Distribution pairing with smooth parameter families

Around a fixed $x_0\in X$ take a closed ball with a slightly larger closed ball still inside $X$. The hypothesis on the larger ball supplies a single compact $K$ for all its slices. Every parameter derivative of $F$ has support in $K$ for parameters in the smaller ball: for $y\mathrm{\notin }K$ the function is identically zero for all parameters in a neighborhood, so all parameter derivatives vanish there. On the smaller ball times $K$, every mixed derivative is uniformly continuous by compactness. Thus $p_m(F(x,\cdot )-F(x_0,\cdot ))\to 0$ for each $m$, and F1 gives continuity of $a$.

Fix a coordinate $i$. For each $\mathrm{\mid }\beta \mathrm{\mid }\le m$, apply F5 on the segment from $0$ to $h$ (reverse its orientation if $h<0$) to the curve $t\mapsto {\mathrm{\partial }}_y^{\beta }F(x+t{e}_i,y)-t{\mathrm{\partial }}_i{\mathrm{\partial }}_y^{\beta }F(x,y)$. The derivative increment is bounded uniformly in $y\in K$ by a modulus of continuity tending to zero with $\mathrm{\mid }h\mathrm{\mid }$. Dividing the resulting inequality by $\mathrm{\mid }h\mathrm{\mid }$ proves [step 1.1, F1, F2, F5] $p_m(\frac{F(x+h{e}_i,\cdot )-F(x,\cdot )}{h}-{\mathrm{\partial }}_{i}F(x,\cdot ))⟶0.$ The F1 estimate passes this limit through $u$. Apply step 1.1 to the parameter derivative slices for continuity of the resulting derivative, and repeat for every multi-index. This proves smoothness and the derivative formula.

For this clause assume Countable Choice and use F7 for Lebesgue measure. Fix compact $E\subseteq X$. By F4 choose a smooth cutoff $\theta =1$ near $E$ with compact parameter support in $X$. Extend $\tilde{F}=\theta F$ by zero to all parameter space. It is smooth, and the support hypothesis on $\mathrm{supp}\theta$ gives a common compact $K$ for all its slices and their $y$ derivatives. Choose a closed box $Q$ whose interior contains that parameter support. At level $j$ divide each side into $2^j$ equal pieces, disjointify the cells by assigning shared faces in coordinate order, and let $t_C$ be each cell's lower corner. Put [step 2.1, given, F2, F4, F6, F7] $S_j(y)={\sum }_C{\lambda }_r(E\cap C)\tilde{F}(t_C,y)\mathrm{.}$ These are tests supported in $K$. For every $m$, uniform continuity of the finitely many $y$ derivatives through order $m$ on $Q\times K$ supplies a modulus ${\omega }_m(\delta )\to 0$. For each derivative and fixed $y$, F6 bounds the error between the grid sum and its scalar integral over $E$ by ${\lambda }_r(E){\omega }_m({\mathrm{mesh}}_j)$. The finite sum is exactly the integral of the corresponding step function, and its weights are finite because $E$ lies in a bounded box.

Comparing two grid sums via their scalar integrals gives $p_m(S_j-S_k)\le {\lambda }_r(E)({\omega }_m({\mathrm{mesh}}_j)+{\omega }_m({\mathrm{mesh}}_k))$. Thus F3 gives a limit $S\in {\mathcal{D}}_K$. The degree-zero scalar error in step 3.1 identifies $S(y)={\int }_{E}F(x,y)dx=G(y)$, and its higher-degree errors identify every derivative of $S$ with the corresponding integral. By F1, $u(S_j)\to u(G)$. By finite linearity $u(S_j)={\sum }_C{\lambda }_r(E\cap C)u(\tilde{F}(t_C,\cdot ))$, which converges to ${\int }_{E}a(x)dx$ by the same uniform-continuity integral estimate, since step 2.1 makes that scalar function smooth. This proves interchange. Empty or measure-zero $E$ gives zero sums and integrals; empty $Y$ gives zero slices. All tags and grids are specified, not chosen from an infinite family. $\mathrm{\square }$