}/(p^{\prime }-q),\infty ]$, and $\Vert S_{R}^{z,\infty }\Vert
_{p\rightarrow r}\leq c(z)$ for all $R\geq \left\Vert \rho \right\Vert^2 $.

Recall that $\kappa_{R}^{z, \infty}(x)=\kappa_{R}^z(x)$ for every $|x|>1$. Using the estimates of $\kappa _{R}^{z }$ from Lemmata \[infinity\] and \[extracase\], as well as the estimate (\[phi0\]), it follows that $%
\kappa _{R}^{z,\infty }$ is in $L^{q}(X)$ for every $q>2$. Thus, by Young’s inequality, the operator $f\rightarrow |f|\ast \kappa _{R}^{z,\infty }$ maps $L^{p}(X)$, $p\in \lbrack 1,q^{\prime }]$, continuously into $L^{r}(X)$, for every $r\in \lbrack qp^{\prime }/(p^{\prime }-q),\infty ]$.

Further, for $z\geq n-\frac{1}{2}$, in Lemmata \[infinity\] and [extracase]{} the estimates of the kernel $\kappa _{R}^{z,\infty }$ are uniform with respect to $R$. This implies that the norm $\Vert S_{R}^{z,\infty }\Vert
_{p\rightarrow r}$ is bounded by a constant, uniform with respect to $R$.

Proof of Theorem 2 and Theorem 3
================================

In this section we give the proof of Theorem \[Th1\], which deals with the $L^{p}$-continuity of the maximal operator $S_{\ast }^{z}$ associated with the Riesz means. This allows us to deduce the almost everywhere convergence of Riesz means $S_{R}^{z}(f)$ to $f$, as $R\rightarrow +\infty $.

Recall first that $$S_{\ast }^{z}(f)=\sup_{R>\Vert \rho \Vert ^{2}}|S_{R}^{z}(f)|,\;f\in L^{p}({X%
}).  \label{maximal}$$ The following proposition holds true, [@GIUMA Lemma 4.1].

\[S\*L2\] Let 