Is the sum of convolutions of $L^1$ functions also a convolution?

Question

For $f,g\in L^1(\mathbb{R}^d)$, their convolution is defined as $(f*g)(x)=\displaystyle\int_{\mathbb{R}^d}f(y)g(x-y){\rm d}y$, and it is standard that $f*g\in L^1(\mathbb{R}^d)$, so $*$ is an operator $L^1(\mathbb{R}^d)\times L^1(\mathbb{R}^d)\to L^1(\mathbb{R}^d)$. It would be interesting to know some basic properties of the range of convolution, but it seems even not trivial to determine if this set is closed under addition: for $f_1,f_2,g_1,g_2\in L^1(\mathbb{R}^d)$, must there exist $f,g\in L^1(\mathbb{R}^d)$ such that $$f_1*g_1+f_2*g_2 = f*g?$$ I think the tool of Fourier transform may be useful, but it is known that $\mathcal{F}:L^1(\mathbb{R}^d)\to C_0(\mathbb{R}^d)$ is not surjective, and we don't have a characterization of the range of Fourier transform. So could there be anything to say about the range of convolution of $L^1$ functions?

Answer 1

In fact, $$ L^1(\Bbb{R^d}) \ast L^1(\Bbb{R^d}) = L^1(\Bbb{R^d}), $$ which is thus in particular a vector space.

This follows from the Cohen-Hewitt factorization theorem. See here and here for details.