I know it is probably a silly question, but is there anyone could help me to complete of the corollary $3.1$ of that document? I pass a lot of time to try understanding the problem, but I can't advance on the subject.
Corollary $3.1$. If $M_1 \subset M_2$, then $$\lambda(M_2) \leq \lambda(M_1),$$ i.e. the bigger the domain, the smaller the first eigenvalue.
Proof. $$\lambda_1(M_1)=\min \{E(u), u \in C^{\infty}(M_1) \}$$
But then, $u \in C^{\infty}(M_1)$, $u=0$ on $M_2$, and $$\lambda_1(M_2)=\min \{E(u), u \in C^{\infty}(M_2) \}$$
I think it's a bit confusing what they wrote, bordering on incorrect. I'm sure you can find a better reference for these things.
The idea is to use the theorem that $ \lambda_1(M) \leq E(u) / \langle u, u \rangle = \| \nabla u \|^2 / \| u \|^2 $ for any sufficiently nice $ u $ (it does not have to be $ C^\infty $).
Let $ \varphi_1 \in C^\infty(M_1) $ be the eigenvector corresponding to $ \lambda_1(M_1) $. By extending by $ 0 $, $\varphi_1 $ defines a piecewise differentiable function $ \widetilde \varphi_1 $ on $ M_2 $. It is not $ C^\infty $. Now use the theorem: $$ \lambda_1(M_2) \leq \| \nabla \widetilde \varphi_1 \|^2_{M_2} / \| \widetilde \varphi_1 \|^2_{M_2} = \| \nabla \varphi_1 \|^2_{M_1} / \| \varphi_1 \|^2_{M_1}= \lambda_1(M_1) $$ I put subscripts on the norms $ \| \cdot \|_{M_i} $ to indicate in which space it is taken. The equality $ \| \nabla \widetilde \varphi_1 \|^2_{M_2} / \| \widetilde \varphi_1 \|^2_{M_2} = \| \nabla \varphi_1 \|^2_{M_1} / \| \varphi_1 \|^2_{M_1} $ is because $ \widetilde \varphi_1 $ has support in $ M_1 $ and will be obvious if you write out each side in terms of integrals.
EDIT: Fixed a bunch of typos.