[proofplan] We construct the inverse of the weak Dirichlet Laplacian as a [compact operator](/page/Compact%20Operator) $K$ on $L^2(U)$. The weak formulation shows that $K$ is self-adjoint, positive, and injective. The compact self-adjoint spectral theorem then gives an [orthonormal basis](/page/Orthonormal%20Basis) of $L^2(U)$ consisting of eigenvectors of $K$ with positive eigenvalues tending to $0$. Taking reciprocals of those eigenvalues produces the Dirichlet Laplacian eigenvalues, and the weak equation for $K$ gives the displayed eigenfunction identity. [/proofplan] [step:Define the weak resolvent as a compact operator on $L^2(U)$] Let $H$ denote the [Hilbert space](/page/Hilbert%20Space) $L^2(U)$ with [inner product](/page/Inner%20Product) \begin{align*} (f,g)_H := \int_U f g \, d\mathcal{L}^n \end{align*} for $f,g \in L^2(U)$. By the assumed well-posedness of the variational Dirichlet problem, define the solution operator \begin{align*} K_0: L^2(U) \to H^1_0(U) \end{align*} as follows: for $f \in L^2(U)$, $K_0 f$ is the unique element of $H^1_0(U)$ satisfying \begin{align*} \int_U \nabla (K_0 f) \cdot \nabla v \, d\mathcal{L}^n = \int_U f v \, d\mathcal{L}^n \end{align*} for every $v \in H^1_0(U)$. Let \begin{align*} j: H^1_0(U) \to L^2(U) \end{align*} be the inclusion map. Define \begin{align*} K: L^2(U) \to L^2(U) \end{align*} by $K := j \circ K_0$. Since $K_0: L^2(U) \to H^1_0(U)$ is bounded by hypothesis and $j: H^1_0(U) \to L^2(U)$ is compact by hypothesis, the composition $K = j \circ K_0$ is a compact linear operator on $L^2(U)$. [/step] [step:Verify that the resolvent is self-adjoint and positive] Let $f,g \in L^2(U)$. Since $Kf \in H^1_0(U)$ and $Kg \in H^1_0(U)$ as representatives of the corresponding weak solutions, we may use $v = Kg$ in the weak equation for $Kf$ and $v = Kf$ in the weak equation for $Kg$. This gives \begin{align*} \int_U \nabla (Kf) \cdot \nabla (Kg) \, d\mathcal{L}^n = \int_U f Kg \, d\mathcal{L}^n \end{align*} and \begin{align*} \int_U \nabla (Kg) \cdot \nabla (Kf) \, d\mathcal{L}^n = \int_U g Kf \, d\mathcal{L}^n. \end{align*} The two left-hand sides are equal by symmetry of the Euclidean dot product. Hence \begin{align*} (Kf,g)_H = \int_U Kf g \, d\mathcal{L}^n = \int_U f Kg \, d\mathcal{L}^n = (f,Kg)_H. \end{align*} Thus $K$ is self-adjoint on $L^2(U)$. Taking $g=f$ in the weak formulation and using $v=Kf$ gives \begin{align*} (Kf,f)_H = \int_U f Kf \, d\mathcal{L}^n = \int_U |\nabla (Kf)|^2 \, d\mathcal{L}^n \geq 0. \end{align*} Therefore $K$ is positive. [guided] We first prove self-adjointness, because the spectral theorem we will use applies to compact [self-adjoint operators](/page/Self-Adjoint%20Operators) on Hilbert spaces. Let $f,g \in L^2(U)$. The element $Kf$ is the weak solution with source $f$, so it belongs to $H^1_0(U)$ and satisfies \begin{align*} \int_U \nabla (Kf) \cdot \nabla v \, d\mathcal{L}^n = \int_U f v \, d\mathcal{L}^n \end{align*} for every $v \in H^1_0(U)$. Since $Kg \in H^1_0(U)$, we may choose $v=Kg$ and obtain \begin{align*} \int_U \nabla (Kf) \cdot \nabla (Kg) \, d\mathcal{L}^n = \int_U f Kg \, d\mathcal{L}^n. \end{align*} Now apply the same weak formulation with source $g$ and [test function](/page/Test%20Function) $v=Kf$. This gives \begin{align*} \int_U \nabla (Kg) \cdot \nabla (Kf) \, d\mathcal{L}^n = \int_U g Kf \, d\mathcal{L}^n. \end{align*} The left-hand sides are equal because $a \cdot b = b \cdot a$ for vectors $a,b \in \mathbb{R}^n$. Therefore \begin{align*} \int_U f Kg \, d\mathcal{L}^n = \int_U g Kf \, d\mathcal{L}^n. \end{align*} In the Hilbert space notation $(h_1,h_2)_H = \int_U h_1 h_2 \, d\mathcal{L}^n$, this is exactly \begin{align*} (f,Kg)_H = (Kf,g)_H. \end{align*} Thus $K$ is self-adjoint. We also need positivity, because it rules out negative resolvent eigenvalues. Taking the weak equation for $Kf$ and using the test function $v=Kf$ gives \begin{align*} \int_U f Kf \, d\mathcal{L}^n = \int_U |\nabla (Kf)|^2 \, d\mathcal{L}^n. \end{align*} The right-hand side is nonnegative because it is the integral of the nonnegative function $|\nabla(Kf)|^2$ with respect to $\mathcal{L}^n$. Hence \begin{align*} (Kf,f)_H \geq 0. \end{align*} This proves that $K$ is positive. [/guided] [/step] [step:Show that the resolvent has trivial kernel] Let $f \in L^2(U)$ satisfy $Kf=0$ in $L^2(U)$. Since $Kf$ is also the weak solution associated to $f$, the weak formulation gives \begin{align*} \int_U f v \, d\mathcal{L}^n = \int_U \nabla (Kf) \cdot \nabla v \, d\mathcal{L}^n = 0 \end{align*} for every $v \in H^1_0(U)$. The space $C_c^\infty(U)$ is dense in $L^2(U)$, and $C_c^\infty(U) \subset H^1_0(U)$ by the definition of $H^1_0(U)$. Hence $H^1_0(U)$ is dense in $L^2(U)$. Since the continuous linear functional \begin{align*} \ell_f: L^2(U) \to \mathbb{R} \end{align*} defined by \begin{align*} \ell_f(w) := \int_U f w \, d\mathcal{L}^n \end{align*} vanishes on the dense subspace $H^1_0(U)$, it vanishes on all of $L^2(U)$. Taking $w=f$ gives \begin{align*} \int_U f^2 \, d\mathcal{L}^n = 0. \end{align*} Therefore $f=0$ in $L^2(U)$, and so $\ker K = \{0\}$. [/step] [step:Apply the compact self-adjoint spectral theorem to the resolvent] Because $U$ is nonempty and open, there exist $x_0 \in U$ and $r>0$ such that $B(x_0,r) \subset U$. Choosing countably many pairwise disjoint measurable subsets of $B(x_0,r)$ with positive finite $\mathcal{L}^n$-measure, their indicator functions define a countably infinite orthogonal family in $L^2(U)$. Hence $L^2(U)$ is infinite-dimensional. By the compact self-adjoint spectral theorem for Hilbert spaces (citing a result not yet in the wiki: Compact [Spectral Theorem for Compact Self-Adjoint Operators](/theorems/538)), applied to the compact self-adjoint operator $K: L^2(U) \to L^2(U)$, the Hilbert space $L^2(U)$ has an orthonormal basis consisting of eigenvectors of $K$ together with an orthonormal basis of $\ker K$. Since $\ker K=\{0\}$ and $L^2(U)$ is infinite-dimensional, there exists a countably infinite $L^2(U)$-orthonormal basis $(\varphi_k)_{k=1}^\infty$ of $L^2(U)$ and real nonzero eigenvalues $(\mu_k)_{k=1}^\infty$ such that \begin{align*} K\varphi_k = \mu_k \varphi_k \end{align*} for every $k \in \mathbb{N}$. Since $K$ is positive, each $\mu_k$ is positive. Indeed, if $K\varphi_k=\mu_k\varphi_k$ and $\|\varphi_k\|_{L^2(U)}=1$, then \begin{align*} \mu_k = (K\varphi_k,\varphi_k)_H \geq 0. \end{align*} Because $\mu_k \neq 0$, this gives $\mu_k>0$. The same spectral theorem says that the nonzero eigenvalues of the compact operator $K$, listed with multiplicity, can accumulate only at $0$. Reorder the basis so that \begin{align*} \mu_1 \geq \mu_2 \geq \mu_3 \geq \cdots > 0. \end{align*} Then \begin{align*} \mu_k \to 0 \end{align*} as $k \to \infty$. [/step] [step:Invert the resolvent eigenvalues to obtain Dirichlet Laplacian eigenvalues] For each $k \in \mathbb{N}$, define \begin{align*} \lambda_k := \frac{1}{\mu_k}. \end{align*} Since \begin{align*} \mu_1 \geq \mu_2 \geq \mu_3 \geq \cdots > 0 \end{align*} and $\mu_k \to 0$, the sequence $(\lambda_k)_{k=1}^\infty$ satisfies \begin{align*} 0 < \lambda_1 \leq \lambda_2 \leq \lambda_3 \leq \cdots \end{align*} and \begin{align*} \lambda_k \to \infty. \end{align*} It remains to verify the weak eigenfunction identity. Since $K\varphi_k=\mu_k\varphi_k$ and $K\varphi_k \in H^1_0(U)$, we have $\mu_k\varphi_k \in H^1_0(U)$. Because $\mu_k>0$, it follows that $\varphi_k \in H^1_0(U)$. The weak equation defining $K\varphi_k$ gives, for every $v \in H^1_0(U)$, \begin{align*} \int_U \nabla (K\varphi_k) \cdot \nabla v \, d\mathcal{L}^n = \int_U \varphi_k v \, d\mathcal{L}^n. \end{align*} Substituting $K\varphi_k=\mu_k\varphi_k$ gives \begin{align*} \mu_k \int_U \nabla \varphi_k \cdot \nabla v \, d\mathcal{L}^n = \int_U \varphi_k v \, d\mathcal{L}^n. \end{align*} Dividing by $\mu_k>0$ and using $\lambda_k=1/\mu_k$ yields \begin{align*} \int_U \nabla \varphi_k \cdot \nabla v \, d\mathcal{L}^n = \lambda_k \int_U \varphi_k v \, d\mathcal{L}^n. \end{align*} Thus each $\varphi_k$ is a weak Dirichlet eigenfunction of $-\Delta$ with eigenvalue $\lambda_k$. [/step] [step:Match multiplicities and complete the proof] If $\mu>0$ is an eigenvalue of $K$, then the preceding calculation shows that every vector in $\ker(K-\mu I)$ is a weak Dirichlet eigenfunction with eigenvalue $1/\mu$. Conversely, if $\varphi \in H^1_0(U)$ is nonzero and satisfies \begin{align*} \int_U \nabla \varphi \cdot \nabla v \, d\mathcal{L}^n = \lambda \int_U \varphi v \, d\mathcal{L}^n \end{align*} for every $v \in H^1_0(U)$ with $\lambda>0$, then the uniqueness part of the weak Dirichlet problem implies \begin{align*} K\varphi = \frac{1}{\lambda}\varphi. \end{align*} Indeed, $K\varphi$ is the unique element $u \in H^1_0(U)$ such that \begin{align*} \int_U \nabla u \cdot \nabla v \, d\mathcal{L}^n = \int_U \varphi v \, d\mathcal{L}^n \end{align*} for every $v \in H^1_0(U)$, and $u=\lambda^{-1}\varphi$ satisfies this same identity. Therefore the eigenspaces of the weak Dirichlet Laplacian and the nonzero eigenspaces of $K$ correspond exactly by $\lambda = 1/\mu$, with the same dimensions. Listing the $\lambda_k$ according to the multiplicities inherited from the compact self-adjoint spectral decomposition of $K$, the sequence $(\varphi_k)_{k=1}^\infty$ is an $L^2(U)$-orthonormal basis of eigenfunctions in $H^1_0(U)$ and satisfies the asserted weak eigenvalue equation. This proves the theorem. [/step]