From the article Weyl's law of Matt Stevenson page $3$, there's a lemma where I'd like some help to understand the proof.
Lemma. (Domain Monotonicity) If $\Omega_1 \subset \Omega_2$, then $\mu_n(\Omega_2) \leq \mu_n (\Omega_1)$.
Pf : Remark that $H_0^1 (\Omega_1) \subset H_0^1 (\Omega_2)$, so $\Phi_n(H_0^1 (\Omega_1)) \subset \Phi_n(H_0^1 (\Omega_2)).$ Then by the version $2$ of the Minimax Principle, we conclude that $\mu_n (\Omega_2) \leq \mu_n (\Omega_1)$
I know that we could extend continuously every eigenfunction of $\Omega_1$ by vanishing on $\Omega_2$, i.e
$$ w'(x) = \begin{cases} w(x) & x \in \Omega_1 \\ 0 & x \in \Omega_2 - \Omega_1 \\ \end{cases}. $$
However, I do not see the order relation between $\mu_n(\Omega_2) = \inf_{X \in \Phi_n(V)} \sup_{u \in X} \frac{\|\nabla u(\Omega_2)\|^2}{\|u(\Omega_2)\|^2}$ and $\mu_n(\Omega_1) = \inf_{X \in \Phi_n(V)} \sup_{u \in X} \frac{\|\nabla u(\Omega_1)\|^2}{\|u(\Omega_1)\|^2}$ and how to explain that.
Edit :
Let $$R_n(H^1_0) = \{\sup_{u\in X^n}\rho(u)\ |\ X\in\Phi_n(H^1_0)\}.$$ Notice that $R$ is the image of a function $\Phi_n(H)\to\mathbb{R}$ taking $X\mapsto\sup_{u\in X}\mathcal{R}(u)$. As $\Phi_n(H^1_0 (\Omega_1))\subset\Phi_n(H^1_0 (\Omega_2))$, we have that $R_n(H^1_0 (\Omega_1))\subseteq R_n(H^1_0 (\Omega_2))$. From this fact, by the comment of Renart, we are sur that $\inf (R_n(H^1_0 (\Omega_1))) \geq \inf(R_n(H^1_0 (\Omega_2)))$. By the Minimax principle, it is clear that $\mu_n(\Omega_1) \geq \mu_n(\Omega_2)$.
Question : Could anyone be able to explain rigorously that relation?
Question : If we proceed by contradiction, why is it not possible to get $\mu_n(\Omega_2) = \inf_{X \in \Phi_n(V)} \sup_{u \in X} \frac{\|\nabla u(\Omega_2)\|^2}{\|u(\Omega_2)\|^2} > \mu_n(\Omega_1) = \inf_{X \in \Phi_n(V)} \sup_{u \in X} \frac{\|\nabla u(\Omega_1)\|^2}{\|u(\Omega_1)\|^2}$ ?