 L1 convolution algebra

With left Haar measure, $(f\ast g)(x)={\int }_{G}f(y)g(y^{-1}x)dm(y)$ makes $L^1(G)$ a Banach algebra, since $\Vert f\ast g{\Vert }_1\le \Vert f{\Vert }_1\Vert g{\Vert }_1$. For an Abelian group it is commutative, and the involution is $f^{\ast }(x)=\overline{f(x^{-1})}$. Its nonzero characters of an algebra are exactly integration against inverses of continuous unitary group characters.

// Target: geometry-and-topology.bigb