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So my conjecture was not true, I am sorry. However, the initial (main) question remains open: if somebody could explain how to separate "regular states" from "irregular" using only the properties of $M(G)$, that would be great.

So my conjecture was wrong, I am sorry. However, the initial (main) question remains open: if somebody could explain how to separate "regular states" from "irregular" using only the properties of $M(G)$, that would be great.

P.S. Excuse me, I understood that the state (3) satisfies (4), because this functional acts only on the discrete part $\alpha^d$ of the measure $\alpha$ $$ \sigma(\alpha)=\sigma(\alpha^d) $$ and this means that we can consider the discrete topology on the group $G$, and with this topology $G$ becomes again a locally compact Abelian group $G^d$, and $\sigma$ can be considered as a state on the algebra $M(G^d)$, and we again can apply the Bochner theorem.

So my conjecture was not true, I am sorry. However, the initial (main) question remains open: if somebody could explain how to separate "regular states" from "irregular" using only the properties of $M(G)$, that would be great.

In other words if we have a state $\sigma:M(G)\to{\mathbb C}$, do we have a possibility to understand if it can be represented in the form (2) if we look only at the properties of $\sigma$ as a functional on $M(G)$? (We can't restore the topology of $G$ from $M(G)$, so our answer can't be like this: $\sigma$ must be generated by some continuous positive-definite function $f$.)

In other words if we have a state $\sigma:M(G)\to{\mathbb C}$, do we have a possibility to understand if it can be represented in the form (2) if we look only at the properties of $\sigma$ as a functional on $M(G)$? (We can't restore the topology of $G$ from $M(G)$, so our answer can't be like "$\sigma$ must be generated by some continuous positive-definite function $f$". Equally, we can't say "$\sigma$ must be $C_0(G)$-weakly continuous", because we can't restore $C_0(G)$ from $M(G)$.)