Let $\lambda_1$ the first eigenvalue of the Laplace operator with Dirichlet boundary conditions in $\Omega \subset \mathbb{R}^n$ a bounded domain. Consider the set:
$$A:=\{u\in C^\infty(\overline{\Omega})\::\: u>0 \text{ in }\Omega \text{, and }u=0\text{ in }\partial \Omega\}. $$
Prove that... $$\lambda_1=\sup_{u\in U} \inf_{x\in \Omega} \frac{-\Delta u(x)}{u(x)}.$$
My ideas: I know that I must use that the first proper function $f_1$ of the Laplace operator is in $A$, but I've seen other formulations similar to this problem using a particular Sobolev space $W_0^{1,p}$ but the characterization of $\lambda_1$ is in terms of the infimum in the Sobolev space of $\frac{||u||_{1,p}^p}{||u||_p^p}$´, but I don't know how to give the characterization of supreme in $u$ and infimum in $\Omega$.
Any idea, help or commentary will be appreciated♥
If $v\in A$ is an eigenfunction of $-\Delta$ to the eigenvalue $\lambda_1$, then $$ \inf_{x\in\Omega}\frac{-\Delta v(x)}{v(x)}=\inf_{x\in\Omega}\lambda_1\frac{v(x)}{v(x)}=\lambda_1. $$ Thus $\lambda_1\leq\sup_{u\in A}\inf_{x\in\Omega}\frac{-\Delta u(x)}{u(x)}$.
On the other hand, if $u\in A$, then $\inf_{x\in\Omega}\frac{-\Delta u(x)}{u(x)}\leq\lambda_1$: For if not, then $-\Delta u(x)>\lambda_1 u(x)$ for all $x\in\Omega$, which implies $$ \int u(x)v(x)\,dx<-\frac 1 {\lambda_1}\int u(x)\Delta v(x)\,dx=-\frac 1{\lambda_1}\int \Delta u(x) v(x)\,dx=\int u(x)v(x)\,dx, $$ a contradiction. Hence $\sup_{u\in A}\inf_{x\in\Omega}\frac{-\Delta u(x)}{u(x)}\leq\lambda_1$.