Convolution of L1 & L2 function: definition

Question

A book that I'm reading makes the following statement that I'm not sure how to understand: On $\mathbb R^n$, if $f\in L1$ and $g\in L2$, we have: $$\widehat{f*g}=\hat f \hat g$$ How do I read it? I know that the assumptions imply $\hat f$ is continuous and bounded, $\hat g\in L2$ so that the product on the right-hand side is a well-defined $L2$-function. Hence it is the Fourier transform of an $L2$-function $(\hat f \hat g\check{)}$ which, apparently, is equal to $f*g$. For all I know it could be the definition of $f*g$ in this case. But is $f*g$ actually defined by an a.e. convergent integral as in the case $f,g\in L1$?

Answer 1

Hint: notice that in order to define $f\ast g$ it is enough to assume $f$ and $g$ measurable such that $y\mapsto f(x-y)g(y)$ is in $L^1$. (Of course it is sufficient to assume $f,g\in L^1$ by Young's theorem, but is not necessary.)