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Ground eigenvalue as the form variational infimum

Statement

Assume Countable Choice and use Q, qh, and the operator h from Self-adjoint unbounded polynomial interaction at fixed regulator. Let λ1 be the least eigenvalue of h. Then inf⁡{qh(u):u∈Q, ∥u∥2=1}=λ1. The infimum is attained by every normalized vector in the lowest eigenspace; no simplicity or nondegeneracy is assumed.

Facts & Assumptions

Given: Countable Choice, the form domain Q, the form qh, and the complete eigenfamily of h.

[A1]

The Axiom of Countable Choice (ACω): countable eigenbasis choices used in the form realization are permitted.

[F1]

Self-adjoint unbounded polynomial interaction at fixed regulator: the eigenvalues λn are real, ordered increasingly with multiplicity, have a least value λ1, and for u∈Q, qh(u)=∑n≥1λn∣⟨u,ψn⟩∣2.

[F2]

Fourier expansion in a Hilbert space: ∑n∣⟨u,ψn⟩∣2=∥u∥22.

Proof

technique · direct

Given: u∈Q and the ordered real eigenvalue list.

1.1A1F1F2algebra

Since λn≥λ1 for every n, [A1], [F1] and [F2] imply qh(u)−λ1∥u∥22=∑n(λn−λ1)∣⟨u,ψn⟩∣2≥0. Thus every unit vector in Q has form value at least λ1.

2.1F1F2step 1.1algebra∎

A normalized eigenvector ψ1 belongs to D(h)⊆Q by Self-adjoint unbounded polynomial interaction at fixed regulator, has norm one, and its sole nonzero coefficient is at an eigenvalue λ1. Therefore qh(ψ1)=λ1, so the lower bound is attained and equals the infimum. If the lowest eigenspace has dimension greater than one, the same coefficient calculation applies to every normalized vector in it.

Depends on

Dependency tree · 2 levels

Sources