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In the book by Bensoussan and Lions, they introduce the weighted spaces with exponentially decaying weights to study elliptic equations with bounded coefficients on the whole space $\mathbb{R}^n$. They mentioned that there are classical regularity results based on these spaces.

For example, for each $p\in [1,\infty)$, the weighted $L^p_\mu(\mathbb{R}^d)$ space on $\mathbb{R}^d$ is defined to be the set of Lebesgue measurable functions such that $f\omega_\mu(x)\in L^p(\mathbb{R}^n)$, i.e., $$\|f\|_{L^p_\mu}=\int_{\mathbb{R}^d}|f|^p\omega^p_\mu(x)\,dx< \infty,$$ where $\omega_\mu(x)=\exp(-\mu\sqrt{1+|x|^2})$ for $\mu>0$, and the weighted sobolev space $W^{1,p}_\mu(\mathbb{R}^d)$ is defined to be the space of functions such that $u\omega_\mu\in L^p(\mathbb{R}^n)$ and $\partial_{x_i} u\omega_\mu\in L^p(\mathbb{R}^n)$, where $\partial_{x_i}$ denotes the weak derivative in the distribution sense. Similarly we define the high-order sobolev space $W^{2,p}_\mu(\mathbb{R}^d)$ such that $\partial_{x_ix_j}u\omega_\mu\in L^p$ for all $i,j$.

I am interested in a reference on the embedding properties between spaces of different orders. For example, it is pointed out (without a proof) in the book that the injection $$ W^{2,p}_\mu\hookrightarrow W^{1,p}_\nu \tag{1}$$ with $\nu>\mu$ is compact. Could you briefly outline or provide a reference for a proof of this statement? Does it follow from the results for the classical sobolev space?


Note that Corollary 3.3 in Hooton's paper implies the injection $W^{2,p}_\mu\hookrightarrow W^{1,p}_\mu \tag{2}$ is not compact.

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This question has an open bounty worth +50 reputation from John ending in 6 days.

Looking for an answer drawing from credible and/or official sources.

  • $\begingroup$ I believe that you can just take a compact exhaustion $\Omega_i$ of $\mathbb{R}^n$. On every $\Omega_i$, the weighted norm is equivalent to the usual Soboelv norm. Using Rellich-Kochandrov and a diagonal sequence gives you a converging subsequence, the condition $\mu>\nu$ ensures that the $W_\nu^{1,p}$ norm of the sequence converges to zero outside of the exhaustion. $\endgroup$ – user128470 20 hours ago
  • $\begingroup$ @user128470 Could you please elaborate more on the role of the condition $\mu>\nu$? And do you want to expand the comment into an answer, so that I can accept it? $\endgroup$ – John 19 hours ago
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Expanding on my comment, let $u_i\in W^{2,p}_\mu(\mathbb{R}^d)$ be a bounded sequence, that is, $$ |u_i|_{W^{2,p}_\mu(\mathbb{R}^d)}\leq C $$ and consider the balls $B_k(0)$, $k\in\mathbb{N}$. Let $W^{2,p}(B_k(0))$ be the usual Sobolev space. It is easy to see that $$ |u_i|_{W^{2,p}({B_k(0)})}\leq C\exp(\mu\sqrt{1+k^2})=:C_k. $$ By the Rellich-Kochandrov Theorem we may find a subsequence (again labeled $u_i$) such that $u_i\to v_1$ in $W^{1,p}(B_1(0))$ and pointwise almost everywhere. We may then choose another subsequence such that $u_i\to v_2$ to $W^{1,p}(B_2(0))$. By pointwise convergence it follows that $v_1=v_2$ almost everywhere in $B_1(0)$. Continuing this procedure and taking a diagonal subsequence we find a well-defined function $v$ defined by $v(x)=v_k(x)$ where $k$ is chosen such that $x\in B_k(0)$ such that $u_i\to v$ in $W^{1,p}(B_k(0))$ for every $k\in\mathbb{N}$ and pointwise almost everywhere. We may also assume that all partial derivatives converge pointwise almost everywhere. By point-wise convergence and Fatou's lemma it follows that $v\in W^{1,p}_\mu(\mathbb{R}^d)$. Now let $\nu>\mu$ and $\epsilon>0$. As $v\in W^{1,p}_\mu(\mathbb{R}^d)$ we may choose $K\in\mathbb {N}$ such that $$ \bigg(\int_{\mathbb{R}^d\setminus B_K(0)}\omega_\nu^p(|v|^p+|\nabla v|^p)\bigg)^{1/p}\leq \bigg(\int_{\mathbb{R}^d\setminus B_K(0)}\omega_\mu^p(|v|^p+|\nabla v|^p)\bigg)^{1/p}\leq \frac{\epsilon}{3}. $$ On the other hand, $$ \bigg(\int_{\mathbb{R}^d\setminus B_K(0)}\omega_\nu^p(|u_i|^p+|\nabla u_i|^p)\bigg)^{1/p}\leq \exp((\nu-\mu)\sqrt{1+K^2})\bigg(\int_{\mathbb{R}^d\setminus B_K(0)}\omega_\mu^p(|u_i|^p+|\nabla u_i|^p)\bigg)^{1/p}\leq \frac{\epsilon}{3}, $$ provided $K$ is large enough such that $\exp((\nu-\mu)\sqrt{1+K^2})C\leq \frac{\epsilon}{3}$ (here we use $\nu>\mu$). Finally, as $u_i\to v$ in $W^{1,p}(B_K(0))$ it follows that $$ \bigg(\int_{B_K(0)} |u_i-v|^p+|\nabla u_i-\nabla v|^p\bigg)\leq \frac{\epsilon}{3} $$ for all $i\geq I$, where $I$ depends on $\epsilon$ and $K$. Consequently, $$ |u_i-v|_{W_\nu^{1,p}(\mathbb{R}^d}\leq 3\frac{\epsilon}{3}=\epsilon. $$

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