Fractional weak Green identity and interior flux cancellation

Choose a whole-space extension $Q\in H^r({\mathbb{R}}^d;{\mathbb{R}}^d)$ of $q$, componentwise by [F1]. For $a$ in the interior of $K$ and $0<t<1$, set $A_{t}x=a+t(x-a)$. For every Schwartz vector field, the affine change-of-variables formula in [F5] and dominated convergence give $Q\circ A_t\to Q$ in $H^r({\mathbb{R}}^d)$ as $t\uparrow 1$; the dilation operators are uniformly bounded for $1\mathrm{/}2\le t\le 1$ because their Fourier norm is $t^{-d}\int ⟨t\xi {⟩}^{2r}\mathrm{\mid }\widehat{Q}(\xi ){\mathrm{\mid }}^{2}d\xi \le 2^d\mathrm{\parallel }Q{\mathrm{\parallel }}_{H^r}^2$. Density of Schwartz space in the Bessel completion extends this convergence to every $Q$.

Let $f_0$ be the zero extension of $\mathrm{div}q$ from $K$ to ${\mathbb{R}}^d$. Strong continuity of affine dilations in $L^2$, first for smooth compactly supported functions and then by density and the uniform change-of-variables bound in [F5], gives $t{f}_0\circ A_t\to f_0$ in $L^2({\mathbb{R}}^d)$. Since $A_t(\bar{K})$ is compactly contained in $K$, there is a bounded open neighborhood $N_t\supset \bar{K}$ with $A_t(N_t)\subset K$. On $N_t$, $Q\circ A_t=q\circ A_t$ almost everywhere, because [F5] sends the null set where $Q$ and $q$ differ to a null preimage under the invertible affine map. This divergence identity follows directly without assuming that $q$ has all first weak derivatives in $L^2$: for $\varphi \in C_c^{\mathrm{\infty }}(N_t)$ put $\psi =\varphi \circ A_t^{-1}\in C_c^{\mathrm{\infty }}(K)$. Apply [F5] to the positive and negative parts of each integrable component product and use $\mathrm{\nabla }\varphi =t(\mathrm{\nabla }\psi )\circ A_t$ to obtain ${\int }_{N_t}(Q\circ A_t)\cdot \mathrm{\nabla }\varphi =t^{1-d}{\int }_{K}q\cdot \mathrm{\nabla }\psi =-t^{1-d}{\int }_K(\mathrm{div}q)\psi =-{\int }_{N_t}t(\mathrm{div}q)\circ A_t\varphi$. Thus $\mathrm{div}(Q\circ A_t)=t(\mathrm{div}q)\circ A_t$ distributionally.

For fixed $t<1$, let $q_{t,ϵ}:={\rho }_{ϵ}\ast (Q\circ A_t)$ for a nonnegative smooth unit-mass mollifier supported in the unit ball. Since $Q\circ A_t\in H^r\subset {\mathcal{S}}^{\mathrm{\prime }}$ and ${\rho }_{ϵ}\in \mathcal{S}$, [F6] gives $\widehat{q_{t,ϵ}}=\widehat{Q\circ A_t}\widehat{{\rho }_{ϵ}}$. Nonnegativity and unit mass give $\mathrm{\mid }\widehat{{\rho }_{ϵ}}(\xi )\mathrm{\mid }\le 1$, while its defining integral and dominated convergence give $\widehat{{\rho }_{ϵ}}(\xi )\to 1$ for each $\xi$. The exact weighted Fourier characterization in [F1] and dominated convergence therefore give $q_{t,ϵ}\to Q\circ A_t$ in $H^r({\mathbb{R}}^d)$. Choose $N_t$ bounded with $\bar{K}\subset N_t$ and $A_t(N_t)\subset K$, as in step 2.1, and put $f_t:=t(\mathrm{div}q)\circ A_t\in L^2(N_t)$. If $ϵ<\mathrm{dist}(\bar{K},N_t^c)$, then for $x\in K$ the convolution kernel $y\mapsto {\rho }_{ϵ}(x-y)$ is supported in $N_t$. Differentiating the smooth convolution under the integral and using the distributional identity ${\int }_{N_t}(Q\circ A_t)\cdot \mathrm{\nabla }\varphi =-{\int }_{N_t}{f}_t\varphi$ with $\varphi (y)={\rho }_{ϵ}(x-y)$ gives, componentwise and after summing, $\mathrm{div}{q}_{t,ϵ}(x)=-{\int }_{N_t}(Q\circ A_t)(y)\cdot {\mathrm{\nabla }}_y{\rho }_{ϵ}(x-y)dy={\int }_{N_t}{f}_t(y){\rho }_{ϵ}(x-y)dy$. Extend $f_t$ by zero from $N_t$ to ${\mathbb{R}}^d$; [F5] gives ${\rho }_{ϵ}\ast f_t\to f_t$ in $L^2({\mathbb{R}}^d)$, so these divergences converge to $f_t$ in $L^2(K)$. Choose $t_n\uparrow 1$ and then sufficiently small ${ϵ}_n$ so that $q_n:=q_{t_n,{ϵ}_n}$ is smooth and $q_n\to q$ in $H^r(K)$ while $\mathrm{div}{q}_n\to \mathrm{div}q$ in $L^2(K)$, using steps 1.1 and 2.1 for the two limits. The countable selection is permitted by [A1].

For a smooth $v$ on a neighbourhood of $\bar{K}$, apply the classical affine-simplex Gauss–Green formula [F3] to each smooth field $q_{n}v$. The volume terms converge because $q_n\to q$ in $L^2(K)$ and $\mathrm{div}{q}_n\to \mathrm{div}q$ in $L^2(K)$; the facet terms converge because the $H^r$ trace in [F1] is continuous for $r>1\mathrm{/}2$. This proves the displayed identity for such smooth $v$.

For arbitrary $v\in H^1(K)$, affine pullback of the published smooth-density result in [F2] gives smooth $v_m\to v$ in $H^1(K)$. The volume integrals in the identity for $v_m$ converge by $L^2$ Cauchy–Schwarz, and their facet traces converge in $L^2(F)$ by the endpoint estimate in [F2]. Passing to the limit proves the identity for every $H^1$ test.

For the global case, take a whole-space extension $U\in H^{1+r}({\mathbb{R}}^d)$ of $u$; [F1]'s Fourier multiplier bound gives $\mathrm{\nabla }U\in H^r({\mathbb{R}}^d)$. Thus both cell traces of $\mathrm{\nabla }u$ on an interior facet are restrictions of the same hyperplane trace, so they agree as vectors and their opposite outward normals cancel. For a global $w\in H^1(\mathrm{\Omega })$, its two value traces agree on each interior facet: on every compact subfacet choose a ball whose closure lies in $\mathrm{\Omega }$ and meets no other facets, multiply $w$ by a smooth cutoff equal to one near that subfacet, extend by zero, and mollify with radii tending to zero below the cutoff margin. [F5] applied to the cutoff function and its weak gradient, with the interior derivative-commutation identity, gives convergence of these smooth approximants in $H^1$ on the smaller neighborhood. Their traces therefore converge on each cell by [F2], while the smooth approximants have identical restrictions from both sides. Exhaust the facet by compact subfacets; its lower-dimensional rim has surface measure zero. Sum the first identity over all cells with $q=\mathrm{\nabla }u$ and test $w$. The interior fluxes cancel by the common vector trace and the interior test traces agree; the volume divergence is $\mathrm{\Delta }u=-f$. Rearranging gives the global formula.