Let $\Omega\subset \mathbb{R}^N$ a bounded domain. By definition the space of radial functions in $H^1_0(\Omega)$ is $$ H^1_{0, rad}(\Omega) = \{u \in H^1_0(\Omega) : u = u \circ R, \forall R \in O(N)\}. $$ I'm trying to find a dense subspace of it. I suspect that the required subspace is $$ C^{\infty}_{0,rad}(\Omega) = \{u \in C^\infty_{0}(\Omega) : u = u \circ R, \forall R \in O(N)\}. $$ I know how to show that $\overline{C^{\infty}_{0,rad}(\Omega)}^{H^1_0(\Omega)} \subset H^1_{0, rad}(\Omega)$.
Now I will write my attempt to prove the reverse inclusion. From the Theorem 2 of Partial Differential Equations, 2010, in the page 265 we know that given $v \in W^{1,2}(\Omega)$ there exists $v_{\epsilon} \in C^\infty(\Omega) \cap W^{1,2}(\Omega)$ such that $\|v_\epsilon - v\|_{W^{1,2}(\Omega)} \to 0$ as $\epsilon \to 0$. Precisely, the function $v_\epsilon$ is of the form $v_\epsilon = \eta_{\epsilon} \ast v$. By construction the function $\eta_{\epsilon}$ is radial.
Now, if $u \in H^{1}_0(\Omega)$ is radial, in particular $u \in W^{1,2}(\Omega)$ and $u_\epsilon$ is radial (convolution of radial is radial, we also assume that $\Omega$ is a radial domain to addapt the proof of this). Therefore $u_\epsilon$ is a smooth function which is radial and converges to $u$ in $W^{1,2}(\Omega)$.
It remains to prove that $u_\epsilon$ has compact support in $\Omega $using the fact that $u \in H^1_0(\Omega)$. However I don't think it is correct. If this is true I would have that $u_\epsilon \in C^\infty_0(\Omega)$. However I got stuck. Any help is welcome.