Suppose $f,g \in \mathcal{S}(\mathbb{R})$, where $\mathcal{S}(\mathbb{R})$ denotes the Schwarz space on $\mathbb{R}$. Then we need to show that $$ x (f *g) = (xf)*g \;\;\;\; \mbox{ a.e. }$$ So, can anyone help me in proving this?
P.S. : Convolution of 2 functions is defined as $$ (f*g )(x) = \int_{-\infty}^{\infty} f(x-y) g(y) \; dy $$