> [!definition] > > Let $U \subset \real^n$ be an [[Open Set|open set]], $\phi \in \cd(U)$ be a [[Space of Test Functions|test function]], $T \in \cd'(U)$ be a [[Distribution|distribution]], and $\alpha$ be a [[Multi-Index|multi-index]]. The **distributional derivative** $D^\alpha T$ is the distribution > $ > \angles{D^\alpha T, \phi} = (-1)^{\alpha}\angles{T, D^\alpha \phi} > $ > The vector $\nabla T = (D^{e_1}T, \cdots, D^{e_d}T)$ is the **distributional gradient** of $T$, where > $ > \angles{\nabla T, \phi} = (\angles{D^{e_1}T, \phi}, \cdots, \angles{D^{e_d}T, \phi}) > $ > > *Proof*. Let $\seq{\phi_n} \subset \cd(U)$ such that $\phi_n \to \phi \in \cd(U)$. Let $K \subset U$ be [[Compactness|compact]] such that $\supp{\phi_n}, \supp{\phi} \subset K$ for each $n \in \nat$, then $\norm{\phi_n - \phi}_{K, \beta} \to 0$ for all $\beta$. > > From here, $\supp{D^\alpha \phi_n}, \supp{D^\alpha \phi} \subset K$ for each $n \in \nat$, so > $ > \norm{D^\alpha \phi_n - D^\alpha \phi}_{K, \beta} = \norm{\phi_n - \phi}_{K, \beta + \alpha} \to 0 > $ > for all $\beta$. So by continuity of $T$, > $ > \angles{D^\alpha T, \phi_n} = \angles{T, D^\alpha \phi_n} \to \angles{T, D^\alpha \phi} = \angles{D^\alpha T, \phi} > $ # Weak Derivatives > [!theorem] > > Let $U \subset \real^n$ be an open subset, $f \in C^1(U)$ and $\phi \in C^\infty_c(U)$, then > $ > \int_U f \cdot \frac{\partial \phi}{\partial x_i} dx = - \int_U \frac{\partial f}{\partial x_i} \cdot \phi dx > $ > If $\alpha$ is a [[Multi-Index|multi-index]] and $f \in C^{\abs{\alpha}}$, then > $ > \int_U f \cdot D^\alpha \phi dx = (-1)^{\abs{\alpha}}\int_U D^\alpha f \cdot \phi dx > $ > [!definition] > > Let $u, v \in \loci$ be [[Locally Integrable|locally integrable]] functions. If for every $\phi \in C_c^\infty(U)$, > $ > \int_U uD^\alpha \phi = (-1)^{\abs{\alpha}}\int v\phi \quad \angles{u, D^\alpha \phi} = (-1)^{\alpha}\angles{D^\alpha u, \phi} > $ > then $v$ is the $\alpha$-th **weak partial derivative** of $u$, and $D^\alpha u = v$, which is unique [[Almost Everywhere|almost everywhere]]. > > *Proof*. Let $v_1, v_2 \in \loci$ such that $D^\alpha u = v_1$ and $D^\alpha u = v_2$, then > $ > \int_U(v_1 - v_2)\phi = 0 \quad \forall \phi \in C_c^\infty(U) > $ > Hence $v_1 - v_2 = 0$ almost everywhere. > [!theorem] > > Let $u, v \in W^{k, p}(U)$ and $\abs{\alpha} \le k$, then > 1. $D^\alpha u \in W^{k - \abs{\alpha}, p}(U)$ and $D^\beta D^\alpha u = D^\alpha D^\beta u = D^{\alpha + \beta} u$ for all $\alpha, \beta$ with $\abs{\alpha} + \abs{\beta} \le k$. > 2. For each $\lambda \in \real$, $\lambda u + v \in W^{k, p}(U)$ and $D^\alpha(\lambda u + v) = \lambda D^\alpha u + D^\alpha v$. > 3. If $V \subset U$ is open, then $u \in W^{k, p}(V)$. > 4. If $\zeta \in C_c^\infty(U)$, then $\zeta u \in W^{k, p}(U)$ and > > $ > D^\alpha(\zeta u) = \sum_{\beta \le \alpha}{\alpha \choose \beta}D^\beta \zeta \cdot D^{\alpha - \beta}u > $ > *Proof*. Let $\phi \in C^\infty(U)$, then $D^\beta \phi \in C^\infty(U)$ as well and > $ > \begin{align*} > \int_U (D^{\alpha}u) \cdot (D^\beta \phi) &= (-1)^{\abs{\alpha}}\int_U u \cdot (D^{\alpha + \beta}\phi) \\ > &= (-1)^{\abs{\alpha} + \abs{\alpha} + \abs{\beta}}\int_U D^{\alpha + \beta}u \cdot \phi \\ > &= (-1)^{\abs{\beta}}\int_U D^{\alpha + \beta}u \cdot \phi > \end{align*} > $ > Hence $D^{\alpha}u \in W^{k - \abs{\alpha}, p}(U)$ with the partials adding up. > > For $(4)$, first suppose that $\abs{\alpha} = 1$. Let $\phi \in C_c^\infty$, then by expanding through the product rule, > $ > \begin{align*} > \int_U \zeta u D^\alpha \phi &= \int uD^\alpha(\zeta \phi) - u (D^\alpha \zeta)\phi \\ > &= -\int_U \zeta \phi D^\alpha u - \int_U u (D^{\alpha}\zeta) \phi \\ > &= -\int_U [\zeta D^\alpha u + uD^\alpha \zeta] \phi > \end{align*} > $ > > Now suppose that the proposition holds for all $\alpha$ with $\abs{\alpha} \le l$. Let $\beta$ be a multi-index with $\abs{\beta} = l + 1$, and let $\alpha$ and $e_j$ such that $\beta = \alpha + e_j$. Let $\phi \in C_c^\infty$, then > $ > \begin{align*} > \int_U \zeta u D^{\alpha}\phi = \int_U \zeta u D^\alpha(D^{e_j}\phi) &= (-1)^{\abs{\alpha}}\int_U\sum_{\sigma \le \alpha}{\alpha \choose \sigma}D^{\sigma}\zeta D^{\alpha - \sigma} u(D^{e_j}\phi) \\ > &= (-1)^{\abs{\alpha} + 1}\int_U\sum_{\sigma \le \alpha}{\alpha \choose \sigma}D^{e_j}\braks{D^{\sigma}\zeta D^{\alpha - \sigma} u}\phi > \end{align*} > $ > by the inductive assumption. Expanding the base case through [[Multi-Index Binomial Detail]] yields > $ > \begin{align*} > \int_U \zeta u D^{\alpha}\phi &= (-1)^{\abs{\alpha} + 1}\int_U\sum_{\sigma \le \alpha}{\alpha \choose \sigma}\braks{D^{\sigma + e_j}\zeta D^{\alpha - \sigma} u + D^{\sigma}\zeta D^{(\alpha + e_j) - \sigma} u}\phi \\ > &= (-1)^{\abs{\alpha} + 1}\int_U\sum_{\sigma \le \alpha}{\alpha\choose \sigma}\paren{D^{\sigma + e_j}\zeta D^{\alpha - \sigma} u}\phi \\ > &+ (-1)^{\abs{\alpha} + 1}\int_U\sum_{\sigma \le \alpha}{\alpha\choose \sigma}\paren{D^{\sigma}\zeta D^{(\alpha + e_j) - \sigma} u}\phi \\ > &= (-1)^{\alpha + e_j}\int_U \sum_{\sigma \le \alpha + e_j}{\alpha\choose \sigma}\paren{D^{\sigma}\zeta D^{(\alpha + e_j) - \sigma} u}\phi > \end{align*} > $