Matching NURBS patch traces give global H1 conformity

Fix a component $q\in \{1,\dots ,m\}$, a test function $\varphi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and a coordinate $j$. The finite physical partition and face-equality convention are those of [F1]. On each cell $K$, the vector field $G_K:=v_q\varphi e_j$ is $C^1$ up to the cell boundary. By [F3], ${\int }_K{v}_q{\mathrm{\partial }}_j\varphi +{\mathrm{\partial }}_j{v}_q\varphi ={\int }_{\mathrm{\partial }K}{v}_q\varphi n_{K,j}dS\mathrm{.}$ Summing this identity over the finitely many cells is legitimate because both volume terms are integrable by the hypotheses and [F4]. On a shared face the two normal fluxes add to $(v_{q,K}-v_{q,K^{\mathrm{\prime }}})\varphi n_{K,j}$, which is zero by the matching-trace assumption; edge overlaps have surface measure zero by [F2]. The exterior boundary term is zero because $\varphi$ is compactly supported in $\mathrm{\Omega }$. Therefore ${\int }_{\mathrm{\Omega }}{v}_q{\mathrm{\partial }}_j\varphi =-{\sum }_K{\int }_K{\mathrm{\partial }}_j{v}_q\varphi \mathrm{.}$

For every component $q$ and coordinate $j$, define $g_{qj}$ to equal the classical ${\mathrm{\partial }}_j{v}_q$ on each open cell and arbitrarily on the cell faces. The faces have $d$-dimensional measure zero under the finite piecewise $C^1$ presentation, and the finite sum of cellwise $L^2$ norms is finite, so $g_{qj}\in L^2(\mathrm{\Omega })$. Step 1.1 gives the weak-derivative identity for every test function, hence $D_j{v}_q=g_{qj}$ weakly. Each component of $v$ is also square integrable on the finite bounded cell partition; [F4] therefore gives $v\in H^1(\mathrm{\Omega };{\mathbb{R}}^m)$ with the asserted componentwise gradient.