[proofplan] We prove the weak derivative identity against an arbitrary [test function](/page/Test%20Function) $\phi \in C_c^\infty(U)$. Since $\phi$ has compact support in $U$, we localize the computation to finitely many rectangular boxes compactly contained in $U$, where the iterated-integral form of [Fubini's theorem](/theorems/2961) and one-dimensional [integration by parts](/theorems/210) are legitimate. Each [integration by parts](/theorems/2098) transfers one classical derivative from $u$ to $\phi$, and all boundary terms vanish because the localized test functions have compact support inside the relevant boxes. Summing the localized identities gives the desired identity on $U$. [/proofplan] [step:Localize the test function to boxes compactly contained in $U$] Fix $\phi \in C_c^\infty(U)$. Let $\mathcal{L}^n$ denote $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}^n$. Define the compact support of $\phi$ by \begin{align*} K := \operatorname{supp}\phi \subset U. \end{align*} Since $K$ is compact and $U$ is open, there are finitely many open rectangular boxes $Q_1,\dots,Q_N \subset \mathbb{R}^n$ such that $\overline{Q_j} \subset U$ for each $j \in \{1,\dots,N\}$ and \begin{align*} K \subset \bigcup_{j=1}^N Q_j. \end{align*} Apply the support-contained form of the [Existence of Partitions of Unity](/theorems/57) to the open cover $Q_1,\dots,Q_N,U\setminus K$ of $U$; the theorem applies because $U \subset \mathbb{R}^n$ is open and these sets form an open cover. Let $\rho_1,\dots,\rho_N,\rho_0 \in C^\infty(U)$ be the resulting smooth [partition of unity](/page/Partition%20of%20Unity), with $\operatorname{supp}\rho_j \subset Q_j$ for $j \in \{1,\dots,N\}$ and $\operatorname{supp}\rho_0 \subset U\setminus K$. Since $\operatorname{supp}\rho_0$ is disjoint from $K$, the function $\rho_0$ vanishes on a neighbourhood of $K$, and therefore \begin{align*} \sum_{j=1}^N \rho_j(x) = 1 \end{align*} for every $x$ in a neighbourhood of $K$. For each $j$, define the function $\phi_j: U \to \mathbb{R}$ by \begin{align*} \phi_j(x) := \rho_j(x)\phi(x) \end{align*} for $x \in U$. Then $\phi_j \in C_c^\infty(Q_j)$ after extending by zero outside $Q_j$, and \begin{align*} \phi = \sum_{j=1}^N \phi_j \end{align*} on $U$. [guided] The goal is to prove the weak derivative identity for the arbitrary test function $\phi \in C_c^\infty(U)$, where $\phi: U \to \mathbb{R}$ is smooth with compact support. Let $\mathcal{L}^n$ denote $n$-dimensional Lebesgue measure on $\mathbb{R}^n$. Define \begin{align*} K := \operatorname{supp}\phi. \end{align*} The set $K$ is compact and satisfies $K \subset U$. Since $U$ is open, every point of $K$ lies in an open rectangular box whose closure is contained in $U$; compactness of $K$ gives finitely many such boxes $Q_1,\dots,Q_N \subset \mathbb{R}^n$ with $\overline{Q_j} \subset U$ and \begin{align*} K \subset \bigcup_{j=1}^N Q_j. \end{align*} This localization lets us use product-coordinate integration by parts while keeping all derivatives of $u$ defined on neighbourhoods of the closed boxes. Apply the support-contained form of the [Existence of Partitions of Unity](/theorems/57) to the open cover $Q_1,\dots,Q_N,U\setminus K$ of $U$. The hypotheses of the theorem are satisfied: $U \subset \mathbb{R}^n$ is open, each $Q_j$ is open, $U\setminus K$ is open because $K$ is closed, and the family covers $U$ because the boxes cover $K$. Hence there are smooth functions $\rho_1,\dots,\rho_N,\rho_0 \in C^\infty(U)$ forming a partition of unity subordinate to this cover. In particular, $\operatorname{supp}\rho_j \subset Q_j$ for $j \in \{1,\dots,N\}$ and $\operatorname{supp}\rho_0 \subset U\setminus K$. Since $\operatorname{supp}\rho_0$ is disjoint from $K$, the function $\rho_0$ vanishes on a neighbourhood of $K$, so \begin{align*} \sum_{j=1}^N \rho_j(x) = 1 \end{align*} for every $x$ in a neighbourhood of $K$. For each $j$, define the function $\phi_j: U \to \mathbb{R}$ by \begin{align*} \phi_j(x) := \rho_j(x)\phi(x) \end{align*} for $x \in U$. Then $\phi_j$ is smooth, compactly supported in $Q_j$, and extends by zero to an element of $C_c^\infty(Q_j)$. Since $\sum_{j=1}^N \rho_j = 1$ near $K$ and $\phi$ vanishes outside $K$, we have \begin{align*} \phi = \sum_{j=1}^N \phi_j \end{align*} on all of $U$. Now fix $j \in \{1,\dots,N\}$ and write \begin{align*} Q_j = (a_1,b_1) \times \cdots \times (a_n,b_n). \end{align*} Because $\phi_j \in C_c^\infty(Q_j)$ and $u \in C^{|\alpha|}(U)$, the functions $uD^\alpha\phi_j$ and $D^\alpha u\,\phi_j$ are continuous on the compact support of $\phi_j$ and vanish outside a compact subset of $Q_j$. Hence they are integrable with respect to $\mathcal{L}^n$ on $Q_j$, and the standard iterated-integral form of [Fubini's theorem](/theorems/2961) applies on the rectangle $Q_j$. Applying one-dimensional [integration by parts](/theorems/210) successively in each coordinate direction transfers exactly $\alpha_i$ derivatives in the $x_i$ direction from $\phi_j$ onto $u$. Every boundary term at $x_i=a_i$ and $x_i=b_i$ is zero because $\phi_j$ and its derivatives are compactly supported inside $Q_j$, so they vanish near the boundary hyperfaces. The total number of integrations by parts is $|\alpha|$, and therefore the accumulated sign is $(-1)^{|\alpha|}$: \begin{align*} \int_{Q_j} u(x)D^\alpha \phi_j(x)\,d\mathcal{L}^n(x) = (-1)^{|\alpha|} \int_{Q_j} D^\alpha u(x)\phi_j(x)\,d\mathcal{L}^n(x). \end{align*} Since $\phi_j$ and $D^\alpha\phi_j$ vanish outside $Q_j$, the same identity may be written over $U$: \begin{align*} \int_U u(x)D^\alpha \phi_j(x)\,d\mathcal{L}^n(x) = (-1)^{|\alpha|} \int_U D^\alpha u(x)\phi_j(x)\,d\mathcal{L}^n(x). \end{align*} Summing this identity over $j \in \{1,\dots,N\}$ and using linearity of the [Lebesgue integral](/page/Lebesgue%20Integral) gives \begin{align*} \int_U u(x)D^\alpha\left(\sum_{j=1}^N \phi_j\right)(x)\,d\mathcal{L}^n(x) = (-1)^{|\alpha|} \int_U D^\alpha u(x)\left(\sum_{j=1}^N \phi_j(x)\right)\,d\mathcal{L}^n(x). \end{align*} Using $\sum_{j=1}^N \phi_j = \phi$ on $U$, we obtain \begin{align*} \int_U u(x)D^\alpha \phi(x)\,d\mathcal{L}^n(x) = (-1)^{|\alpha|} \int_U D^\alpha u(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} This is the defining identity for $D^\alpha u$ to be the [weak derivative](/page/Weak%20Derivative) of $u$ of order $\alpha$. [/guided] [/step] [step:Prove the identity on one rectangular box by repeated integration by parts] Fix $j \in \{1,\dots,N\}$, and write \begin{align*} Q_j = (a_1,b_1) \times \cdots \times (a_n,b_n). \end{align*} Since $\phi_j \in C_c^\infty(Q_j)$ and $u \in C^{|\alpha|}(U)$, all products appearing below are integrable with respect to $\mathcal{L}^n$ on $Q_j$: the support of $\phi_j$ is compactly contained in $Q_j$, $D^\beta\phi_j$ is bounded for every multi-index $\beta$ with $|\beta| \le |\alpha|$, and $D^\gamma u$ is continuous on neighbourhoods of that compact support for every multi-index $\gamma$ with $|\gamma| \le |\alpha|$. Hence the iterated-integral form of [Fubini's theorem](/theorems/2961) applies on the rectangle $Q_j$. We then apply one-dimensional [integration by parts](/theorems/210) successively in the coordinate directions. For each coordinate $x_i$, every relevant slice has compact support away from the endpoints $a_i$ and $b_i$, so the boundary contribution at $x_i=a_i$ and $x_i=b_i$ is zero. After applying integration by parts $\alpha_i$ times in the $x_i$ direction for each $i \in \{1,\dots,n\}$, the total number of transferred derivatives is $|\alpha|$, and hence the accumulated sign is $(-1)^{|\alpha|}$. Therefore \begin{align*} \int_{Q_j} u(x)D^\alpha \phi_j(x)\,d\mathcal{L}^n(x) = (-1)^{|\alpha|} \int_{Q_j} D^\alpha u(x)\phi_j(x)\,d\mathcal{L}^n(x). \end{align*} [/step] [step:Sum the localized identities and recover the integral over $U$] Because $\phi_j$ and $D^\alpha\phi_j$ vanish outside $Q_j$, we may write the previous identity as an identity over $U$: \begin{align*} \int_U u(x)D^\alpha \phi_j(x)\,d\mathcal{L}^n(x) = (-1)^{|\alpha|} \int_U D^\alpha u(x)\phi_j(x)\,d\mathcal{L}^n(x). \end{align*} Summing over $j \in \{1,\dots,N\}$ and using the linearity of the Lebesgue integral gives \begin{align*} \int_U u(x)D^\alpha\left(\sum_{j=1}^N \phi_j\right)(x)\,d\mathcal{L}^n(x) = (-1)^{|\alpha|} \int_U D^\alpha u(x)\left(\sum_{j=1}^N \phi_j(x)\right)\,d\mathcal{L}^n(x). \end{align*} Since $\sum_{j=1}^N \phi_j = \phi$ on $U$, this becomes \begin{align*} \int_U u(x)D^\alpha \phi(x)\,d\mathcal{L}^n(x) = (-1)^{|\alpha|} \int_U D^\alpha u(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} This is exactly the defining identity for $D^\alpha u$ to be the [weak derivative](/page/Weak%20Derivative) of $u$ of order $\alpha$. [/step]