Let $\rho\in\mathscr{D}$, $0\leq\rho\leq 1, \rho(0) = 1$, and $\sum_{n\in\mathbb{Z}^d}$ $\rho(x-n) = 1$. Denote, $\rho_{n,\epsilon}() = \tau_n\rho(\frac{x}{\epsilon})$, where $\tau$ is the translation by $n$. If $\phi\in\mathscr{S}$ and $\psi\in\mathscr{D}$, then the following sum converges to $\phi*\psi$ in Schwartz as $\epsilon\searrow 0$: $$\sum_n {(\tau_{n\epsilon}\phi)\psi(n\epsilon)\int \rho_{n,\epsilon}(x)dx}$$.
Any idea how to get started on this problem? The only thing I know in this context is that the convolution of a test function and an approximate identity converges to the test function in $\mathscr{D}$, but this does not seem to be helpful.
$\int \rho_{n,\epsilon}(x)dx = \int \rho_{\epsilon}(x)dx=\epsilon^{d}\int\rho(x)dx$ and the latter integral is considered of limit $\frac{1}{2^{j}}\searrow 0$ of Riemann summation of which the partition is equilateral cube of length $\frac{1}{2^{j}}$. That is, $$\int \rho(x)dx=lim_{j\searrow \infty}\frac{1}{2^{jd}}\sum_{m\in\frac{\mathbb{Z}^d}{2^j}}\rho(m)=lim_{j\searrow \infty}\frac{1}{2^{jd}}\sum_{0\leq k_{l}s\leq 2^{j}-1}\,\,\sum_{n\in\mathbb{Z}^d}\rho[(\frac{k_{1}}{2^{j}},...,\frac{k_{d}}{2^{j}})+n]=lim_{j\searrow \infty}\frac{1}{2^{jd}}\sum_{0\leq k_{l}s\leq 2^{j}-1}1=lim_{j\searrow \infty}\frac{1}{2^{jd}}2^{jd}=1$$ where $\frac{\mathbb{Z}^d}{2^j}$ means vertices of the partition of equilateral cube of length $\frac{1}{2^j}$ and $0\leq k_{l}s\leq 2^{j}-1$ are integers. Then you can easily recognize $$\sum_n {(\tau_{n\epsilon}\phi)\psi(n\epsilon)\int \rho_{n,\epsilon}(x)dx}=\epsilon^{d}\sum_n {(\tau_{n\epsilon}\phi)\psi(n\epsilon)}$$ is also the Riemann summation of $\phi*\psi$ and the summation is finite because $\psi\in\mathscr{D}$. Then split the integral $\phi*\psi$ into summation of integrals of equilateral cube bases of length $\epsilon$ (of which the number of summation is also finite) and subtract the corresponding ones of the Riemann summation $\epsilon^{d}\sum_n {(\tau_{n\epsilon}\phi)\psi(n\epsilon)}$. The difference of $\phi$ and $\tau_{n\epsilon}\phi$ on the same cube is arbitrarily small as $\epsilon\searrow 0$ and so are $y^{m}\phi(y-x)$ and $y^{m}\tau_{n\epsilon}\phi(y)$ where $x$ belongs to that cube, because $y^{m}\phi(y-x)$ is uniformly continuous. $\psi$ is also uniformly continuous and has a compact support, we can make the total difference arbitrarily small. (Surely $y^{m}$ means the $m$ products of the coordinates $y_{1},...,y_{d}$ where $y=(y_{1},...,y_{d})$)