Tensor-product tests are dense in smooth compact kernel topology

Fix $\mathrm{\Phi }\in \mathcal{D}(X^n)$ and regard $X^n$ as ${\mathbb{R}}^D$, $D=dn$. The coordinate projections of $\mathrm{supp}\mathrm{\Phi }$ lie in a compact $K\subset X$. By [F3] choose $\chi \in \mathcal{D}(X)$ equal to one near $K$. Choose $L>0$ so large that every coordinate of $\mathrm{supp}\chi$ lies strictly inside $(-L\mathrm{/}2,L\mathrm{/}2)$. Extend $\mathrm{\Phi }$ by zero and then $L$-periodically in each of the $D$ coordinates; because its support is interior to the period cube, the resulting function is smooth on the compact $D$-torus.

For $k\in {\mathbb{Z}}^D$, put $e_k(x)=\exp (2\pi ik\cdot x\mathrm{/}L)$, $Q=[-L\mathrm{/}2,L\mathrm{/}2{]}^D$, and $\widehat{\mathrm{\Phi }}(k)=L^{-D}{\int }_Q\mathrm{\Phi }(x)\overline{e_k(x)}dx$. These are Riemann integrals by [F7]. Periodic integration by parts in one coordinate has no boundary term: the endpoint product agrees by periodicity, and [F8] converts the integral of a derivative to that endpoint difference. Writing $\mathrm{\Delta }={\sum }_{j=1}^D{\mathrm{\partial }}_j^2$, applying this identity to $(1-\mathrm{\Delta }{)}^q\mathrm{\Phi }$, and using the character derivative from [F6], gives $\widehat{(1-\mathrm{\Delta }{)}^q\mathrm{\Phi }}(k)=(1+(2\pi \mathrm{/}L{)}^2\mathrm{\mid }k{\mathrm{\mid }}^2{)}^q\widehat{\mathrm{\Phi }}(k)$. The left side is bounded by the volume of $Q$ times the supremum of the continuous function $(1-\mathrm{\Delta }{)}^q\mathrm{\Phi }$, so $\mathrm{\mid }\widehat{\mathrm{\Phi }}(k)\mathrm{\mid }\le C_{\mathrm{\Phi },q}(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{-q}\mathrm{.}$ For integer $m\ge 1$, the number of lattice points in $m\le \mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}<m+1$ is at most $(2m+3{)}^D-(2m-1{)}^D\le C_D(m+1{)}^{D-1}$, by expanding the two powers; the shell $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}<1$ is finite. Thus, for every multi-index $\alpha$, choosing $q$ with $2q>D+\mathrm{\mid }\alpha \mathrm{\mid }$ makes ${\sum }_k\mathrm{\mid }k{\mathrm{\mid }}^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }\widehat{\mathrm{\Phi }}(k)\mathrm{\mid }$ finite. The Fourier series and each derivative series therefore converge absolutely and uniformly on the torus. Integrating each uniformly convergent derivative series along a coordinate segment and using [F8] shows it is the corresponding derivative of the uniformly convergent Fourier sum; iterating over coordinates gives the same assertion for every multi-index.

The uniform sum has the same Fourier coefficients as $\mathrm{\Phi }$, since uniform convergence allows integration term by term and distinct integer characters have zero cross integral by [F6]–[F8]. Their difference is a continuous $D$-periodic function with every multi-index Fourier coefficient zero. For $D=1$, all its Fejer means are finite sums of those zero coefficients, so [F5] makes the function zero. For $D>1$, fix the first frequency $k_1$ and integrate in the first coordinate to obtain a continuous $(D-1)$-periodic coefficient function; continuity follows from [F4] and the compact integral bound [F7]. Its remaining Fourier coefficients are the corresponding coefficients of the original difference by [F7], so induction makes each such coefficient function zero. Applying the one-dimensional case to each fixed choice of the remaining coordinates shows the original difference is zero. Hence the Fourier series sums to $\mathrm{\Phi }$, with all derivatives uniformly.

Let $S_M\mathrm{\Phi }$ be the finite Fourier sum over $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}\le M$. On ${\mathbb{R}}^D$, put $P_M(x_1,\dots ,x_n)=({\prod }_{j=1}^n\chi (x_j))S_M\mathrm{\Phi }(x_1,\dots ,x_n)\mathrm{.}$ Each Fourier character factors over the $n$ blocks $x_j\in {\mathbb{R}}^d$; after multiplication by the product cutoff, each term is an elementary tensor of tests in $\mathcal{D}(X)$. Also $({\prod }_j\chi (x_j))\mathrm{\Phi }=\mathrm{\Phi }$, since every coordinate projection of its support lies in $K$. The product rule [F8] and the derivative convergence from step 2.1 show $P_M\to \mathrm{\Phi }$ uniformly with every derivative, with all supports in the fixed compact $(\mathrm{supp}\chi {)}^n$. By [F1] this is convergence in the LF topology. Applying the finite symmetrization operator preserves convergence and turns the approximants into symmetric finite tensor sums whenever $\mathrm{\Phi }$ is symmetric.

Suppose ${\sum }_{n=0}^N\int f_n(x_1,\dots ,x_n){\prod }_j\varphi (x_j)dx=0$ for every real smooth $\varphi$, with symmetric test kernels $f_n$ as in [F9]. Replacing $\varphi$ by $t\varphi$ makes a scalar polynomial in $t$ vanish for every real $t$, so each homogeneous term $\int f_n{\varphi }^{\otimes n}$ is zero for every $\varphi$. For fixed real test functions $h_1,\dots ,h_n$, substitute $\varphi ={\sum }_j{t}_j{h}_j$. The coefficient of $t_1\cdots t_n$ is $n!\int f_n(x_1,\dots ,x_n){\prod }_j{h}_j(x_j)dx$, by symmetry, and must vanish. Complex product tests are finite linear combinations of real product tests by separating real and imaginary parts. Step 3.1 makes their span dense, so choose approximants ${\psi }_M\to \overline{f_n}$ with support in a fixed compact rectangle $C$. The componentwise convention [F7, F9] and the Riemann-sum triangle inequality give $\mid \int f_n({\psi }_M-\overline{f_n})\mid \le \mathrm{vol}(C)\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{\psi }_M-\overline{f_n}{\mathrm{\parallel }}_{\mathrm{\infty }}⟶0.$ Thus $\int \mathrm{\mid }{f}_n{\mathrm{\mid }}^2=0$. If $f_n$ were nonzero at some point, its support constraint and continuity give a closed nondegenerate rectangle $Q\subset \mathrm{int}C$ and $c>0$ with $\mathrm{\mid }{f}_n\mathrm{\mid }\ge c$ on $Q$. Insert the faces of $Q$ into a Darboux grid of $C$. Since $\mathrm{\mid }{f}_n{\mathrm{\mid }}^2\ge 0$ everywhere and $\mathrm{\mid }{f}_n{\mathrm{\mid }}^2\ge c^2$ on $Q$, the lower Darboux sum is at least $c^2\mathrm{vol}(Q)>0$, so the real Riemann integral is positive, a contradiction. Hence $f_n=0$. The constant term is zero by taking $\varphi =0$. This proves uniqueness and the well-definedness claim used by the definition.