[proofplan] We split the eigenfunction into its positive and negative parts. Testing the weak eigenvalue equation against these truncations shows that each nonzero part also attains the first Rayleigh quotient, hence is itself a nonnegative first eigenfunction. The assumed weak strong maximum principle then forces every nonzero part to be positive almost everywhere. Since the positive and negative parts have disjoint supports, both cannot be nonzero; therefore the original eigenfunction has one fixed sign almost everywhere. [/proofplan]

[step:Split the eigenfunction into positive and negative parts] Define the positive and negative parts of $u$ by the maps \begin{align*} u^+: U &\to [0,\infty) \end{align*} and \begin{align*} u^+(x) &= \max\{u(x),0\} \end{align*} for $x \in U$, and \begin{align*} u^-: U &\to [0,\infty) \end{align*} and \begin{align*} u^-(x) &= \max\{-u(x),0\} \end{align*} for $x \in U$. For each measurable set $A \subset U$, let $\mathbb{1}_A:U\to\{0,1\}$ denote the indicator function of $A$. The Sobolev truncation property gives $u^+,u^- \in H^1_0(U)$, and their weak gradients satisfy \begin{align*} \nabla u^+ = \mathbb{1}_{\{u>0\}}\nabla u \end{align*} and \begin{align*} \nabla u^- = -\mathbb{1}_{\{u<0\}}\nabla u \end{align*} $\mathcal{L}^n$-a.e. in $U$. Also $u=u^+-u^-$ and $u^+u^-=0$ $\mathcal{L}^n$-a.e. in $U$. [/step]

[step:Show that every nonzero sign part is a first eigenfunction]Use $u^+ \in H^1_0(U)$ as a [test function](/page/Test%20Function) in the weak eigenvalue equation for $u$. Since $\nabla u \cdot \nabla u^+ = |\nabla u^+|^2$ $\mathcal{L}^n$-a.e. and $u u^+ = |u^+|^2$ $\mathcal{L}^n$-a.e., we obtain \begin{align*} \int_U |\nabla u^+|^2\, d\mathcal{L}^n(x) = \lambda_1 \int_U |u^+|^2\, d\mathcal{L}^n(x). \end{align*} Thus, if $u^+ \neq 0$ in $L^2(U)$, then $u^+$ attains the Rayleigh quotient value $\lambda_1$. Similarly, using $-u^- \in H^1_0(U)$ as a test function gives \begin{align*} \int_U |\nabla u^-|^2\, d\mathcal{L}^n(x) = \lambda_1 \int_U |u^-|^2\, d\mathcal{L}^n(x). \end{align*} Hence, if $u^- \neq 0$ in $L^2(U)$, then $u^-$ also attains the Rayleigh quotient value $\lambda_1$.[/step]

[guided]The purpose of introducing $u^+$ and $u^-$ is that they isolate the regions where $u$ is positive and negative while remaining admissible test functions. The Sobolev truncation property ensures that both $u^+$ and $u^-$ belong to $H^1_0(U)$, so they can be inserted into the weak formulation. Testing with $u^+$ gives \begin{align*} \int_U \nabla u \cdot \nabla u^+\, d\mathcal{L}^n(x) = \lambda_1 \int_U u u^+\, d\mathcal{L}^n(x). \end{align*} On the set $\{x\in U:u(x)>0\}$, we have $u^+=u$ and $\nabla u^+=\nabla u$. On the set $\{x\in U:u(x)\leq 0\}$, we have $u^+=0$ and $\nabla u^+=0$ $\mathcal{L}^n$-a.e. Therefore \begin{align*} \nabla u \cdot \nabla u^+ = |\nabla u^+|^2 \end{align*} and \begin{align*} u u^+ = |u^+|^2 \end{align*} $\mathcal{L}^n$-a.e. in $U$. Substituting these identities gives \begin{align*} \int_U |\nabla u^+|^2\, d\mathcal{L}^n(x) = \lambda_1 \int_U |u^+|^2\, d\mathcal{L}^n(x). \end{align*} If $u^+$ is not the zero element of $L^2(U)$, this equality says precisely that the Rayleigh quotient of $u^+$ equals $\lambda_1$. For the negative part, we test with $-u^- \in H^1_0(U)$ because $u=-u^-$ on the set where $u<0$. The weak equation gives \begin{align*} \int_U \nabla u \cdot \nabla(-u^-)\, d\mathcal{L}^n(x) = \lambda_1 \int_U u(-u^-)\, d\mathcal{L}^n(x). \end{align*} Since $\nabla u = -\nabla u^-$ and $u=-u^-$ on $\{u<0\}$, while both $u^-$ and $\nabla u^-$ vanish $\mathcal{L}^n$-a.e. on $\{u\geq 0\}$, this becomes \begin{align*} \int_U |\nabla u^-|^2\, d\mathcal{L}^n(x) = \lambda_1 \int_U |u^-|^2\, d\mathcal{L}^n(x). \end{align*} Thus every nonzero sign part of $u$ is itself a minimizer for the first Rayleigh quotient.[/guided]

[step:Apply the weak strong maximum principle to the nonzero sign parts] Let $w \in \{u^+,u^-\}$ be nonzero in $L^2(U)$. The previous step shows that \begin{align*} \int_U |\nabla w|^2\, d\mathcal{L}^n(x) = \lambda_1 \int_U |w|^2\, d\mathcal{L}^n(x). \end{align*} Define $A:H^1_0(U)\times H^1_0(U)\to\mathbb{R}$ and $B:H^1_0(U)\times H^1_0(U)\to\mathbb{R}$ by \begin{align*} A[v,z] = \int_U \nabla v \cdot \nabla z\, d\mathcal{L}^n(x) \end{align*} and \begin{align*} B[v,z] = \int_U vz\, d\mathcal{L}^n(x). \end{align*} Because $w\neq 0$ in $L^2(U)$, $B[w,w]>0$. For a fixed $\varphi\in H^1_0(U)$, define $q_\varphi:I_\varphi\to\mathbb{R}$ on an open interval $I_\varphi\subset\mathbb{R}$ containing $0$ and small enough that $B[w+t\varphi,w+t\varphi]>0$ by \begin{align*} q_\varphi(t)=\frac{A[w+t\varphi,w+t\varphi]}{B[w+t\varphi,w+t\varphi]}. \end{align*} Since $\lambda_1$ is the minimum of the Rayleigh quotient on $H^1_0(U)\setminus\{0\}$ and $q_\varphi(0)=\lambda_1$, the real differentiable function $q_\varphi$ has a minimum at $0$, so $q_\varphi'(0)=0$. Differentiating the quotient gives \begin{align*} 0=q_\varphi'(0)=\frac{2A[w,\varphi]B[w,w]-2A[w,w]B[w,\varphi]}{B[w,w]^2}. \end{align*} Using $A[w,w]=\lambda_1 B[w,w]$ and $B[w,w]>0$, this identity reduces to \begin{align*} A[w,\varphi]=\lambda_1 B[w,\varphi]. \end{align*} Thus \begin{align*} \int_U \nabla w \cdot \nabla \varphi\, d\mathcal{L}^n(x) = \lambda_1 \int_U w\varphi\, d\mathcal{L}^n(x) \end{align*} for every $\varphi \in H^1_0(U)$. Also $w\geq 0$ $\mathcal{L}^n$-a.e. in $U$ and $w\neq 0$. By the assumed weak strong maximum principle, $w>0$ $\mathcal{L}^n$-a.e. in $U$. [/step]

[step:Exclude coexistence of positive and negative parts] Suppose both $u^+$ and $u^-$ are nonzero in $L^2(U)$. Applying the previous step to each of them gives \begin{align*} u^+>0 \quad \mathcal{L}^n\text{-a.e. in } U \end{align*} and \begin{align*} u^->0 \quad \mathcal{L}^n\text{-a.e. in } U. \end{align*} But $u^+u^-=0$ $\mathcal{L}^n$-a.e. in $U$, so the two inequalities cannot both hold. Therefore exactly one of $u^+$ and $u^-$ is nonzero. If $u^+\neq 0$ and $u^-=0$, then $u=u^+$ $\mathcal{L}^n$-a.e. in $U$, and the weak strong maximum principle gives \begin{align*} u>0 \quad \mathcal{L}^n\text{-a.e. in } U. \end{align*} If $u^-=0$ is false, then $u^+=0$ and $u=u^-(-1)$ $\mathcal{L}^n$-a.e. in $U$, so \begin{align*} u<0 \quad \mathcal{L}^n\text{-a.e. in } U. \end{align*} This proves that a nonzero first Dirichlet eigenfunction has a fixed strict sign almost everywhere in $U$. [/step]