\subset \CySpT^{m, \, 2}$, given $\uv_h \in \CySpT_h^{m, \, 2}$, we have that $$\begin{aligned}
    \norm{\uv_h}_\rho^2 & =  \norm{\uvt_{\rho z, \, h}}_\rho^2 +
    \norm{\dfrac{1}{m}\parT{\rho
            \,\ut_{\theta, \, h} + \ut_{\rho, \,
                h}}}_\rho^2\\ 
    & \leq 3R\,\norm{\uvt_{\rho z, \, h}}^2 +
    2R^3\,\norm{\ut_{\theta, \, h}}^2 < \infty 
    \end{aligned}$$ and $$\begin{aligned}
    \norm{\divergtm \uv_h}_\rho^2 & = \norm{\divergtsm
        \uvt_h}_\rho^2 \leq
    2\,\norm{\divergt
        \uvt_{\rho z, \, h}}_\rho^2
    + 2\,\norm{\ut_{\theta, \,
            h}}_\rho^2 \\ 
    & \leq 2R\,\norm{\divergt \uvt_{\rho z, \, h}}^2 +
    2R\,\norm{\ut_{\theta, \, h}}^2 < \infty 
    \end{aligned}$$ since $\uvt_{\rho z, \, h} \in \CaSpB^{1*}$ and $\ut_{\theta,
        \, h} \in \CaSpB^2$.\
Finally, $\CySpT_h^{m, \, 3} \subset \CySpT^{m, \, 3}$, given $u_h \in \CySpT_h^{m, \, 3}$, we have that $$\begin{aligned}
    \norm{u_h}_\rho^2 & = \norm{\ut_h}_\rho^2 \leq R\,
    \norm{\ut_h}^2 < \infty 
    \end{aligned}$$ since $\ut_h \in \CaSpB^2$.

The projectors $\Pibr^{m, \, k}: \CySpTT^{m, \, k} \rightarrow
\CySpT_h^{m, \, k}$ are defined as in [@oh:2015], exploiting those used in Cartesian coordinates –, that is $$\label{eq:cyl_interp}
\begin{alignedat}{3}
&\Pibr^0 u && = \parT{\eta_{m, \, 0}^{-1} \comp \Pi^0
    \comp \eta_{m, \, 0}}\parT{u}~,  &&\quad u \in \CySpTT^{m,
    \, 0}~,\\ 
&\Pibr^1 \uv && = \parT{\eta_{m, \, 1}^{-1}
    \comp \parT{\Pi^1 \times \Pi^0}\comp \eta_{m, \,
        1}}\parT{\uv}~,  && \quad \uv \in \CySpTT^{m, \, 1}~,\\ 
&\Pibr^2 \uv && = \parT{\eta_{m, \, 2}^{-1}
    \comp \parT{\Pi^{1*}\times \Pi^2}\comp \eta_{m, 