$T$ is a tempered distribution on $\mathbb R$, $f$ is a Schwartz functions on $\mathbb R$. We define $T\ast f$ as $(T\ast f) (l)$=$T(f(-x)\ast l)$ for all $l$ Schwartz function, where the last $\ast$ is convolution. Asume $f(-x)\ast l$ is a Schwartz function and $T\ast f$ is a tempered distribution. Show that $F(T\ast f)$=$(1/(2\pi)) F(T)F(f)$ where $F$ is Fourier transform.
My idea: $F(T\ast f) (l) = T\ast f (F(l))$ Fourier transform of a tempered distribution, then $T\ast f (F(l))=T(f(-x)\ast F(l))$ definition of $T\ast f$ Now How to proceed?