Form min-max below the essential spectrum

By Choice [A1], [F1] and the spectral calculus give, for $u\in Q(A)$, $q_A[u]=\int \lambda d{E}_u(\lambda )$; the lower bound ensures this integral is finite after the stated shift. The form norm in the Statement is exactly $\mathrm{\parallel }(A+(c+1)I{)}^{1\mathrm{/}2}u\mathrm{\parallel }$.

Fix $t<\alpha$. Since $A\ge -cI$, the spectral points in $(-\mathrm{\infty },t]$ lie in the compact interval $[-c,t]$. If this interval contained infinitely many spectral points, compactness of the interval and closedness of $\sigma (A)$ would give an accumulation point $s\le t$ in $\sigma (A)$. Such an accumulation point is essential by [F2], contradicting $s\le t<\alpha$. Thus there are finitely many spectral points below $t$, each isolated with finite multiplicity. By [F3], $E_A((-\mathrm{\infty },t))$ is supported on their eigenspaces, so it has finite rank.

Now assume the form-domain inclusion is compact. The map $(A+(c+1)I{)}^{-1}:H\to Q(A)$ is bounded because $\mathrm{\parallel }(A+(c+1)I{)}^{1\mathrm{/}2}(A+(c+1)I{)}^{-1}f\mathrm{\parallel }\le \mathrm{\parallel }f\mathrm{\parallel }$ by [F3]. Composing with the compact inclusion $Q(A)\hookrightarrow H$ shows that this resolvent is compact by [F5]. It is positive, self-adjoint and injective. By [F4], its nonzero eigenspaces have finite dimension and span $H$. On each such subspace, $A=(A+(c+1)I)-(c+1)I$ has eigenvalue $r^{-1}-(c+1)$ when the resolvent eigenvalue is $r>0$. The compact spectral theorem says these $r$ can accumulate only at zero; therefore the resulting eigenvalues of $A$ have finite multiplicity and can accumulate only at $+\mathrm{\infty }$. This proves the compact-embedding conclusion.

Define ${\nu }_k=\inf \{t\in \mathbb{R}:\mathrm{rank}{E}_A((-\mathrm{\infty },t))\ge k\}$, with $\inf \mathrm{\varnothing }=+\mathrm{\infty }$. If there are at least $k$ eigenvalues below $\alpha$, their finite-rank spectral projections show that ${\nu }_k$ is the $k$th such eigenvalue counted with multiplicity. If fewer exist, no $t<\alpha$ has rank at least $k$. When $\alpha <\mathrm{\infty }$, closedness of ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)$ gives $\alpha \in {\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)$, so every $t>\alpha$ contains an interval about $\alpha$ inside $(-\mathrm{\infty },t)$ whose spectral projection has infinite rank by [F2]; hence ${\nu }_k=\alpha$. When $\alpha =+\mathrm{\infty }$, the spectrum consists only of discrete eigenvalues; if fewer than $k$ exist, their finite-dimensional eigenspaces span $H$ by [F3], so $\dim H<k$, no $k$-dimensional subspace exists, and ${\nu }_k=+\mathrm{\infty }$. Thus ${\nu }_k$ has exactly the target value in every case.

Let $L\subset Q(A)$ have dimension $k$, and set $m_L={\sup }_{u\in L,\mathrm{\parallel }u\mathrm{\parallel }=1}{q}_A[u]$. By finite-dimensional Gram--Schmidt, choose an orthonormal basis $e_1,\dots ,e_k$ of $L$. For a unit vector $u={\sum }_i{a}_i{e}_i$, each $\mathrm{\mid }{a}_i\mathrm{\mid }\le 1$ and $q_A[u]={\sum }_{i,j}{a}_i\overline{a_j}{q}_A(e_i,e_j)$, so $\mathrm{\mid }{q}_A[u]\mathrm{\mid }\le {\sum }_{i,j}\mathrm{\mid }{q}_A(e_i,e_j)\mathrm{\mid }<\mathrm{\infty }$ uniformly on the unit sphere. Thus $m_L<\mathrm{\infty }$. For any $t>m_L$, if $\mathrm{rank}{E}_A((-\mathrm{\infty },t))<k$, the projection restricted to $L$ has a nonzero kernel; choose a unit $u\in L$ with $E_A((-\mathrm{\infty },t))u=0$. Its spectral measure is supported in $[t,\mathrm{\infty })$, so [F1] gives $q_A[u]\ge t>m_L$, a contradiction. Hence $\mathrm{rank}{E}_A((-\mathrm{\infty },t))\ge k$, so ${\nu }_k\le t$ for every $t>m_L$ and therefore ${\nu }_k\le m_L$. Taking the infimum over $L$ yields ${\nu }_k\le {\lambda }_k(A)$.

If ${\nu }_k<\mathrm{\infty }$ and $t>{\nu }_k$, monotonicity of the interval projections gives some $s<t$ with $\mathrm{rank}{E}_A((-\mathrm{\infty },s))\ge k$. Choose a $k$-dimensional subspace $L$ of this range. Its spectral support is bounded below by $-c$ and above by $s$, hence $L\subset Q(A)$; [F1] also gives $q_A[u]\le s\mathrm{\parallel }u{\mathrm{\parallel }}^2$ on $L$. Thus ${\lambda }_k(A)\le s<t$; letting $t$ decrease to ${\nu }_k$ gives ${\lambda }_k(A)\le {\nu }_k$. If ${\nu }_k=+\mathrm{\infty }$, Step 2.1 shows $\dim H<k$, so the variational infimum is also $+\mathrm{\infty }$. Combining with Step 3.1 gives ${\lambda }_k(A)={\nu }_k$, proving the min-max identity and its spectral interpretation.

Steps 1.1-4.1 establish the form min-max characterization, including the essential-spectrum threshold and the empty-spectrum convention; Step 1.3 proves the complete discrete eigenbasis under compact form-domain embedding.