# I would like to find some references about the demonstration of this theorem that the book doesn't cite anyone

In Ambrosetti, A. and Arcoya, D.,it An Introduction to Nonlinear Functional Analysis and Elliptic Problems there is a theorem that the autors doesn't prove and dont let any reference about it, the theorem is

Let, $$\label{equ1.4.1} \left\{ \begin{array}{lrl} -\Delta u = \lambda m(x)u,& &\,\, x\in \Omega\\ \hspace{0.6cm}\, u = 0, & &\,\, x\in\partial\Omega \end{array} \right.$$ Assume that $\Omega\subset{R^{N}}$ is bounded and open, $r\in \left( \frac{N}{2}, +\infty \right)\cap \left(1,+\infty \right)$, and $m\in L^{r}(\Omega)$. Consider the set \begin{eqnarray*} \Omega_{+} & = & \left\{ x\in\Omega ;, m(x) > 0 \right\} \end{eqnarray*} The following assertions hold.

• $0$ is not an eigenvalue of the equation above.

• If the Lebesgue measure $\left|\Omega_{+} \right|$ of $\Omega_{+}$ is zero, then the equation has no positive eigenvalue.

• If $\left|\Omega_{+} \right| > 0$, then the positive eigenvalues of the equation above deﬁne a nondecreasing unbounded sequence $\left\{ \lambda_{n} \right\}_{n\in \mathbb{N}} \subset (0, +\infty)$. In addition, $\lambda_{n}$ is characterized by $$\frac{1}{\lambda_{n}} = \sup_{F\in \mathcal{F}_{n}}\inf\left\{ \displaystyle\int m(x)u^{2}(x)dx : \, \displaystyle\int \left|\nabla u(x)\right|^{2} = 1, \,\,\,\,u\in F \right\}$$ where\, ${\mathcal{F}}_{n} = \left\{ F\subset H:\, F\,\,\mbox{is a subspace with}\,\,\dim F = n \right\}.$

Has someone already seen this theorem proved anywhere?

Ad 1. Eigenvalue $0$ would mean that there is a nonzero function $u$ for which $$\left\{ \begin{array}{lrl} -\Delta u = 0& &\,\, x\in \Omega,\\ \hspace{0.6cm}\, u = 0 & &\,\, x\in\partial\Omega. \end{array} \right.$$ By the maximum principle for harmonic functions, such $u$ has to be zero.
Ad 2. You're not mentioning the space you're working in, but I assume it's $W^{1,2}(\Omega)$. By Sobolev embedding, we have $u \in L^{\frac{2n}{n-2}}$. The exponent $r$ is chosen so that the equation can be tested with $u$ itself (by Holder's inequality), resulting in $$\int_\Omega |\nabla u|^2 = - \int_\Omega u \Delta u = \lambda \int_\Omega m u^2.$$ The condition $|\Omega_+| = 0$ means $m \le 0$ a.e. in $\Omega$. If we assume that $\lambda \ge 0$, the RHS is nonpositive, while the LHS is nonnegative. Thus, $\nabla u \equiv 0$ in $\Omega$ and in consequence $u \equiv 0$.
Ad 3. This requires some more work and knowledge. Are you familiar with the case $m=1$, i.e. with the usual spectral theory for $-\Delta$?
Chapter 4 of this thesis deals with eigenvalues for the weighted Laplacian. It has a few references, so it is a starting point. link This paper deals with the $p$-laplacian but again the references could be useful. link2