Let $d\in \mathbb{N}$, $B_1=\{x\in \mathbb{R}^d;|x|\leq 1\}$,
$f\in L^1(B_1)$ be a radial function (i.e. $f(x)=f(|x|)$) satisfying $||f-f_n||_{L^1(B_1)}\to 0 \ (n \to \infty)$ for some $(f_n)_{n\in \mathbb{N}} \subset C(B_1)$, and $f_n(x)=0$ if $\ |x|=1$.
Then, is there exist a $(g_n)_{n\in \mathbb{N}} \subset C(B_1)$ such that $g_n(x)=0$ if $\ |x|=1$, $||f-g_n||_{L^1(B_1)}\to 0 \ (n \to \infty)$, and $g_n$ is a radial?
Say $O$ is the orthogonal group, with Haar measure $dT$. Define $$g_n(x)=\int_Of_n\circ T\,dT.$$
Then $g_n$ is radial and $$f(x)-g_n(x)=\int_O(f(Tx)-f_n(Tx))\,dT,$$ since $f$ is radial. So $$||f-g_n||=\left|\left|\int_O(f_n\circ T-g_n\circ T)\right|\right|\le\int_O||f\circ T-f_n\circ T||\,dT=||f-f_n||.$$