DF93]: A. Defant, K. Floret, Tensor Norms and Operator Ideals (1993), Mathematics Studies 176, North-Holland.