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 with the usual $\sup$-norm, and $M(G)$ the space of complex Radon measures 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 $f:G\to{\mathbb C}$ 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, 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 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 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.

  • $\begingroup$ I suspect your "regular" means "weak* continuous" and "irregular" means "not weak* continuous". $\endgroup$
    – Nik Weaver
    Jul 25, 2022 at 16:13
  • $\begingroup$ @NikWeaver but what is "weak*", if we don't know how to restore $C_0(G)$? $\endgroup$ Jul 25, 2022 at 16:16
  • $\begingroup$ Weak* relative to the duality pairing between $C_0(G)$ and $M(G)$. $\endgroup$
    – Nik Weaver
    Jul 25, 2022 at 18:07
  • $\begingroup$ @NikWeaver the question is how to separate "good" states from "bad" states without $C_0(G)$. It is supposed that we know only the properties of $M(G)$ as a Banach algebra with involution, we don't know that it has a predual space and we can't restore this predual (unless you can suggest a construction that recovers $C_0(G)$ from $M(G)$). $\endgroup$ Jul 25, 2022 at 18:16

1 Answer 1


Your definition (2) explicitly uses the Fourier transform, so let's think about this.

For any locally compact group $G$ we can turn both $L^1(G)$ and $M(G)$ into Banach $*$-algebras, as you do. Then we can consider $*$-representations on Hilbert spaces, say $\pi:L^1(G) \rightarrow B(H)$. These are always norm-decreasing, so by taking a direct sum (say over all cyclic representations, so the sum is not too big) we can find a maximal $*$-representation, which induces a maximal $C^*$-algebra norm on $L^1(G)$. The completion is $C^*(G)$ the group $C^*$-algebra.

$L^1(G)$ is an ideal in $M(G)$ (identified as the continuous measures with respect to the Haar measure) and in fact $M(G)$ is the multiplier algebra of $L^1(G)$. Given any $\pi:L^1(G) \rightarrow B(H)$ we view $H$ as a left $L^1(G)$ module which is essential, as $\pi$ is non-degenerate. By Cohen-Hewitt Factorization every $\xi\in H$ is of the form $\pi(f)\eta$ for some $f\in L^1(G), \eta\in H$. We can hence extend $\pi$ to $\tilde\pi:M(G)\rightarrow B(H)$ by defining $\tilde\pi(\mu) \xi = \tilde\pi(\mu)\pi(f)\eta = \pi(\mu \star f)\eta$. This can be checked to be well-defined and to give a $*$-homomorphism.

When $G$ is abelian (more generally, amenable) there is no need to consider arbitrary $\pi$ but just the left-regular representation $\lambda:L^1(G)\rightarrow B(L^2(G))$, and so we identify $C^*(G)$ with $C^*_r(G)$ the reduced group $C^*$-algebra acting on $L^2(G)$. The Fourier transform $\mathcal F:L^2(G) \rightarrow L^2(\hat G)$ intertwines $C^*_r(G)$ and the action of $C_0(\hat G)$, giving an isomorphism.

The Fourier transform $\mathcal F:L^1(G) \rightarrow C_0(\hat G)$ has image the Fourier Algebra $A(\hat G)$ which is dense in $C_0(\hat G)$. This is compatible with the isomorphism between $C^*_r(G)$ and $C_0(\hat G)$.

We extend $\lambda$ to $M(G) \rightarrow B(L^2(G))$ which maps into the $C^*$-algebraic multiplier algebra $M(C^*_r(G))$. The Fourier transform identifies $M(C^*_r(G))$ with $M(C_0(\hat G)) = C^b(\hat G)$ and is compatible with $\mathcal F:M(G)\rightarrow B(\hat G)\subseteq C^b(\hat G)$ where $B(\hat G)$ is the Fourier-Stieljtes algebra.

So you are interested in states on $M(G)$ which arise by integrating a $B(\hat G)$ function against a probability measure in $M(\hat G) = C_0(\hat G)^*$. Un-winding the isomorphisms, these are the states on $M(G)$ which come from a state $\mu$ on $C^*(G)$. We first extend $\mu$ to the multiplier algebra $M(C^*(G))$, then take our mapping $M(G) \rightarrow M(C^*(G))$, and then apply the extension of the state.

You hence ask: which states on $M(G)$ arise in this way? This could be asked for any locally compact group.

The answer is: there must be a state $\mu$ on $L^1(G)$ whose canonical extension to the multiplier algebra $M(L^1(G)) = M(G)$ gives the state you started with. To show this:

  • If we start with a state on $C^*(G)$, then restrict to $L^1(G)$. The extensions to the relevant multiplier algebras are the same.
  • Conversely, if we have a state $\mu$ on $L^1(G)$ which extends to your state on $M(G)$ then we can perform the GNS construction to find $\pi:L^1(G)\rightarrow B(H)$ and $\xi\in H$ with $\mu(f) = (\xi|\pi(f)\xi)$ for each $f\in L^1(G)$. Hence $\mu$ extends to $C^*(G)$.

So how do we detect when a state $\mu$ on $M(G)$ arises from the extension of a state on $L^1(G)$? Let $(e_i)$ be a bounded approximate identity for $L^1(G)$. Then the extension $\tilde\mu$ satisfies (or is defined by) $$ \tilde\mu(\alpha) = \lim_i \mu(\alpha\star e_i) \qquad (\alpha\in M(G)). $$ As we want $\mu = \tilde\mu$, we need exactly that $\mu(\alpha) = \lim_i \mu(\alpha \star e_i)$ for each $\alpha\in M(G)$.

For example, your "bad" state $\mu:\alpha\mapsto\alpha(\{e\})$ actually annihilates $L^1(G)$, and so cannot be an extension of a state on $L^1(G)$.

Conclusion: You need to know not just $M(G)$ but also how $L^1(G)$ sits inside $M(G)$.

About extensions: let $A$ be a $C^*$-algebra, and let $\mu\in A^*$. As $A^*$ is an essential $A$-module, by Cohen-Hewitt factorisation, there are $b\in A, \lambda\in A^*$ with $\mu=a\cdot\lambda$. Then we may define $\tilde\mu:M(A)\rightarrow\mathbb C$ by $\tilde\mu(x) = \lambda(xa)$. Let $(e_i)$ be a bounded approximate identity for $A$, so that for $x\in M(A)$ we have that $xa = \lim_i e_i x e_i a$ and so $$ \tilde\mu(x) = \lambda(xa) = \lim_i \lambda(e_i x e_i a) = \lim_i \mu(e_i x e_i). $$ As we can choose $e_i$ self-adjoint, it follows that $\mu$ positive implies $\tilde\mu$ positive. If $\mu$ is a state (meaning $\|\mu\|=1$ and $\mu$ is positive) then by the norm estimates we get from Cohen-Hewitt, also $\|\tilde\mu\|=1$ so $\tilde\mu$ is a state.

  • 1
    $\begingroup$ @SergeiAkbarov For $G=\mathbb{R}$ (reals) it is possible to restore $L^1(G)$ from $M(G)$: It is the largest separable ideal in $M(G)$. $\endgroup$
    – user95282
    Jul 26, 2022 at 10:27
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    $\begingroup$ By "separable" I mean "separable in norm" --- it has a countable dense subset. A measure $\mu\in M(\mathbb{R})$ is in $L^1(\mathbb{R})$ if and only if its principal ideal $M(\mathbb{R})\ast\mu$ is separable. $\endgroup$
    – user95282
    Jul 26, 2022 at 11:00
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    $\begingroup$ R. Larsen, Proc. Amer. Math. Soc. 19 (1968), 569--572, proves that if a measure on a second countable LC group has a separable orbit then it is absolutely continuous. $\endgroup$
    – user95282
    Jul 26, 2022 at 15:11
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    $\begingroup$ For another way of recognizing $L^1(G)$ inside $M(G)$, see B.E. Johnson, Proc. Amer. Math. Soc. 15 (1964), 186--188. $\endgroup$
    – user95282
    Jul 26, 2022 at 15:36
  • 1
    $\begingroup$ Ah, sorry, a typo on my part. You should have $\tilde\mu(x) = \lambda(xa)$. $\endgroup$ Jul 28, 2022 at 19:42

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