Let $\Omega = B_1(0)$ the ball centered at $0$ with radius equal 1 and $u \in H^1_0(\Omega).$ Also consider $f \in C^\infty(S^{N-1})$. I'm trying to prove the function $g(x) := u(x) f(x/|x|)$ is in $H^1_0(\Omega)$. The difficulties for me are in the fact that $x/|x|$ has no weak derivative. I took a sequence $u_n \in C^1_0(\Omega)$ such that $$||u_n - u||_{H^1(\Omega)} \rightarrow 0.$$ My first attempt was to prove that $g_n(x) := u_n(x) f(x/|x|)$ is in $C^1_0(\Omega)$. I don't think this is true.
Any help is welcome.
I don't think that this is true for $n = 2$. Let us identify $f$ with a function on $[0,2\pi]$. Suppose that $u \equiv 1$ on a small ball with radius $R$ around the origin. Then, we get $$ |\nabla g(x)| = \frac1{r(x)} |f'(\theta(x))| $$ for $0 < |x| < R$ with the usual polar coordinates $(r(x), \theta(x))$. That is, $g$ is differentiable on $B_R \setminus \{0\}$. However, $$ \int_{B_R} |\nabla g(x)|^2 \mathrm{d}x = \int_0^{2\pi}\int_0^R \frac1{r^2} |f'(\theta)|^2 r \mathrm{d}r \mathrm{d}\theta = \int_0^{2\pi}|f'(\theta)|^2 \mathrm{d}\theta \int_0^R \frac1{r}\mathrm{d}r $$ and this is $+\infty$ (unless $f$ is constant).