Rectangular continuous divergence inverse on mean-zero tests

For the given rectangle, fix once for all inputs and support sets an affine copy of the bump in [A1] on each side interval. Every derivative on $\mathrm{\mid }s\mathrm{\mid }<1$ is the exponential times a finite sum of rational terms with poles of finite order at $\mathrm{\mid }s\mathrm{\mid }=1$; $e^{-1\mathrm{/}(1-s^2)}$ tends to zero faster than every such power, so extension by zero is smooth. By [F1] its continuous compact-interval integral exists; positivity makes it nonzero, so normalization gives ${\theta }_i$. For $d=1$, put $(Rh)(x)={\int }_{a_1}^{x}h(s)ds$. The primitive difference quotient differs from $h(x)$ by at most $\sup \{\mathrm{\mid }h(s)-h(x)\mathrm{\mid }:s\text{ lies between }x\text{ and }x+\delta \}$, which tends to zero by continuity. Its derivative is therefore $h$, and it vanishes near both endpoints because $h$ is compactly supported and has total integral zero. Its support is compact inside the interval, and its $j$th derivative is $h^{(j-1)}$ for $j\ge 1$; the zeroth seminorm is bounded by $(b_1-a_1)\sup \mathrm{\mid }h\mathrm{\mid }$.

Here and below the coordinate integrals of smooth compactly supported functions may be computed on the closed coordinate intervals after extension by zero. This extension is smooth on a neighborhood of the closed box: compact support leaves a neighborhood of every boundary face where the function is zero. By [F1] the function and every coordinate section are Riemann integrable; [F2], applied to successive coordinate blocks, identifies all iterated orders with the same box integral. In particular, the marginal $H(x^{\mathrm{\prime }})={\int }_{a_1}^{b_1}h(s,x^{\mathrm{\prime }})ds$ has total integral equal to ${\int }_{\mathrm{\Omega }}h$.

To differentiate a marginal, fix all other coordinates and restrict the differentiated coordinate to a compact interval inside its open side interval. The integrand and each parameter derivative are continuous on the resulting compact two-variable rectangle, so repeated [F3] gives ${\mathrm{\partial }}_{x^{\mathrm{\prime }}}^{\beta }H(x^{\mathrm{\prime }})={\int }_{a_1}^{b_1}{\mathrm{\partial }}_{x^{\mathrm{\prime }}}^{\beta }h(s,x^{\mathrm{\prime }})ds,\sup \mathrm{\mid }{\mathrm{\partial }}^{\beta }H\mathrm{\mid }\le (b_1-a_1)\sup \mathrm{\mid }{\mathrm{\partial }}_{x^{\mathrm{\prime }}}^{\beta }h\mathrm{\mid }\mathrm{.}$ The inequality follows from [F4] with the second integrand zero. Continuity of these derivatives jointly in the remaining parameters follows from the same bound applied to their differences and uniform continuity of the integrand derivative on each compact box.

The same reasoning differentiates an integral with upper endpoint $x_1$ in its other parameters. For its $x_1$ derivative, the difference quotient of ${\int }_{a_1}^{x_1}f(s,x^{\mathrm{\prime }})ds$ differs from $f(x_1,x^{\mathrm{\prime }})$ by at most ${\sup }_{s\text{ between }{x}_1\text{ and }{x}_1+\delta }\mathrm{\mid }f(s,x^{\mathrm{\prime }})-f(x_1,x^{\mathrm{\prime }})\mathrm{\mid }$, which tends to zero by continuity; higher endpoint derivatives then follow by ordinary differentiation. This proves the finite-integral identities used in the recursion.

For $d>1$, let $x^{\mathrm{\prime }}=(x_2,\dots ,x_d)$ and $H(x^{\mathrm{\prime }})={\int }_{a_1}^{b_1}h(s,x^{\mathrm{\prime }})ds$. By step 1.2, $\int H=\int h=0$, and steps 1.3–2.1 show that $H$ and the displayed primitives are smooth. Assuming the construction in dimension $d-1$, define $(Rh{)}_1={\int }_{a_1}^{x_1}(h(s,x^{\mathrm{\prime }})-{\theta }_1(s)H(x^{\mathrm{\prime }}))ds,(Rh{)}_j={\theta }_1(x_1)(R^{\mathrm{\prime }}H{)}_{j-1}(x^{\mathrm{\prime }})(j\ge 2)\mathrm{.}$ Differentiating gives $\mathrm{div}(Rh)=h-{\theta }_{1}H+{\theta }_{1}H=h$. The first component vanishes both before the support and after its total first-coordinate integral, which is zero; the other components are compactly supported by induction.

If inputs are supported in fixed compact $K$, each coordinate projection $K_i$ is compactly contained in $(a_i,b_i)$. Choose closed intervals $J_i⋐(a_i,b_i)$ containing $K_i$ and the support of the already fixed bump ${\theta }_i$. Their product is a common compact output-support set $K^{\mathrm{\prime }}={\prod }_i{J}_i$: before the first-coordinate support the primitive is zero, and after it the primitive is zero because its full first-coordinate integral is zero; the other components have the required support by induction.

Write $p_m(f)={\max }_{\mathrm{\mid }\alpha \mathrm{\mid }\le m}\sup \mathrm{\mid }{\mathrm{\partial }}^{\alpha }f\mathrm{\mid }$, taking the maximum also over components for vector fields, and put ${\mathrm{\ell }}_1=b_1-a_1$. Step 1.3 gives $p_m(H)\le {\mathrm{\ell }}_1{p}_m(h)$. For the first component in step 3.1, a derivative with no $x_1$ differentiation is an integral of ${\mathrm{\partial }}_{x^{\mathrm{\prime }}}^{\beta }h-{\theta }_1{\mathrm{\partial }}_{x^{\mathrm{\prime }}}^{\beta }H$ and is bounded by ${\mathrm{\ell }}_1(1+{\mathrm{\ell }}_1\mathrm{\parallel }{\theta }_1{\mathrm{\parallel }}_{\mathrm{\infty }})p_m(h)$. If the derivative differentiates $x_1$ at least once, the endpoint identity in step 2.1 gives a derivative of $h$ minus a fixed derivative of ${\theta }_1$ times a derivative of $H$, again bounded by a constant times $p_m(h)$. For each remaining component ${\theta }_1(x_1)(R^{\mathrm{\prime }}H{)}_{j-1}(x^{\mathrm{\prime }})$, the product derivative is bounded by a fixed bump-derivative constant times $p_m(R^{\mathrm{\prime }}H)\le C_m^{\mathrm{\prime }}{p}_m(H)$ by induction. The dimension-one bounds in step 1.1 start this induction. Thus $p_m(Rh)\le C_m{p}_m(h)(m\ge 0),$ where $C_m$ depends only on the fixed rectangle, bumps, dimension and $m$, not on $h$. These are all fixed-support seminorm estimates, proving the full stated continuity. Every displayed construction is linear in $h$, and the bumps were fixed once independently of $h$ and $K$, so the same map $R$ serves all compact input supports.