= Radial Schwartz approximation of tempered distributions
Choose a radial <mollifier> $\rho\in C_c^\infty$ with integral one and support in the unit ball, and a radial <cutoff function> $\chi\in C_c^\infty$ equal to one there. For a <radial tempered distribution> $T$, the functions
$$
f_j(x)=\chi(x/j)(T*\rho_{1/j})(x),\qquad \rho_\varepsilon(x)=\varepsilon^{-n}\rho(x/\varepsilon),
$$
are radial <Schwartz functions>. Smoothness comes from <convolution of a tempered distribution with a Schwartz function>, and compact support comes from the cutoff. Pairing with $\varphi$ gives $\langle f_j,\varphi\rangle=\langle T,\check\rho_{1/j}*(\chi(\cdot/j)\varphi)\rangle$. With $q_m(\varphi)=\max_{|\beta|\leq m}\sup_x(1+|x|)^m|\partial^\beta\varphi(x)|$, the cutoff tail and the <mean value theorem> give
$$
q_m\!\left(\check\rho_{1/j}*(\chi(\cdot/j)\varphi)-\varphi\right)\leq C_mj^{-1}q_{m+1}(\varphi).
$$
The continuity estimate for $T$ therefore proves $f_j\to T$ even in the <strong dual topology>. Inserting a cutoff is essential because convolution alone need not give rapid decay.
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