0}$, in fact $$\begin{aligned}
    \norm{u_h}_\rho^2 = \norm{\rho \ut_h}_\rho^2 \leq R^3
    \,\norm{\ut_h}^2 < \infty 
    \end{aligned}$$ and $$\begin{aligned}
    \norm{\gradtm u_h}_\rho^2 & = \norm{\gradtsm
        \ut_h}_\rho^2 \leq R \,
    \norm{\gradtsm \ut_h}^2\\ 
    & \leq 3R\, \norm{\ut_h}^2 + 2R^3 \, \norm{\gradt \ut_h}^2 <
    \infty 
    \end{aligned}$$ since $\ut_h \in \CaSpB^0$ . The proofs of the other cases are similar. For $\CySpT_h^{m,
        \, 1} \subset \CySpT^{m, \, 1}$, given $\uv_h \in
    \CySpT_h^{m, \, 1}$, we have that $$\begin{aligned}
    \norm{\uv_h}_\rho^2 & =  \norm{\dfrac{1}{m}\parT{\rho \,
            \ut_{\rho, \, h} - \ut_{\theta, \,
                h}}}_\rho^2 +
    \norm{\dfrac{\rho}{m}\ut_{z, \,
            h}}_\rho^2 + \norm{\ut_{\theta, \,
            h}}_\rho^2\\ 
    & \leq 2 R^3\,\norm{\ut_{\rho, \, h}}^2 + R^3\,\norm{\ut_{z,
            \, h}}^2 +  3R\,\norm{\ut_{\theta, \, h}}^2 < \infty 
    \end{aligned}$$ and $$\begin{aligned}
    \norm{\curltm\uv_h}_\rho^2 & =
    \norm{\curltsm\uvt_h}_\rho^2
    \\ 
    & =\norm{\uvt_{\rho z, \, h} + \gradt \ut_{\theta, \,
            h}}_\rho^2 \\ 
    & + \norm{\dfrac{1}{m}\parT{-\rho\,\curltsc \uvt_{\rho z, \,
                h} - \ut_{z, \, h} - \pd{z} \ut_{\theta, \,h}}}_\rho^2 \\ 
    & \leq 6\, \norm{\uvt_{\rho z, \, h}}_\rho^2 + 4
    \,\norm{\rho\,\curltsc \uvt_{\rho z, \, h}}_\rho^2 +
    4\,\norm{\gradt \ut_{\theta, \, h}}_\rho^2\\ 
    & \leq 6R\, \norm{\uvt_{\rho z, \, h}}^2 + 4R^3\,
    \norm{\curltsc \uvt_{\rho z, \, h}}^2 + 4R\, \norm{\gradt
        \ut_{\theta, \, h}}^2 < \infty 
    \end{aligned}$$ since $\uvt_{\rho z, \, h} \in \CaSpB^1$ and $\ut_{\theta, \,
        h} \in \CaSpB^0$.\
For $\CySpT_h^{m, \, 2} 