W^{1,p}(Omega) is reflexive for 1<p<infinity

The gradient embedding. Write ${\mathcal{A}}_1=\{0,e_1,\dots ,e_n\}$ and define $\mathrm{\Phi }(u):=(D^{\alpha }u{)}_{\alpha \in {\mathcal{A}}_1}$ for $u\in W^{1,p}(\mathrm{\Omega })$, regarded as an element of the real vector space $Y:={\prod }_{\alpha \in {\mathcal{A}}_1}{L}^p(\mathrm{\Omega })$ equipped with the ${\mathrm{\ell }}^p$-sum norm $\mathrm{\parallel }(w_{\alpha }){\mathrm{\parallel }}_Y:=({\sum }_{\alpha }\mathrm{\parallel }{w}_{\alpha }{\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}^p{)}^{1\mathrm{/}p}$. By [F1] the map $\mathrm{\Phi }$ is linear and $\mathrm{\parallel }\mathrm{\Phi }(u){\mathrm{\parallel }}_Y=\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}$ for every $u$; hence $\mathrm{\Phi }$ is a linear isometry onto its image $S:=\mathrm{\Phi }(W^{1,p}(\mathrm{\Omega }))$, and $Y$ is a Banach space (a finite ${\mathrm{\ell }}^p$-sum of the Banach spaces $L^p(\mathrm{\Omega })$).

$W^{1,p}(\mathrm{\Omega })$ is complete. By [F2] the space $W^{1,p}(\mathrm{\Omega })$ is a complete normed space, the Countable Choice needed there being supplied by DC.

Finite sums of reflexive spaces are reflexive. Each factor $L^p(\mathrm{\Omega })$ is reflexive by [F3]; we show that a finite ${\mathrm{\ell }}^p$-sum of reflexive Banach spaces is reflexive. For two factors $X,Z$: a bounded sequence in $X{\oplus }_{p}Z$ has bounded coordinate sequences, so two successive extractions using [F4] give a subsequence $(x_k,z_k)$ with $x_k\rightharpoonup x$ in $X$ and $z_k\rightharpoonup z$ in $Z$; every bounded linear functional on $X{\oplus }_{p}Z$ has the form $(x,z)\mapsto f(x)+g(z)$ with $f\in X^{\ast }$, $g\in Z^{\ast }$ bounded by the norm of the functional (restrict the functional to each factor), so $f(x_k)+g(z_k)\to f(x)+g(z)$ and the subsequence converges weakly in $X{\oplus }_{p}Z$; [F4] then makes $X{\oplus }_{p}Z$ reflexive. Induction over the finitely many factors gives the reflexivity of $Y$.

$S$ is closed. Since $\mathrm{\Phi }$ is a surjective isometry from the complete space $W^{1,p}(\mathrm{\Omega })$ onto $S$ by steps 1.1 and 1.2, the space $S$ is complete, and a complete subset of the normed space $Y$ is closed.

$Y$ is reflexive. By step 2.1 the finite ${\mathrm{\ell }}^p$-sum $Y$ of the reflexive spaces $L^p(\mathrm{\Omega })$ is reflexive.

$S$ is reflexive. By steps 2.2 and 3.1, $S$ is a closed linear subspace of the reflexive Banach space $Y$; [F5] therefore makes $S$, with the restricted norm, reflexive.

Reflexivity transfers to $W^{1,p}(\mathrm{\Omega })$. The isometry $\mathrm{\Phi }:W^{1,p}(\mathrm{\Omega })\to S$ satisfies ${\mathrm{\Phi }}^{\ast \ast }\circ J_{W^{1,p}(\mathrm{\Omega })}=J_S\circ \mathrm{\Phi }$ for the canonical maps: both sides send $u$ to the functional $s^{\ast }\mapsto s^{\ast }(\mathrm{\Phi }(u))$ on $S^{\ast }$. If $\psi \in W^{1,p}(\mathrm{\Omega }{)}^{\ast \ast }$ is given, then ${\mathrm{\Phi }}^{\ast \ast }\psi \in S^{\ast \ast }$ and, $S$ being reflexive by step 4.1, ${\mathrm{\Phi }}^{\ast \ast }\psi =J_S(s)$ for some $s\in S$; writing $s=\mathrm{\Phi }(u)$ gives $J_S(\mathrm{\Phi }(u))={\mathrm{\Phi }}^{\ast \ast }(J_{W^{1,p}(\mathrm{\Omega })}(u))$, hence $J_{W^{1,p}(\mathrm{\Omega })}(u)=\psi$ because ${\mathrm{\Phi }}^{\ast \ast }$ is injective. So the canonical map of $W^{1,p}(\mathrm{\Omega })$ is surjective, that is, $W^{1,p}(\mathrm{\Omega })$ is reflexive.