Let $G$ be a locally compact Abelian group (we can think that $G={\mathbb R}$). Let $C_0(G)$ be the space of continuous functions $u:G\to{\mathbb C}$ [vanishing at infinity][1] with the usual $\sup$-norm, and $M(G)$ the space of [complex Radon measures][2] on $G$, understood as a Banach space dual to the Banach space $C_0(G)$:
$$
M(G)=C_0(M)^*.
$$ 
The Banach space $M(G)$ is naturally endowed with the structure of a Banach algebra with the convolution of measures as the multiplication
$$
\alpha * \beta,\qquad \alpha,\beta\in M(G),
$$
and, moreover, it also has a natural involution:
$$
\alpha^\bullet(u)=\overline{\alpha(\overline{\widetilde{u}})},\qquad \alpha\in M(G),\quad u\in C_0(G),
$$
where $\overline{z}$ is the usual involition of the complex number $z\in{\mathbb C}$, and $\widetilde{u}$ is the antipode of the function $u\in C_0(G)$:
$$
\widetilde{u}(t)=u(t^{-1}),\qquad t\in G.
$$

So $M(G)$ is a *Banach algebra with involution*. And formally we can define *states* on $M(G)$, as linear continuous functionals $\sigma:M(G)\to{\mathbb C}$ with the properties
$$
\sigma(\alpha^\bullet*\alpha)\ge 0, \qquad \sigma(\delta^e)=1,\qquad \alpha\in M(G),
$$
where $e\in G$ is the unit in $G$ and $\delta^e$ is the delta-measure supported in $e$.

It is more or less obvious, that each normalized *continuous* [positive-definite function][3] $f:G\to{\mathbb C}$ defines a state on $M(G)$ by the formula
$$
\sigma(\alpha)=\int_G f(t)\ \alpha(d t) \tag{1}
$$
and by the [Bochner theorem][4], we can represent the action of $\sigma$ as the integral by some positive measure $\mu$ on the Pontryagin dual group $\widehat{G}$:
$$
\sigma(\alpha)=\int_{\widehat{G}} {\mathcal F}(\alpha)(\chi)\ \mu(d\chi) \tag{2}
$$
where 
$$
{\mathcal F}(\alpha)(\chi)=\int_G \chi(t)\ \alpha(d t),\qquad \chi\in\widehat{G}
$$
is the Fourier transform of the measure $\alpha$.

I would call such states $\sigma$ on $M(G)$ *"regular states"*, since intuitively they are what we expect them to be.

But strange thing, there are some other, *"irregular states"* on $M(G)$. Namely, the function
$$
f(t)=\begin{cases}1, & t=e\\ 0, & t\ne e\end{cases}
$$
(where $e\in G$ is the unit of the group $G$) gives another state by the same formula (1), which can be simplified in this case as follows:
$$
\sigma(\alpha)=\int_G f(t)\ \alpha(d t)=\alpha(\{e\}). \tag{3}
$$
And if $G$ is not discrete, then $f$ cannot be a continuous function on $G$, and in this case the state (3) cannot be represented in the form (2) (i.e., as an integral of the Fourier transform by some measure $\mu\ge 0$ on $\widehat{G}$).

And I deduce from this that there is a qualitative difference between the states on $M(G)$, which can be represented in the form (2) and those that cannot. My question is

> what is this qualitative difference between the states (2) and (3) in terms of the properties of $M(G)$ considering it as the Banach algebra with involution?

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 $M(G)$.)

My conjecture is that perhaps the difference between (2) and (3) is that these states relate differently to the monotonicity conditions. Using the [Beppo Levi lemma][5] one can show that if a sequence of measures $\alpha_n\in M(G)$ tends to a measure $\alpha\in M(G)$ monotonously with respect to the *preorder on $M(G)$ generated by the involution*,
$$
0\le\alpha_1\le\alpha_2\le...\le\alpha_n\le...\le\sup_n\alpha_n=\alpha
$$
then the values of the state $\sigma$ of the form (2) tend to what we need:
$$
\sigma(\alpha_n)\underset{n\to\infty}{\longrightarrow}\sigma(\alpha) \tag{4}
$$
But if we consider the state $\sigma$ of the form (3), we cannot use the Bochner theorem, cannot represent $\sigma$ in the form (2), and because of that I am not sure that this state satisfies the condition (4). So a simplified version of my question is the following:

> is the state $\sigma$ of the form (3) on $M(G)$ monotone in the sense that it preserves the monotonous convergence of sequences? 

I think this must be simple, but I stuck in this, and if somebody could clarify this to me, I would be very grateful. 

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.


  [1]: https://en.wikipedia.org/wiki/Vanish_at_infinity
  [2]: https://www.wiley.com/en-us/Real+Analysis%3A+Modern+Techniques+and+Their+Applications%2C+2nd+Edition-p-9780471317166
  [3]: https://en.wikipedia.org/wiki/Positive-definite_function_on_a_group
  [4]: https://en.wikipedia.org/wiki/Bochner%27s_theorem
  [5]: https://en.wikipedia.org/wiki/Monotone_convergence_theorem#Beppo_Levi's_lemma