Proof of Approximation of a Twice Continuously Differentiable Function with Bounded First and Second Derivatives by Test Functions

Each result cited is universally quantified over the data in its own statement.

Step 0 (growth of $f$). By Gradients of Functions with Bounded Derivatives Belong to the Tangent Space; Constants Are Tangent; the Score Identity; the Score Has Mean Zero; Translation of Tangent Fields §tangent, read with its dimension equal to $q$ (its measure plays no role in that inequality), $|f(x)|\le|f(0_{\mathbb{R}^{q}})|+\sqrt{q}\,M\lVert x\rVert$ for every $x\in\mathbb{R}^{q}$.

Step 1 (cutting off). Let $\chi$, $\chi_{R}$ and the constants $M_{1},M_{2}$ be as in Scaled Cutoffs and the Second-Moment Test Functions: Uniform Derivative Bounds and Agreement on a Ball §cutoff. For $n\in\mathbb{N}$, read as a positive real number, let $h_{n}=\chi_{n}f$. It is of class $C^{2}$ by Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set, vanishes at every $x$ with $\lVert x\rVert\ge2n$, so that $h_{n}$, and each of its first partial derivatives (which vanish on the open set $\{\lVert x\rVert>2n\}$), is compactly supported by Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set; by the product rule

The function $\chi_{n}$ is constant on the open sets $\{\lVert x\rVert<n\}$ (value $1$) and $\{\lVert x\rVert>2n\}$ (value $0$), so its first and second partial derivatives vanish there, by Partial Derivative on a Euclidean Open Set. At the remaining points, $n\le\lVert x\rVert\le2n$ and Step 0 gives $|f(x)|\le(|f(0_{\mathbb{R}^{q}})|+2\sqrt{q}M)\,n=:A\,n$. Hence, for all $x$, using $1\le n$, $0\le\chi_{n}\le1$ and the bounds of Scaled Cutoffs and the Second-Moment Test Functions: Uniform Derivative Bounds and Agreement on a Ball §cutoff,

Let $M'$ be the larger of these two bounds. Moreover, if $\lVert x\rVert<n$, then $h_{n}=f$ on the open ball $\{\lVert y\rVert<n\}$ around $x$, so $\partial_{i}h_{n}(x)=\partial_{i}f(x)$ and $\partial_{j}\partial_{i}h_{n}(x)=\partial_{j}\partial_{i}f(x)$.

Step 2 (mollifying). Let $\rho$ be a mollifier kernel of radius $1$ on $\mathbb{R}^{q}$ (Existence of Mollifier Kernels of Every Radius) and, for $\varepsilon>0$, $\rho_{\varepsilon}(y)=(\varepsilon^{-1})^{q}\rho(\varepsilon^{-1}y)$, a mollifier kernel of radius $\varepsilon$ by Rescaling a Mollifier Kernel. With $\Omega=\mathbb{R}^{q}$ in Convolution of a Continuous Function with a Compactly Supported Continuous Kernel one has $\Omega^{\varepsilon}=\mathbb{R}^{q}$, so for every continuous $g:\mathbb{R}^{q}\to\mathbb{R}$ the convolution $g*\rho_{\varepsilon}$ is defined on $\mathbb{R}^{q}$; if $|g|\le K$ then $|g*\rho_{\varepsilon}|\le K\int\rho_{\varepsilon}\,d\lambda_{q}=K$ by monotonicity of the integral (Linearity and Monotonicity of the Lebesgue Integral) and $0\le\rho_{\varepsilon}$. Let $g$ be of class $C^{1}$ and compactly supported, and let $i\in[q]$ and $x\in\mathbb{R}^{q}$. By Differentiating a Convolution through the Kernel, claim 2, $\partial_{i}(g*\rho_{\varepsilon})(x)=\int g(x-y)\,\partial_{i}\rho_{\varepsilon}(y)\,dy$. The function $y\mapsto g(x-y)$ is of class $C^{1}$ with $i$th partial derivative $-(\partial_{i}g)(x-y)$, by claim 1 of A Composition of $C^k$ Maps Between Euclidean Open Sets is of Class $C^k$, the map $y\mapsto x-y$ being smooth by claims 2 and 3 of Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set; and $\rho_{\varepsilon}$ is of class $C^{1}$ and compactly supported, as it vanishes whenever $\lVert y\rVert>\varepsilon$ (Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set). So Integration by Parts on Euclidean Space Against a Compactly Supported Function of Class $C^{1}$ §parts gives

Applying (D) to $g=h_{n}$ and then to $g=\partial_{i}h_{n}$ (of class $C^{1}$ and compactly supported) gives $\partial_{i}(h_{n}*\rho_{\varepsilon})=(\partial_{i}h_{n})*\rho_{\varepsilon}$ and $\partial_{j}\partial_{i}(h_{n}*\rho_{\varepsilon})=(\partial_{j}\partial_{i}h_{n})*\rho_{\varepsilon}$, both bounded by $M'$ by Step 1. For $0<\varepsilon\le1$ the function $h_{n}*\rho_{\varepsilon}$ is smooth by claim 2 of Convolution with a $C^k$ Kernel is of Class $C^k$, and it vanishes whenever $\lVert x\rVert>2n+1$, since then the integrand $h_{n}(x-y)\rho_{\varepsilon}(y)$ vanishes for every $y$: either $\lVert y\rVert>\varepsilon$ and $\rho_{\varepsilon}(y)=0$, or $\lVert y\rVert\le1$ and $\lVert x-y\rVert\ge\lVert x\rVert-\lVert y\rVert>2n$ by the triangle inequality (claim 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$), so $h_{n}(x-y)=0$; so it is compactly supported by Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set, hence a test function.

Step 3 (choice of the radius). Let $K_{n}=\bar{B}(0_{\mathbb{R}^{q}},2n+2)$, compact by A Closed Euclidean Ball is Convex and Compact. Each of the finitely many continuous functions $\partial_{i}h_{n}$ and $\partial_{j}\partial_{i}h_{n}$ satisfies Mollification Converges Uniformly on Compact Subsets (with $\Omega=\mathbb{R}^{q}$, $\delta=1$, $K=K_{n}$ and $\eta=n^{-1}$). Let $\varepsilon_{n}$ be half the least of the finitely many radii it provides and of $1$; then $\varepsilon_{n}\in(0,1]$ lies strictly below each of those radii, so that theorem applies with $\varepsilon=\varepsilon_{n}$ and gives

for all $x\in K_{n}$ and $i,j\in[q]$. Let $\psi_{n}=h_{n}*\rho_{\varepsilon_{n}}\in C_{c}^{\infty}(\mathbb{R}^{q})$. By Step 2, its first and second partial derivatives are bounded by $M'$.

Step 4 (convergence). Fix $x\in\mathbb{R}^{q}$ and $i,j\in[q]$. For every $n>\lVert x\rVert$ (such $n$ exist by The Archimedean Property of the Real Numbers, and all larger $n$ qualify) one has $x\in K_{n}$, and by Steps 1–3

and likewise for the second derivatives. Since $n^{-1}\to0$, the stated limits hold by Limit of a Sequence of Real Numbers. $\blacksquare$