I am trying understand why if $u$ is an eigenfunction for the first eigenvalue $\sigma_1$, then $|u|$ is one also.
Theorem 8.38. Let $L$ be a self-adjoint operator satisfying $(8.5)$ and $(8.6)$. Then the minimum eigenvalue is simple and has a positive eigenfunction.
Proof. If $u$ is an eigenfunction of $\sigma_1$, then it follows from the formula $(8.92)$ that $|u|$ is one also.
Relevant informations:
$(8.92)$ is the variational characterization of $\sigma_1$. More precisely, if the operator $L$ has the form
$$Lu = D_i(a^{ij}D_j u + b^i u) - b^i D_i u + c u,$$ where $[a^{ij}]$ is symmetric, then The associated quadratic form on $H = W_0^{1,2}(\Omega)$ is then given by $$\mathscr{L}(u,u) = \int_{\Omega} (a^{ij} D_i u D_j u + 2 b^i u D_i u + c u^2) dx.$$ The ratio $$J(u) = \frac{\mathscr{L}(u,u)}{(u,u)}, u \not\equiv 0, u \in H$$ is called the Rayleigh quotient. $$(8.92) \ \sigma = \inf_H J.$$
The conditions satisfied by $L$ are $$ \quad a^{i j}(x) \xi_{i} \xi_{j} \geqslant \lambda|\xi|^{2}, \quad \forall x \in \Omega, \xi \in \mathbb{R}^{n} \tag{8.5}$$ We also assume (unless stated otherwise) that $L$ has bounded coefficients; that is for some constants $\Lambda$ and $v \geqslant 0$ we have for all $x \in \Omega$
$$ \sum\left|a^{i j}(x)\right|^{2} \leqslant \Lambda^{2}, \quad \lambda^{-2} \sum\left(\left|b^{i}(x)\right|^{2}+\left|c^{i}(x)\right|^{2}\right)+\lambda^{-1}|d(x)| \leqslant v^{2} \tag{8.6}$$ Transcribed from this screenshot.
Thanks in advance!
Let’s try to evaluate $\mathscr{L}(|u|,|u|)$ keeping in mind that $|u| = u^+ + u^-$ where:
$u^+ (x) = \max \{ u(x), 0\}$ and $u^- (x) := \max \{-u(x), 0\}$ for $x \in \Omega$,
$u^\pm \geq 0$ a.e. in $\Omega^\pm := \{ x \in \Omega : \pm u(x) \geq 0\}$,
$u^\mp = 0$ in $\Omega^\pm$,
and $u = u^+ - u^-$;
we get:
\begin{split} \mathscr{L}(|u|,|u|) &= \mathscr{L}(u^+ + u^-, u^+ + u^-) \\ &= \int_{\Omega^+} (a^{ij} D_i u^+ D_j u^+ + 2 b^i u^+ D_i u^+ + c (u^+)^2) dx + \int_{\Omega^-} (a^{ij} D_i u^- D_j u^- + 2 b^i u^- D_i u^- + c (u^-)^2) dx \\ &= \int_{\Omega^+} (a^{ij} D_i u^+ D_j u^+ + 2 b^i u^+ D_i u^+ + c (u^+)^2) dx + \int_{\Omega^-} (a^{ij} D_i (-u^-) D_j (-u^-) + 2 b^i (-u^-)D_i (-u^-) + c (-u^-)^2) dx \\ &= \mathscr{L}(u^+ - u^-, u^+ - u^-) \\ &= \mathscr{L} (u,u)\;. \end{split}
Since $\| |u|\|_2^2 = ( |u|, |u|) = (u,u) = \|u\|^2$, we have equality between the Rayleigh quotients $\frac{\mathscr{L} (|u|, |u|)}{\| |u|\|^2}$ and $\frac{\mathscr{L} (u,u)}{\|u\|^2}$, therefore if $u$ is a first eigenfunction then $|u|$ also does.