Group algebra of a locally compact group

Remarks

With these operations $L^1(G)$ is a Banach algebra with an isometric involution, that is, a Banach $*$-algebra: $\norm{f * h}_1 \le \norm{f}_1 \norm{h}_1$, $(f * h)^* = h^* * f^*$ and $\norm{f^*}_1 = \norm{f}_1$.

Example: $G = \Z$ with counting measure. Then $L^1(\Z) = \ell^1(\Z)$, $\Delta \equiv 1$, the product is $(f * h)(n) = \sum_m f(m) h(n - m)$ and $f^*(n) = \overline{f(-n)}$. The sequence $\delta_0$ is a unit: $(\delta_0 * h)(n) = h(n)$. Here $\ell^1(\Z)$ contains the group algebra $\C[\Z]$ of finitely supported sequences as a dense subalgebra.

Non-example of a unit: $L^1(\R)$ has no unit. If $e * f = f$ for all $f \in L^1(\R)$, then taking Fourier transforms gives $\hat e = 1$ wherever some $\hat f$ is nonzero, so $\hat e \equiv 1$, which contradicts the Riemann–Lebesgue lemma $\hat e(\xi) \to 0$. In general $L^1(G)$ has a unit exactly when $G$ is discrete, but it always has a bounded approximate unit.

$L^1(G)$ is commutative if and only if $G$ is abelian.