[proofplan] We prove each of the four properties directly from the definition of the [weak derivative](/page/Weak%20Derivative). Part (i) shows that iterated weak derivatives commute by transferring all derivatives onto the test function, where classical commutativity applies, and tracking signs via $(-1)^{|\alpha|}$. Part (ii) is immediate from linearity of the integral. Part (iii) uses the zero extension of test functions from $V$ to $U$. Part (iv) establishes the Leibniz formula by induction on $|\alpha|$: the base case uses the classical product rule for $\zeta \phi$ to rearrange the weak-derivative integral, and the inductive step applies the base case to each term of the sum. [/proofplan] [step:Show iterated weak derivatives commute: $D^\beta(D^\alpha u) = D^{\alpha+\beta} u$] Let $\phi \in C_c^\infty(U)$. By the definition of the [weak derivative](/page/Weak%20Derivative) $D^\alpha u$: \begin{align*} \int_U (D^\alpha u)\, D^\beta \phi \, d\mathcal{L}^n = (-1)^{|\alpha|} \int_U u\, D^\alpha(D^\beta \phi) \, d\mathcal{L}^n. \end{align*} Since $\phi \in C_c^\infty(U)$, classical partial derivatives commute: $D^\alpha(D^\beta \phi) = D^{\alpha+\beta}\phi$. Substituting: \begin{align*} \int_U (D^\alpha u)\, D^\beta \phi \, d\mathcal{L}^n = (-1)^{|\alpha|} \int_U u\, D^{\alpha+\beta} \phi \, d\mathcal{L}^n. \end{align*} Since $u \in W^{k,p}(U)$ and $|\alpha| + |\beta| \le k$, the weak derivative $D^{\alpha+\beta}u$ exists in $L^p(U)$ and satisfies: \begin{align*} \int_U u\, D^{\alpha+\beta} \phi \, d\mathcal{L}^n = (-1)^{|\alpha+\beta|} \int_U (D^{\alpha+\beta} u)\, \phi \, d\mathcal{L}^n. \end{align*} Combining and using $(-1)^{|\alpha|} \cdot (-1)^{|\alpha+\beta|} = (-1)^{|\alpha| + |\alpha| + |\beta|} = (-1)^{2|\alpha| + |\beta|} = (-1)^{|\beta|}$: \begin{align*} \int_U (D^\alpha u)\, D^\beta \phi \, d\mathcal{L}^n = (-1)^{|\beta|} \int_U (D^{\alpha+\beta} u)\, \phi \, d\mathcal{L}^n. \end{align*} This is the defining relation for $D^\beta(D^\alpha u) = D^{\alpha+\beta} u$, and the derivative lies in $L^p(U)$ since $D^{\alpha+\beta}u \in L^p(U)$. [guided] The idea is to move all derivatives onto the smooth test function $\phi$, where classical commutativity applies, then move them back. Let $\phi \in C_c^\infty(U)$. We start from the integral that defines $D^\beta(D^\alpha u)$: \begin{align*} \int_U (D^\alpha u)\, D^\beta \phi \, d\mathcal{L}^n. \end{align*} We apply the definition of the weak derivative $D^\alpha u$ to transfer $\alpha$ derivatives onto the test function $D^\beta \phi$: \begin{align*} \int_U (D^\alpha u)\, D^\beta \phi \, d\mathcal{L}^n = (-1)^{|\alpha|} \int_U u\, D^\alpha(D^\beta \phi) \, d\mathcal{L}^n. \end{align*} Now $\phi$ is smooth, so $D^\alpha(D^\beta \phi) = D^{\alpha+\beta}\phi$ by commutativity of classical partial derivatives. Using the definition of $D^{\alpha+\beta}u$: \begin{align*} (-1)^{|\alpha|} \int_U u\, D^{\alpha+\beta}\phi \, d\mathcal{L}^n = (-1)^{|\alpha|} \cdot (-1)^{|\alpha+\beta|} \int_U (D^{\alpha+\beta}u)\, \phi \, d\mathcal{L}^n. \end{align*} For the sign: $(-1)^{|\alpha|} \cdot (-1)^{|\alpha+\beta|} = (-1)^{2|\alpha| + |\beta|} = (-1)^{|\beta|}$, since $(-1)^{2|\alpha|} = 1$. Therefore \begin{align*} \int_U (D^\alpha u)\, D^\beta \phi \, d\mathcal{L}^n = (-1)^{|\beta|} \int_U (D^{\alpha+\beta}u)\, \phi \, d\mathcal{L}^n, \end{align*} which is precisely the statement that $D^\beta(D^\alpha u) = D^{\alpha+\beta}u$ in the weak sense. [/guided] [/step] [step:Verify linearity of the weak derivative] Let $\phi \in C_c^\infty(U)$. By linearity of the integral and the definition of the weak derivative: \begin{align*} \int_U (\lambda u + \mu v)\, D^\alpha \phi \, d\mathcal{L}^n &= \lambda \int_U u\, D^\alpha \phi \, d\mathcal{L}^n + \mu \int_U v\, D^\alpha \phi \, d\mathcal{L}^n \\ &= \lambda(-1)^{|\alpha|} \int_U (D^\alpha u)\, \phi \, d\mathcal{L}^n + \mu(-1)^{|\alpha|} \int_U (D^\alpha v)\, \phi \, d\mathcal{L}^n \\ &= (-1)^{|\alpha|} \int_U (\lambda D^\alpha u + \mu D^\alpha v)\, \phi \, d\mathcal{L}^n. \end{align*} This shows $D^\alpha(\lambda u + \mu v) = \lambda D^\alpha u + \mu D^\alpha v$, and the right-hand side lies in $L^p(U)$ since $L^p(U)$ is a vector space. [/step] [step:Restrict weak derivatives to open subsets] Let $V \subset U$ be open and let $\phi \in C_c^\infty(V)$. Define the zero extension: \begin{align*} \tilde{\phi}: U \to \mathbb{R}, \quad \tilde{\phi}(x) = \begin{cases} \phi(x) & x \in V \\ 0 & x \in U \setminus V \end{cases}. \end{align*} Since $\phi$ has compact support in $V$, the extension $\tilde{\phi}$ belongs to $C_c^\infty(U)$. Using $u \in W^{k,p}(U)$: \begin{align*} \int_U u\, D^\alpha \tilde{\phi} \, d\mathcal{L}^n = (-1)^{|\alpha|} \int_U (D^\alpha u)\, \tilde{\phi} \, d\mathcal{L}^n. \end{align*} Since $\operatorname{supp}(\tilde{\phi}) \subset V$, both integrals restrict to $V$: \begin{align*} \int_V u\, D^\alpha \phi \, d\mathcal{L}^n = (-1)^{|\alpha|} \int_V (D^\alpha u)\, \phi \, d\mathcal{L}^n. \end{align*} This shows that $D^\alpha u|_V$ is the weak derivative of $u|_V$ on $V$, so $u \in W^{k,p}(V)$. [/step] [step:Establish the Leibniz formula by induction on $|\alpha|$] **Base case ($|\alpha| = 1$).** Let $\alpha = e_i$ for some $i \in \{1, \ldots, n\}$. Let $\phi \in C_c^\infty(U)$. Since $\zeta \in C_c^\infty(U)$, the product $\zeta\phi \in C_c^\infty(U)$. The classical product rule gives $\zeta\, \partial_{x_i}\phi = \partial_{x_i}(\zeta\phi) - (\partial_{x_i}\zeta)\,\phi$. We compute: \begin{align*} \int_U (\zeta u)\, \partial_{x_i}\phi \, d\mathcal{L}^n &= \int_U u\,(\zeta\, \partial_{x_i}\phi) \, d\mathcal{L}^n \\ &= \int_U u\, \partial_{x_i}(\zeta\phi) \, d\mathcal{L}^n - \int_U u\,(\partial_{x_i}\zeta)\,\phi \, d\mathcal{L}^n. \end{align*} Applying the definition of the weak derivative $\partial_{x_i}u$ to the first integral with test function $\zeta\phi \in C_c^\infty(U)$: \begin{align*} \int_U u\, \partial_{x_i}(\zeta\phi) \, d\mathcal{L}^n = -\int_U (\partial_{x_i}u)\,(\zeta\phi) \, d\mathcal{L}^n = -\int_U (\zeta\, \partial_{x_i}u)\,\phi \, d\mathcal{L}^n. \end{align*} Combining: \begin{align*} \int_U (\zeta u)\, \partial_{x_i}\phi \, d\mathcal{L}^n = -\int_U (\zeta\, \partial_{x_i}u + u\,\partial_{x_i}\zeta)\,\phi \, d\mathcal{L}^n. \end{align*} This shows $D_i(\zeta u) = \zeta\, D_i u + u\, D_i\zeta$, which is the Leibniz formula for $|\alpha| = 1$. [guided] The idea is to isolate the weak derivative of $\zeta u$ by rearranging the integral $\int (\zeta u)\,\partial_{x_i}\phi \, d\mathcal{L}^n$. We cannot directly apply the definition of the weak derivative to $\zeta u$ because we do not yet know what $D_i(\zeta u)$ is -- that is what we are trying to determine. Instead, we use a trick: rewrite $\zeta\, \partial_{x_i}\phi$ using the classical product rule $\zeta\,\partial_{x_i}\phi = \partial_{x_i}(\zeta\phi) - (\partial_{x_i}\zeta)\,\phi$. This is valid because $\zeta$ and $\phi$ are both smooth. Since $\zeta\phi \in C_c^\infty(U)$, we can use it as a test function for the weak derivative $\partial_{x_i} u$: \begin{align*} \int_U u\, \partial_{x_i}(\zeta\phi) \, d\mathcal{L}^n = -\int_U (\partial_{x_i}u)\,(\zeta\phi) \, d\mathcal{L}^n. \end{align*} Putting it together: \begin{align*} \int_U (\zeta u)\, \partial_{x_i}\phi \, d\mathcal{L}^n &= \int_U u\,\partial_{x_i}(\zeta\phi) \, d\mathcal{L}^n - \int_U u\,(\partial_{x_i}\zeta)\,\phi \, d\mathcal{L}^n \\ &= -\int_U (\zeta\, \partial_{x_i}u)\,\phi \, d\mathcal{L}^n - \int_U (u\,\partial_{x_i}\zeta)\,\phi \, d\mathcal{L}^n \\ &= -\int_U (\zeta\, \partial_{x_i}u + u\,\partial_{x_i}\zeta)\,\phi \, d\mathcal{L}^n. \end{align*} Since this holds for all $\phi \in C_c^\infty(U)$, we conclude $D_i(\zeta u) = \zeta\, D_i u + u\, D_i \zeta$, confirming the Leibniz formula $D^{e_i}(\zeta u) = \sum_{\beta \le e_i} \binom{e_i}{\beta} D^\beta\zeta\, D^{e_i - \beta}u$ (the sum has two terms: $\beta = 0$ and $\beta = e_i$). **Inductive step.** Assume the Leibniz formula holds for all multi-indices of order $\le m$, and let $\gamma$ satisfy $|\gamma| = m + 1$. Write $\gamma = \alpha + e_i$ where $|\alpha| = m$. By part (i), $D^\gamma(\zeta u) = D_i(D^\alpha(\zeta u))$. The inductive hypothesis gives: \begin{align*} D^\alpha(\zeta u) = \sum_{\beta \le \alpha} \binom{\alpha}{\beta} D^\beta\zeta\, D^{\alpha - \beta}u. \end{align*} Each term $D^\beta\zeta\, D^{\alpha-\beta}u$ is a product of a smooth function ($D^\beta\zeta \in C_c^\infty(U)$) with an $L^p$ function ($D^{\alpha-\beta}u$), so we can apply the base case ($|\text{order}| = 1$) product rule and linearity (part (ii)) to obtain: \begin{align*} D_i(D^\beta\zeta\, D^{\alpha-\beta}u) = (D^{\beta+e_i}\zeta)(D^{\alpha-\beta}u) + (D^\beta\zeta)(D^{\alpha-\beta+e_i}u). \end{align*} Summing over $\beta \le \alpha$ with coefficients $\binom{\alpha}{\beta}$: \begin{align*} D^\gamma(\zeta u) = \sum_{\beta \le \alpha} \binom{\alpha}{\beta} (D^{\beta+e_i}\zeta)(D^{\alpha-\beta}u) + \sum_{\beta \le \alpha} \binom{\alpha}{\beta} (D^\beta\zeta)(D^{\gamma-\beta}u). \end{align*} In the first sum, substitute $\sigma = \beta + e_i$ (so $\sigma$ ranges over multi-indices with $e_i \le \sigma \le \alpha + e_i = \gamma$, and $\binom{\alpha}{\beta} = \binom{\alpha}{\sigma - e_i}$). In the second sum, set $\sigma = \beta$ (so $0 \le \sigma \le \alpha$). The overlapping range $e_i \le \sigma \le \alpha$ contributes from both sums with total coefficient $\binom{\alpha}{\sigma - e_i} + \binom{\alpha}{\sigma}$. By the Pascal identity for multi-index binomial coefficients, $\binom{\alpha}{\sigma - e_i} + \binom{\alpha}{\sigma} = \binom{\alpha + e_i}{\sigma} = \binom{\gamma}{\sigma}$. The boundary terms ($\sigma = 0$ from the second sum only, $\sigma = \gamma$ from the first sum only) each contribute $\binom{\gamma}{\sigma}$ as well. Combining: \begin{align*} D^\gamma(\zeta u) = \sum_{\sigma \le \gamma} \binom{\gamma}{\sigma} D^\sigma\zeta\, D^{\gamma - \sigma}u, \end{align*} which completes the induction. [/guided] **Inductive step.** Assume the formula holds for all multi-indices of order $\le m$. Let $\gamma$ be a multi-index with $|\gamma| = m + 1$. Write $\gamma = \alpha + e_i$ for some $\alpha$ with $|\alpha| = m$. By part (i), $D^\gamma(\zeta u) = D_i(D^\alpha(\zeta u))$. By the inductive hypothesis: \begin{align*} D^\alpha(\zeta u) = \sum_{\beta \le \alpha} \binom{\alpha}{\beta} D^\beta\zeta\, D^{\alpha - \beta}u. \end{align*} Applying $D_i$ to each term using linearity (part (ii)) and the base case product rule: \begin{align*} D_i(D^\beta\zeta\, D^{\alpha-\beta}u) = (D^{\beta+e_i}\zeta)(D^{\alpha-\beta}u) + (D^\beta\zeta)(D^{\alpha-\beta+e_i}u). \end{align*} Therefore: \begin{align*} D^\gamma(\zeta u) = \sum_{\beta \le \alpha} \binom{\alpha}{\beta} (D^{\beta+e_i}\zeta)(D^{\alpha-\beta}u) + \sum_{\beta \le \alpha} \binom{\alpha}{\beta} (D^\beta\zeta)(D^{\alpha+e_i-\beta}u). \end{align*} Substituting $\sigma = \beta + e_i$ in the first sum (so $\sigma$ ranges over $e_i \le \sigma \le \alpha + e_i$) and $\sigma = \beta$ in the second (so $0 \le \sigma \le \alpha$), then combining using the Pascal identity for multi-indices $\binom{\alpha}{\sigma - e_i} + \binom{\alpha}{\sigma} = \binom{\alpha + e_i}{\sigma} = \binom{\gamma}{\sigma}$: \begin{align*} D^\gamma(\zeta u) = \sum_{\sigma \le \gamma} \binom{\gamma}{\sigma} D^\sigma\zeta\, D^{\gamma - \sigma}u. \end{align*} [/step]