# Convolution between Tempered distribution and schwartz function

$T$ 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?

• I think it is ok now Oct 14, 2014 at 2:49
• Does your comment indicate that you completed the proof? So why not answer your own question and let us see if it is correct.
– Vobo
Oct 16, 2014 at 16:16

Let $(T|f)$ be the pairing between tempered distributions and Schwartz functions. If $f$ is Schwartz function then set $\check f(x) = f(-x)$. Also let $F$ denote the Fourier transform. We normalize so that $F^4 = I$ on the Schwartz space. If $T$ is a tempered distribution then $F(T)$ is defined by $(T|F^{-1}(f))=(F(T)|f)$. We note that if $f$ is a Schwartz function and $T$ is a tempered distribution then $T*f$ is defined by $(T|\check f *g)=(T*f | g)$. Finally here is the calculation: $$(F(T * f)|g)= (T * f | F^{-1}(g)) = (T|\check f * F^{-1}(g)) =(T|F^{-1}(F(\check f) g)) =(F(T)|F(\check f)g)$$ So the formula is $F(T*f)=F(\check f) F(T)$.