Characterisation of positive elements in l¹(Z) Consider the Banach $^* $-algebra $\ell^1(\mathbb Z)$ with multiplication given by convolution and involution given by $a^*(n)=\overline{a(-n)}$.
I would like to find nice necessary and sufficient conditions for an element $b\in\ell^1(\mathbb Z)$ to be positive, that is, to be of the form $a^* * a$ for some $a\in\ell^1(\mathbb Z)$.
By now, I have found two necessary conditions. Namely, if $b\in\ell^1(\mathbb Z)$ is positive, then $$b(-n)=\overline{b(n)}$$ and $$\lvert b(n)\rvert\leq b(0)$$ for every $n\in\mathbb Z$.
Edit: As t3suji states in his comment below both conditions follow from the more general fact that $a$ is a positive-definite function.

Question: Is this condition also sufficient for positivity? If not, what to I have to add?

Good references would also be great.
Motivation: In the end I want to investigate the (failure of) the Gelfand–Naimark theorem for the above non-C*-algebra.
 A: Although I am not sure that answering this question will help all that much with your original motivating question/problem, I may as well post a link to Bochner's theorem (see these remarks on Wikipedia).
The passage I have in minds says:

Positive-definiteness arises naturally in the theory of the Fourier transform; it is easy to see directly that to be positive-definite is a necessary condition ... to be the Fourier transform of a function $g$ on the real line with $g(y) \geq 0$.
The converse result is Bochner's theorem, stating that a continuous positive-definite function on the real line is the Fourier transform of a (positive) measure.

A: There are four facts which clarify things a little bit:
1) The inclusion $\ell^1 {\mathbb Z} \subset C(S^1)$ preserves the spectrum. (That is Wiener's Theorem)
2) If $f = \sum_i a_i z^{i}\in \ell^1 {\mathbb Z}$ has non-negative coefficients, then $\|f\| = \|f\|_{C(S^1)}$.
3) If $f(z)>0$ for all $z \in S^1$, then $f$ has a square-root in $\ell^1 {\mathbb Z}$ by holomorphic functional calculus.
4) $2 - z-z^{-1}$ is non-negative, but has no square-root in $\ell^1{\mathbb Z}$.
