[proofplan]
We first prove the reflexivity of $L^p(\Omega)$ from the standard $L^p$ duality theorem: for $1<p<\infty$, the dual of $L^p$ is represented by $L^{p'}$, and applying the same theorem to $L^{p'}$ makes the canonical embedding into the bidual surjective. For the [Sobolev space](/page/Sobolev%20Space), we identify $W^{1,p}(\Omega)$ with the graph of the weak-gradient operator inside the finite product $L^p(\Omega)^{n+1}$. The product is reflexive because finite products of reflexive Banach spaces are reflexive, and the graph is closed because strong $L^p$ limits preserve the [distributional derivative](/page/Distributional%20Derivative) identity. Finally, a closed linear subspace of a reflexive [Banach space](/page/Banach%20Space) is reflexive.
[/proofplan]
[step:Identify the bidual of $L^p(\Omega)$ using $L^p$ duality]
Let $p'=\frac{p}{p-1}$ denote the Hölder conjugate exponent of $p$, so that $1<p'<\infty$ and
\begin{align*}
\frac{1}{p}+\frac{1}{p'}=1.
\end{align*}
We regard $L^p(\Omega)$ as $L^p(\Omega,\mathcal B(\Omega),\mathcal L^n)$. Since $\Omega\subset\mathbb R^n$ is open, the [measure space](/page/Measure%20Space) $(\Omega,\mathcal B(\Omega),\mathcal L^n)$ is $\sigma$-finite.
By the standard $L^p$ duality theorem for $\sigma$-finite measure spaces, applied with exponent $p$, every bounded linear functional on $L^p(\Omega)$ is represented uniquely by an element of $L^{p'}(\Omega)$ through the pairing
\begin{align*}
\Lambda_g(f)=\int_\Omega f(x)g(x)\,d\mathcal L^n(x), \qquad f\in L^p(\Omega),
\end{align*}
where $g\in L^{p'}(\Omega)$. This is a cited result not yet resolved to a wiki theorem ID in this run: $L^p$ duality theorem.
Apply the same duality theorem with exponent $p'$. Since the Hölder conjugate of $p'$ is $p$, every bounded linear functional on $L^{p'}(\Omega)$ is represented uniquely by an element of $L^p(\Omega)$. Therefore every element of $(L^p(\Omega))^{**}$ is evaluation against some $f\in L^p(\Omega)$ under the above duality pairing. Equivalently, the canonical embedding
\begin{align*}
J_p:L^p(\Omega)\to (L^p(\Omega))^{**}
\end{align*}
is surjective. Since $J_p$ is always an isometric linear embedding, $L^p(\Omega)$ is reflexive.
[/step]
[step:Embed $W^{1,p}(\Omega)$ into a finite product of $L^p$ spaces]
Define the Banach space
\begin{align*}
E=L^p(\Omega)^{n+1}
\end{align*}
with product norm
\begin{align*}
\|(f_0,f_1,\dots,f_n)\|_E=\left(\sum_{i=0}^n \|f_i\|_{L^p(\Omega)}^p\right)^{1/p}.
\end{align*}
Since each factor $L^p(\Omega)$ is reflexive by the previous step, the standard Banach-space fact that finite products of reflexive Banach spaces are reflexive implies that $E$ is reflexive. This is a cited result not yet resolved to a wiki theorem ID in this run: finite products of reflexive Banach spaces are reflexive.
Define
\begin{align*}
T:W^{1,p}(\Omega)\to E,\qquad T(u)=(u,\partial_{x_1}u,\dots,\partial_{x_n}u).
\end{align*}
Here $\partial_{x_i}u\in L^p(\Omega)$ denotes the $i$th [weak derivative](/page/Weak%20Derivative) of $u$. The weak derivative is linear, so $T$ is linear. By the definition of the Sobolev norm,
\begin{align*}
\|T(u)\|_E=\|u\|_{W^{1,p}(\Omega)}
\end{align*}
for every $u\in W^{1,p}(\Omega)$. Thus $T$ is a linear isometry onto its range. Therefore it remains to prove that $T(W^{1,p}(\Omega))$ is closed in $E$.
[/step]
[step:Prove that the Sobolev graph is closed in the product space]
Let $(u_k)_{k=1}^\infty$ be a sequence in $W^{1,p}(\Omega)$ such that $T(u_k)$ converges in $E$ to some element $(v_0,v_1,\dots,v_n)\in E$. This means
\begin{align*}
u_k\to v_0 \quad \text{in } L^p(\Omega)
\end{align*}
and, for each $i\in\{1,\dots,n\}$,
\begin{align*}
\partial_{x_i}u_k\to v_i \quad \text{in } L^p(\Omega).
\end{align*}
Fix $i\in\{1,\dots,n\}$ and let $\varphi\in C_c^\infty(\Omega)$ be a [test function](/page/Test%20Function). Since $\varphi$ and $\partial_{x_i}\varphi$ are smooth with compact support in $\Omega$, both belong to $L^{p'}(\Omega)$. For each $k\in\mathbb N$, the definition of the weak derivative gives
\begin{align*}
\int_\Omega u_k(x)\partial_{x_i}\varphi(x)\,d\mathcal L^n(x)=-\int_\Omega \partial_{x_i}u_k(x)\varphi(x)\,d\mathcal L^n(x).
\end{align*}
Passing to the limit uses Hölder's inequality with conjugate exponents $p$ and $p'$. Indeed,
\begin{align*}
\left|\int_\Omega (u_k(x)-v_0(x))\partial_{x_i}\varphi(x)\,d\mathcal L^n(x)\right|\le \|u_k-v_0\|_{L^p(\Omega)}\|\partial_{x_i}\varphi\|_{L^{p'}(\Omega)}\to 0,
\end{align*}
and
\begin{align*}
\left|\int_\Omega (\partial_{x_i}u_k(x)-v_i(x))\varphi(x)\,d\mathcal L^n(x)\right|\le \|\partial_{x_i}u_k-v_i\|_{L^p(\Omega)}\|\varphi\|_{L^{p'}(\Omega)}\to 0.
\end{align*}
Therefore
\begin{align*}
\int_\Omega v_0(x)\partial_{x_i}\varphi(x)\,d\mathcal L^n(x)=-\int_\Omega v_i(x)\varphi(x)\,d\mathcal L^n(x).
\end{align*}
Since this holds for every $\varphi\in C_c^\infty(\Omega)$ and every $i\in\{1,\dots,n\}$, the function $v_i$ is the $i$th weak derivative of $v_0$. Hence $v_0\in W^{1,p}(\Omega)$ and
\begin{align*}
T(v_0)=(v_0,v_1,\dots,v_n).
\end{align*}
Thus $T(W^{1,p}(\Omega))$ is closed in $E$.
[guided]
The goal is to show that no [limit point](/page/Limit%20Point) of the graph escapes the graph. Suppose a sequence of graph elements $T(u_k)$ converges in the product space $E=L^p(\Omega)^{n+1}$ to an element $(v_0,v_1,\dots,v_n)$. By the definition of the product norm, convergence in $E$ means convergence in each component:
\begin{align*}
u_k\to v_0 \quad \text{in } L^p(\Omega),
\end{align*}
and, for each $i\in\{1,\dots,n\}$,
\begin{align*}
\partial_{x_i}u_k\to v_i \quad \text{in } L^p(\Omega).
\end{align*}
What must be proved? We must show that $(v_0,v_1,\dots,v_n)$ is again of the form $T(v_0)$, which means precisely that $v_i$ is the $i$th weak derivative of $v_0$ for every $i$. The definition of weak derivative is distributional: for every test function $\varphi\in C_c^\infty(\Omega)$, we must prove
\begin{align*}
\int_\Omega v_0(x)\partial_{x_i}\varphi(x)\,d\mathcal L^n(x)=-\int_\Omega v_i(x)\varphi(x)\,d\mathcal L^n(x).
\end{align*}
Fix $i\in\{1,\dots,n\}$ and fix a test function $\varphi\in C_c^\infty(\Omega)$. Since $\varphi$ and $\partial_{x_i}\varphi$ are smooth with compact support in $\Omega$, they are bounded and supported on a compact subset of $\Omega$ with finite $\mathcal L^n$-measure. Hence
\begin{align*}
\varphi\in L^{p'}(\Omega) \quad \text{and} \quad \partial_{x_i}\varphi\in L^{p'}(\Omega).
\end{align*}
For each $k\in\mathbb N$, because $\partial_{x_i}u_k$ is the weak derivative of $u_k$, the defining identity gives
\begin{align*}
\int_\Omega u_k(x)\partial_{x_i}\varphi(x)\,d\mathcal L^n(x)=-\int_\Omega \partial_{x_i}u_k(x)\varphi(x)\,d\mathcal L^n(x).
\end{align*}
We now pass to the limit on both sides. For the left side, apply Hölder's inequality with the conjugate exponents $p$ and $p'$, pairing $u_k-v_0\in L^p(\Omega)$ with $\partial_{x_i}\varphi\in L^{p'}(\Omega)$. This gives
\begin{align*}
\left|\int_\Omega (u_k(x)-v_0(x))\partial_{x_i}\varphi(x)\,d\mathcal L^n(x)\right|\le \|u_k-v_0\|_{L^p(\Omega)}\|\partial_{x_i}\varphi\|_{L^{p'}(\Omega)}.
\end{align*}
The first factor tends to $0$ by the assumed $L^p$ convergence of $u_k$ to $v_0$, while the second factor is fixed and finite. Therefore
\begin{align*}
\int_\Omega u_k(x)\partial_{x_i}\varphi(x)\,d\mathcal L^n(x)\to \int_\Omega v_0(x)\partial_{x_i}\varphi(x)\,d\mathcal L^n(x).
\end{align*}
For the right side, apply Hölder's inequality again, now pairing $\partial_{x_i}u_k-v_i\in L^p(\Omega)$ with $\varphi\in L^{p'}(\Omega)$. We obtain
\begin{align*}
\left|\int_\Omega (\partial_{x_i}u_k(x)-v_i(x))\varphi(x)\,d\mathcal L^n(x)\right|\le \|\partial_{x_i}u_k-v_i\|_{L^p(\Omega)}\|\varphi\|_{L^{p'}(\Omega)}.
\end{align*}
The first factor tends to $0$ by the assumed $L^p$ convergence of $\partial_{x_i}u_k$ to $v_i$, and the second factor is fixed and finite. Hence
\begin{align*}
\int_\Omega \partial_{x_i}u_k(x)\varphi(x)\,d\mathcal L^n(x)\to \int_\Omega v_i(x)\varphi(x)\,d\mathcal L^n(x).
\end{align*}
Taking limits in the weak derivative identity for $u_k$ gives
\begin{align*}
\int_\Omega v_0(x)\partial_{x_i}\varphi(x)\,d\mathcal L^n(x)=-\int_\Omega v_i(x)\varphi(x)\,d\mathcal L^n(x).
\end{align*}
Because the test function $\varphi\in C_c^\infty(\Omega)$ was arbitrary, this proves that $v_i$ is the $i$th weak derivative of $v_0$. Because $i\in\{1,\dots,n\}$ was arbitrary, all first weak derivatives of $v_0$ belong to $L^p(\Omega)$. Thus $v_0\in W^{1,p}(\Omega)$ and
\begin{align*}
T(v_0)=(v_0,v_1,\dots,v_n).
\end{align*}
So the limit point lies in the range of $T$, which proves that $T(W^{1,p}(\Omega))$ is closed in $E$.
[/guided]
[/step]
[step:Transfer reflexivity from the closed product subspace to $W^{1,p}(\Omega)$]
The range $T(W^{1,p}(\Omega))$ is a closed linear subspace of the reflexive Banach space $E$. By the standard Banach-space theorem that closed linear subspaces of reflexive Banach spaces are reflexive, $T(W^{1,p}(\Omega))$ is reflexive. This is a cited result not yet resolved to a wiki theorem ID in this run: closed subspaces of reflexive Banach spaces are reflexive.
Since $T:W^{1,p}(\Omega)\to T(W^{1,p}(\Omega))$ is a surjective linear isometry, reflexivity transfers through this isometric isomorphism. Therefore $W^{1,p}(\Omega)$ is reflexive. Together with the first step, this proves both asserted reflexivity statements.
[/step]
