Is it true that $\frac{dU}{dt}*V = U*\frac{dV}{dt}$ where not both $U,V$ are distributions and $*$ denotes convolution?

Question

A theorem in my textbook states that

Let $U,V$ be distributions and $*$ denote a convolution. Then $$\frac{dU}{dt}*V = U*\frac{dV}{dt}$$

Does this theorem hold if only $U$ is a distribution but $V$ is a "normal function" like $\sin(t)$? Are regular functions also distributions?

Answer 1

You can easily answer to this question by taking the Fourier transform. You will get $$ {\cal F}\left[\int_{-\infty}^\infty dt'\frac{dV}{dt}(t-t')U(t')\right]=i\omega V(\omega)U(\omega) $$ and $$ {\cal F}\left[\int_{-\infty}^\infty dt'\frac{dU}{dt}(t-t')V(t')\right]=i\omega U(\omega)V(\omega). $$