Schwartz approximate identities converge in the sense of tempered distributions

The reflected approximate identity converges in Schwartz space. Put $\check{\mathrm{\Phi }}(u)=\mathrm{\Phi }(-u)$ and $K_t={\check{\mathrm{\Phi }}}_t\ast \psi$, so that $K_t(y)={\int }_{{\mathbb{R}}^n}{\mathrm{\Phi }}_t(x-y)\psi (x)dx={\int }_{{\mathbb{R}}^n}\check{\mathrm{\Phi }}(w)\psi (y-tw)dw$, the last form by the substitution $x=y-tw$. Then $K_t-\psi =\int \check{\mathrm{\Phi }}(w)(\psi (\cdot -tw)-\psi (\cdot ))dw$ because $\int \check{\mathrm{\Phi }}=\int \mathrm{\Phi }=1$ by [F1], [F4]. Fix multi-indices $\alpha ,\beta$ and $t\le 1$. Differentiating under the integral and applying the mean value theorem along the segment from $y$ to $y-tw$ gives $\mathrm{\mid }{\mathrm{\partial }}^{\beta }{K}_t(y)-{\mathrm{\partial }}^{\beta }\psi (y)\mathrm{\mid }\le {\int }_{{\mathbb{R}}^n}\mathrm{\mid }\check{\mathrm{\Phi }}(w)\mathrm{\mid }t\mathrm{\mid }w\mathrm{\mid }{\int }_0^1\mid \mathrm{\nabla }{\mathrm{\partial }}^{\beta }\psi (y-stw)\mid dsdw\mathrm{.}$ Since $\mathrm{\mid }{y}^{\alpha }\mathrm{\mid }\le {\sum }_{\gamma \le \alpha }\left(\genfrac{}{}{0px}{}{\alpha }{\gamma }\right)\mathrm{\mid }(y-stw{)}^{\gamma }\mathrm{\mid }(t\mathrm{\mid }w\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }-\mathrm{\mid }\gamma \mathrm{\mid }}\le (1+t\mathrm{\mid }w\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}{\sum }_{\gamma \le \alpha }\left(\genfrac{}{}{0px}{}{\alpha }{\gamma }\right)\mathrm{\mid }(y-stw{)}^{\gamma }\mathrm{\mid }$, taking the supremum in $y$ and using $t\le 1$ yields $p_{\alpha \beta }(K_t-\psi )\le t{C}_{\alpha }{\int }_{{\mathbb{R}}^n}\mathrm{\mid }\check{\mathrm{\Phi }}(w)\mathrm{\mid }(1+\mathrm{\mid }w\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }+1}dw\cdot {\max }_{i\le n}{\max }_{\gamma \le \alpha }{p}_{\gamma ,\beta +e_i}(\psi ),$ with $C_{\alpha }=\sqrt{n}{\sum }_{\gamma \le \alpha }\left(\genfrac{}{}{0px}{}{\alpha }{\gamma }\right)$, and the integral is finite by [F3]. Hence $p_{\alpha \beta }(K_t-\psi )\to 0$ as $t\downarrow 0$: this is convergence in every Schwartz seminorm, i.e. $K_t\to \psi$ in $\mathcal{S}$.

The pairing identity. For every $t>0$, $⟨{\mathrm{\Phi }}_t\ast f,\psi ⟩=⟨f,K_t⟩$. Indeed, the parameter-pairing lemma applied to $H(x)=\psi (x){\mathrm{\Phi }}_t(x-\cdot )$ and $u=f$ gives $⟨f,\int H(x)dx⟩=\int ⟨f,{\mathrm{\Phi }}_t(x-\cdot )⟩\psi (x)dx=\int ({\mathrm{\Phi }}_t\ast f)(x)\psi (x)dx$, and $\int H(x)dx=K_t$ by the computation of step 1.1; the seminorm majorants required by that lemma are supplied by [F1] and [F3], since $\mathrm{\mid }\psi (x)\mathrm{\mid }(1+\mathrm{\mid }x\mathrm{\mid }{)}^N$ is integrable for every $N$.

Conclusion. By step 1.1 $K_t\to \psi$ in $\mathcal{S}$, so the finite-seminorm bound of [F2] gives $⟨f,K_t⟩\to ⟨f,\psi ⟩$; step 2.1 identifies this with $⟨{\mathrm{\Phi }}_t\ast f,\psi ⟩\to ⟨f,\psi ⟩$ as $t\downarrow 0$, which is convergence ${\mathrm{\Phi }}_t\ast f\to f$ in ${\mathcal{S}}^{\mathrm{\prime }}$ by the definition of that convergence. Sequences and the dyadic scale $t=2^{-j}$ are instances. For the convolution form, $\mathrm{\Phi }\ast \mathrm{\Phi }\in \mathcal{S}$ by Schwartz convolution and product laws, $\int (\mathrm{\Phi }\ast \mathrm{\Phi })=(\int \mathrm{\Phi }{)}^2=1$, and the substitution $z=tw$ gives $(\mathrm{\Phi }\ast \mathrm{\Phi }{)}_t={\mathrm{\Phi }}_t\ast {\mathrm{\Phi }}_t$, so the statement applies to the Schwartz function $\mathrm{\Phi }\ast \mathrm{\Phi }$. This proves the lemma.