Is it true that $f\ast g\in L^2(G)$, when $f$ and $g$ be in $L^2(G)$?

Question

Let $G$ be a unimodular locally compact group and $L^2(G)$ be Banach algebra with convolution product, that is, if $f$ and $g$ be in $L^2(G)$ then $f\ast g$ defined as following: $$f\ast g(x)=\int f(y)g(y^{-1}x)dy=\int f(xy)g(y^{-1})dy$$ Is it true that $f\ast g\in L^2(G)$?

For the case that $G$ be a copmact group, it is follows from invariance of Harr integrals under translations and inversions.

Answer 1

The Answer is No. In fact, $L^2(G) \ast L^2 (G) = A(G)$ where $A(G)$ is a Fourier algebra.