[proofplan] We first introduce the positive part of the given function and verify, using the [Sobolev lattice property](/page/Sobolev%20Space), that it is an admissible nonnegative [test function](/page/Test%20Function) in $H^1_0(U)$. Testing the weak inequality with that function converts the hypothesis into a quadratic energy inequality supported on the positivity set. The Rayleigh quotient definition of the first Dirichlet eigenvalue $\lambda_1(U)$ then gives a strictly positive lower bound for the same energy unless the positive part vanishes. Hence the function has no positive part in $L^2(U)$, which is exactly the assertion that $u \le 0$ almost everywhere. [/proofplan] [step:Use the positive part as an admissible test function] Define the positive part map $P: \mathbb{R} \to \mathbb{R}$ by $P(t)=\max\{t,0\}$ for every $t \in \mathbb{R}$. Set $u^+ := P \circ u$. Define the measurable set $\{u>0\}:=\{x\in U:u(x)>0\}$, and let $\mathbb{1}_{\{u>0\}}:U\to\{0,1\}$ denote its indicator function, so $\mathbb{1}_{\{u>0\}}(x)=1$ for $x\in\{u>0\}$ and $\mathbb{1}_{\{u>0\}}(x)=0$ for $x\in U\setminus\{u>0\}$. Since $P$ is Lipschitz, $P(0)=0$, and $u \in H^1_0(U)$, the [Sobolev lattice property](/page/Sobolev%20Space) gives $u^+ \in H^1_0(U)$, $u^+ \ge 0$ $\mathcal{L}^n$-a.e. in $U$, and \begin{align*} \nabla u^+(x) = \mathbb{1}_{\{u>0\}}(x)\nabla u(x) \end{align*} for $\mathcal{L}^n$-a.e. $x \in U$. Thus $v=u^+$ is an admissible nonnegative test function in the hypothesis. Substituting $v=u^+$ gives \begin{align*} \int_U \nabla u(x)\cdot \nabla u^+(x)\,d\mathcal{L}^n(x) - \lambda \int_U u(x)u^+(x)\,d\mathcal{L}^n(x) \le 0. \end{align*} On $\{u>0\}$ we have $u=u^+$ and $\nabla u^+=\nabla u$ a.e.; on $\{u\le 0\}$ we have $u^+=0$ and $\nabla u^+=0$ a.e. Therefore \begin{align*} \nabla u(x)\cdot \nabla u^+(x) = |\nabla u^+(x)|^2 \end{align*} for $\mathcal{L}^n$-a.e. $x \in U$, and \begin{align*} u(x)u^+(x) = |u^+(x)|^2 \end{align*} for $\mathcal{L}^n$-a.e. $x \in U$. Hence \begin{align*} \int_U |\nabla u^+(x)|^2\,d\mathcal{L}^n(x) - \lambda \int_U |u^+(x)|^2\,d\mathcal{L}^n(x) \le 0. \end{align*} [guided] The hypothesis allows us to test only against nonnegative functions in $H^1_0(U)$, so the natural choice is the part of $u$ that could violate the desired conclusion. Define the map $P: \mathbb{R} \to \mathbb{R}$ by $P(t)=\max\{t,0\}$ for every $t \in \mathbb{R}$. Then $u^+ := P \circ u$ records exactly where $u$ is positive. Define the measurable set $\{u>0\}:=\{x\in U:u(x)>0\}$, and let $\mathbb{1}_{\{u>0\}}:U\to\{0,1\}$ denote its indicator function. Thus $\mathbb{1}_{\{u>0\}}(x)=1$ on the positive set and $\mathbb{1}_{\{u>0\}}(x)=0$ off the positive set. Since $P$ is Lipschitz and satisfies $P(0)=0$, the [Sobolev lattice property](/page/Sobolev%20Space) applies to $u \in H^1_0(U)$ and gives $u^+ \in H^1_0(U)$. It also gives the weak gradient formula \begin{align*} \nabla u^+(x) = \mathbb{1}_{\{u>0\}}(x)\nabla u(x) \end{align*} for $\mathcal{L}^n$-a.e. $x \in U$. Because $u^+ \ge 0$ almost everywhere, $u^+$ is an admissible test function. Substituting $v=u^+$ into the assumed weak inequality gives \begin{align*} \int_U \nabla u(x)\cdot \nabla u^+(x)\,d\mathcal{L}^n(x) - \lambda \int_U u(x)u^+(x)\,d\mathcal{L}^n(x) \le 0. \end{align*} We now rewrite both integrands in terms of $u^+$. On the set $\{u>0\}$, the positive part satisfies $u^+=u$ and the gradient identity gives $\nabla u^+=\nabla u$ a.e. On the set $\{u\le 0\}$, the function $u^+$ is zero, and the same gradient identity gives $\nabla u^+=0$ a.e. Therefore \begin{align*} \nabla u(x)\cdot \nabla u^+(x) = |\nabla u^+(x)|^2 \end{align*} for $\mathcal{L}^n$-a.e. $x \in U$. Similarly, since $u^+=u$ on $\{u>0\}$ and $u^+=0$ on $\{u\le 0\}$, we have \begin{align*} u(x)u^+(x) = |u^+(x)|^2 \end{align*} for $\mathcal{L}^n$-a.e. $x \in U$. Substituting these two identities into the tested inequality yields \begin{align*} \int_U |\nabla u^+(x)|^2\,d\mathcal{L}^n(x) - \lambda \int_U |u^+(x)|^2\,d\mathcal{L}^n(x) \le 0. \end{align*} This is the key reduction: the original one-sided weak inequality has become a coercive quadratic inequality for the positive part alone. [/guided] [/step] [step:Use the Rayleigh quotient gap to force the positive part to vanish] Since $U\subset\mathbb{R}^n$ is nonempty, bounded, and open, the [Rayleigh quotient characterization of the first Dirichlet eigenvalue](/theorems/533) for the Dirichlet Laplacian, applied to $L=-\Delta$, gives that every nonzero $w \in H^1_0(U)$ satisfies \begin{align*} \int_U |\nabla w(x)|^2\,d\mathcal{L}^n(x) \ge \lambda_1(U)\int_U |w(x)|^2\,d\mathcal{L}^n(x). \end{align*} The same inequality also holds for $w=0$, since both sides are then equal to $0$. Applying this with $w=u^+$ gives \begin{align*} \int_U |\nabla u^+(x)|^2\,d\mathcal{L}^n(x) - \lambda \int_U |u^+(x)|^2\,d\mathcal{L}^n(x) \ge (\lambda_1(U)-\lambda)\int_U |u^+(x)|^2\,d\mathcal{L}^n(x). \end{align*} Combining this lower bound with the inequality from the previous step yields \begin{align*} 0 \ge (\lambda_1(U)-\lambda)\int_U |u^+(x)|^2\,d\mathcal{L}^n(x). \end{align*} Since $\lambda_1(U)-\lambda>0$ and the integral of $|u^+|^2$ is nonnegative, it follows that \begin{align*} \int_U |u^+(x)|^2\,d\mathcal{L}^n(x)=0. \end{align*} Thus $u^+=0$ in $L^2(U)$. [/step] [step:Translate vanishing of the positive part into the maximum principle] Since $u^+=0$ in $L^2(U)$ and $u^+ \ge 0$ $\mathcal{L}^n$-a.e. in $U$, we have $u^+(x)=0$ for $\mathcal{L}^n$-a.e. $x \in U$. By the definition $u^+(x)=\max\{u(x),0\}$, this means $u(x)\le 0$ for $\mathcal{L}^n$-a.e. $x \in U$. This is the desired weak maximum principle below the first Dirichlet eigenvalue. [/step]