RT0 facet-flux unisolvence and global H(div) conformity

On the reference simplex ${\widehat{S}}_d=\{x_i>0,{\sum }_i{x}_i<1\}$, each coordinate facet has outward normal $-e_i$ and area $1\mathrm{/}(d-1)!$, while the top facet has outward normal $(1,\dots ,1)\mathrm{/}\sqrt{d}$ and area $\sqrt{d}\mathrm{/}(d-1)!$. For $\widehat{v}=a+cx$, the constant-vector facet fluxes sum to zero by [F3], and the flux sum of $cx$ is $c\mathrm{/}(d-1)!=dc\mathrm{\mid }{\widehat{S}}_d\mathrm{\mid }$. If all facet fluxes vanish then $c=0$; the coordinate-facet fluxes then force each $a_i=0$. The flux map is injective between spaces of dimension $d+1$, hence bijective.

For any physical simplex, [F2] transports the reference flux degrees of freedom bijectively, so the same unisolvence holds. On each facet, $(a+cx)\cdot n$ is constant because $x\cdot n$ is constant on its supporting hyperplane; equal integrated fluxes therefore give equal normal components.

Let $v$ be a piecewise RT0 field with matching shared-facet fluxes. For $\varphi \in C_c^{\mathrm{\infty }}(D)$, apply [F3] to $\varphi v$ cellwise and sum: interior facet terms cancel by step 2.1 and the common value of $\varphi$, and boundary terms vanish because $\varphi$ is compactly supported. Thus ${\int }_{D}v\cdot \mathrm{\nabla }\varphi =-{\sum }_K{\int }_K(\mathrm{div}v)\varphi$. The cellwise divergence is constant and in $L^2$, so $v\in H(\mathrm{div};D)$ by definition.

Conversely, let $v$ be a piecewise RT0 field in H(div), with global $L^2$ divergence $g$. For every test supported in one cell, the global distributional identity and cellwise Gauss–Green give zero pairing with $g$ minus that cell's polynomial divergence. This difference is in $L^2(K)$ and hence in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(K)$ by [F7]; extending from real to complex tests by linearity, injectivity [F7] makes the difference zero almost everywhere on the cell. For arbitrary $\varphi \in C_c^{\mathrm{\infty }}(D)$, summing cellwise Green formulas and comparing with the global distributional identity yields ${\sum }_{F=K^{+}\cap K^{-}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{o}\mathrm{r}}{\int }_F\varphi (v_{K^{+}}\cdot n_{K^{+}}+v_{K^{-}}\cdot n_{K^{-}})=0$. Each normal sum is constant on its facet. If one were nonzero, choose a nonnegative smooth test supported in a small ball around a relative-interior point of that facet, meeting no other facet, and positive on a subpatch; the sum would be a nonzero constant times a positive surface integral, a contradiction. Hence all interior fluxes match.

Suppose the interior fluxes match. For $w\in H^1(D)$, approximate each restriction $w{\mathrm{\mid }}_K$ in H1 by smooth-neighborhood functions using [F5]. The trace bound [F4] and volume convergence pass cellwise Gauss–Green to each restriction. Interior traces of $w$ agree by [F4], and the normal components of $v$ agree by step 2.1, so interior terms cancel. If $c_F$ denotes the constant outward normal component on a boundary facet $F$, the resulting identity is ${\int }_D(v\cdot \mathrm{\nabla }w+\mathrm{div}vw)={\sum }_{F\subset \mathrm{\partial }D}{c}_F{\int }_F{\gamma }_{F}w$. By [F6] this is the weak normal-trace functional; in particular, if every boundary flux is zero then $v\in H_0(\mathrm{div};D)$.

Conversely, if $v\in H_0(\mathrm{div};D)$, its zero trace and the identity in step 4.1 give ${\sum }_{F\subset \mathrm{\partial }D}{c}_F{\int }_F{\gamma }_{F}w=0$ for every $w\in H^1(D)$. For any selected boundary facet, choose a smooth function on ${\mathbb{R}}^d$ supported in a small ball about a relative-interior point of that facet, positive on a subpatch, and meeting no other boundary facet; its restriction lies in $H^1(D)$. The sum then reduces to the selected $c_F$ times a positive surface integral, so $c_F=0$. Thus every boundary facet flux vanishes.