Consider $N = n+ k$ and in $\mathbb{R}^N$ and also take the variable $(x,y)$. In the article https://doi.org/10.1017/prm.2022.43 is defined for $\gamma \geq 0$ the space $$ H^{1,2}_{\gamma}(\mathbb{R}^N) = \{u\in L^2(\mathbb{R}^N): \partial_{x_i}u,|x|^{\gamma}\partial_{y_i}u \in L^2(\mathbb{R}^N)\}, $$ and it is said that it is a Hilbert space with the norm $$ \|u\|^2 = \|u\|_{L^2(\mathbb{R}^N)}^2 +\sum_{i=1}^{n} \|\partial_{x_i}u\|_{L^2(\mathbb{R}^N)}^2 +\sum_{i=1}^{k} \||x|^{\gamma}\partial_{y_i}u\|_{L^2(\mathbb{R}^N)}^2. $$ How to prove it? I could prove it, but with some restriction on $\gamma$, which doesn't seem to be necessary.
The proof of this fact can be seem in Tri, Nguyen Minh. Sobolev imbedding theorems for degenerate metrics. International Centre for Theoretical Physics, 1995. , theorem 4. However I can not understand all the proof. Let me put here the ideia. Taking $u_n$ a Cauchy sequence in $H^{1,2}_\gamma(\mathbb{R}^N)$, one can see that $u_n, \partial_{x_i} u_n$ and $|x|^\gamma \partial_{y_j} u_n$ are Cauchy sequences in $L^2(\mathbb{R}^N)$. Let $u, v_i, w_j$ the limits in $L^2(\mathbb{R}^N)$ of $u_n, \partial_{x_i} u_n$ and $|x|^\gamma \partial_{y_j} u_n$ respectively. I know how to prove that $u$ has $x_i-$ weak derivate and $\partial_{x_i} u = v_i$. Now given $\varphi \in C^\infty_0(\mathbb{R}^N)$, notice that \begin{align} \int_{\mathbb{R}^N} (|x|^\gamma u) \partial_{y_i} \varphi dz & = \lim_{n \to +\infty} \int_{\mathbb{R}^N} (|x|^\gamma u_n) \partial_{y_i} \varphi dz \\ & = - \lim_{n \to +\infty} \int_{\mathbb{R}^N} \partial_{y_i} (|x|^\gamma u_n) \varphi dz \\ &= - \lim_{n \to +\infty} \int_{\mathbb{R}^N} |x|^\gamma \partial_{y_i}(u_n) \varphi dz \\ & = - \int_{\mathbb{R}^N} w_j \varphi dz. \end{align} This shows $|x|^\gamma u$ has $y_j-$weak derivative and $\partial_{y_j} (|x|^\gamma \partial_{y_j} u) = w_j$. Now I would like to use a product rule to show that $\partial_{y_j} (|x|^\gamma \partial_{y_j} u) = |x|^\gamma \partial_{y_j} u$. I dont't know how guarantee the function $u$ has $y_j-$weak derivative. I suspect it has to be $\frac{w_j}{|x|^\gamma}$. In this case, how to proove that $\frac{w_j}{|x|^\gamma} \in L^1_{loc}(\mathbb{R}^N)$ for all $\gamma > 0$?
Here is the proof from the cited book(in dimension 2):