Tensor products of tempered distributions

By continuity [F3], choose a basic neighbourhood $V_u=\{f:p_{{\alpha }_j{\beta }_j}(f)<{\epsilon }_j,1\le j\le k\}$ on which $\mathrm{\mid }u(f)\mathrm{\mid }<1$. Intersecting with $p_{00}(f)<1$ if necessary ensures $k\ge 1$. Set $M_u={\max }_{1\le j\le k}\{\mathrm{\mid }{\alpha }_j\mathrm{\mid },\mathrm{\mid }{\beta }_j\mathrm{\mid }\}$, $P_{M_u}(f)={\max }_{\mathrm{\mid }\alpha \mathrm{\mid },\mathrm{\mid }\beta \mathrm{\mid }\le M_u}{p}_{\alpha \beta }(f)$, and $c_u=2{\max }_j{\epsilon }_j^{-1}$. For $D(f)={\max }_j(p_{{\alpha }_j{\beta }_j}(f)\mathrm{/}{\epsilon }_j)>0$, the test $f\mathrm{/}(2D(f))$ lies in $V_u$, so $\mathrm{\mid }u(f)\mathrm{\mid }\le 2D(f)\le c_u{P}_{M_u}(f)$. If $D(f)=0$, every positive scalar multiple of $f$ lies in $V_u$, forcing $u(f)=0$ and the same bound. Applying this argument to $v$ gives finite $M_v,c_v$ with $\mathrm{\mid }v(g)\mathrm{\mid }\le c_v{P}_{M_v}(g)$.

For fixed $\mathrm{\Phi }\in \mathcal{S}({\mathbb{R}}^{p+q})$, the map $x\mapsto \mathrm{\Phi }(x,\cdot )$ is smooth with values in $\mathcal{S}({\mathbb{R}}^q)$: Taylor's formula in the $x$ coordinates, with its remainder differentiated in $y$, tends to zero in each $p_{\gamma \delta }$ on every compact $x$ set because every mixed Schwartz seminorm of $\mathrm{\Phi }$ is finite. Continuity of $v$ therefore makes $h(x):=v(\mathrm{\Phi }(x,\cdot ))$ smooth and gives ${\mathrm{\partial }}_x^{\beta }h(x)=v({\mathrm{\partial }}_x^{\beta }\mathrm{\Phi }(x,\cdot ))$. For every $\alpha ,\beta$, [F3] and step 1.1 yield ${\sup }_x\mathrm{\mid }{x}^{\alpha }{\mathrm{\partial }}_x^{\beta }h(x)\mathrm{\mid }\le c_v{\max }_{\mathrm{\mid }\gamma \mathrm{\mid },\mathrm{\mid }\delta \mathrm{\mid }\le M_v}{\sup }_{x,y}\mathrm{\mid }{x}^{\alpha }{y}^{\gamma }{\mathrm{\partial }}_x^{\beta }{\mathrm{\partial }}_y^{\delta }\mathrm{\Phi }(x,y)\mathrm{\mid }<\mathrm{\infty }\mathrm{.}$ Thus $h\in \mathcal{S}({\mathbb{R}}^p)$.

Define $w(\mathrm{\Phi })=u(h)$. Applying the bound in step 1.1 to the finitely many $x$ seminorms with indices at most $M_u$, and then the displayed estimate in step 2.1, bounds $\mathrm{\mid }w(\mathrm{\Phi })\mathrm{\mid }$ by a finite constant times the maximum of finitely many product Schwartz seminorms of $\mathrm{\Phi }$. Hence $w$ is continuous on $\mathcal{S}({\mathbb{R}}^{p+q})$. For $\mathrm{\Phi }=\phi \otimes \psi$, the inner pairing is $\phi (x)v(\psi )$, giving the claimed product value; on compactly supported tests the formula is the same iterated pairing as [F6].

Finite sums of Schwartz product tests are dense in $\mathcal{S}({\mathbb{R}}^{p+q})$. Indeed, choose $\chi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^p)$ and $\eta \in C_c^{\mathrm{\infty }}({\mathbb{R}}^q)$ equal to one near the unit balls by [F4], and set ${\chi }_R(x)=\chi (x\mathrm{/}R)$ and ${\eta }_R(y)=\eta (y\mathrm{/}R)$. The product rule and rapid decay show ${\chi }_R(x){\eta }_R(y)\mathrm{\Phi }(x,y)\to \mathrm{\Phi }(x,y)$ in every product Schwartz seminorm: the undifferentiated tails lie outside balls of radius proportional to $R$, and each differentiated cutoff contributes a factor $R^{-j}$ on the same tails. For fixed $R$, [F5] approximates this compactly supported smooth function by finite sums of product tests with common compact support; on that support, uniform convergence of all derivatives implies convergence in every Schwartz seminorm. Thus any two continuous functionals with the same product values agree on a dense subspace and hence everywhere. The same density argument, iterated over finitely many factors, shows that different parenthesizations of the constructed tensor product agree; this also agrees with the compact-test associativity in [F6].