\overline{ f(y)K(y^{-1}x)}\,dy\right)dx \\
& = \int_G 
 \left(\int_G g(x) K(x^{-1}y)\,dx\right) \overline{ f(y)}\,dy\\
& =\langle{ g\ast K},{f}\rangle_{L_2},\end{aligned}$$ where $\overline{K(y^{-1}x)}=K(x^{-1}y)$. Note that we can interchange the integral by Fubini’s theorem since $$\int_G \lvert{g(x)}\rvert
 \left(\int_G \lvert{ f(y)K(y^{-1}x)}\rvert\,dy\right)dx \leq \lVert{g}\rVert_{L_{r',m^{-1}}}\cdot
 \lVert{\lvert{f}\rvert*\lvert{K}\rvert}\rVert_{L_{r,m}}$$ and $\lvert{f}\rvert*\lvert{K}\rvert\in L_{r,m}(G) $ by Young’s inequality  with $q=1$ and $p=r$, $f\in L_{1,m}(G)$ and $g=K\in L_{r,w}(G)$. Note that Fubini’s theorem shows that $$\int |f(y)| \cdot (|g| \ast |K|)(y) \, dy
      < \infty \, .$$ Since this holds for any $f \in C_c (G)$, we see $|g| \ast |K| < \infty$ almost everywhere, so that $g$ and $K$ are convolvable. By density of $C_c(G)$ in $ L_{r,m}(G) $ we get that $RC_K^*g=g\ast K$, so that $g\ast K\in  L_{r',m^{-1}}(G) $. Hence the convolution operator acts continuously on $ L_{r',m^{-1}}(G) $ and it coincides with $RC_K^*$.

To show the second claim, observe first that for any $f\in
C_c(G)\subset \mathcal{T}_w\subset  L_{r',m^{-1}}(G) $, since $K\in\mathcal{T}_w$, both $\lvert{f}\rvert*\lvert{K}\rvert$ and $(\lvert{f}\rvert*\lvert{K}\rvert)\ast\lvert{K}\rvert$ exist, so that by (77d) of [@dadedelastte17] the convolution is associative and $$RC_K^2 f = (f\ast K)\ast K = f\ast (K\ast K)= f \ast K = RC_K f.$$ By density, and since $RC_K$ is bounded on $ L_{r,m}(G) $ by assumption, we get that $RC^2_K=RC_K$ and hence $\operatorname{Ran}RC_K\subset \mathcal M_{r,m}$. 