[proofplan] The proof has three components. **Completeness:** given a Cauchy sequence $(u_k) \subset W^{1,p}(\Omega)$, the components $(u_k)$ and $(D_{e_i} u_k)$ are Cauchy in $L^p(\Omega)$, so by completeness of $L^p$ they converge to limits $u$ and $v_i$. We then identify $v_i = D_{e_i} u$ in the weak sense by passing to the limit in the integration-by-parts identity defining the weak derivative. **Reflexivity** for $1 < p < \infty$ and **separability** for $1 \leq p < \infty$ follow by realising $W^{1,p}(\Omega)$ as a closed subspace of the product space $L^p(\Omega)^{n+1}$ via the map $u \mapsto (u, D_{e_1} u, \dots, D_{e_n} u)$, since closed subspaces inherit reflexivity and separability. The Hilbert structure for $p = 2$ is read off from the inner product $(u, v)_{H^1} = (u, v)_{L^2} + \sum_i (D_{e_i} u, D_{e_i} v)_{L^2}$. [/proofplan]

[step:Take a Cauchy sequence and extract $L^p$ limits of the components]Fix $1 \leq p \leq \infty$. Let $(u_k)_{k \geq 1} \subset W^{1,p}(\Omega)$ be a Cauchy sequence. By definition of the $W^{1,p}$ norm, \begin{align*} \|u_k - u_\ell\|_{W^{1,p}}^p = \|u_k - u_\ell\|_{L^p}^p + \sum_{i=1}^n \|D_{e_i}(u_k - u_\ell)\|_{L^p}^p \quad \text{for } 1 \leq p < \infty, \end{align*} and analogously with the supremum for $p = \infty$. Hence each of the sequences $(u_k)$ and $(D_{e_i} u_k)$ for $i = 1, \dots, n$ is Cauchy in $L^p(\Omega)$. By the [Completeness of $L^p$ Spaces](/theorems/???), there exist $u, v_1, \dots, v_n \in L^p(\Omega)$ such that \begin{align*} u_k \xrightarrow{L^p} u, \qquad D_{e_i} u_k \xrightarrow{L^p} v_i \quad (i = 1, \dots, n). \end{align*}[/step]

[guided]We are given a sequence $(u_k)$ that is Cauchy in $W^{1,p}(\Omega)$ and we want to produce a $W^{1,p}$-limit. The strategy is to first extract limits of the function values and of each weak partial derivative separately in $L^p$, then verify that these limits are consistent — that is, that the limit of $D_{e_i} u_k$ really is the weak derivative of the limit of $u_k$. The first step uses only completeness of $L^p$. By definition of the Sobolev norm, \begin{align*} \|u_k - u_\ell\|_{W^{1,p}}^p = \|u_k - u_\ell\|_{L^p}^p + \sum_{i=1}^n \|D_{e_i}(u_k - u_\ell)\|_{L^p}^p \end{align*} for $1 \leq p < \infty$, with the analogous statement for $p = \infty$ where the maximum replaces the sum. The Cauchy assumption $\|u_k - u_\ell\|_{W^{1,p}} \to 0$ as $k, \ell \to \infty$ therefore forces each of the $n+1$ summands on the right to go to zero, so each component sequence is Cauchy in $L^p$. We now invoke the [Completeness of $L^p$ Spaces](/theorems/???). The hypotheses are that $(\Omega, \mathcal{B}(\Omega), \mathcal{L}^n)$ is a measure space and $1 \leq p \leq \infty$ — both granted. The conclusion produces limits $u, v_1, \dots, v_n \in L^p(\Omega)$ with \begin{align*} u_k \xrightarrow{L^p} u, \qquad D_{e_i} u_k \xrightarrow{L^p} v_i \quad (i = 1, \dots, n). \end{align*} The danger now is that the $v_i$ are merely $L^p$ limits of weak derivatives — there is no a priori reason for $v_i$ to be the weak derivative of $u$. The next step verifies this.[/guided]

[step:Identify $v_i$ as the weak partial derivative $D_{e_i} u$]Fix $i \in \{1, \dots, n\}$ and a test function $\phi \in C_c^\infty(\Omega)$. By the [Definition of Weak Derivative](/page/Weak%20Derivative), each $u_k \in W^{1,p}(\Omega)$ satisfies \begin{align*} \int_\Omega u_k(x)\, D_{e_i}\phi(x) \, d\mathcal{L}^n(x) = -\int_\Omega (D_{e_i} u_k)(x)\, \phi(x) \, d\mathcal{L}^n(x). \end{align*} We pass to the limit $k \to \infty$ on both sides. By Hölder's inequality with exponents $p$ and $p' = p/(p-1)$ (with the convention $1' = \infty$, $\infty' = 1$), \begin{align*} \left|\int_\Omega (u_k - u)\, D_{e_i}\phi \, d\mathcal{L}^n\right| \leq \|u_k - u\|_{L^p(\Omega)} \|D_{e_i}\phi\|_{L^{p'}(\Omega)} \xrightarrow{k \to \infty} 0, \end{align*} since $D_{e_i}\phi \in C_c^\infty(\Omega) \subset L^{p'}(\Omega)$ and $u_k \to u$ in $L^p$. Identically, \begin{align*} \left|\int_\Omega (D_{e_i} u_k - v_i)\, \phi \, d\mathcal{L}^n\right| \leq \|D_{e_i} u_k - v_i\|_{L^p(\Omega)} \|\phi\|_{L^{p'}(\Omega)} \xrightarrow{k \to \infty} 0. \end{align*} Passing to the limit therefore yields \begin{align*} \int_\Omega u\, D_{e_i}\phi \, d\mathcal{L}^n = -\int_\Omega v_i\, \phi \, d\mathcal{L}^n \quad \text{for every } \phi \in C_c^\infty(\Omega). \end{align*} This is precisely the definition of $v_i$ being the $i$-th weak partial derivative of $u$, so $v_i = D_{e_i} u$ in $L^p(\Omega)$.[/step]

[guided]We have $L^p$ limits $u$ and $v_i$ but need to show $v_i = D_{e_i} u$ in the weak sense. The only tool tying weak derivatives to function values is the integration-by-parts identity that defines the weak derivative, so the strategy is to write that identity for each $u_k$ and pass to the limit. Fix $i \in \{1, \dots, n\}$ and a test function $\phi \in C_c^\infty(\Omega)$. The [Definition of Weak Derivative](/page/Weak%20Derivative) applied to $u_k \in W^{1,p}(\Omega)$ gives \begin{align*} \int_\Omega u_k(x)\, D_{e_i}\phi(x) \, d\mathcal{L}^n(x) = -\int_\Omega (D_{e_i} u_k)(x)\, \phi(x) \, d\mathcal{L}^n(x). \end{align*} We want to send $k \to \infty$. Each side is a duality pairing of an $L^p$ function with the test function (or its derivative); these are $L^{p'}$ functions where $p'$ is the [Hölder Conjugate](/theorems/???) of $p$. The natural tool is Hölder's inequality. For the left-hand side, we apply Hölder with exponents $p$ and $p'$ (where $1/p + 1/p' = 1$, with $1' = \infty$ and $\infty' = 1$): \begin{align*} \left|\int_\Omega (u_k - u)\, D_{e_i}\phi \, d\mathcal{L}^n\right| \leq \|u_k - u\|_{L^p(\Omega)} \|D_{e_i}\phi\|_{L^{p'}(\Omega)}. \end{align*} We must check that $D_{e_i}\phi \in L^{p'}(\Omega)$. Since $\phi \in C_c^\infty(\Omega)$, the derivative $D_{e_i}\phi$ is also smooth and compactly supported, hence in $L^q(\Omega)$ for every $q \in [1, \infty]$ — in particular for $q = p'$. The first factor $\|u_k - u\|_{L^p}$ tends to $0$ by the previous step, so the entire bound vanishes. The right-hand side argument is symmetric, using $\phi \in L^{p'}(\Omega)$ and $D_{e_i} u_k \to v_i$ in $L^p$: \begin{align*} \left|\int_\Omega (D_{e_i} u_k - v_i)\, \phi \, d\mathcal{L}^n\right| \leq \|D_{e_i} u_k - v_i\|_{L^p(\Omega)} \|\phi\|_{L^{p'}(\Omega)} \xrightarrow{k \to \infty} 0. \end{align*} Passing to the limit on both sides of the integration-by-parts identity yields \begin{align*} \int_\Omega u\, D_{e_i}\phi \, d\mathcal{L}^n = -\int_\Omega v_i\, \phi \, d\mathcal{L}^n \quad \text{for every } \phi \in C_c^\infty(\Omega). \end{align*} This is exactly the defining property of $v_i$ as the $i$-th weak partial derivative of $u$, so $v_i = D_{e_i} u$ as elements of $L^p(\Omega)$. Why does this argument work for *every* $1 \leq p \leq \infty$? Because Hölder pairs $L^p$ with $L^{p'}$, and test functions are in every $L^q$. For $p = 1$ the conjugate is $p' = \infty$ and $\phi$ is bounded; for $p = \infty$ we have $p' = 1$ and $\phi$ is integrable; for $p \in (1, \infty)$ both factors are between these extremes. The argument is uniform across the range.[/guided]

[step:Conclude $u \in W^{1,p}(\Omega)$ and $u_k \to u$ in $W^{1,p}$] Combining the two previous steps, $u \in L^p(\Omega)$ has weak partial derivatives $D_{e_i} u = v_i \in L^p(\Omega)$ for $i = 1, \dots, n$, so $u \in W^{1,p}(\Omega)$. The Sobolev convergence \begin{align*} \|u_k - u\|_{W^{1,p}}^p = \|u_k - u\|_{L^p}^p + \sum_{i=1}^n \|D_{e_i} u_k - D_{e_i} u\|_{L^p}^p \xrightarrow{k \to \infty} 0 \end{align*} follows immediately because each summand tends to zero by Step 1. Hence $W^{1,p}(\Omega)$ is a Banach space. [/step]

[step:Embed $W^{1,p}(\Omega)$ as a closed subspace of $L^p(\Omega)^{n+1}$]Define the linear map \begin{align*} T: W^{1,p}(\Omega) &\to L^p(\Omega)^{n+1} \\ u &\mapsto (u, D_{e_1} u, \dots, D_{e_n} u), \end{align*} where $L^p(\Omega)^{n+1}$ is endowed with the norm $\|(f_0, \dots, f_n)\|^p = \sum_{i=0}^n \|f_i\|_{L^p}^p$ for $1 \leq p < \infty$, and the maximum norm for $p = \infty$. By construction $\|Tu\| = \|u\|_{W^{1,p}}$, so $T$ is an isometric embedding. Its image $T(W^{1,p}(\Omega))$ is closed in $L^p(\Omega)^{n+1}$: if $(Tu_k) \to (f_0, \dots, f_n)$ in $L^p(\Omega)^{n+1}$, then $(u_k)$ is Cauchy in $W^{1,p}(\Omega)$, by Step 3 there is $u \in W^{1,p}(\Omega)$ with $u_k \to u$ in $W^{1,p}$, and uniqueness of $L^p$ limits forces $f_0 = u$ and $f_i = D_{e_i} u$, so $(f_0, \dots, f_n) = Tu$ lies in the image.[/step]

[guided]We want to bootstrap structural properties (reflexivity, separability) from $L^p$ to $W^{1,p}$. The cleanest way is to realise $W^{1,p}$ as a closed subspace of a finite product of $L^p$ spaces — which is the same kind of space — and then use the fact that closed subspaces inherit both properties. Define \begin{align*} T: W^{1,p}(\Omega) &\to L^p(\Omega)^{n+1} \\ u &\mapsto (u, D_{e_1} u, \dots, D_{e_n} u). \end{align*} The codomain is the product space $L^p(\Omega)^{n+1}$, which is itself a Banach space under the norm $\|(f_0, \dots, f_n)\|^p = \sum_{i=0}^n \|f_i\|_{L^p}^p$ for $1 \leq p < \infty$ (with the maximum for $p = \infty$). With this choice of norm, \begin{align*} \|Tu\|^p = \|u\|_{L^p}^p + \sum_{i=1}^n \|D_{e_i} u\|_{L^p}^p = \|u\|_{W^{1,p}}^p, \end{align*} so $T$ is an isometric embedding. In particular it is injective and continuous. We now show the image is closed. Suppose $(Tu_k)$ converges to $(f_0, \dots, f_n)$ in $L^p(\Omega)^{n+1}$. Componentwise this says $u_k \to f_0$ and $D_{e_i} u_k \to f_i$ in $L^p$. Convergent sequences are Cauchy, so $(Tu_k)$ is Cauchy in $L^p(\Omega)^{n+1}$, and by the isometry $(u_k)$ is Cauchy in $W^{1,p}$. By Step 3 there is $u \in W^{1,p}(\Omega)$ with $u_k \to u$ in $W^{1,p}$, hence $Tu_k \to Tu$ in $L^p(\Omega)^{n+1}$. By uniqueness of $L^p$ limits, $Tu = (f_0, \dots, f_n)$, so the image is closed. The image is therefore a closed subspace of $L^p(\Omega)^{n+1}$, and $T$ is a bijection between $W^{1,p}(\Omega)$ and its image. Since $T$ is an isometry, $W^{1,p}(\Omega)$ inherits any property of closed subspaces of $L^p(\Omega)^{n+1}$ that is preserved under isometric isomorphism — in particular reflexivity and separability.[/guided]

[step:Deduce reflexivity for $1 < p < \infty$] For $1 < p < \infty$, the space $L^p(\Omega)$ is reflexive by [Reflexivity of $L^p$ Spaces](/theorems/???). A finite product of reflexive Banach spaces is reflexive (the dual of $X \oplus Y$ identifies with $X^* \oplus Y^*$ and the canonical map $X \oplus Y \to (X \oplus Y)^{**}$ decomposes as the direct sum of the canonical maps), so $L^p(\Omega)^{n+1}$ is reflexive. By the [Closed Subspace of a Reflexive Space is Reflexive](/theorems/???) theorem, the closed subspace $T(W^{1,p}(\Omega))$ is reflexive, and reflexivity is preserved under isometric isomorphism, so $W^{1,p}(\Omega)$ is reflexive. [/step]

[step:Deduce separability for $1 \leq p < \infty$] For $1 \leq p < \infty$, the space $L^p(\Omega)$ is separable by the [Separability of $L^p$ Spaces](/theorems/548). Theorem 548 requires the underlying measure space to be $\sigma$-finite with countably generated $\sigma$-algebra; since $\Omega \subset \mathbb{R}^n$ is open, $\mathcal{L}^n$ is $\sigma$-finite on $\Omega$ (e.g. $\Omega = \bigcup_{m=1}^\infty K_m$ with $K_m = \{x \in \Omega : |x| \leq m,\ \operatorname{dist}(x, \partial\Omega) \geq 1/m\}$, each of finite measure) and $\mathcal{B}(\Omega)$ is generated by the countable collection of rational balls. A finite product of separable metric spaces is separable, so $L^p(\Omega)^{n+1}$ is separable. Subspaces of separable metric spaces are separable, so the closed subspace $T(W^{1,p}(\Omega))$ is separable, and via the isometry $T$ so is $W^{1,p}(\Omega)$. The hypothesis $p < \infty$ is consumed here: $L^\infty(\Omega)$ is not separable (the indicator functions $\{\mathbb{1}_{B(x, 1)} : x \in \Omega\}$ form an uncountable set with pairwise distance $1$ in $L^\infty$), and correspondingly $W^{1,\infty}(\Omega)$ is not separable. [/step]

[step:Verify the Hilbert structure for $p = 2$] For $p = 2$ the space $L^2(\Omega)$ is a Hilbert space with inner product $(f, g)_{L^2} = \int_\Omega f\, \bar{g}\, d\mathcal{L}^n$. Define \begin{align*} (\cdot, \cdot)_{H^1}: H^1(\Omega) \times H^1(\Omega) &\to \mathbb{R} \quad (\text{or } \mathbb{C}) \\ (u, v) &\mapsto (u, v)_{L^2} + \sum_{i=1}^n (D_{e_i} u, D_{e_i} v)_{L^2}. \end{align*} This is bilinear (sesquilinear in the complex case), symmetric (Hermitian), and positive definite: $(u, u)_{H^1} = \|u\|_{H^1}^2 \geq 0$ with equality iff $u = 0$ in $H^1$. The induced norm $\|u\|_{H^1} = (u, u)_{H^1}^{1/2}$ coincides with the $W^{1,2}$ norm. By Step 3 this norm is complete, so $H^1(\Omega)$ is a Hilbert space; by Step 6 (with $p = 2 < \infty$) it is separable. This completes the proof. [/step]