TITLE: Compactness Sobolev embedding for radial functions on $\mathbb{R}^N$
QUESTION [7 upvotes]: I want to show:
Let $N\geq 2$ and $2< q <2^\ast$. Then the embedding \begin{align}
H^1_{\text{rad}}(\mathbb{R}^N)\hookrightarrow L^q(\mathbb{R}^N)
\end{align}
is compact.
I was able to show that \begin{align}|u(r)|\leq C R^{\frac{-(N-1)}{2}} \|\nabla u\|_2^{\frac{1}{2}} \|u\|_2^{\frac{1}{2}}\leq \hat C R^{\frac{-(N-1)}{2}} \|u\|_{H^1} \end{align} holds almost everywhere for $r\geq R$.
How can I conclude now? I think I should somehow use the Rellich-Kondrachov compact embedding for bounded domains.

REPLY [10 votes]: Raise both sides of your inequality to the $q-2$ power and then multiply the resulting by $|u(r)|^2r^{N-1}$ to get
$$
|u(r)|^qr^{N-1}\leq C \| u\|_{H^1}^{q-2}R^{(1-N)(q-2)/2}|u(r)|^2r^{N-1}.
$$
Integrating this over the set $\{ r>R\}$ and calling $A_R=\{ |x|>R\}$, we get
$$
\| u\|_{L^q(A_R)} \leq C\| u\|_{H^1} R^{(1-N)(q-2)/2}.
$$
Now take a bounded sequence $(u_k)$ in $H^1_r(\mathbb{R}^N)$, using a diagonal argument and Rellich's theorem in the ball $B_R$, we can take a sequence $R_m \to \infty$ so that (calling the subsequence the same as the original sequence) $u_k\to u$ strongly in $L^q(B_{R_m})$ for every $m$. Then
$$
\| u_k-u_l\|_{L^q(\mathbb{R}^N)} \leq \| u_k-u_l\|_{L^q(B_{R_m})} + \| u_k-u_l\|_{L^q(A_{R_m})}.
$$
The secodn term can be made arbitrarily small (uniformly in $k,l$) by choosing $m$ big enough, and then for such an $m$ fixed, the first term can be made small too by choosing $k,l$ big enough.