[proofplan] We prove completeness directly from the definition of weak derivatives. A [Cauchy sequence](/page/Cauchy%20Sequence) in $W^{k,p}(U)$ gives, for every multi-index $\alpha$ with $|\alpha| \le k$, a Cauchy sequence of weak derivatives in $L^p(U)$; completeness of $L^p$ supplies candidate limits $v_\alpha$. The only substantive point is to show that these limits are compatible, namely that $v_\alpha$ is the [weak derivative](/page/Weak%20Derivative) $D^\alpha v_0$. Once this is proved, convergence in the Sobolev norm is immediate, and the Hilbert statement follows by identifying the displayed [inner product](/page/Inner%20Product) with the $W^{1,2}$ norm. [/proofplan] [step:Take componentwise limits in $L^p(U)$] Let $L^p(U)$ denote the Lebesgue space $L^p(U,\mathcal{B}(U),\mathcal{L}^n)$, where $\mathcal{B}(U)$ is the Borel $\sigma$-algebra on $U$ and $\mathcal{L}^n$ is $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) restricted to $U$. Let $\mathbb{N}_0 := \mathbb{N} \cup \{0\}$. For a multi-index $\alpha = (\alpha_1,\dots,\alpha_n) \in \mathbb{N}_0^n$, define $|\alpha| := \alpha_1+\cdots+\alpha_n$ and define the weak derivative notation $D^\alpha u := \partial_{x_1}^{\alpha_1}\cdots\partial_{x_n}^{\alpha_n}u$ whenever this weak derivative exists. Let $(u_j)_{j=1}^{\infty}$ be a Cauchy sequence in $W^{k,p}(U)$. For each multi-index $\alpha = (\alpha_1,\dots,\alpha_n) \in \mathbb{N}_0^n$ with $|\alpha| \le k$, the sequence \begin{align*} (D^\alpha u_j)_{j=1}^{\infty} \end{align*} is Cauchy in $L^p(U)$, because the Sobolev norm controls each derivative component. By completeness of $L^p(U)$ for $1 \le p \le \infty$ (citing a result not yet in the wiki: Completeness of $L^p$ Spaces), there exists a function $v_\alpha \in L^p(U)$ such that \begin{align*} D^\alpha u_j \to v_\alpha \quad \text{in } L^p(U) \end{align*} as $j \to \infty$. In particular, for the zero multi-index $0 = (0,\dots,0)$, there is a function $u := v_0 \in L^p(U)$ such that \begin{align*} u_j \to u \quad \text{in } L^p(U). \end{align*} [/step] [step:Pass to the limit in the weak derivative identity] We prove that $v_\alpha = D^\alpha u$ weakly on $U$ for every multi-index $\alpha$ with $|\alpha| \le k$. Fix such an $\alpha$, and let $\phi \in C_c^\infty(U)$ be a real-valued [test function](/page/Test%20Function) \begin{align*} \phi: U \to \mathbb{R}. \end{align*} For every $j \in \mathbb{N}$, since $D^\alpha u_j$ is the weak derivative of $u_j$, the weak derivative identity gives \begin{align*} \int_U u_j D^\alpha \phi\,d\mathcal L^n = (-1)^{|\alpha|}\int_U D^\alpha u_j \phi\,d\mathcal L^n. \end{align*} Let $K := \operatorname{supp}\phi$, which is compact and contained in $U$. For any $1 \le q \le \infty$, let $L^q(U)$ denote $L^q(U,\mathcal{B}(U),\mathcal{L}^n)$. Since $\phi$ and $D^\alpha\phi$ are smooth with compact support, both belong to $L^q(U)$ for every $1 \le q \le \infty$. Let $q$ denote the conjugate exponent to $p$, with $q=\infty$ when $p=1$ and $q=1$ when $p=\infty$. Hölder's inequality (citing a result not yet in the wiki: Hölder's Inequality) gives \begin{align*} \left|\int_U (u_j-u)D^\alpha\phi\,d\mathcal L^n\right| \le \|u_j-u\|_{L^p(U)}\|D^\alpha\phi\|_{L^q(U)}. \end{align*} The right-hand side tends to $0$ as $j \to \infty$. Likewise, \begin{align*} \left|\int_U (D^\alpha u_j-v_\alpha)\phi\,d\mathcal L^n\right| \le \|D^\alpha u_j-v_\alpha\|_{L^p(U)}\|\phi\|_{L^q(U)}, \end{align*} which also tends to $0$. Passing to the limit in the weak derivative identity yields \begin{align*} \int_U u D^\alpha \phi\,d\mathcal L^n = (-1)^{|\alpha|}\int_U v_\alpha \phi\,d\mathcal L^n. \end{align*} Thus $v_\alpha$ is the weak derivative $D^\alpha u$. [guided] The central issue is compatibility: the functions $v_\alpha$ were obtained independently as $L^p$ limits of the derivative sequences $(D^\alpha u_j)_{j=1}^{\infty}$. We must show that they are not arbitrary limits, but the actual weak derivatives of the same limiting function $u = v_0$. Fix a multi-index $\alpha = (\alpha_1,\dots,\alpha_n)$ with $|\alpha| \le k$, and fix a real-valued test function \begin{align*} \phi: U \to \mathbb{R} \end{align*} with $\phi \in C_c^\infty(U)$. Because $u_j \in W^{k,p}(U)$, the weak derivative $D^\alpha u_j$ exists in $L^p(U)$, and the definition of weak derivative gives \begin{align*} \int_U u_j D^\alpha \phi\,d\mathcal L^n = (-1)^{|\alpha|}\int_U D^\alpha u_j \phi\,d\mathcal L^n. \end{align*} We want to let $j \to \infty$ on both sides. The left side contains $u_j$, and we know $u_j \to u$ in $L^p(U)$. The right side contains $D^\alpha u_j$, and we know $D^\alpha u_j \to v_\alpha$ in $L^p(U)$. To justify passing limits through the integrals, let $q$ be the conjugate exponent to $p$, so that $1/p+1/q=1$, with the conventions $q=\infty$ for $p=1$ and $q=1$ for $p=\infty$. Here $L^q(U)$ denotes $L^q(U,\mathcal{B}(U),\mathcal{L}^n)$, the Lebesgue space over $U$ with respect to $n$-dimensional Lebesgue measure. Since $\phi$ is smooth with compact support, both $\phi$ and $D^\alpha\phi$ belong to $L^q(U)$. Hölder's inequality (citing a result not yet in the wiki: Hölder's Inequality), applied to the product $(u_j-u)D^\alpha\phi$, gives \begin{align*} \left|\int_U (u_j-u)D^\alpha\phi\,d\mathcal L^n\right| \le \|u_j-u\|_{L^p(U)}\|D^\alpha\phi\|_{L^q(U)}. \end{align*} The factor $\|D^\alpha\phi\|_{L^q(U)}$ is finite and independent of $j$, while $\|u_j-u\|_{L^p(U)} \to 0$. Hence \begin{align*} \int_U u_j D^\alpha\phi\,d\mathcal L^n \to \int_U u D^\alpha\phi\,d\mathcal L^n. \end{align*} Applying the same inequality to the product $(D^\alpha u_j-v_\alpha)\phi$ gives \begin{align*} \left|\int_U (D^\alpha u_j-v_\alpha)\phi\,d\mathcal L^n\right| \le \|D^\alpha u_j-v_\alpha\|_{L^p(U)}\|\phi\|_{L^q(U)}. \end{align*} Again the second factor is finite and independent of $j$, and the first factor tends to $0$. Therefore \begin{align*} \int_U D^\alpha u_j \phi\,d\mathcal L^n \to \int_U v_\alpha \phi\,d\mathcal L^n. \end{align*} Taking the limit in the weak derivative identity for $u_j$ gives \begin{align*} \int_U u D^\alpha \phi\,d\mathcal L^n = (-1)^{|\alpha|}\int_U v_\alpha \phi\,d\mathcal L^n. \end{align*} This is precisely the definition that $v_\alpha$ is the weak derivative $D^\alpha u$ on $U$. Since $\alpha$ was arbitrary with $|\alpha|\le k$, every weak derivative of $u$ up to order $k$ exists and belongs to $L^p(U)$. [/guided] [/step] [step:Conclude convergence in the Sobolev norm] From the previous step, $D^\alpha u = v_\alpha$ for every multi-index $\alpha$ with $|\alpha|\le k$. Therefore $u \in W^{k,p}(U)$. Moreover, for every such $\alpha$, \begin{align*} \|D^\alpha u_j-D^\alpha u\|_{L^p(U)} = \|D^\alpha u_j-v_\alpha\|_{L^p(U)} \to 0. \end{align*} If $1 \le p < \infty$, summing the finitely many derivative norms in the definition of the $W^{k,p}$ norm gives \begin{align*} \|u_j-u\|_{W^{k,p}(U)} \to 0. \end{align*} If $p=\infty$, taking the maximum over the same finite set of multi-indices gives the same conclusion. Thus every Cauchy sequence in $W^{k,p}(U)$ converges in $W^{k,p}(U)$, so $W^{k,p}(U)$ is complete. Hence $W^{k,p}(U)$ is a [Banach space](/page/Banach%20Space). [/step] [step:Identify $H^1(U)$ as a Hilbert space] Now take $p=2$ and $k=1$. For $u,v \in H^1(U)$, define \begin{align*} (u,v)_{H^1(U)} := \int_U u v\,d\mathcal L^n + \sum_{i=1}^n \int_U \partial_{x_i}u\,\partial_{x_i}v\,d\mathcal L^n. \end{align*} Bilinearity and symmetry follow from the corresponding properties of the [Lebesgue integral](/page/Lebesgue%20Integral). If $(u,u)_{H^1(U)}=0$, then \begin{align*} \int_U |u|^2\,d\mathcal L^n + \sum_{i=1}^n \int_U |\partial_{x_i}u|^2\,d\mathcal L^n = 0. \end{align*} Each term is non-negative, so $\|u\|_{L^2(U)}=0$ and hence $u=0$ in $L^2(U)$. Thus the displayed formula is an inner product. Its induced norm is \begin{align*} \|u\|_{H^1(U)} = \left(\|u\|_{L^2(U)}^2+\sum_{i=1}^n\|\partial_{x_i}u\|_{L^2(U)}^2\right)^{1/2}, \end{align*} which is exactly the standard $W^{1,2}(U)$ norm. Since $W^{1,2}(U)$ is complete by the Banach-space result already proved, $H^1(U)$ is complete with respect to the norm induced by this inner product. Therefore $H^1(U)$ is a [Hilbert space](/page/Hilbert%20Space). [/step]