Compact embedding in $L^1$

Question

Is it true that the space $H^1(\mathbb{R}^2)\cap L^2(\mathbb{R}^2,(1+|x|^2)^2dx)$ is compactly embedded in $L^1(\mathbb{R}^2)?$ I don't think that the dimension $2$ is important. Here $H^1$ is the usual Sobolev space and $L^2(\mathbb{R}^2,(1+|x|^2)^2dx)$ is the weighted $L^2$ space, i.e. $\int_{\mathbb{R}^2}|f|^2(1+|x|^2)^2dx<+\infty.$ It seems to be similar to the previous question here Proving a subset of $H^1(\mathbb{R}^d)$ is compactly embedded in $L^2(\mathbb{R}^d)$., but I cannot adapt it to $L^1$. Thank you.

Answer 1

Note that it suffices to show that if $f_n \rightharpoonup f$ in $H^1\cap L^2((1+|x|^2)^2dx)$, then it converges strongly in $L^1$. Indeed, suppose that this is true. Then, denote by $T:H^1\cap L^2((1+|x|^2)^2dx) \to L^1$ the injection map. For any sequence $T(f_n)$ with $f_n \in B_1$, we use Banach Alaoglu (since $H^1\cap L^2((1+|x|^2)^2dx)$ is a Hilbert space) to extract a weakly converging subsequence $f_{n_j}$, which then yields that $Tf_{n_j}$ is strongly converging by assumption, showing compactness.

To this end, note that $$ \|f_n-f\|_{L^1}=\|f_n-f\|_{L^1(|x|\leq R)}+\|f_n-f\|_{L^1(|x|> R)} $$ for any fixed positive $R$. For any fixed $R$, the first term tends to zero by Rellich, so we just need to ensure we can make the last term sufficiently small. Indeed, it is bounded above by $\|f_n\|_{L^1(|x|> R)}+\|f\|_{L^1(|x|> R)}$, which may be bounded as follows: By weak convergence, there exists some constant $C>0$ so that $\|f_n\|_{L^2((1+|x|^2)^2)},\|f\|_{L^2((1+|x|^2)^2)}\leq C$. We thus have by Cauchy-Schwarz $$ \int_{|x|\geq R}|f(x)|dx \leq (\int_{|x|\geq R}|f(x)|^2 (1+|x|^2)^2dx)^\frac{1}{2}(\int_{|x|\geq R}(1+|x|^2)^{-2}dx)^\frac{1}{2} \leq C^\frac{1}{2} f(R) $$ where $f(R)=(\int_{|x|\geq R}(1+|x|^2)^{-2}dx)^\frac{1}{2} \to 0$ as $R \to \infty$ by dominated convergence, which completes the proof.