Continuous density of the weak surface divergence for a C1 tangent field

Fix a chart $X:V\to S$. Its coordinate tangent vectors are independent by [F2], so there are unique functions $K^a$ with $K=K^a{\mathrm{\partial }}_{a}X$; their $C^1$ regularity is part of the stated tangent-field convention. The chart area density $J_X$ is positive and $C^1$ by [F1]. Therefore $g_X(X(u)):=J_X(u{)}^{-1}{\sum }_{a=1}^2{\mathrm{\partial }}_a(J_X{K}^a)(u)$ is continuous on the chart.

Let $\varphi \in C^1(S)$ have compact support in this chart. By the definition of the surface gradient, $K\cdot {\mathrm{\nabla }}_S\varphi =K^a{\mathrm{\partial }}_a(\varphi \circ X)$. The surface-area formula and [F3] give $-{\int }_{S}K\cdot {\mathrm{\nabla }}_S\varphi dS=-{\int }_V{J}_X{K}^a{\mathrm{\partial }}_a(\varphi \circ X)du={\int }_V{\mathrm{\partial }}_a(J_X{K}^a)(\varphi \circ X)du={\int }_S{g}_X\varphi dS\mathrm{.}$ Here the repeated $a$ is summed over $1,2$; compact support lets each product $J_X{K}^a(\varphi \circ X)$ extend by zero as a $C^1$ function, and integration by parts on a rectangle containing the support has no boundary term. Thus $g_X$ from step 1.1 represents ${\mathrm{Div}}_{S}K$ on this chart.

If $S$ is empty, the unique empty function is the asserted density and every chart assertion is vacuous; hence assume $S$ is nonempty. Write the finite chart cover in [F1] as $S\cap O_j$, with $O_j\subset {\mathbb{R}}^3$ open and $S\cap O_j$ the corresponding chart image. All balls $B(x,r)$ with $x\in S$ and $\bar{B}(x,2r)\subset O_j$ for some $j$ cover $S$. Take a finite subcover, assign a chart to each of its finitely many balls, and apply [F4] to each closed inner ball in its assigned $O_j$. This gives finitely many nonnegative ambient cutoffs; group their sums as $b_j$ by chart and put $b={\sum }_j{b}_j$. Then $b>0$ on a neighborhood of $S$, and ${\chi }_j=b_j\mathrm{/}b$ there restricts to a $C^1(S)$ function, with support compactly contained in its chart, such that ${\sum }_j{\chi }_j=1$. Each ${\chi }_j{g}_j$ extends continuously by zero off its chart, so $g:={\sum }_j{\chi }_j{g}_j$ is continuous on $S$. For every $\varphi \in C^1(S)$, applying step 2.1 to ${\chi }_j\varphi$ and summing gives ${\int }_{S}g\varphi dS=-{\sum }_j{\int }_{S}K\cdot {\mathrm{\nabla }}_S({\chi }_j\varphi )dS=-{\int }_{S}K\cdot {\mathrm{\nabla }}_S\varphi dS$, because ${\sum }_j{\mathrm{\nabla }}_S{\chi }_j=0$. Thus $g$ represents ${\mathrm{Div}}_{S}K$. The construction makes only finitely many choices.

Fix any chart image $U=S\cap O$ and $x\in U$. By [F4], choose $\eta \in C_c^{\mathrm{\infty }}(O)$ equal to one near $x$. The function $h=\eta (g_X-g)$ on $U$, extended by zero to $S\setminus U$, is continuous on $S$, since its support is compactly contained in $U$. For every ambient test $\psi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$, the restriction of $\eta \psi$ has compact support in $U$. Steps 2.1 and 3.1 therefore give ${\int }_{S}h\psi dS={\int }_U(g_X-g)\eta \psi dS=0$. By [F5], $h=0$ on $S$, and $\eta (x)=1$ gives $g_X(x)=g(x)$. Since the chart and point were arbitrary, the global density has the stated formula in every chart.

If $g_1,g_2$ are continuous representatives, then ${\int }_S(g_1-g_2)\varphi dS=0$ for every $\varphi \in C^1(S)$, hence for every restriction of an ambient test in $C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$. Applying [F5] to $g_1-g_2$ gives pointwise equality. Together with step 4.1 this proves uniqueness and chart independence of the asserted continuous representative.