Dense subspace of radial functions of $H^1(\mathbb{R}^N)$

Question

By definition the space of radial functions in $H^1(\mathbb{R}^N)$ is $$ H^1_{rad}(\mathbb{R}^N) = \{u \in H^1(\mathbb{R}^N) : u = u \circ R, \forall R \in O(N)\}. $$ I'm trying to find a dense subspace of it. I suspect that the required subspace is $$ C^{\infty}_{0,rad}(\mathbb{R}^N) = \{u \in C^\infty_{0}(\mathbb{R}^N) : u = u \circ R, \forall R \in O(N)\}. $$ I know how to show that $\overline{C^{\infty}_{0,rad}(\mathbb{R}^N)}^{H^1(\mathbb{R}^N)} \subset H^1_{rad}(\mathbb{R}^N)$. To prove the reverse inclusion I tried to use that $C^\infty_0(\mathbb{R}^N)$ is dense in $H^1(\mathbb{R}^N)$. However I got stuck.

Answer 1

The same proof as for Sobolev spaces will work:

1. Take $\phi_k(x):= \phi(x/k)$, for some radial $\phi\in C_c^\infty(\mathbb{R}^N)$ with $\phi\equiv 1$ on the unit ball, supported in $B(0,2)$, and $0\leq \phi \leq 1$ throughout. Set $w_k=w\cdot \phi_k$. Using the product rule you can check that $\| w-w_k\|_{H^1}\to 0$ as $k\to \infty$.

2. By 1, compactly supported, radial, $H^1$ functions are dense. To finish, approximate these by convolving with a radial mollifier (convolution of radial functions is radial).