Involution on L1 of a locally compact group

Assume AC. Let $G$ be an LCH group with a fixed left Haar measure $\mu$, let ${\mathrm{\Delta }}_G$ be the modular function of $G$ (Modular function of a locally compact group), and write $L^1(G):=L^1(G,\mu ;\mathbb{C})$ with norm $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$ (Complex Haar L^p spaces and compactly supported functions). For a complex-valued function $f$ on $G$ put $f^{\ast }(x):={\mathrm{\Delta }}_G(x^{-1})\overline{f(x^{-1})}(x\in G),$ and define the involution of $L^1(G)$ to be the map $f\mapsto f^{\ast }$ that assigns to the class of $f$ the class of $f^{\ast }$.

Thus the involution is the composition of inversion, complex conjugation and multiplication by the continuous positive function ${\mathrm{\Delta }}_G\circ \mathrm{inv}$, where $\mathrm{inv}(x)=x^{-1}$. On a unimodular group ${\mathrm{\Delta }}_G\equiv 1$ (Unimodular locally compact group) and the involution reduces to the naive inversion $f^{\ast }(x)=\overline{f(x^{-1})}$.

Well-definedness. The map $\mathrm{inv}$ is a homeomorphism of $G$ (Left and right translations and inversion in a topological group are homeomorphisms), so $\mathrm{inv}$ and its inverse carry Borel sets to Borel sets and a Borel function $f$ has $f\circ \mathrm{inv}$ Borel; ${\mathrm{\Delta }}_G\circ \mathrm{inv}$ is continuous, hence Borel (The modular function is a continuous homomorphism, Continuous functions on Euclidean spaces are Borel measurable), and complex conjugation is continuous. So $f^{\ast }$ is measurable whenever $f$ is. Integrability and norm: the change of variables under inversion (Haar change of variables under inversion) applied to the nonnegative Borel function $H(t):={\mathrm{\Delta }}_G(t)\mathrm{\mid }f(t)\mathrm{\mid }$ gives ${\int }_G\mathrm{\mid }{f}^{\ast }\mathrm{\mid }d\mu ={\int }_{G}H(x^{-1})d\mu (x)={\int }_G{\mathrm{\Delta }}_G(x^{-1})H(x)d\mu (x)={\int }_G\mathrm{\mid }f\mathrm{\mid }d\mu <\mathrm{\infty },$ because ${\mathrm{\Delta }}_G(x^{-1}){\mathrm{\Delta }}_G(x)=1$, so $f^{\ast }$ is $\mu$-integrable with $\mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_1=\mathrm{\parallel }f{\mathrm{\parallel }}_1$. Independence of the representative: applying the same identity to the indicator of a Borel set $E$ gives $\mu (E^{-1})={\int }_E{\mathrm{\Delta }}_G(x^{-1})d\mu (x)$, and the density ${\mathrm{\Delta }}_G\circ \mathrm{inv}$ is strictly positive, so $\mu (E^{-1})=0$ exactly when $\mu (E)=0$; hence $f=0$ a.e. implies $f^{\ast }=0$ a.e., and two representatives of one class of $L^1(G)$ give two representatives of one class. The assignment is therefore a well-defined map $L^1(G)\to L^1(G)$.

Normalisation. The factor ${\mathrm{\Delta }}_G(x^{-1})$ is the one that makes the involution isometric, $\mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_1=\mathrm{\parallel }f{\mathrm{\parallel }}_1$, and later makes it reverse convolution, $(f\ast g{)}^{\ast }=g^{\ast }\ast f^{\ast }$ (The L1 involution is isometric, involutive and reverses convolution); without it the map $f\mapsto \overline{f(x^{-1})}$ is not isometric on a nonunimodular group (Naive inversion is not the L1 involution on a nonunimodular group ↗).