mappings $$y \mapsto f_n(y) K(y^{-1}x), \qquad y \mapsto g(y) K(y^{-1}x)$$ are integrable. Then, given $x\in G \setminus F$, for all $y\in G \setminus E$ $$\begin{aligned}
  |f_n(y) K(y^{-1}x) |\leq |g(y) K(y^{-1}x)|,   \quad \lim_{n\to\infty} f_n(y) K(y^{-1}x) = f(y) K(y^{-1}x) .\end{aligned}$$ For $x \in G \setminus F$, the function $y\mapsto g(y) K(y^{-1}x)$ is integrable, so that by dominated convergence we obtain $$g(x)=\lim_{n\to\infty} \int_G f_n(y) K(y^{-1}x)\,dy = \int_G f(y) K(y^{-1}x)\,dy = f\ast K (x),$$ so $RC_K$ is indeed closed.

\[prop:kernel\_property\] Denote by $r'$ the dual exponent $1/r+1/r'=1$. Assume that the right convolution operator $$RC_K:  L_{r,m}(G) \to  L_{r,m}(G) , \qquad RC_Kf= f\ast K$$ is bounded, then

i)  the right convolution operator is bounded on $ L_{r',m^{-1}}(G) $ and it coincides with the adjoint of $RC_K$;

ii) the operator $RC_K$ is a projection from $ L_{r,m}(G) $ onto the reproducing kernel Banach space $\mathcal M_{r,m}$.

Here and in the following, the duality pairing is the sesqui-linear form $$\langle{f},{g}\rangle_{L_2}= \int_G f(x) \overline{g(x)}\, dx, \qquad f\in L_{r,m}(G) ,\, g\in L_{r',m^{-1}}(G) .$$

Since $RC_K$ is a bounded operator on $L_{rm}(G)$, the adjoint is a bounded operator on $ L_{r,m}(G) '$. Take $g\in L_{r',m^{-1}}(G) $ and $f\in C_c(G)\subset L_{r,m}(G) $, then $$\begin{aligned}
{1}
 \langle{RC_K^*g},{f}\rangle_{L_2} & =  \langle{g},{RC_Kf}\rangle_{L_2} =\int_G g(x)
 \left(\int_G 