Let $u_1, \dots, u_d, w \in H^1(\mathbb{R}^d).$ We set $u:=(u_1, \dots, u_d)^{\top}.$ Is the following identity true for the whole space? (If we assume the divergence theorem for Sobolev functions on bounded smooth-boundary sets.) $$ \int_{\mathbb{R}^d} \mathrm{div}(u)w \, \mathrm{d}x = -\int_{\mathbb{R}^d} u^{\top} \nabla w \, \mathrm{d}x. $$ I think so because: \begin{align*} \int_{\mathbb{R}^d} \mathrm{div}(u)w \, dx &= \lim\limits_{R \to +\infty} \int_{B_R(0)} \mathrm{div}(u)w \, dx = \lim\limits_{R \to +\infty} \left[-\int_{B_R(0)} u^{\top} \nabla w \, dx + \int_{\partial B_R(0)} (u^{\top} n) w\, ds \right] \\ &= -\int_{\mathbb{R}^d} u^{\top} \nabla w \, dx + \lim\limits_{R \to +\infty} \int_{\partial B_R(0)} (u^{\top} n) w\, ds. \end{align*}
In order for the boundary integral to vanish, the corresponding integrand must converge to $0$ fast enough. Adequate conditions must be placed on the functions $u$ and $v$ for this. A sufficient condition would be $|u(x)||v(x)| \in o(\Vert x \Vert^{1-d})$ for $\Vert x \Vert \to + \infty,$ because for the $d$-dimensional volume of the unit sphere $\mathrm{vol}_d( \partial B_R(0)) = c_d R^{d-1}$ (the real constant $c_d > 0$ depends only on the dimension $d \in \mathbb{N}$), this is certainly for Schwartz -Functions $u,v \in \mathcal{S}({\mathbb{R}^d})$ fulfilled. However, we are dealing with Sobolev functions, fortunately the boundary integral also vanishes if $u,v \in H^1(\mathbb{R}^d)$ holds, since in this case we could argue by density of the Schwartz space in $H^1(\mathbb{R}^d)$.
Could anyone confirm this?