Min-max principle below the essential spectrum

Suppose first H is nonzero. For any real r<Lambda with r>=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) intersect [c,r] is finite. These points are discrete eigenvalues of finite multiplicity by [A1]. Since E is carried by sigma(A) intersect [c,infinity), $E((-\mathrm{\infty },r])$ is the finite sum of their singleton projections. For r<c it is zero. In particular the total multiplicity below each r<Lambda is finite. The eigenvalues below Lambda can therefore be ordered from below: if any remain, choose one such u; the nonempty finite set of remaining values <=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<Lambda. A countable increasing sequence of bounds r approaching Lambda (or infinity) 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<=n-1), and alpha_n^D for the first over D(A). If dim L=n and dim F=r<=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 intersect F-perp. Thus the supremum on L is at least the infimum on Q(A) intersect F-perp. For each F take the infimum over L, then the supremum over F, giving beta_n<=alpha_n; this remains true when there is no L because alpha_n=+infinity. The trial-space inclusion gives alpha_n<=alpha_n^D.

If 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 Lambda+1/m and use closedness. Hence every interval around Lambda has infinite-rank projection by [A1]. If Lambda=+infinity and only k eigenvalues with multiplicity occur, there is no other spectrum: every finite spectral point lies below 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<Lambda, choose orthonormal eigenvectors phi_1,...,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((-infinity,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,infinity), so q_A[x]>=E_n for unit x in Q(A) intersect F_0-perp. Equality holds at phi_n. Consequently E_n<=beta_n<=alpha_n<=alpha_n^D<=E_n by step 1.2. This proves the formulas, the D(A) version and both attainments in this case.

If E_n=Lambda is finite, exactly k<n eigenvalues occur below 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<=n-1. By support, vectors perpendicular to it have no spectral mass below Lambda, so the inner infimum is at least Lambda and beta_n>=Lambda. For any epsilon>0, the range of E((Lambda-epsilon,Lambda+epsilon)) is infinite dimensional by step 2.1. Choose n independent vectors there and use finite Gram-Schmidt to get an n-dimensional subspace L_epsilon in the same range. Bounded spectral support puts it in D(A) and gives q_A[x]<=Lambda+epsilon on its unit sphere. Hence alpha_n^D<=Lambda+epsilon. Let epsilon decrease to zero in the inequalities of step 1.2 to conclude beta_n=alpha_n=alpha_n^D=Lambda. No form-cross-term estimate or Weyl-sequence approximation is needed.

If E_n=Lambda=+infinity, 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=+infinity. F=H has dimension k<=n-1 and is admissible for beta_n. Its orthogonal complement contains no unit vector, so its inner infimum is +infinity and beta_n=+infinity. This includes the zero Hilbert space for every n>=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) intersect 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.