Distributional derivative vs standard derivative of mollified distribution.

Question

Define

• $\check{f}(y):=f(-y)$
• $\tau_x f(y):=f(y-x)$
• $T_f(\phi):=\int f(x)\phi(x)dx$
• $T\ast f(x):=T(\tau_x \check{f})$ (Convolution of a distribution with a function which returns a function)

I was wondering if the distributional derivative of $S$ is the same as convolving with a standard sequence of mollifiers, deriving and then taking the distributional limit. In other words, given $S$ a distribution,

$T_{(S\ast \rho_n)'}\xrightarrow{D'} S'\quad ?$

where $S'$ is the distributional derivative of $S$, $(S\ast\rho_n)'$ is the standard derivative of the function $S\ast\rho_n$, "$\xrightarrow{D'}$" denotes the convergence in the sense of distributions and $(\rho_n)_{n\ge 1}$ is a standard sequence of mollifiers.

Answer 1

Using two facts valid in $\mathcal{D}',$

1. $(S*\rho)' = S'*\rho$ for all $S\in\mathcal{D}'$ and all $\rho\in C^\infty_c$,
2. $S*\rho_n \to S$ for all $S\in\mathcal{D}'$ and all mollifiers $(\rho_n),$

we get that $(S*\rho_n)' = S'*\rho_n \to S'$ in $\mathcal{D}'.$