= c^{-1} \langle RC_K h, g \rangle_{L_2}
      = c^{-1} \langle h, RC_K g \rangle_{L_2}
      = c^{-1} \langle h, g \rangle_{L_2}
      = c^{-1} \|g\|_{L_{r',m^{-1}}} \, .$$ Hence, $c^{-1} \cdot \|g\|_{L_{r',m^{-1}}} \leq \|\Gamma_g\| \leq \|g\|_{L_{r',m^{-1}}}$.

We now prove that the map $g\mapsto \Gamma_g$ is surjective. Take $\Gamma$ in the anti-linear dual of $\mathcal M_{r,m}$. Since $\mathcal M_{r,m}$ is a subspace of $ L_{r,m}(G) $ there exists $g'\in L_{r',m^{-1}}(G) $ such that $\Gamma(f)=\overline{\langle{f},{g'}\rangle_{L_2}}$ for all $f\in \mathcal
M_{r,m}$. By setting $g=RC_Kg'\in \mathcal M_{r',m^{-1}}$, as above $$\Gamma(f)=\overline{\langle{f},{g'}\rangle_{L_2}}=
\overline{\langle RC_K f, g \rangle_{L_2}}=\overline{\langle{f},{g}\rangle_{L_2}}=\Gamma_g(f) \quad \text{for all}\,f\in \mathcal M_{r,m},$$ thus $\Gamma=\Gamma_g$.

Now we can prove that in the following setting Assumption \[assumption1\] is fulfilled.

\[lem:kernel\_continuity\] Fix $r\in (1,\infty)$ and assume that the right convolution operator $RC_K$ is bounded on $ L_{r,m}(G) $. Then the sets $\operatorname{span}\{\pi(x)u\}_{x\in G}$ and $\operatorname{span}\{\lambda(x)K\}_{x\in G}$ are dense in $\operatorname{Co}(L_{r,m})$ and $\mathcal M_{r,m}$, respectively. Thus, Assumption \[assumption1\] is fulfilled.

By the correspondence principle, it is enough to show the second claim. Let $\Gamma\in \mathcal M'_{r,m}$ be such that for all $x\in   G$, $$\Gamma(\lambda(x)K)=0.$$ By the above corollary, there exists $g \in \mathcal{M}_{r',m^{-1}}$ such that $\Gamma(f) = \langle g,f \rangle_{L_2}$ for all $f \in \mathcal{M}_{r,m}$. In particular, $$0
      = \Gamma \big(\lambda(x) K \big)
      