**(i) Iterated Derivatives** We prove that $D^\beta(D^\alpha u) = D^{\alpha+\beta}u$ in the weak sense. Let $\phi \in C_c^\infty(U)$ be an arbitrary test function. By definition, we must evaluate the integral of the proposed derivative against the test function: \begin{align*} \int_U (D^\alpha u) D^\beta \phi \, dx. \end{align*} Using the definition of the weak derivative $D^\alpha u$, we transfer the derivative $\alpha$ onto the test function $D^\beta \phi$: \begin{align*} \int_U (D^\alpha u) D^\beta \phi \, dx = (-1)^{|\alpha|} \int_U u D^\alpha(D^\beta \phi) \, dx. \end{align*} Since $\phi$ is smooth ($C_c^\infty$), classical partial derivatives commute. Thus $D^\alpha(D^\beta \phi) = D^{\alpha+\beta} \phi$. Substituting this into the integral: \begin{align*} (-1)^{|\alpha|} \int_U u D^{\alpha+\beta} \phi \, dx. \end{align*} We manipulate the sign factor. Note that $(-1)^{|\alpha|} = (-1)^{|\alpha|} \cdot 1 = (-1)^{|\alpha|} (-1)^{|\alpha+\beta|} (-1)^{|\alpha+\beta|}$. Actually, a simpler manipulation is to recognize that we want to match the definition of $D^{\alpha+\beta} u$. The definition requires: \begin{align*} \int_U u D^{\alpha+\beta} \phi \, dx = (-1)^{|\alpha+\beta|} \int_U (D^{\alpha+\beta} u) \phi \, dx. \end{align*} Substituting this back into our expression: \begin{align*} (-1)^{|\alpha|} \left[ (-1)^{|\alpha+\beta|} \int_U (D^{\alpha+\beta} u) \phi \, dx \right] &= (-1)^{|\alpha| + |\alpha| + |\beta|} \int_U (D^{\alpha+\beta} u) \phi \, dx \\ &= (-1)^{2|\alpha| + |\beta|} \int_U (D^{\alpha+\beta} u) \phi \, dx. \end{align*} Since $2|\alpha|$ is even, $(-1)^{2|\alpha|} = 1$. Thus, we arrive at: \begin{align*} \int_U (D^\alpha u) D^\beta \phi \, dx = (-1)^{|\beta|} \int_U (D^{\alpha+\beta} u) \phi \, dx. \end{align*} This is exactly the defining relation for $D^\beta(D^\alpha u)$ being equal to $D^{\alpha+\beta} u$. Since $D^{\alpha+\beta} u \in L^p(U)$ (by hypothesis that $u \in W^{k,p}$ and total order is $\le k$), the derivative exists in $L^p$.

**(ii) Linearity** Let $\phi \in C_c^\infty(U)$. We compute the distributional derivative of the linear combination: \begin{align*} \int_U (\lambda u + \mu v) D^\alpha \phi \, dx &= \lambda \int_U u D^\alpha \phi \, dx + \mu \int_U v D^\alpha \phi \, dx \quad (\text{Linearity of integral}) \\ &= \lambda (-1)^{|\alpha|} \int_U (D^\alpha u) \phi \, dx + \mu (-1)^{|\alpha|} \int_U (D^\alpha v) \phi \, dx \quad (\text{Def. of weak derivative}) \\ &= (-1)^{|\alpha|} \int_U (\lambda D^\alpha u + \mu D^\alpha v) \phi \, dx. \end{align*} Thus, $\lambda D^\alpha u + \mu D^\alpha v$ satisfies the definition of the weak derivative $D^\alpha(\lambda u + \mu v)$.

**(iii) Restriction** Let $V \subset U$ be open. Let $\phi \in C_c^\infty(V)$. We define the zero extension $\tilde{\phi}$ on $U$ by: \begin{align*}

\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 $V$, $\tilde{\phi} \in C_c^\infty(U)$. Since $u \in W^{k,p}(U)$, we have: \begin{align*} \int_U u D^\alpha \tilde{\phi} \, dx = (-1)^{|\alpha|} \int_U (D^\alpha u) \tilde{\phi} \, dx. \end{align*} Restricting the integrals to the support of $\phi$ (which is inside $V$): \begin{align*} \int_V u D^\alpha \phi \, dx = (-1)^{|\alpha|} \int_V (D^\alpha u) \phi \, dx. \end{align*} Thus, $D^\alpha u |_V$ is the weak derivative of $u|_V$ on the set $V$. **(iv) Leibniz' Formula** We proceed by induction on the order of the derivative $|\alpha|$. **Base Case ($|\alpha|=1$):** Let $\alpha = e_i$. Let $\phi \in C_c^\infty(U)$. We want to find $w$ such that $\int \zeta u \phi_{x_i} = - \int w \phi$. Note that for smooth functions, $(\zeta \phi)_{x_i} = \zeta_{x_i} \phi + \zeta \phi_{x_i}$, which implies $\zeta \phi_{x_i} = (\zeta \phi)_{x_i} - \zeta_{x_i} \phi$. Since $\zeta \in C_c^\infty(U)$, the product $\zeta \phi$ is also in $C_c^\infty(U)$. \begin{align*} \int_U (\zeta u) \phi_{x_i} \, dx &= \int_U u (\zeta \phi_{x_i}) \, dx \\ &= \int_U u [(\zeta \phi)_{x_i} - \zeta_{x_i} \phi] \, dx \\ &= \int_U u (\zeta \phi)_{x_i} \, dx - \int_U u \zeta_{x_i} \phi \, dx. \end{align*} Applying the definition of the weak derivative $u_{x_i}$ to the first term (with test function $\psi = \zeta \phi$): \begin{align*} \int_U u (\zeta \phi)_{x_i} \, dx = - \int_U u_{x_i} (\zeta \phi) \, dx = - \int_U (\zeta u_{x_i}) \phi \, dx. \end{align*} Substituting this back: \begin{align*} \int_U \zeta u \phi_{x_i} \, dx &= - \int_U (\zeta u_{x_i}) \phi \, dx - \int_U (u \zeta_{x_i}) \phi \, dx \\ &= - \int_U (\zeta u_{x_i} + u \zeta_{x_i}) \phi \, dx. \end{align*} Thus $D_i(\zeta u) = \zeta D_i u + u D_i \zeta$, which matches the formula for $|\alpha|=1$. **Induction Step:** Assume the formula holds for all multiindices $\alpha$ with $|\alpha| \le m$. Let $\gamma$ be a multiindex with $|\gamma| = m+1$. We can 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 induction hypothesis: \begin{align*} D^\alpha(\zeta u) = \sum_{\beta \le \alpha} \binom{\alpha}{\beta} D^\beta \zeta D^{\alpha-\beta} u. \end{align*} Now we apply $D_i$ to this sum. By linearity (part ii) and the base case product rule (since the terms in the sum are products of functions in $W^{k,p}$ and $C_c^\infty$): \begin{align*} D^\gamma(\zeta u) &= \sum_{\beta \le \alpha} \binom{\alpha}{\beta} D_i \left( D^\beta \zeta D^{\alpha-\beta} u \right). \end{align*} Applying the base case product rule to each term $A \cdot B$ where $A = D^\beta \zeta$ and $B = D^{\alpha-\beta} u$: \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*} Substituting this back into the sum: \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*} We shift the index in the first sum. Let $\sigma = \beta + e_i$. As $\beta$ ranges over $0 \dots \alpha$, $\sigma$ ranges over $e_i \dots \alpha+e_i$. \begin{align*} \text{Sum 1} = \sum_{e_i \le \sigma \le \alpha+e_i} \binom{\alpha}{\sigma-e_i} D^\sigma \zeta D^{\alpha+e_i-\sigma} u. \end{align*} In the second sum, let $\sigma = \beta$. \begin{align*} \text{Sum 2} = \sum_{0 \le \sigma \le \alpha} \binom{\alpha}{\sigma} D^\sigma \zeta D^{\alpha+e_i-\sigma} u. \end{align*} We combine these sums to form the sum over $0 \le \sigma \le \alpha+e_i = \gamma$. We use the Pascal identity for multiindices: $\binom{\alpha}{\sigma} + \binom{\alpha}{\sigma-e_i} = \binom{\alpha+e_i}{\sigma} = \binom{\gamma}{\sigma}$. (Note: If $\sigma = 0$, the first term is zero. If $\sigma = \gamma$, the second term is zero. The identity holds essentially everywhere). \begin{align*} D^\gamma(\zeta u) = \sum_{\sigma \le \gamma} \binom{\gamma}{\sigma} D^\sigma \zeta D^{\gamma-\sigma} u. \end{align*} This completes the induction.