Derivative of Convolution of Distribution and Test Function on $\mathbb R$ Let $T\in\mathcal D'(\mathbb R)$ be a distribution and let $f\in C^\infty_c(\mathbb R)$ be a test function.
How may I show that
$$(T*f)'=T'*f=T*f'\tag*{?}$$
This is what I did:
$$\begin{align*}(T*f)'(x)&=\lim_{h\to0}\frac{T*f(x+h)-T*f(x)}{h}\\&=\lim_{h\to0}\frac{\int T(y)f(x+h-y)-\int T(y)f(x-y)}{h}\\&=\lim_{h\to0}\frac{\int T(y)(f(x-y+h)-f(x-y))}{h}\\&=\int T(y)\lim_{h\to0}\frac{f(x-y+h)-f(x-y)}{h}\tag*{DCT}\\&=\int T(y)f'(x-y)\\&=T*f'(x)\end{align*}$$
Since convolution is commutative, one can similarly show that $(T*f)'=T'*f$.
Therefore, $?$ holds.
However, I feel that there is something off about my argument. For example, I did not use the fact that $f$ is compactly supported. Moreover, is $T$ not supposed to take arguments in $C_c^\infty(\mathbb R)$ and not in $\mathbb R$ since $T$ is a linear functional on $C_c^\infty(\mathbb R)$?
 A: There's a major technicality in your argument which involves the commuting of limits, which requires you to comment on the different topologies in play (the weak-* topology on $\mathcal{D}'(\mathbb{R})$ compared to the natural projective limit topology of $C^{\infty}_c$). Furthermore, the proof won't be symmetric in $T$ and $g$. Instead, here's how I'd prove it:
For $g\in C^{\infty}(\mathbb{R})$, denote $\tilde{g}(x)=g(-x)$. Then, by definition, for any test-function $\varphi\in C_c^{\infty}(\mathbb{R})$,
$$
\langle T*f,\varphi\rangle= \langle T,\varphi\;*\tilde{f}\rangle
$$
Thus, by definition of the weak derivative and the fact that $g'*\tilde{f}=g*\tilde{f}'$ (since they are both $C^1$ in the usual sense),
$$
\langle (T*f)',g\rangle=-\langle T*f,g'\rangle=-\langle T,g'*\tilde{f}\rangle=-\langle T,g* \tilde{f}'\rangle=\langle T,g * \tilde{(f')}\rangle=\langle T*f',g\rangle
$$
For the other identity, we simply follow the first three equalities above and then realise, again since $f$ and $g$ are $C^1$, that
$$
-\langle T,g*\tilde{f}'\rangle=-\langle T,(g*\tilde{f})'\rangle=\langle T',g*\tilde{f}\rangle=\langle T'*f,g\rangle
$$
