constant-sign solution of with $f{\left(}x,u{\right)}=\left|u\right|^{p^*-2}u$. In particular, every extremal function of or satisfies the a priori estimates in Theorems \[Th1\], \[Th2\], and \[Th3\].

As a remark, due to the scaling law , every extremal function of or generates in fact an infinite family of extremal functions.

Preliminary to the proof of Theorem \[Th5\], we prove the following result.

\[Pr1\] Let $\overrightarrow{\theta}={\left(}\theta_1,\dotsc,\theta_n{\right)}$ be such that $\theta_i>0$ for all $i=1,\dotsc,n$ and $\sum_{i=1}^n1/\theta_i=n/p$. Then the following assertions hold true.

1.  For any extremal function $u$ of , $u\circ\tau_{\overrightarrow{\theta}}$ is an extremal function of , where $$\label{Pr1Eq1}
    \tau_{\overrightarrow{\theta}}{\left(}x{\right)}:=\left(\lambda_{\overrightarrow{\theta},1}x_1,\dotsc,\lambda_{\overrightarrow{\theta},n}x_n\right),\quad\lambda_{\overrightarrow{\theta},i}:=\theta_i^{1/\theta_i}\prod_{j=1}^ n\theta_j^{-p/{\left(}n\theta_i\theta_j{\right)}}\,,$$ for all $x\in{\mathbb{R}}^n$ and $i=1,\dotsc,n$.

2.  For any extremal function $u$ of , $u\circ\sigma_{\overrightarrow{\theta},u}\circ\tau_{\overrightarrow{\theta}}^{-1}$ is an extremal function of , where $\tau_{\overrightarrow{\theta}}$ is as in and $$\label{Pr1Eq2}
    \sigma_{\overrightarrow{\theta},u}{\left(}x{\right)}:=\left(\mu_{\overrightarrow{\theta},1}{\left(}u{\right)}x_1,\dotsc,\mu_{\overrightarrow{\theta},n}{\left(}u{\right)}x_n\right),\quad\mu_{\overrightarrow{\theta},i}{\left(}u{\right)}:=\frac{\overset{n}{\underset{j=1}{\prod}}{\left(}\int_{{\mathbb{R}}^n}\left|\partial_{x_j}u\right|^{p_j}dx{\right)}^{p/{\left(}n\theta_ip_j{\right)}}}{{\left(}\int_{{\mathbb{R}}^n}\left|\partial_{x_i}u\right|^{p_i}dx{\right)}^{1/p_i}}\,,$$ for all $x\in{\mathbb{R}}^n$ and $i=1,\dotsc,n$.

We begin with proving Point (i). We fix an extremal function $u_0$ of . Since $\sum_{i=1}^n1/\theta_i=n/p$, we obtain $$\label{Pr1Eq3}
\prod_{i=1}^n{\left(}\int_{{\mathbb{R}}^n}\big|\partial_{x_i}\big(u_0\circ\tau_{\overrightarrow{\theta}}\big)\big|^{p_i}dx{\right)}^{p/{\left(}np_i{\right)}}\le\frac{p}{n}\sum_{i=1}^n\frac{1}{\theta_i}{\left(}\int_{{\mathbb{R}}^n}\big|\partial_{x_i}\big(u_0\circ\tau_{\overrightarrow{\theta}}\big)\big|^{p_i}dx{\right)}^{\theta_i/p_i}.$$ For any function $u\in D^{1,\operatorname{\overrightarrow{p}}}{\left(}{\mathbb{R}}^n{\right)}$, simple calculations give $$\label{Pr1Eq4}
\sum_{i=1}^n\frac{1}{\theta_i}{\left(}\int_{{\mathbb{R}}^n}\big|\partial_{x_i}\big(u\circ\tau_{\overrightarrow{\theta}}\big)\big|^{p_i}dx{\right)}^{\theta_i/p_i}={\left(}\prod_{j=1}^ n\theta_j^{-p/{\left(}n\theta_j{\right)}}{\right)}\sum_{i=1}^n{\left(}\int_{{\mathbb{R}}^n}\left|\partial_{x_i}u\right|^{p_i}dx{\right)}^{\theta_i/p_i}$$ and $$\label{Pr1Eq5}
\int_{{\mathbb{R}}^n}\big|u\circ\tau_{\overrightarrow{\theta}}\big|^{p^*}dx=\int_{{\mathbb{R}}^n}\left|u\right|^{p^*}dx\,.$$ By invertibility of $\tau_{\overrightarrow{\theta}}$ and since $u_0$ is an extremal function of , it follows from and that $$\label{Pr1Eq6}
\frac{\overset{n}{\underset{i=1}{\sum}}\frac{1}{\theta_i}{\left(}\int_{{\mathbb{R}}^n}\big|\partial_{x_i}\big(u_0\circ\tau_{\overrightarrow{\theta}}\big)\big|^{p_i}dx{\right)}^{\theta_i/p_i}}{\big(\int_{{\mathbb{R}}^n}\big|u_0\circ\tau_{\overrightarrow{\theta}}\big|^{p^*}dx\big)^{p/p^*}}=\inf_{\underset{u\ne0}{u\in D^{1,\operatorname{\overrightarrow{p}}}{\left(}{\mathbb{R}}^n{\right)}}}\frac{\overset{n}{\underset{i=1}{\sum}}\frac{1}{\theta_i}{\left(}\int_{{\mathbb{R}}^n}\left|\partial_{x_i}u\right|^{p_i}dx{\right)}^{\theta_i/p_i}}{\left(\int_{{\mathbb{R}}^n}\left|u\right|^{p^*}dx\right)^{p/p^*}}\,.$$ Now, we claim that $$\label{Pr1Eq7}
\inf_{\underset{u\ne0}{u\in D^{1,\operatorname{\overrightarrow{p}}}{\left(}{\mathbb{R}}^n{\right)}}}\frac{\overset{n}{\underset{i=1}{\sum}}\frac{1}{\theta_i}{\left(}\int_{{\mathbb{R}}^n}\left|\partial_{x_i}u\right|^{p_i}dx{\right)}^{\theta_i/p_i}}{\left(\int_{{\mathbb{R}}^n}\left|u\right|^{p^*}dx\right)^{p/p^*}}\le\frac{n}{p}\cdot\inf_{\underset{u\ne0}{u\in D^{1,\operatorname{\overrightarrow{p}}}{\left(}{\mathbb{R}}^n{\right)}}}\frac{\overset{n}{\underset{i=1}{\prod}}{\left(}\int_{{\mathbb{R}}^n}\left|\partial_{x_i}u\right|^{p_i}dx{\right)}^{p/{\left(}np_i{\right)}}}{\left(\int_{{\mathbb{R}}^n}\left|u\right|^{p^*}dx\right)^{p/p^*}}\,.$$ We prove this claim. For any function $u\in D^{1,\operatorname{\overrightarrow{p}}}{\left(}{\mathbb{R}}^n{\right)}$, 