Is it possible to construct a vector space on convolution

Question

Let $S$ be a set of functions on $R$ such that for any two functions $f$ and $g$ in $S$, the convolution: $$ (f{\ast}g)(x)={\int}f(y)\ g(x-y)\ dy $$ exists. Since the Dirac delta is technically not a function, exclude it from any consideration. Is it possible to define an identity element for convolution that is also a member of $S$? And, is it possible to define a unique inverse element for each member of $S$ such that the convolution produces the identity element? If not can the set of functions, $S$, be restricted to some special subset such that an identity element and unique inverse elements are possible under convolution?

Answer 1

Partial answer: If functions in $S$ are integrable and if $S$ contains a function like $e^{-x^{2}}$ of $e^{-|x|}$ or $\frac 1 {1+x^{2}}$ or any function which Fourier transform never vanishes then you cannot have an identity element in $S$. This is because If $g$ is an identity element then $\hat {f} \hat {g}=\hat{f}$ and choosing $f$ to be one those special functions you get $\hat {g}\equiv 1$. But this cannot be true for any integrable function $g$.