The L1 involution is isometric, involutive and reverses convolution

Conjugate-linearity. Let $f,g\in L^1(G)$ and $\alpha ,\beta \in \mathbb{C}$. For every $x$ the defining formula of [F1] gives $(\alpha f+\beta g{)}^{\ast }(x)={\mathrm{\Delta }}_G(x^{-1})\overline{\alpha f(x^{-1})+\beta g(x^{-1})}=\bar{\alpha }{f}^{\ast }(x)+\bar{\beta }{g}^{\ast }(x)$, as conjugation is additive and $\overline{\alpha \beta }=\bar{\alpha }\bar{\beta }$; since the identity holds pointwise it holds for the classes, so $(\alpha f+\beta g{)}^{\ast }=\bar{\alpha }{f}^{\ast }+\bar{\beta }{g}^{\ast }$.

Isometry. For $f\in L^1(G)$ the modulus of $f^{\ast }$ is $\mathrm{\mid }{f}^{\ast }(x)\mathrm{\mid }={\mathrm{\Delta }}_G(x^{-1})\mathrm{\mid }f(x^{-1})\mathrm{\mid }$, the factor being positive. Apply the change of variables [F3] (whose choice hypothesis is discharged by [A1]) to the nonnegative Borel function $H(t):={\mathrm{\Delta }}_G(t)\mathrm{\mid }f(t)\mathrm{\mid }$. Then ${\int }_G\mathrm{\mid }{f}^{\ast }\mathrm{\mid }d\mu ={\int }_{G}H(x^{-1})d\mu (x)={\int }_G{\mathrm{\Delta }}_G(x^{-1})H(x)d\mu (x)={\int }_G\mathrm{\mid }f\mathrm{\mid }d\mu$, using ${\mathrm{\Delta }}_G(x^{-1}){\mathrm{\Delta }}_G(x)=1$. Hence $\mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_1=\mathrm{\parallel }f{\mathrm{\parallel }}_1$.

Involutivity. For $f\in L^1(G)$ and $x\in G$, the definition gives $(f^{\ast }{)}^{\ast }(x)={\mathrm{\Delta }}_G(x^{-1})\overline{f^{\ast }(x^{-1})}={\mathrm{\Delta }}_G(x^{-1})\overline{{\mathrm{\Delta }}_G(x)\stackrel{‾}{f(x)}}={\mathrm{\Delta }}_G(x^{-1}){\mathrm{\Delta }}_G(x)f(x)=f(x)$, where ${\mathrm{\Delta }}_G$ is real-valued and ${\mathrm{\Delta }}_G(x){\mathrm{\Delta }}_G(x^{-1})=1$ by [F2]; hence $(f^{\ast }{)}^{\ast }=f$.

The involution preserves $C_c(G)$. If $f\in C_c(G)$ then $f^{\ast }$ is continuous by [F1] and [F8], and its support is $\mathrm{supp}(f^{\ast })=\mathrm{inv}(\mathrm{supp}f)$, a compact set because $\mathrm{inv}$ is a homeomorphism and continuous images of compact sets are compact; thus $f^{\ast }\in C_c(G)$.

Anti-multiplicativity on $C_c(G)$. Let $f,g\in C_c(G)$ and $x\in G$. By [F5] and [F1], $(f\ast g{)}^{\ast }(x)={\mathrm{\Delta }}_G(x^{-1})\overline{{\int }_{G}f(y)g(y^{-1}{x}^{-1})d\mu (y)}={\mathrm{\Delta }}_G(x^{-1}){\int }_G\overline{f(y)}\overline{g(y^{-1}{x}^{-1})}d\mu (y)$, conjugation being continuous. In the other order, [F5] and [F1] give $(g^{\ast }\ast f^{\ast })(x)={\int }_G{g}^{\ast }(y)f^{\ast }(y^{-1}x)d\mu (y)={\int }_G{\mathrm{\Delta }}_G(y^{-1}){\mathrm{\Delta }}_G(x^{-1}y)\overline{g(y^{-1})}\overline{f(x^{-1}y)}d\mu (y)={\mathrm{\Delta }}_G(x^{-1}){\int }_G\overline{g(y^{-1})}\overline{f(x^{-1}y)}d\mu (y)$, where [F2] computes ${\mathrm{\Delta }}_G(y^{-1}){\mathrm{\Delta }}_G(x^{-1}y)={\mathrm{\Delta }}_G(y^{-1}{x}^{-1}y)={\mathrm{\Delta }}_G(x^{-1})$. Substituting $y=xu$ in this last integral and using left invariance [F4] turns it into ${\mathrm{\Delta }}_G(x^{-1}){\int }_G\overline{g(u^{-1}{x}^{-1})}\overline{f(u)}d\mu (u)$, which is the first expression with the order of the two factors interchanged. Hence $(f\ast g{)}^{\ast }(x)=(g^{\ast }\ast f^{\ast })(x)$ for every $x$, so $(f\ast g{)}^{\ast }=g^{\ast }\ast f^{\ast }$ for $f,g\in C_c(G)$.

Anti-multiplicativity on $L^1(G)$. Fix $F,G\in L^1(G)$ and choose $f_n,g_n\in C_c(G)$ with $f_n\to F$ and $g_n\to G$ in $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$, possible by [F7]. By step 1.5, $(f_n\ast g_n{)}^{\ast }=g_n^{\ast }\ast f_n^{\ast }$ for every $n$. The left side converges to $(F\ast G{)}^{\ast }$: indeed $f_n\ast g_n\to F\ast G$ by joint continuity of the extension [F6], and the involution is isometric by step 1.2, hence norm continuous. The right side converges to $G^{\ast }\ast F^{\ast }$ by joint continuity [F6] applied to $g_n^{\ast }\to G^{\ast }$ and $f_n^{\ast }\to F^{\ast }$, again by step 1.2. Since limits in the normed space $L^1(G)$ are unique, $(F\ast G{)}^{\ast }=G^{\ast }\ast F^{\ast }$.

Concatenating: step 1.1 gives conjugate-linearity, step 1.2 isometry, step 1.3 involutivity and step 2.1 the reversal property, for all $f,g\in L^1(G)$. ∎