Is it possible to talk about the spectrum of a function? I know that $$A:=(L^1(\mathbb R^n,\mathbb C),+,*)$$ is a Banach algebra (the product operation ‘$*$’ is the usual convolution product). Therefore, for some function $f\in L^1$, we we can talk about the spectrum $\sigma(f)$, that is, the set of all scalars $\lambda\in \mathbb C$ such that $f-\lambda \mathbf{1}$ is not invertible in $A$.
Does this mean that the spectrum of an $L^1$ random variable kind of corresponds with the values it might take on?
Edit: The Banach algebra $A$ is not unital as pointed out in the comments, but one could in principle consider a unital algebra which contains $A$ (which always exists by general facts). In particular, one could consider the algebra $A_e=A\oplus \mathbb C$ with the obvious norm $$\|(f,z)\|_{A_e}:=\|f\|_{L^1}+|z|$$ (where one morally adds the dirac delta to $L^1(\mathbb R^n)$).
This is a partial answer. I consider the unital B.a. $A_e$ as in the edited question. The spectrum of a function $f\in L^1$ always contains $0$ and it is bounded by the $L^1$ norm of $f$. The question is, does it contain something besides zero?
Consider the Fourier transform $\hat f$ of $f$. On the Fourier side, the operator $f-\lambda \mathbf 1$ is the multiplication operator by the function $\hat f(\xi)-\lambda$. The inverse of this operator is the multiplication operator by the function $(\hat f(\xi)-\lambda)^{-1}$, and we want to know when this is an element of the algebra $A_e$. It is clear that, if $\lambda$ is in the range of $\hat f$, then the function $(\hat f(\xi)-\lambda)^{-1}$ is unbounded, and it cannot be the Fourier transform of some element of $A_e$. This argument implies that $$ \overline{\operatorname{range}(\hat f)}\subseteq\sigma(f). $$ It is possible that the equality holds, it seems likely to me (and it would be quite a nice fact). I suspect this is something other people have done, but I couldn’t find a reference for that.
Edit: I think one can replace the $L^1$ norm by the $L^\infty$ norm of the Fourier transform and run the exact same argument again to obtain the equality. I am not sure on whether this makes sense though, I hope someone else will make this point clear.