[proofplan] The proof is a direct consequence of the spectral representation of the quadratic form. For every admissible vector $\psi$, the spectral measure of $\psi$ is supported in $\sigma(H)\subset [E_0,\infty)$, so the Rayleigh quotient is bounded below by $E_0$. Since $E_0$ is assumed to be an eigenvalue, a corresponding eigenvector attains this lower bound. The finite-dimensional trial-space estimate follows by restricting the infimum to a smaller class of vectors, with compactness giving existence of the minimum. [/proofplan] [step:Represent the quadratic form by the spectral measure of a vector] Let $E_H$ denote the projection-valued spectral measure of the self-adjoint operator $H$. For each $\psi\in\mathcal{H}$, define the finite positive Borel measure $\mu_\psi$ on $\mathbb{R}$ by \begin{align*} \mu_\psi(B):=(E_H(B)\psi,\psi)_{\mathcal{H}} \end{align*} for every Borel set $B\subset\mathbb{R}$. By the [spectral theorem for self-adjoint operators](/theorems/6911), applied to the lower-semibounded operator $H$ (citing a result not yet in the wiki: spectral theorem for lower-semibounded [self-adjoint operators](/page/Self-Adjoint%20Operators)), the form domain is \begin{align*} Q(H)=\left\{\psi\in\mathcal{H}:\int_{\mathbb{R}}|\lambda|\,d\mu_\psi(\lambda)<\infty\right\}, \end{align*} and for every $\psi\in Q(H)$ the associated quadratic form is \begin{align*} q_H[\psi]=\int_{\mathbb{R}}\lambda\,d\mu_\psi(\lambda). \end{align*} Also, \begin{align*} \|\psi\|_{\mathcal{H}}^2=\mu_\psi(\mathbb{R}). \end{align*} [guided] We first translate the quadratic form into spectral-measure language, because the desired lower bound is a statement about where the spectrum lies. Let $E_H$ be the projection-valued spectral measure associated with the self-adjoint operator $H$. For a vector $\psi\in\mathcal{H}$, define a finite positive Borel measure $\mu_\psi$ on $\mathbb{R}$ by \begin{align*} \mu_\psi(B):=(E_H(B)\psi,\psi)_{\mathcal{H}} \end{align*} for each Borel set $B\subset\mathbb{R}$. Positivity follows because $E_H(B)$ is an [orthogonal projection](/theorems/437), and finiteness follows from \begin{align*} \mu_\psi(\mathbb{R})=(E_H(\mathbb{R})\psi,\psi)_{\mathcal{H}}=(\psi,\psi)_{\mathcal{H}}=\|\psi\|_{\mathcal{H}}^2. \end{align*} We now use the spectral theorem for self-adjoint operators, applied in its quadratic-form version to the lower-semibounded operator $H$ (citing a result not yet in the wiki: spectral theorem for lower-semibounded self-adjoint operators). It identifies the form domain as \begin{align*} Q(H)=\left\{\psi\in\mathcal{H}:\int_{\mathbb{R}}|\lambda|\,d\mu_\psi(\lambda)<\infty\right\}, \end{align*} and gives, for every $\psi\in Q(H)$, \begin{align*} q_H[\psi]=\int_{\mathbb{R}}\lambda\,d\mu_\psi(\lambda). \end{align*} This formula is the bridge between the operator-theoretic hypothesis $E_0=\inf\sigma(H)$ and the variational expression $R_H[\psi]$. [/guided] [/step] [step:Bound every Rayleigh quotient from below by the spectral bottom] Fix $\psi\in Q(H)\setminus\{0\}$. Since $E_0=\inf\sigma(H)$, the spectrum satisfies $\sigma(H)\subset [E_0,\infty)$. The spectral measure $\mu_\psi$ is supported on $\sigma(H)$, so $\lambda\geq E_0$ for $\mu_\psi$-almost every $\lambda\in\mathbb{R}$. Therefore \begin{align*} q_H[\psi]=\int_{\mathbb{R}}\lambda\,d\mu_\psi(\lambda)\geq \int_{\mathbb{R}}E_0\,d\mu_\psi(\lambda)=E_0\mu_\psi(\mathbb{R})=E_0\|\psi\|_{\mathcal{H}}^2. \end{align*} Because $\psi\neq 0$, division by $\|\psi\|_{\mathcal{H}}^2>0$ gives \begin{align*} R_H[\psi]\geq E_0. \end{align*} Taking the infimum over $\psi\in Q(H)\setminus\{0\}$ yields \begin{align*} \inf_{\psi\in Q(H)\setminus\{0\}}R_H[\psi]\geq E_0. \end{align*} [/step] [step:Use a ground-state eigenvector to attain the lower bound] Since $E_0$ is an eigenvalue of $H$, there exists $\phi\in D(H)\setminus\{0\}$ such that \begin{align*} H\phi=E_0\phi. \end{align*} Because $D(H)\subset Q(H)$, the vector $\phi$ is admissible. Using the definition of the quadratic form on operator-domain vectors, \begin{align*} q_H[\phi]=(H\phi,\phi)_{\mathcal{H}}=(E_0\phi,\phi)_{\mathcal{H}}=E_0\|\phi\|_{\mathcal{H}}^2. \end{align*} Hence \begin{align*} R_H[\phi]=E_0. \end{align*} Consequently \begin{align*} \inf_{\psi\in Q(H)\setminus\{0\}}R_H[\psi]\leq E_0. \end{align*} Combining this with the lower bound from the previous step proves \begin{align*} E_0=\inf_{\psi\in Q(H)\setminus\{0\}}R_H[\psi]. \end{align*} [/step] [step:Restrict the variational problem to a finite-dimensional trial subspace] Let $L\subset Q(H)$ be a nonzero finite-dimensional linear subspace. Since \begin{align*} L\setminus\{0\}\subset Q(H)\setminus\{0\}, \end{align*} restricting the class of admissible vectors can only increase the infimum: \begin{align*} E_0=\inf_{\psi\in Q(H)\setminus\{0\}}R_H[\psi]\leq \inf_{\psi\in L\setminus\{0\}}R_H[\psi]. \end{align*} It remains only to justify that the infimum on $L\setminus\{0\}$ is a minimum. Define the unit sphere in $L$ by \begin{align*} S_L:=\{\psi\in L:\|\psi\|_{\mathcal{H}}=1\}. \end{align*} Because $L$ is finite-dimensional and nonzero, $S_L$ is compact. The restriction of the quadratic form $q_H$ to $L$ is a finite-dimensional quadratic form and is therefore continuous. On $S_L$ one has $R_H[\psi]=q_H[\psi]$, so $R_H$ is continuous on $S_L$ and attains its minimum there. Since $R_H[c\psi]=R_H[\psi]$ for every $\psi\in L\setminus\{0\}$ and every scalar $c\in\mathbb{C}\setminus\{0\}$, minimizing over $L\setminus\{0\}$ is the same as minimizing over $S_L$. Thus \begin{align*} \inf_{\psi\in L\setminus\{0\}}R_H[\psi]=\min_{\psi\in L\setminus\{0\}}R_H[\psi]. \end{align*} Therefore \begin{align*} E_0\leq \min_{\psi\in L\setminus\{0\}}R_H[\psi], \end{align*} which is the claimed Rayleigh-Ritz upper bound. [/step]