I need a hand with this question:
If $f\in{L_1(\mathbb{R})}$ and $g\in{L_2(\mathbb{R})}$, then prove that $\widehat{f*g}=\hat{f}\cdot \hat{g}$
As a tip, i have been told to prove that:
$L_2(\mathbb{R}^n)\rightarrow L_2(\mathbb{R}^n)$
$h\rightarrow f*h$
and
$L_2(\mathbb{R}^n)\rightarrow L_2(\mathbb{R}^n)$
$h\rightarrow \hat{f}\cdot h$
are continuous.
Thanks a lot for any help.
As a consequence of Fubini's theorem, the result is true when $g\in L^1\cap L^2$. Indeed, when $n=1$, write
$$\widehat{f *g}(x)=\int_{\Bbb R}e^{-itx}\int_{\Bbb R}f(t-s)g(s)dsdt$$ and $e^{-itx}=e^{-i(t-s)x}e^{-isx}$. The case $n$ arbitrary can be done similarily.
The purpose of the hint is to jump from this case to the general case. Indeed, thanks to Plancherel's theorem and the density of $L^1\cap L^2$ in $L^2$ we will be able to conclude. Indeed, let $g\in L^2$ and $(g_k,k\geqslant 1)$ a sequence of function of $L^1\cap L^2$ such that $\lVert g-g_k\rVert_{L^2}\to 0$ and $g_k\to g$ almost everywhere. Then $\widehat{f*g_k}=\widehat f\cdot\widehat{g_k}$ for each $k$. Now take the limit $k\to \infty$.