Min-max principle below the essential spectrum

Suppose first $H$ is nonzero. For any real $r<\mathrm{\Lambda }$ with $r\ge c$, each point of $[c,r]$ has a finite-rank interval neighborhood by [A1]. Use the family of all such intervals and the finite subcover property [A4]. Each chosen interval contains finitely many spectral points, so $\sigma (A)\cap [c,r]$ is finite. These points are discrete eigenvalues of finite multiplicity by [A1]. Since $E$ is carried by $\sigma (A)\cap [c,\mathrm{\infty })$, $E((-\mathrm{\infty },r])$ is the finite sum of their singleton projections. For $r<c$ it is zero. In particular the total multiplicity below each $r<\mathrm{\Lambda }$ is finite. The eigenvalues below $\mathrm{\Lambda }$ can therefore be ordered from below: if any remain, choose one such $u$; the nonempty finite set of remaining values $\le u$ has a smallest member, which is the smallest remaining value altogether. Repeat with its finite multiplicity. No value can be omitted forever, since only finitely many terms lie below any fixed $u<\mathrm{\Lambda }$. A countable increasing sequence of bounds $r$ approaching $\mathrm{\Lambda }$ (or $\mathrm{\infty }$) covers all values; [A5] licenses the associated orthonormal eigenvector choices.

Write ${\alpha }_n$ for the first outer value over $Q(A)$, ${\beta }_n$ for the second ($\dim F\le n-1$), and ${\alpha }_n^D$ for the first over $D(A)$. If $\dim L=n$ and $\dim F=r\le n-1$, choose finite bases and impose the $r$ equations $⟨x,f_j⟩=0$ on $x$ in $L$. Finite-dimensional elimination gives a nonzero solution, since there are fewer equations than unknowns. Normalize it to obtain a unit vector in $L\cap F^{\perp }$. Thus the supremum on $L$ is at least the infimum on $Q(A)\cap F^{\perp }$. For each $F$ take the infimum over $L$, then the supremum over $F$, giving ${\beta }_n\le {\alpha }_n$; this remains true when there is no $L$ because ${\alpha }_n=+\mathrm{\infty }$. The trial-space inclusion gives ${\alpha }_n\le {\alpha }_n^D$.

If $\mathrm{\Lambda }$ is finite, it belongs to the essential spectrum: that set is nonempty in this case, bounded below by $c$, and closed; for every positive integer $m$ choose a point between its infimum and $\mathrm{\Lambda }+1\mathrm{/}m$ and use closedness. Hence every interval around $\mathrm{\Lambda }$ has infinite-rank projection by [A1]. If $\mathrm{\Lambda }=+\mathrm{\infty }$ and only $k$ eigenvalues with multiplicity occur, there is no other spectrum: every finite spectral point lies below $\mathrm{\Lambda }$ and is one of these eigenvalues. Support then makes their eigenspaces sum to all of $H$, so $\dim H=k$ and $D(A)=Q(A)=H$. The same last assertion holds directly with $k=0$ for $H=\{0\}$.

If $E_n<\mathrm{\Lambda }$, choose orthonormal eigenvectors ${\phi }_1,\dots ,{\phi }_n$ ordered with multiplicity as in step 1.1. Distinct eigenspaces are orthogonal because symmetry gives $(E_j-E_k)⟨{\phi }_j,{\phi }_k⟩=0$; within each eigenspace choose an orthonormal basis by finite Gram-Schmidt. For $L_0=\mathrm{span}\{{\phi }_1,\dots ,{\phi }_n\}$, contained in $D(A)$, $q_A[{\sum }_j{a}_j{\phi }_j]={\sum }_j{E}_j\mathrm{\mid }{a}_j{\mathrm{\mid }}^2$, so its unit-sphere supremum is $E_n$. For $F_0=\mathrm{span}\{{\phi }_1,\dots ,{\phi }_{n-1}\}$, all eigenspaces strictly below $E_n$ are contained in $F_0$. The projection $E((-\mathrm{\infty },E_n))$ therefore annihilates every $x$ in $F_0^{\perp }$; repetitions of $E_n$ need not be removed. Its scalar measure is carried on $[E_n,\mathrm{\infty })$, so $q_A[x]\ge E_n$ for unit $x$ in $Q(A)\cap F_0^{\perp }$. Equality holds at ${\phi }_n$. Consequently $E_n\le {\beta }_n\le {\alpha }_n\le {\alpha }_n^D\le E_n$ by step 1.2. This proves the formulas, the $D(A)$ version and both attainments in this case.

If $E_n=\mathrm{\Lambda }$ is finite, exactly $k<n$ eigenvalues occur below $\mathrm{\Lambda }$ with multiplicity. Let $F_0$ be the span of all their orthonormal eigenvectors (zero if $k=0$). It is an admissible space of dimension $k\le n-1$. By support, vectors perpendicular to it have no spectral mass below $\mathrm{\Lambda }$, so the inner infimum is at least $\mathrm{\Lambda }$ and ${\beta }_n\ge \mathrm{\Lambda }$. For any $ϵ>0$, the range of $E((\mathrm{\Lambda }-ϵ,\mathrm{\Lambda }+ϵ))$ is infinite dimensional by step 2.1. Choose $n$ independent vectors there and use finite Gram-Schmidt to get an $n$-dimensional subspace $L_{ϵ}$ in the same range. Bounded spectral support puts it in $D(A)$ and gives $q_A[x]\le \mathrm{\Lambda }+ϵ$ on its unit sphere. Hence ${\alpha }_n^D\le \mathrm{\Lambda }+ϵ$. Let $ϵ$ decrease to zero in the inequalities of step 1.2 to conclude ${\beta }_n={\alpha }_n={\alpha }_n^D=\mathrm{\Lambda }$. No form-cross-term estimate or Weyl-sequence approximation is needed.

If $E_n=\mathrm{\Lambda }=+\mathrm{\infty }$, step 2.1 gives $\dim H=k<n$, $D(A)=Q(A)=H$, with $k$ the exhausted total multiplicity. There is no $n$-dimensional trial space, so ${\alpha }_n={\alpha }_n^D=+\mathrm{\infty }$. $F=H$ has dimension $k\le n-1$ and is admissible for ${\beta }_n$. Its orthogonal complement contains no unit vector, so its inner infimum is $+\mathrm{\infty }$ and ${\beta }_n=+\mathrm{\infty }$. This includes the zero Hilbert space for every $n\ge 1$.

Finally suppose $Q(A)$ has a subspace of dimension $n-1$. Any finite-dimensional $F$ contained in $Q(A)$ with $\dim F<n-1$ can be enlarged inside $Q(A)$ to dimension $n-1$: as long as its dimension is smaller, choose a vector from the given $(n-1)$-dimensional subspace not in the current span and adjoin it. Enlarging $F$ shrinks $Q(A)\cap F^{\perp }$, so its inner infimum cannot decrease. Taking suprema shows that restricting the second outer family to exact dimension $n-1$ leaves ${\beta }_n$ unchanged; the reverse inequality is family inclusion. When $n=1$ the only space is $F=\{0\}$. When $H$ is finite dimensional and $n$ exceeds $\dim H+1$, the at-most convention remains necessary. The real lower bound $c$ was never shifted away, so negative eigenvalues and threshold zero require no special argument. AC is used exactly in [A5] and the infimum-approaching sequence in step 2.1.