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I've seen it stated in several sources and lecture notes for Abstract Harmonic Analysis that for a locally compact group $G$, $L^{1}(G)$ is unital if and only if $G$ is discrete.

What about the locally compact group $\mathbb{T} = \{\lambda\in\mathbb{C}: |\lambda| = 1\}$, which is not discrete because the arclength measure of a point on the unit circle is $0$.

But since it is compact, the constant function $1\in L^{1}(G)$.

roo
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    But the multiplication in $L^1$ is not pointwise multiplication (the pointwise product is generally not in $L^1$), but convolution. The unit for convolution is the Dirac measure in the unit of $G$. – Daniel Fischer Nov 04 '13 at 01:18
  • Thank you! Is that the map $\delta_{1} = \chi_{{1}}$? – roo Nov 04 '13 at 01:26
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    If $G$ is discrete, it is represented by $\chi_{{1}}$, otherwise, you can't represent it by a function, since it's not absolutely continuous wrt the Haar measure. – Daniel Fischer Nov 04 '13 at 01:29
  • Where Can I get a proof for it? – budi Sep 26 '19 at 08:53

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The multiplication in the algebra $L^1(G)$ is convolution, not the pointwise product. For groups like $\mathbb{T}$ or $\mathbb{R}^n$, the pointwise product of $L^1$ functions is generally not in $L^1$.

You can extend the convolution to the space of Borel measures of bounded variation, and then you get a unital algebra, where the unit is the Dirac measure (point mass) in the unit of $G$. You can represent a point mass as an $L^1$ function with respect to the Haar measure only if singletons have positive measure. That is the case only if $G$ is discrete.

Daniel Fischer
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  • Can you say a bit more about why isn't a point mass isn't an $L^1$ function? It seems measurable, since the pre-image of {0}, {1} are both measurable (albeit the latter with 0 measure) and the integral would evaluate to 0 (0 * measure of domain w value 0 = 01, plus 1 measure of domain w value 1 = 1*0), hence is finite? – lukemassa Apr 25 '21 at 17:25
  • Oh wait, is the issue that L^1 isn't all the integrable functions, but rather equivalence classes of integrable functions, and the point mass function when points don't have measure is equivalent to 0? – lukemassa Apr 25 '21 at 17:31
  • If $G$ is not discrete, then the point mass $\delta_e\not\in L^1(G)$, but one would still have to show that there is no other possible unit of $L^1(G)$ (which is not a unit of $M(G)$). – Sha Vuklia Jan 11 '24 at 19:44