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.

  • $\begingroup$ Just for sake of confirmation, is $H_{rad}$, space of radially symmetric functions? $\endgroup$
    – Harish
    Feb 21, 2015 at 20:16
  • $\begingroup$ Yes, this is true. $\endgroup$
    – Peter
    Feb 21, 2015 at 22:17

1 Answer 1


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.


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