Theorem $2$ (Variational principle for the principal eigenvalue)

I'm trying understand the Theorem $2$ (Variational principle for the principal eigenvalue) in the section of the eigenvalues of the elliptic symmetric operator on chapter $6$ of the Evans' book and I'm stuck in steps $5$, $6$, $7$ and $8$ of the proof. I will put the whole proof and the motivation of the weak solution below.

My doubts are

$1.$ Why $\lambda_1$ has finite multiplicity imply that $u = \sum_\limits{k=1}^m (u,w_k) w_k$ in the step $5$?

$2.$ Why $B[u^+,u^+] = \lambda_1 ||u^+||^2_{L^2(U)}$ and $B[u^-,u^-] = \lambda_1 ||u^-||^2_{L^2(U)}$? I can't see why the previous equality ensures this.

$3.$ Why $(19)$ and $(20)$ holds by the step $5$? This step ensures the result follows for normalized functions with respect to the norm of $L^2(U)$ and we assume $||u||_{L^2(U)}^2 = 1$, but not assume this for $u^+$ and $u^-$.

$4.$ Why exist $\chi \in \mathbb{R}$ with the property mentioned in the step $8$? I would understand if the $u$ and $\tilde{u}$ would be constants, but they are functions. How to ensure the existence of this constant in this case?

Thanks in advance!

1. Equation $(8)$ states that $$ u = \sum_{k=1}^\infty d_kw_k = \sum_{k=1}^\infty (u,w_k) w_k. $$ The line before says that $d_k = 0$ if $\lambda_k>\lambda_1$, which means that the only terms that preserves in the infinite sum are the terms $d_kw_k$ with $w_k$ the eigenfunctions corresponding to $\lambda_1$. Since $\lambda_1$ has multiplicity, say $m$, it follows that there are $m$ eigenfunctions corresponding to $\lambda_1$ and so $$ u = \sum_{k=1}^\infty (u,w_k)w_k = \sum_{k=1}^m (u,w_k)w_k. $$

2. On the equality just before the claim, note that we use the \emph{Rayleigh formula} (See the equivalent statement of this in the Remarks), which asserts that $$ \lambda_1\le \frac{B[v,v]}{\|v\|_{L^2(U)}^2} \ \ \textrm{ for any $v\in H_0^1(U), v\neq 0$}. $$ In particular, \begin{align*} \lambda_1 = B[u,u] & = \color{red}{B[u^+,u^+]} + B[u^-,u^-] \\ & \ge \color{red}{\lambda_1\|u^+\|_{L^2(U)}^2} + \color{blue}{B[u^-,u^-]} \\ & \ge \lambda_1\|u^+\|_{L^2(U)}^2 + \color{blue}{\lambda_1\|u^-\|_{L^2(U)}^2} \\ & = (\alpha + \beta)\lambda_1 = \lambda_1. \end{align*} Since we start with $\lambda_1$ and end up with $\lambda_1$, this means that every inequality that we used must in fact be an equality, i.e. the blue expressions must equal to each other and similarly for red.

3. You are correct that $u^+$ and $u^-$ are not normalised so you can't really invoke the claim in Step $5$ (called this Claim $1$). But one can actually prove the following similar claim by adapting the proof of Claim $1$: If $u\in H_0^1(U)\setminus 0$, then $u$ is a weak solution of \begin{alignat*}{2} Lu & = \lambda_1u && \ \ \textrm{ in $U$} \\ u & = 0 && \ \ \textrm{ on $\partial U$}, \end{alignat*} if and only if $\lambda_1\|u\|_{L^2(U)}^2 = B[u,u]$.

4. In Step $8$, since we assume that $u$ and $\tilde u$ are two nontrivial (i.e. $u, \tilde u\neq 0$) weak solutions of $(10)$, it follows from $(16)$ and $(17)$ that either $u>0$ in $U$ or $u<0$ in $U$, and similarly for $\tilde u$. In particular, this means that $$ A\colon = \int_U u\, dx \neq 0 \ \ \textrm{ and } \ \ B\colon = \int_U \tilde u\, dx \neq 0. $$ But these integrals $A$ and $B$ are just real numbers (well, convince yourself that these integrals are finite number). The existence of $\chi$ is clear here, since one is really asking: can I find $\chi$ such that $A-\chi B = 0$? Yes, pick $\chi = A/B$ (which is well-defined since $B\neq 0$).