= Solution
The <multiplication of a distribution by a smooth function> is $\langle\psi T,\varphi\rangle=\langle T,\psi\varphi\rangle$. If $\psi$ is a <Schwartz function>, the <Leibniz rule> shows that $\varphi\mapsto\psi\varphi$ is a continuous map $\mathcal S\to\mathcal S$, so $\psi T$ is a <tempered distribution>. When both $\psi$ and $T$ are radial, $\psi\circ R^t=\psi$, and
$$
\langle(\psi T)\circ R,\varphi\rangle
=\langle T,\psi(\varphi\circ R^t)\rangle
=\langle T,(\psi\varphi)\circ R^t\rangle
=\langle T,\psi\varphi\rangle.
$$
Thus \b[Multiplication by a radial <Schwartz function> preserves <radial tempered distributions>].
For the <convolution of a tempered distribution with a Schwartz function>, set
$$
(T*\psi)(x)=\langle T_y,\psi(x-y)\rangle.
$$
Translations of $\psi$ depend smoothly on $x$ in the <Schwartz space>, so this is a <smooth function> with $\partial^\alpha(T*\psi)(x)=\langle T_y,\partial^\alpha\psi(x-y)\rangle$. The bound for $T$ by finitely many <seminorms>, together with $1+|y|\leq(1+|x|)(1+|x-y|)$, proves
$$
|\partial^\alpha(T*\psi)(x)|\leq C_\alpha(1+|x|)^m.
$$
Thus the function also defines a <tempered distribution>. For a <radial function> $\psi$, put $g_x(y)=\psi(x-y)$. Then $\psi(Rx-y)=\psi(x-R^ty)=g_x(R^ty)$, so
$$
(T*\psi)(Rx)=\langle T,g_x\circ R^t\rangle
=\langle T\circ R,g_x\rangle=(T*\psi)(x).
$$
Hence \b[<Convolution> with a radial <Schwartz function> preserves <radial tempered distributions>].
Smoothing alone need not give a <Schwartz function>: for the constant <tempered distribution> $T=1$ and a <Schwartz function> with integral one, $T*\psi=1$. The <radial Schwartz approximation of tempered distributions> therefore combines smoothing with a large-radius cutoff. Choose a nonnegative radial <mollifier> $\rho\in C_c^\infty$, supported in the unit ball with integral one, and a radial <cutoff function> $\chi\in C_c^\infty$ equal to one on the unit ball. Put
$$
\rho_\varepsilon(x)=\varepsilon^{-n}\rho(x/\varepsilon),\qquad
\chi_j(x)=\chi(x/j),\qquad
\boxed{f_j(x)=\chi_j(x)(T*\rho_{1/j})(x).}
$$
Each $f_j$ is a <smooth function> of <compact support>, hence a <Schwartz function>, and the two invariance calculations above make it radial.
It remains to prove convergence, including the simultaneous changes of both scales. With $\check\rho(x)=\rho(-x)$, the distributional <convolution> pairing is
$$
\langle f_j,\varphi\rangle
=\left\langle T,\check\rho_{1/j}*(\chi_j\varphi)\right\rangle.
$$
For every fixed $m$, the <Leibniz rule>, rapid decay outside the radius-$j$ ball, and the <chain rule> for $\chi(x/j)$ give
$$
q_m((\chi_j-1)\varphi)\leq \frac{C_m}{j}q_{m+1}(\varphi).
$$
Convolution by $\check\rho_\varepsilon$ is uniformly bounded in $q_m$ for $0<\varepsilon\leq1$, because its shifts have size at most one. The <mean value theorem> applied to $\varphi(x-y)-\varphi(x)$ similarly gives
$$
q_m(\check\rho_\varepsilon*\varphi-\varphi)\leq C_m\varepsilon q_{m+1}(\varphi).
$$
Splitting the error into the convolved cutoff error and the <approximate identity> error proves
$$
q_m\!\left(\check\rho_{1/j}*(\chi_j\varphi)-\varphi\right)
\leq \frac{C_m'}j q_{m+1}(\varphi).
$$
If $|\langle T,\theta\rangle|\leq Cq_m(\theta)$, this yields
$$
|\langle f_j-T,\varphi\rangle|\leq \frac{C''}j q_{m+1}(\varphi).
$$
Consequently
$$
\boxed{f_j\longrightarrow T\quad\text{in }\mathcal S'}
$$
weakly, and even in the <strong dual topology>, since $q_{m+1}$ is uniformly bounded on every subset of the <Schwartz space> that is a <bounded set in a topological vector space>. No assertion that $T*\rho_{1/j}$ itself is rapidly decreasing is needed.
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