The distributional gradient of a locally integrable function smooth off one point

Set $F:={\mathrm{\partial }}_{i}u\cdot \psi$ and $H:=u\cdot {\mathrm{\partial }}_i\psi$ on ${\mathbb{R}}^n$. By [F4] replace $u$ and each ${\mathrm{\partial }}_{i}u$ at $p$ by $0$ without changing their local integrability classes or integrals. On $O:={\mathbb{R}}^n\setminus \{p\}$ these functions, and the real and imaginary parts of $F,H$, are continuous. For any one of these real functions $v$ extended by $0$ at $p$ and every real $a$, the set $\{v>a\}\cap O$ is open in $O$, hence open in ${\mathbb{R}}^n$; the full superlevel set is this open set with $\{p\}$ added when $0>a$. It is therefore Borel, so the real and imaginary parts of $F,H$ are Borel measurable. Choose a ball containing the compact support of $\psi$ and its derivatives. On it $\mathrm{\mid }F\mathrm{\mid }\le \mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }{\mathrm{\partial }}_{i}u\mathrm{\mid }$ and $\mathrm{\mid }H\mathrm{\mid }\le \mathrm{\parallel }{\mathrm{\partial }}_i\psi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }u\mathrm{\mid }$, and both products vanish outside it; monotonicity and homogeneity yield $F,H\in L^1({\lambda }_n)$.

Let $\pi :{\mathbb{R}}^n\to {\mathbb{R}}^n$ be the coordinate permutation moving the $i$-th coordinate to the last position, $\pi (x)=(x_1,\dots ,x_{i-1},x_{i+1},\dots ,x_n,x_i)$. Then $\mathrm{\parallel }\pi (x){\mathrm{\parallel }}^2={\sum }_j{x}_j^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2$, so $\pi$ is a linear isometry, hence an orthogonal operator with $\mathrm{\mid }\det D\pi \mathrm{\mid }=\mathrm{\mid }\det \pi \mathrm{\mid }=1$ ([F9]); $\pi$ is a $C^1$ diffeomorphism of ${\mathbb{R}}^n$ onto itself. Applying the change-of-variables identity of [F9] to $F$ and to $H$, and using $\mathrm{\mid }\det D\pi \mathrm{\mid }=1$, gives ${\int }_{{\mathbb{R}}^n}Fd{\lambda }_n={\int }_{{\mathbb{R}}^n}F\circ {\pi }^{-1}d{\lambda }_n,{\int }_{{\mathbb{R}}^n}Hd{\lambda }_n={\int }_{{\mathbb{R}}^n}H\circ {\pi }^{-1}d{\lambda }_n\mathrm{.}$ The compositions $F\circ {\pi }^{-1}$ and $H\circ {\pi }^{-1}$ are Borel and in $L^1({\lambda }_n)$ because $\int \mathrm{\mid }F\circ {\pi }^{-1}\mathrm{\mid }d{\lambda }_n=\int \mathrm{\mid }F\mathrm{\mid }d{\lambda }_n<\mathrm{\infty }$ and likewise for $H$.

Identify ${\mathbb{R}}^n$ with ${\mathbb{R}}^{n-1}\times \mathbb{R}$ by the standard ordering, so that a point with last coordinate $t$ and remaining coordinates $y$ is written $(y,t)$; the permutation of step 2.1 then satisfies $\pi (x)=(y(x),x_i)$ where $y(x)$ lists the coordinates other than the $i$-th. Define $G(y,t):=F(x(y,t))$ and $K(y,t):=H(x(y,t))$, where $x(y,t)\in {\mathbb{R}}^n$ is the point with $i$-th coordinate $t$ and remaining coordinates $y$. Then $G=F\circ {\pi }^{-1}$ and $K=H\circ {\pi }^{-1}$ under the identification, so step 2.1 reads $\int Fd{\lambda }_n=\int Gd{\lambda }_n$ and $\int Hd{\lambda }_n=\int Kd{\lambda }_n$; by [F7] and [F8] these equal the repeated integrals ${\int }_{{\mathbb{R}}^{n-1}}({\int }_{\mathbb{R}}G(y,t)dt)dy$ and ${\int }_{{\mathbb{R}}^{n-1}}({\int }_{\mathbb{R}}K(y,t)dt)dy$, the inner sections being integrable for almost every $y$.

Fix $y\in {\mathbb{R}}^{n-1}$ such that the line $L_y=\{x(y,t):t\in \mathbb{R}\}$ does not pass through $p$; the excluded set of $y$ is the single point $y^{\ast }$ listing the coordinates of $p$ other than $p_i$, which is a degenerate box and hence ${\lambda }_{n-1}$-null ([F4]). For such $y$, the map $t\mapsto u(x(y,t))$ is $C^1$ on $\mathbb{R}$ with derivative $t\mapsto {\mathrm{\partial }}_{i}u(x(y,t))$ ([F11]), and $t\mapsto \psi (x(y,t))$ is smooth with compact support ([F3]). Choose $a<b$ with the support of $t\mapsto \psi (x(y,t))$ contained in $(a,b)$. Then $v(t):=u(x(y,t))\psi (x(y,t))$ is differentiable at every point of $[a,b]$ with continuous derivative $v^{\mathrm{\prime }}={\mathrm{\partial }}_{i}u\psi +u{\mathrm{\partial }}_i\psi$ there, so $v^{\mathrm{\prime }}$ is integrable in the Darboux sense ([F10]); applying the real integration-by-parts formula of [F10] to the real and imaginary parts of both factors on $[a,b]$, combining the resulting identities bilinearly and using that the boundary terms vanish gives ${\int }_{\mathbb{R}}G(y,t)dt=-{\int }_{\mathbb{R}}K(y,t)dt,$ where the Darboux integrals have been replaced by the equal Lebesgue integrals of the continuous compactly supported integrands ([F10]).

The equality of step 4.1 holds for every $y$ outside the null set $\{y^{\ast }\}$ ([F4]), hence almost everywhere, and the section functions are integrable there by step 3.1. Integrating the identity over ${\mathbb{R}}^{n-1}$ and using the linearity of the integral gives $\int Fd{\lambda }_n=\int Gd{\lambda }_n=-\int Kd{\lambda }_n=-\int Hd{\lambda }_n$. Therefore, by [F1] and [F2], $⟨{\mathrm{\partial }}_i{T}_u,\psi ⟩=-⟨T_u,{\mathrm{\partial }}_i\psi ⟩=-{\int }_{{\mathbb{R}}^n}u{\mathrm{\partial }}_i\psi d{\lambda }_n=-{\int }_{{\mathbb{R}}^n}Hd{\lambda }_n={\int }_{{\mathbb{R}}^n}Fd{\lambda }_n={\int }_{{\mathbb{R}}^n}{\mathrm{\partial }}_{i}u\psi d{\lambda }_n=⟨T_{{\mathrm{\partial }}_{i}u},\psi ⟩\mathrm{.}$ Since $\psi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ was arbitrary, ${\mathrm{\partial }}_i{T}_u=T_{{\mathrm{\partial }}_{i}u}$ in ${\mathcal{D}}^{\mathrm{\prime }}({\mathbb{R}}^n)$, as claimed.