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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 $C_c^\infty(\mathbb R)$?

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  • $\begingroup$ The easiest way is to look at $1_{x\in [0,b]} \ast T \ast f'= T \ast (f(x)-f(x-b))$, and $T' \ast f = T \ast f'$ is by definition of the distributional derivative. Otherwise let $g_a(x) = \frac{f(x+a)-f(x)}{a}$ then $\lim_{a \to 0} g_a = f'$ where the convergence is in $D(\mathbb{R})$. Therefore $(T \ast f)' = \lim_{a \to 0} T \ast g_a = T \ast f'$ $\endgroup$ – reuns Oct 15 '17 at 21:04
  • $\begingroup$ The meaning of $(T\ast f)(x)$ is by definition $T(y\mapsto f(x-y))$. Try to apply this and the fact that the sequence of functions $\{ (y\mapsto [f(x+h-y)-f(x-y)]/h) \vert h\to 0\}$ converges to $(y\mapsto f(x-y))$ in the space $\mathcal{D}$ (why?) ! And now you use the continuity of $T$. $\endgroup$ – Vobo Oct 20 '17 at 18:50

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