[proofplan] We test the weak Dirichlet eigenvalue equation with the eigenfunction itself. This expresses $\lambda$ as the quotient of the Dirichlet energy of $u$ by its nonzero $L^2$ mass, so first $\lambda \geq 0$. To rule out $\lambda = 0$, we use the [Poincaré Inequality](/theorems/53) on the zero-trace [Sobolev space](/page/Sobolev%20Space) $H_0^1(U)$: zero Dirichlet energy forces the $L^2$ norm of $u$ to vanish, contradicting that an eigenfunction is nonzero. [/proofplan] [step:Fix the measure convention and the $L^2$ spaces over $U$] Throughout the proof, let $\mathcal{L}^n$ denote $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on the ambient Euclidean space $\mathbb{R}^n$; integrals over $U$ are taken with respect to the restriction of $\mathcal{L}^n$ to $U$. Let $\mathcal{B}(U)$ denote the Borel $\sigma$-algebra of the subspace $U$. Define $L^2(U)$ to be the real [Hilbert space](/page/Hilbert%20Space) $L^2(U, \mathcal{B}(U), \mathcal{L}^n; \mathbb{R})$ of square-integrable real-valued [measurable functions](/page/Measurable%20Functions) on $U$, modulo $\mathcal{L}^n$-a.e. equality. Define $L^2(U;\mathbb{R}^n)$ to be the real Hilbert space of square-integrable measurable maps from $U$ to $\mathbb{R}^n$, again modulo $\mathcal{L}^n$-a.e. equality. [/step] [step:Test the weak eigenvalue equation with the eigenfunction] Since $u$ belongs to the zero-trace [Sobolev space](/page/Sobolev%20Space) $H_0^1(U)$, it is an admissible [test function](/page/Test%20Function) in the weak Dirichlet eigenvalue equation stated in the theorem. Taking $v = u$ gives \begin{align*} \int_U \nabla u(x) \cdot \nabla u(x)\, d\mathcal{L}^n(x) = \lambda \int_U u(x)^2\, d\mathcal{L}^n(x). \end{align*} Equivalently, \begin{align*} \int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x) = \lambda \int_U u(x)^2\, d\mathcal{L}^n(x). \end{align*} The quantity \begin{align*} M := \int_U u(x)^2\, d\mathcal{L}^n(x) \end{align*} is strictly positive because $u$ is nonzero in $L^2(U)$. Indeed, if $u = 0$ in $L^2(U)$, then all weak first derivatives of $u$ are zero as distributions, so $u$ represents the zero element of $H_0^1(U)$, contrary to the definition of an eigenfunction. Hence \begin{align*} \lambda = \frac{\int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x)}{M}. \end{align*} [guided] The theorem statement defines the weak Dirichlet eigenvalue equation as the identity that holds for every test function $v$ in the zero-trace [Sobolev space](/page/Sobolev%20Space) $H_0^1(U)$. Since the eigenfunction itself satisfies $u \in H_0^1(U)$, we may choose $v = u$. Substituting this choice into the weak eigenvalue equation 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). \end{align*} Using the Euclidean identity $\nabla u(x) \cdot \nabla u(x) = |\nabla u(x)|^2$ for $\mathcal{L}^n$-a.e. $x \in U$, and $u(x)u(x)=u(x)^2$, this becomes \begin{align*} \int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x) = \lambda \int_U u(x)^2\, d\mathcal{L}^n(x). \end{align*} Define the $L^2$ mass of $u$ by \begin{align*} M := \int_U u(x)^2\, d\mathcal{L}^n(x). \end{align*} Because $u$ is nonzero as an eigenfunction in $H_0^1(U)$, it is also nonzero as an element of $L^2(U)$: if $u = 0$ in $L^2(U)$, then the weak first derivatives of $u$ are all zero distributions, so the Sobolev class of $u$ is the zero element of $H_0^1(U)$. Thus its $L^2$ norm is nonzero, so $M > 0$. Therefore division by $M$ is valid, and we obtain \begin{align*} \lambda = \frac{\int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x)}{M}. \end{align*} This quotient formula is the main point of the proof: the numerator is an energy and the denominator is a nonzero mass. [/guided] [/step] [step:Deduce nonnegativity from the energy identity] The integrand $|\nabla u(x)|^2$ is nonnegative for $\mathcal{L}^n$-a.e. $x \in U$, so \begin{align*} \int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x) \geq 0. \end{align*} Since $M > 0$, the quotient obtained above gives $\lambda \geq 0$. [/step] [step:Exclude the zero eigenvalue using the Dirichlet boundary condition] Assume for contradiction that $\lambda = 0$. The energy identity gives \begin{align*} \int_U |\nabla u(x)|^2\, d\mathcal{L}^n(x) = 0. \end{align*} Since $|\nabla u|^2 \geq 0$ $\mathcal{L}^n$-a.e., this implies $\nabla u = 0$ in $L^2(U;\mathbb{R}^n)$. Because $U$ is a bounded open subset of $\mathbb{R}^n$ and $u$ belongs to the zero-trace [Sobolev space](/page/Sobolev%20Space) $H_0^1(U)$, we use the version of the [Poincaré Inequality](/theorems/53) valid for arbitrary bounded open sets and zero-trace functions. The boundedness of $U$ is exactly the domain hypothesis required for this version of the inequality. It gives a constant $C_U > 0$, depending only on $U$, such that \begin{align*} \|u\|_{L^2(U)} \leq C_U \|\nabla u\|_{L^2(U)}. \end{align*} Therefore \begin{align*} \|u\|_{L^2(U)} = 0. \end{align*} Thus $u = 0$ in $L^2(U)$, contradicting the hypothesis that $u$ is nonzero as an eigenfunction. Hence $\lambda \neq 0$. [/step] [step:Conclude strict positivity] We have proved $\lambda \geq 0$ and $\lambda \neq 0$. Since $\lambda \in \mathbb{R}$, it follows that $\lambda > 0$. This proves the claimed positivity of every weak Dirichlet Laplacian eigenvalue. [/step]