[proofplan] We prove that the standard mollifier $\rho$ belongs to $\mathcal{D}(\mathbb{R}^n)$ in three stages. First, we show the auxiliary function $h(t) = e^{-1/t}\mathbb{1}_{t>0}$ is $C^\infty(\mathbb{R})$ with all derivatives vanishing at $t = 0$, using the fact that $e^{-1/t}$ decays faster than any power of $t$ near $0^+$. Second, we lift this to $\mathbb{R}^n$ by showing the composition $h \circ g$ with $g(x) = 1 - |x|^2$ is smooth across the boundary $|x| = 1$. Third, we verify the normalisation constant is finite and positive. [/proofplan] [step:Show $h \in C^\infty(\mathbb{R})$ with $h^{(k)}(0) = 0$ for all $k$] On $(-\infty, 0)$, $h \equiv 0$ is $C^\infty$. On $(0, \infty)$, $h(t) = e^{-1/t}$ is $C^\infty$ as a composition of smooth functions. By induction, for $t > 0$, $h^{(k)}(t) = P_k(1/t) \, e^{-1/t}$ for a polynomial $P_k$. The base case $k = 0$ holds with $P_0 = 1$. If $h^{(k)}(t) = P_k(1/t)e^{-1/t}$, then differentiating using the product and chain rules gives $h^{(k+1)}(t) = P_{k+1}(1/t)e^{-1/t}$ where $P_{k+1}$ is a polynomial. For the right-hand limits at $t = 0$: for every polynomial $P$ and every $m \geq 0$, \begin{align*} \lim_{t \to 0^+} P(1/t) \, e^{-1/t} &= \lim_{s \to +\infty} P(s) \, e^{-s} = 0, \end{align*} since the exponential $e^{-s}$ decays faster than any polynomial grows. We verify $h^{(k)}(0) = 0$ for all $k$ by induction. The base case: $h(0) = 0$ by definition, and \begin{align*} \lim_{t \to 0^+} \frac{h(t) - h(0)}{t} &= \lim_{t \to 0^+} \frac{e^{-1/t}}{t} = \lim_{s \to \infty} s \, e^{-s} = 0. \end{align*} The left derivative is $0$ (since $h \equiv 0$ on $(-\infty, 0]$), so $h'(0) = 0$. For the inductive step, if $h^{(k)}(0) = 0$ and $h^{(k)}(t) = P_k(1/t)e^{-1/t}$ for $t > 0$: \begin{align*} \lim_{t \to 0^+} \frac{h^{(k)}(t) - h^{(k)}(0)}{t} &= \lim_{t \to 0^+} \frac{P_k(1/t) \, e^{-1/t}}{t} = \lim_{s \to \infty} s \, P_k(s) \, e^{-s} = 0. \end{align*} The left derivative is again $0$, so $h^{(k+1)}(0) = 0$. [/step] [step:Show the composition $h \circ g$ with $g(x) = 1 - |x|^2$ is $C^\infty(\mathbb{R}^n)$] Partition $\mathbb{R}^n$ into three regions. On $B(0,1)$: $g(x) > 0$, so $(h \circ g)(x) = e^{-1/g(x)}$ is $C^\infty$ as a composition of smooth functions. On $\mathbb{R}^n \setminus \overline{B}(0,1)$: $g(x) < 0$, so $(h \circ g)(x) = 0$ is $C^\infty$. On the boundary $\{|x| = 1\}$: by the Faa di Bruno formula, every partial derivative $\partial^\alpha(h \circ g)(x)$ on $B(0,1)$ is a sum of terms $h^{(j)}(g(x)) \cdot \prod_{\ell} \partial^{\beta_\ell} g(x)$. Each factor $\partial^{\beta_\ell} g(x)$ is a polynomial in $x$ (since $g$ is a polynomial), hence bounded near $|x| = 1$. The factor $h^{(j)}(g(x)) = P_j(1/g(x))e^{-1/g(x)} \to 0$ as $x \to x_0$ with $|x_0| = 1$ (since $g(x) \to 0^+$ and $h^{(j)}(0) = 0$ by the previous step). On $\mathbb{R}^n \setminus \overline{B}(0,1)$, all partial derivatives are identically $0$. Therefore $\partial^\alpha(h \circ g)(x) \to 0$ as $x \to x_0$ from both sides, and the derivative extends continuously with value $0$ at $x_0$. Since this holds for every multi-index $\alpha$, $h \circ g \in C^\infty(\mathbb{R}^n)$. [/step] [step:Verify the normalisation constant is well-defined] The function $h(1 - |x|^2)$ is non-negative everywhere and strictly positive on $B(0,1)$ (since $1 - |x|^2 > 0$ there and $e^{-1/t} > 0$ for $t > 0$). It vanishes outside $\overline{B}(0,1)$. Therefore \begin{align*} \int_{\mathbb{R}^n} h(1 - |x|^2) \, d\mathcal{L}^n(x) &= \int_{B(0,1)} \exp\!\left(\frac{-1}{1 - |x|^2}\right) d\mathcal{L}^n(x) > 0, \end{align*} with strict positivity following from the fact that the integrand is continuous and positive on $B(0,1)$, which has positive $\mathcal{L}^n$-measure. The integral is finite because the integrand is bounded (continuous on $\overline{B}(0,1)$, vanishing on the boundary) and $B(0,1)$ has finite measure. Setting $c := 1/\int h(1 - |x|^2) \, d\mathcal{L}^n$ gives $c > 0$ and $\int \rho \, d\mathcal{L}^n = 1$. Radial symmetry is immediate: $\rho(x) = c \, h(1 - |x|^2)$ depends on $x$ only through $|x|$. [/step]