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The Gelfand-Naimark theorem establishes that a complex commutative Banach algebra $A$ with an identity and an involution $x\to x^*$ satisfying $\|x x^*\|=\|x\|^2$ is (isometrically isomorphic to) a $C(K)$ space where $K$ is the compact space consisting of the maximal ideals in $A$.

Is there an analogous characterization of the complex commutative Banach algebras with an identity and an involution which are a convolution algebra $L_1(G)$ with $G$ a locally compact abelian group?

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Rieffel has given in his Ph.D. thesis a fairly satisfactory characterisation of commutative $L_1(G)$-algebras. This requires some terminology.

Let $A$ be a commutative (possibly non-unital) Banach algebra and let $f\in A^*$ be a (norm-one) character. Then $f$ is termed $L^\prime$-inducing, when $A$ is an abstract $L$-space under the ordering given by the cone

$$\{x\in A\colon \|x\| = f(x)\}$$

and $| x\cdot y | \leqslant |x| \cdot |y|$ in the sense of the above-introduced ordering (here $|x|$ stands for the Banach-lattice modulus). Rieffel then offers the following characterisation:

Theorem. Let $A$ be a commutative, semi-simple Banach algebra. Suppose that every norm-one character on $A$ is $L^\prime$-inducing and $A$ is Tauberian. Then $A$ is Banach-algebra isometrically isomorphic to $L_1(G)$ for some locally compact group $G$.

It is to be noted that there exist characterisations of $L_1(S)$-algebras, where $S$ is a locally compact semigroup, that are in a spirit similar to the above result. There is a recent (2016) result by Lau and Hung extending this further to general Fourier-Stieltjes algebras.

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This is a partial answer for the case where $G$ is discrete (equivalently, $L^1(G)$ is unital). First note that if $G$ is a locally compact group then $G$ is discrete if and only if its dual group $\widehat{G}$ is compact. Furthermore, in this case $L^1(G)$ is isomorphic to $C(\widehat{G})$.

Edit: This last assertion is wrong. However, $\widehat{G}$ is still the spectrum of $L^1(G)$, and so the Gelfand transform $L^1(G) \longrightarrow C(\widehat{G})$ exhibits $C(\widehat{G})$ as the $\mathrm{C}^*$-algebra generated from $L^1(G)$ (i.e, every $*$-algebra map $L^1(G) \longrightarrow A$ to a $\mathrm{C}^*$-algebra $A$ factors through a unique $\mathrm{C}^*$-algebra map $C(\widehat{G}) \longrightarrow A$). Consequently, the observations below (which I still believe are correct), do not seem to be very relevant to the question.

Now commutative unital $\mathrm{C}^*$-algebras of the form $C(\widehat{G})$ for $\widehat{G}$ a compact (Hausdorff) abelian group are exactly those commutative unital $\mathrm{C}^*$-algebras which carry a compatible Hopf algebra structure (where the tensor product of $\mathrm{C}^*$-algebras is the completed tensor product). This just follows from the fact that the functor $K \mapsto C(K)$ is a contravariant equivalence, and that the coproduct in the category of commutative unital $\mathrm{C}^*$-algebras coincides with the completed tensor product.

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    $\begingroup$ Can $L^1(G)$ be isomorphic to $C(\hat G)$ when $G$ is infinite discrete? $\endgroup$
    – M.González
    Commented Sep 15, 2015 at 19:39
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    $\begingroup$ The third sentence is, for infinite G, incorrect. As noted many times on MO, $L^1(\Omega)$ cannot be isomorphic as a Banach space to $C(K)$ if $K$ is infinite. $\endgroup$
    – Yemon Choi
    Commented Sep 15, 2015 at 19:44
  • $\begingroup$ Talk of Hopf algebras etc should instead characterize $L^1(G)$ for G locally compact (not nec abelian) as the predual of a commutative Kac algebra. In my view this is really not in the spirit of the original question, but perhaps @M.González can comment on this? $\endgroup$
    – Yemon Choi
    Commented Sep 15, 2015 at 19:47
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    $\begingroup$ I was thinking about a practical criterium. For example, it obviously follows from the Gelfand-Naimark theorem that the space $Bo[0,1]$ of bounded Borel functions on $[0,1]$ is a $C(K)$ space. Moreover I am not familiar with Hopf algebras and Kac algebras. $\endgroup$
    – M.González
    Commented Sep 16, 2015 at 7:13
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    $\begingroup$ Unfortunately your guess is still incorrect, in general, since $L^1(G)$ always has trivial Jacobson radical (this works for all locally compact $G$; for LCA $G$ this comes out of Fourier analysis applied to the Gelfand transform, as shown in e.g. Rudin's book Fourier Analysis on Groups). So when $G$ is infinite, $L^1(G)\to C_0(\widehat{G})$ is never surjective. In the case $G={\bf Z}$ this boils down to the concrete statement: there exist continuous functions on the circle whose Fourier series are not absolutely summable $\endgroup$
    – Yemon Choi
    Commented Sep 16, 2015 at 16:18

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