Naive inversion is not the L1 involution on a nonunimodular group

The modulus of $Jf$ and the inversion formula. For a nonnegative $f\in C_c(G)$, also $Jf\in C_c(G)$ because inversion is a homeomorphism, and $\mathrm{\mid }Jf(x)\mathrm{\mid }=f(x^{-1})$. Thus [F2] gives $\mathrm{\parallel }Jf{\mathrm{\parallel }}_1={\int }_{G}f(x^{-1})d\mu (x)={\int }_G{\mathrm{\Delta }}_G(x^{-1})f(x)d\mu (x)$, while $\mathrm{\parallel }f{\mathrm{\parallel }}_1={\int }_{G}f(x)d\mu (x)$; both integrals are finite by [F4].

A witness in the region $a>1$. Take the open rectangle $U:=(1,2)\times (0,1)$ and a compact rectangle $K:=[3\mathrm{/}2,7\mathrm{/}4]\times [1\mathrm{/}4,3\mathrm{/}4]\subseteq U$, and let $f\in C_c(G)$ be given by [F5] under the Dependent Choice derived from the assumed AC, with ${\mathbf{1}}_K\le f\le {\mathbf{1}}_U$. Then $f\ge 0$, $f\equiv ̸0$, and on $\mathrm{supp}f\subseteq U$ one has $a>1$, hence $a^{-1}>a^{-2}$ everywhere on $(1,2)$.

Comparison of norms. Since $f\ge 0$ and $\mathrm{supp}f\subseteq \{a>1\}$, step 1.1 gives $\mathrm{\parallel }Jf{\mathrm{\parallel }}_1-\mathrm{\parallel }f{\mathrm{\parallel }}_1={\int }_G({\mathrm{\Delta }}_G(x^{-1})-1)f(x)d\mu (x)={\int }_G(a-1)f(a,b)a^{-2}dadb$, where $a-1>0$ on the support of $f$ and $f>0$ on the nonempty open set where $f\ge {\mathbf{1}}_K=1$, namely on the interior of $K$. The integrand is nonnegative continuous with a strictly positive value on an open subset of $G$, and $a^{-2}dadb$ gives positive measure to every nonempty open set, so $\mathrm{\parallel }Jf{\mathrm{\parallel }}_1-\mathrm{\parallel }f{\mathrm{\parallel }}_1>0$: the norms differ.

Since $J$ changes the $L^1$ norm of the nonzero compactly supported function $f$ of step 1.2, it cannot define an isometry of $L^1(G)$; the modular involution [F3], by contrast, is isometric. The naive inversion $f\mapsto \overline{f(x^{-1})}$ is therefore not the $L^1$ involution on this nonunimodular group. ∎