Corollary

The algebra $L^1(G, \nu )$ is semisimple.

This follows from Prop. 21 of I, p. 125 and Prop. 22.

One may therefore identify $L^1(G, \nu )$ with a dense involutive subalgebra of Stell(G), and the canonical injection of $L^1(G, \nu )$ into Stell(G) is then continuous.

Suppose now that the group G is unimodular (INT, VII, §1, No.$^o3$, Def. 3). One may then repeat the same arguments starting from the right regular representation $(f, \mu)\mapsto \boldsymbol{\delta }(\mu)(f) =f*\check{\mu}$ of $L^2(G, \nu )\times \mathscr{M}^1(G)$ in $L^2(G, \nu )$. One then defines a morphism $\boldsymbol{\delta }'$ of Stell(G) into $\mathscr{L}(L^2(G, \nu ))$ such that $\boldsymbol{\delta }=\boldsymbol{\delta }'\circ j$, and one writes $\boldsymbol{\delta }'(\varphi )(f) =f*\varphi$ for $f\in L^2(G, \nu )$ and $\varphi \in$ Stell(G).

For $\varphi , \psi \in$ Stell(G), one has $\boldsymbol{\delta }'(\psi )\circ \boldsymbol{\gamma }'(\varphi ) =\boldsymbol{\gamma }'(\psi )\circ \boldsymbol{\delta }'(\varphi )$, that is to say

for every $f\in L^2(G, \nu )$. In fact, this formula is true for $\varphi , \psi \in$ $L^1(G, \nu )$, and the mappings $(\varphi , \psi )\mapsto (\varphi *f)*\psi$ and $(\varphi , \psi )\mapsto \varphi *(f*\psi )$ are continuous bilinear mappings of Stell(G) $\times$ Stell(G) into $L^2(G, \nu )$.