[proofplan] Choose an orthonormal eigenbasis for the finite-dimensional spectral subspace below and at $E_k$, with eigenvalues repeated according to multiplicity. The [spectral theorem for self-adjoint operators](/theorems/6911) gives two decisive estimates: on the span of the first $k$ eigenvectors the Rayleigh quotient is at most $E_k$, while on the orthogonal complement of the first $k-1$ eigenvectors it is at least $E_k$. Elementary dimension counting then forces every $k$-dimensional trial space to contain a vector in that orthogonal complement, and every codimension-$k-1$ constraint to meet the span of the first $k$ eigenvectors. [/proofplan]

[step:Choose eigenvectors and record the spectral estimates]Let $u_1,\dots,u_k\in D(H)\subset Q(H)$ be orthonormal eigenvectors corresponding to the listed eigenvalues $E_1,\dots,E_k$, so that \begin{align*} Hu_j=E_j u_j \end{align*} for each $j\in\{1,\dots,k\}$. Define the finite-dimensional subspaces \begin{align*} V_k := \operatorname{span}\{u_1,\dots,u_k\}\subset Q(H) \end{align*} and, for $k\geq 2$, \begin{align*} V_{k-1} := \operatorname{span}\{u_1,\dots,u_{k-1}\}\subset Q(H). \end{align*} For $k=1$, set $V_0:=\{0\}$. Since $H$ is self-adjoint and bounded below, it has an associated closed quadratic form \begin{align*} h:Q(H)\times Q(H)\to\mathbb{C} \end{align*} whose Rayleigh quotient is $R_H[\psi]=h[\psi,\psi]/\|\psi\|_{\mathcal{H}}^2$ for nonzero $\psi\in Q(H)$. Let $\Sigma_{\mathrm{ess}}:=\inf\sigma_{\mathrm{ess}}(H)$ denote the bottom of the essential spectrum of $H$. We use the spectral resolution supplied by the spectral theorem for [self-adjoint operators](/page/Self-Adjoint%20Operators). Let \begin{align*} \mathsf{E}: \mathcal{B}(\mathbb{R})\to \mathcal{L}(\mathcal{H}) \end{align*} be the projection-valued spectral measure of $H$. Since $E_1,\dots,E_k$ are exactly the first $k$ eigenvalues below the bottom of the essential spectrum, counted with multiplicity, the spectral subspace for $(-\infty,E_k)$ is contained in $V_{k-1}$, and the spectral subspace generated by $u_1,\dots,u_k$ is $V_k$ up to the possible omission of further eigenvectors with the same eigenvalue $E_k$. For every $\psi\in V_k$, write \begin{align*} \psi=\sum_{j=1}^k c_j u_j \end{align*} with coefficients $c_1,\dots,c_k\in\mathbb{C}$. Orthogonality of the eigenvectors and the definition of the quadratic form give \begin{align*} h[\psi,\psi]=\sum_{j=1}^k E_j |c_j|^2. \end{align*} Since $E_j\leq E_k$ for every $j\leq k$, we obtain \begin{align*} h[\psi,\psi]\leq E_k \sum_{j=1}^k |c_j|^2 = E_k\|\psi\|_{\mathcal{H}}^2. \end{align*} Thus every nonzero $\psi\in V_k$ satisfies \begin{align*} R_H[\psi]\leq E_k. \end{align*} Now let $\psi\in Q(H)\cap V_{k-1}^{\perp}$ be nonzero, where $V_{k-1}^{\perp}$ denotes the orthogonal complement in $\mathcal{H}$. Since $\psi$ is orthogonal to the spectral subspace for $(-\infty,E_k)$, its scalar spectral measure \begin{align*} \mu_\psi(B):=\|\mathsf{E}(B)\psi\|_{\mathcal{H}}^2 \end{align*} for Borel sets $B\subset\mathbb{R}$ is supported in $[E_k,\infty)$. The spectral representation of the closed quadratic form associated to $H$ gives \begin{align*} h[\psi,\psi]=\int_{[E_k,\infty)} \lambda\, d\mu_\psi(\lambda). \end{align*} Because $\lambda\geq E_k$ on the domain of integration, \begin{align*} h[\psi,\psi]\geq E_k\int_{[E_k,\infty)} 1\, d\mu_\psi(\lambda)=E_k\|\psi\|_{\mathcal{H}}^2. \end{align*} Therefore every nonzero $\psi\in Q(H)\cap V_{k-1}^{\perp}$ satisfies \begin{align*} R_H[\psi]\geq E_k. \end{align*}[/step]

[guided]The goal of this step is to isolate the two spectral facts that drive the entire min-max argument. First, choose orthonormal eigenvectors $u_1,\dots,u_k\in D(H)$ so that $Hu_j=E_j u_j$ for each $j\in\{1,\dots,k\}$. This is possible because the eigenvalues below $\Sigma_{\mathrm{ess}}$ are discrete eigenvalues of finite multiplicity, and they are listed with multiplicity. Define \begin{align*} V_k := \operatorname{span}\{u_1,\dots,u_k\}\subset Q(H). \end{align*} For $k\geq 2$, define \begin{align*} V_{k-1} := \operatorname{span}\{u_1,\dots,u_{k-1}\}\subset Q(H), \end{align*} and for $k=1$ define $V_0:=\{0\}$. Why are these the correct spaces? The space $V_k$ is the trial space that should realize the upper bound in the first min-max formula. The space $V_{k-1}$ is the obstruction space consisting of all directions belonging to lower eigenvalues; once we impose orthogonality to it, the spectral theorem forces the energy to be at least $E_k$. Because $H$ is self-adjoint and bounded below, it has an associated closed quadratic form \begin{align*} h:Q(H)\times Q(H)\to\mathbb{C} \end{align*} and the Rayleigh quotient is well-defined on nonzero vectors in $Q(H)$ by \begin{align*} R_H[\psi]=\frac{h[\psi,\psi]}{\|\psi\|_{\mathcal{H}}^2}. \end{align*} Let $\Sigma_{\mathrm{ess}}:=\inf\sigma_{\mathrm{ess}}(H)$ denote the bottom of the essential spectrum of $H$. We use the spectral resolution supplied by the spectral theorem for self-adjoint operators. Let $\mathcal{B}(\mathbb{R})$ denote the Borel $\sigma$-algebra on $\mathbb{R}$, and let $\mathcal{L}(\mathcal{H})$ denote the space of bounded linear operators from $\mathcal{H}$ to itself. Let \begin{align*} \mathsf{E}: \mathcal{B}(\mathbb{R})\to \mathcal{L}(\mathcal{H}) \end{align*} be the projection-valued spectral measure of $H$. The theorem applies because $H$ is self-adjoint. Since $E_1,\dots,E_k$ are the first $k$ eigenvalues below $\Sigma_{\mathrm{ess}}$, counted with multiplicity, there is no spectral mass below $E_k$ except in the span of eigenvectors belonging to the entries $E_1,\dots,E_{k-1}$. Thus the spectral subspace for $(-\infty,E_k)$ is contained in $V_{k-1}$. We now prove the upper spectral estimate on $V_k$. Take $\psi\in V_k$. Then there are coefficients $c_1,\dots,c_k\in\mathbb{C}$ such that \begin{align*} \psi=\sum_{j=1}^k c_j u_j. \end{align*} Since the eigenvectors are orthonormal and $Hu_j=E_j u_j$, the quadratic form satisfies \begin{align*} h[\psi,\psi]=\sum_{j=1}^k E_j |c_j|^2. \end{align*} Because the eigenvalues are ordered and $j\leq k$, each coefficient is weighted by an eigenvalue no larger than $E_k$. Hence \begin{align*} h[\psi,\psi]\leq E_k \sum_{j=1}^k |c_j|^2. \end{align*} Orthonormality gives \begin{align*} \sum_{j=1}^k |c_j|^2=\|\psi\|_{\mathcal{H}}^2. \end{align*} Therefore \begin{align*} R_H[\psi]=\frac{h[\psi,\psi]}{\|\psi\|_{\mathcal{H}}^2}\leq E_k \end{align*} whenever $\psi\in V_k$ is nonzero. Next, prove the lower spectral estimate on $V_{k-1}^{\perp}$. Let $\psi\in Q(H)\cap V_{k-1}^{\perp}$ be nonzero. Define the scalar spectral measure \begin{align*} \mu_\psi(B):=\|\mathsf{E}(B)\psi\|_{\mathcal{H}}^2 \end{align*} for every Borel set $B\subset\mathbb{R}$. Since $\psi$ is orthogonal to $V_{k-1}$, it has no spectral component in the spectral subspace for $(-\infty,E_k)$. Therefore $\mu_\psi$ is supported in $[E_k,\infty)$. The spectral representation of the closed quadratic form gives \begin{align*} h[\psi,\psi]=\int_{[E_k,\infty)} \lambda\, d\mu_\psi(\lambda). \end{align*} This integral is meaningful for $\psi\in Q(H)$ by the definition of the form domain. Since $\lambda\geq E_k$ throughout the support of $\mu_\psi$, monotonicity of the [Lebesgue integral](/page/Lebesgue%20Integral) gives \begin{align*} h[\psi,\psi]\geq E_k\int_{[E_k,\infty)} 1\, d\mu_\psi(\lambda). \end{align*} By the definition of the scalar spectral measure, \begin{align*} \int_{[E_k,\infty)} 1\, d\mu_\psi(\lambda)=\mu_\psi([E_k,\infty))=\|\psi\|_{\mathcal{H}}^2. \end{align*} Thus \begin{align*} R_H[\psi]\geq E_k. \end{align*} This is the lower-energy barrier after all lower eigendirections have been excluded.[/guided]

[step:Prove the first min-max formula using the span of the first $k$ eigenvectors]Since $V_k\subset Q(H)$ and $\dim V_k=k$, the preceding estimate on $V_k$ gives \begin{align*} \inf_{L\subset Q(H),\ \dim L=k}\ \sup_{\psi\in L\setminus\{0\}} R_H[\psi]\leq \sup_{\psi\in V_k\setminus\{0\}} R_H[\psi]\leq E_k. \end{align*} For the reverse inequality, let $L\subset Q(H)$ be a $k$-dimensional linear subspace. Consider the [linear map](/page/Linear%20Map) \begin{align*} P_{k-1}|_L:L\to V_{k-1} \end{align*} where $P_{k-1}:\mathcal{H}\to V_{k-1}$ is the [orthogonal projection](/theorems/437) in $\mathcal{H}$. Since $\dim L=k$ and $\dim V_{k-1}=k-1$, the kernel of $P_{k-1}|_L$ contains a nonzero vector. Choose $\psi\in L\setminus\{0\}$ with $P_{k-1}\psi=0$. Then $\psi\in Q(H)\cap V_{k-1}^{\perp}$, so the lower spectral estimate gives \begin{align*} R_H[\psi]\geq E_k. \end{align*} Therefore \begin{align*} \sup_{\varphi\in L\setminus\{0\}} R_H[\varphi]\geq E_k. \end{align*} Since this holds for every $k$-dimensional linear subspace $L\subset Q(H)$, \begin{align*} \inf_{L\subset Q(H),\ \dim L=k}\ \sup_{\psi\in L\setminus\{0\}} R_H[\psi]\geq E_k. \end{align*} Combining the two inequalities proves the first formula.[/step]

[guided]We now turn the two spectral estimates into the first variational identity. The upper bound is obtained by testing the variational problem on the specific trial space $V_k$. Since $V_k\subset Q(H)$ and $\dim V_k=k$, the infimum over all $k$-dimensional subspaces is no larger than the value at $V_k$: \begin{align*} \inf_{L\subset Q(H),\ \dim L=k}\ \sup_{\psi\in L\setminus\{0\}} R_H[\psi]\leq \sup_{\psi\in V_k\setminus\{0\}} R_H[\psi]. \end{align*} The spectral estimate already proved on $V_k$ gives \begin{align*} \sup_{\psi\in V_k\setminus\{0\}} R_H[\psi]\leq E_k. \end{align*} Therefore \begin{align*} \inf_{L\subset Q(H),\ \dim L=k}\ \sup_{\psi\in L\setminus\{0\}} R_H[\psi]\leq E_k. \end{align*} For the reverse inequality, take an arbitrary $k$-dimensional linear subspace $L\subset Q(H)$. We must show that $L$ contains at least one direction whose Rayleigh quotient is at least $E_k$. Let \begin{align*} P_{k-1}:\mathcal{H}\to V_{k-1} \end{align*} denote the orthogonal projection in the [Hilbert space](/page/Hilbert%20Space) $\mathcal{H}$, and consider its restriction \begin{align*} P_{k-1}|_L:L\to V_{k-1}. \end{align*} This is a linear map from a $k$-dimensional [vector space](/page/Vector%20Space) into a vector space of dimension $k-1$. By the [rank-nullity theorem](/theorems/916), its kernel has positive dimension, so there exists $\psi\in L\setminus\{0\}$ such that $P_{k-1}\psi=0$. The equality $P_{k-1}\psi=0$ means precisely that $\psi\perp V_{k-1}$ in $\mathcal{H}$. Since $L\subset Q(H)$, this vector belongs to $Q(H)\cap V_{k-1}^{\perp}$. The lower spectral estimate applies to every nonzero vector in $Q(H)\cap V_{k-1}^{\perp}$, hence \begin{align*} R_H[\psi]\geq E_k. \end{align*} Because this particular $\psi$ lies in $L\setminus\{0\}$, the supremum over all nonzero vectors of $L$ is at least as large: \begin{align*} \sup_{\varphi\in L\setminus\{0\}} R_H[\varphi]\geq E_k. \end{align*} This argument works for every $k$-dimensional subspace $L\subset Q(H)$, so taking the infimum over all such $L$ gives \begin{align*} \inf_{L\subset Q(H),\ \dim L=k}\ \sup_{\psi\in L\setminus\{0\}} R_H[\psi]\geq E_k. \end{align*} The upper and lower bounds coincide, proving \begin{align*} E_k=\inf_{L\subset Q(H),\ \dim L=k}\ \sup_{\psi\in L\setminus\{0\}} R_H[\psi]. \end{align*}[/guided]

[step:Prove the dual max-min formula by imposing $k-1$ orthogonality constraints]First take $M=V_{k-1}$. Since every nonzero $\psi\in Q(H)\cap V_{k-1}^{\perp}$ satisfies $R_H[\psi]\geq E_k$, we have \begin{align*} \inf_{\psi\in Q(H)\cap V_{k-1}^{\perp},\ \psi\neq 0} R_H[\psi]\geq E_k. \end{align*} Also $u_k\in Q(H)\cap V_{k-1}^{\perp}$ and $R_H[u_k]=E_k$, because $Hu_k=E_k u_k$ and $\|u_k\|_{\mathcal{H}}=1$. Hence \begin{align*} \inf_{\psi\in Q(H)\cap V_{k-1}^{\perp},\ \psi\neq 0} R_H[\psi]=E_k. \end{align*} Taking the supremum over all $(k-1)$-dimensional subspaces gives \begin{align*} \sup_{M\subset Q(H),\ \dim M=k-1}\ \inf_{\psi\in Q(H)\cap M^{\perp},\ \psi\neq 0} R_H[\psi]\geq E_k. \end{align*} For the opposite inequality, let $M\subset Q(H)$ be any linear subspace with $\dim M=k-1$. Since $\dim V_k=k$, the restriction of the orthogonal projection from $V_k$ onto $M$ cannot be injective. Hence there exists a nonzero vector $\psi\in V_k$ such that $\psi\perp M$. The upper spectral estimate on $V_k$ gives \begin{align*} R_H[\psi]\leq E_k. \end{align*} Therefore \begin{align*} \inf_{\varphi\in Q(H)\cap M^{\perp},\ \varphi\neq 0} R_H[\varphi]\leq E_k. \end{align*} Since this holds for every $(k-1)$-dimensional linear subspace $M\subset Q(H)$, \begin{align*} \sup_{M\subset Q(H),\ \dim M=k-1}\ \inf_{\psi\in Q(H)\cap M^{\perp},\ \psi\neq 0} R_H[\psi]\leq E_k. \end{align*} Combining the two inequalities proves the dual max-min formula and completes the proof.[/step]

[guided]We prove the second variational identity by reversing the viewpoint: instead of choosing a $k$-dimensional trial space, we impose $k-1$ orthogonality constraints and look at the best lower bound that remains. First choose the particular constraint space $M=V_{k-1}$. The lower spectral estimate says that every nonzero vector $\psi\in Q(H)\cap V_{k-1}^{\perp}$ satisfies \begin{align*} R_H[\psi]\geq E_k. \end{align*} Therefore \begin{align*} \inf_{\psi\in Q(H)\cap V_{k-1}^{\perp},\ \psi\neq 0} R_H[\psi]\geq E_k. \end{align*} This lower bound is sharp because $u_k\in Q(H)$, $u_k\perp V_{k-1}$ by orthonormality of the eigenvectors, and $Hu_k=E_k u_k$. Since $\|u_k\|_{\mathcal{H}}=1$, the quadratic form identity for an eigenvector gives \begin{align*} R_H[u_k]=E_k. \end{align*} Hence \begin{align*} \inf_{\psi\in Q(H)\cap V_{k-1}^{\perp},\ \psi\neq 0} R_H[\psi]=E_k. \end{align*} Taking the supremum over all $(k-1)$-dimensional subspaces can only increase the value, so \begin{align*} \sup_{M\subset Q(H),\ \dim M=k-1}\ \inf_{\psi\in Q(H)\cap M^{\perp},\ \psi\neq 0} R_H[\psi]\geq E_k. \end{align*} Now let $M\subset Q(H)$ be any linear subspace with $\dim M=k-1$. We show that the inner infimum cannot exceed $E_k$. Consider the orthogonal projection from $\mathcal{H}$ onto $M$ and restrict it to $V_k$. This restriction is a linear map from the $k$-dimensional space $V_k$ into the $(k-1)$-dimensional space $M$. By rank-nullity, it has a nonzero kernel. Thus there exists $\psi\in V_k\setminus\{0\}$ such that $\psi\perp M$. Since $V_k\subset Q(H)$, this vector lies in $Q(H)\cap M^{\perp}$. The upper spectral estimate on $V_k$ gives \begin{align*} R_H[\psi]\leq E_k. \end{align*} Since the infimum over $Q(H)\cap M^{\perp}$ is at most the value at this particular admissible vector, \begin{align*} \inf_{\varphi\in Q(H)\cap M^{\perp},\ \varphi\neq 0} R_H[\varphi]\leq E_k. \end{align*} This holds for every $(k-1)$-dimensional subspace $M\subset Q(H)$, so taking the supremum over $M$ yields \begin{align*} \sup_{M\subset Q(H),\ \dim M=k-1}\ \inf_{\psi\in Q(H)\cap M^{\perp},\ \psi\neq 0} R_H[\psi]\leq E_k. \end{align*} Combining this upper bound with the lower bound proves \begin{align*} E_k=\sup_{M\subset Q(H),\ \dim M=k-1}\ \inf_{\psi\in Q(H)\cap M^{\perp},\ \psi\neq 0} R_H[\psi]. \end{align*} Together with the first formula, this completes the proof of the min-max principle for the eigenvalue $E_k$.[/guided]