Embedding of weighted sobolev space with exponential weights

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.

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.

• 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. – user128470 20 hours ago
• @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? – John 19 hours ago

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, $$|u_i-v|_{W_\nu^{1,p}(\mathbb{R}^d}\leq 3\frac{\epsilon}{3}=\epsilon.$$