Proof of Jentzsch's Theorem via the Variational Absolute Value Trick #
This file proves Jentzsch's theorem for compact self-adjoint positivity-improving operators on L²(ℝ^n, Lebesgue). The proof avoids the Krein-Rutman / Banach lattice cone machinery and uses only the Rayleigh quotient, Cauchy-Schwarz, and L² lattice structure (absolute value, positive/negative parts).
Proof outline (Courant-Hilbert / Barry Simon) #
- Phase 1: Positivity-preserving operators satisfy |Tf| ≤ T|f| a.e.
- Phase 2: Therefore |⟨f, Tf⟩| ≤ ⟨|f|, T|f|⟩.
- Phase 3: If f is a ground state (Tf = lam₀f), then |f| is also a ground state.
- Phase 4: By positivity-improving, the ground state is strictly positive a.e.
- Phase 5: Every ground state eigenvector has constant sign a.e.
- Phase 6: The ground state eigenvalue lam₀ is simple (multiplicity 1).
- Phase 7: All other eigenvalues satisfy |λ| < lam₀ (strict spectral gap).
References #
- Reed-Simon IV, Theorems XIII.43–44
- Simon, Functional Integration and Quantum Physics (2005), §I.13
- Courant-Hilbert, Methods of Mathematical Physics, Ch. VI
Definitions #
An operator on L²(ℝ^n) is positivity-preserving if it maps nonneg functions to nonneg functions. This is weaker than positivity-improving.
Equations
- IsPositivityPreserving T = ∀ (f : ↥(MeasureTheory.Lp ℝ 2 MeasureTheory.volume)), 0 ≤ f → 0 ≤ T f
Instances For
An operator on L²(ℝ^n) is positivity-improving if it maps nonneg nonzero functions to a.e. strictly positive functions (ae-filter version).
Equations
- IsPositivityImproving T = ∀ (f : ↥(MeasureTheory.Lp ℝ 2 MeasureTheory.volume)), (∀ᵐ (x : Ω), 0 ≤ ↑↑f x) → ¬↑↑f =ᵐ[MeasureTheory.volume] 0 → ∀ᵐ (x : Ω), 0 < ↑↑(T f) x
Instances For
An operator on L²(ℝ^n) is positivity-improving if it maps nonneg nonzero functions to a.e. strictly positive functions (Lp lattice version).
Equations
- IsPositivityImproving' T = ∀ (f : ↥(MeasureTheory.Lp ℝ 2 MeasureTheory.volume)), 0 ≤ f → f ≠ 0 → ∀ᵐ (x : Ω), 0 < ↑↑(T f) x
Instances For
Positivity-improving implies positivity-preserving. If f ≥ 0 and f ≠ 0, then Tf > 0 a.e. ≥ 0. If f = 0, then Tf = 0 ≥ 0.
IsPositivityImproving (ae conditions) implies IsPositivityImproving' (Lp lattice).
The definitions differ only in whether the hypotheses use ae-filter or Lp order/equality:
0 ≤ fin Lp ↔0 ≤ᵐ[μ] f(byLp.coeFn_nonneg)f ≠ 0in Lp ↔¬ f =ᵐ[μ] 0(byLp.eq_zero_iff_ae_eq_zero)
Phase 1: Absolute value inequality #
For a positivity-preserving operator T on L²: |Tf| ≤ T|f| a.e.
Proof: f = f⁺ - f⁻ with f⁺, f⁻ ≥ 0. Since T is positivity-preserving, Tf⁺ ≥ 0 and Tf⁻ ≥ 0. Then: |Tf| = |Tf⁺ - Tf⁻| ≤ Tf⁺ + Tf⁻ = T(f⁺ + f⁻) = T|f|
Phase 1: Absolute value inequality for positivity-preserving operators.
Phase 2: Inner product inequality #
|⟨f, Tf⟩| ≤ ⟨|f|, T|f|⟩
Proof: |⟨f, Tf⟩| = |∫ f(x) (Tf)(x) dx| ≤ ∫ |f(x)| |Tf(x)| dx (integral triangle inequality) ≤ ∫ |f(x)| (T|f|)(x) dx (Phase 1: |Tf| ≤ T|f|, and |f| ≥ 0) = ⟨|f|, T|f|⟩ (since |f| ≥ 0 and T|f| ≥ 0)
Phase 2: Inner product inequality.
Phase 3: |f| is a ground state if f is #
The Rayleigh quotient R(g) = ⟨g, Tg⟩ / ‖g‖² achieves its supremum at the ground state eigenvector. The chain of inequalities:
lam₀ ‖f‖² = ⟨f, Tf⟩ ≤ |⟨f, Tf⟩| ≤ ⟨|f|, T|f|⟩ ≤ lam₀ ‖|f|‖²
Since ‖|f|‖ = ‖f‖, we get equality throughout, so |f| achieves the supremum of R, hence |f| is an eigenvector for lam₀.
Phase 3: If f is an eigenvector for the top eigenvalue lam₀, then |f| is also an eigenvector for lam₀.
Phase 4: Ground state is strictly positive a.e. #
Since |f| ≥ 0 and |f| ≠ 0, and T is positivity-improving, T|f| > 0 a.e. But T|f| = lam₀|f|, so |f| > 0 a.e.
Phase 4: The ground state eigenvector (after taking absolute value) is strictly positive almost everywhere.
Phase 5: Every ground state eigenvector has constant sign #
If f is an eigenvector for lam₀, then |f| is also (Phase 3). Define h = |f| - f. Then h ≥ 0 and h is an eigenvector for lam₀. By Phase 4, either h = 0 a.e. (so f = |f| ≥ 0, hence f > 0 a.e.) or h > 0 a.e. Since h = |f| - f, having h > 0 a.e. means |f(x)| > f(x) a.e., i.e., f(x) < 0 a.e. (strictly).
So every ground state eigenvector is either strictly positive or strictly negative a.e.
Phase 5: Every eigenvector for the top eigenvalue has constant sign.
Phase 6: The top eigenvalue lam₀ is simple #
If u, v are two orthogonal eigenvectors for lam₀ with ⟨u, v⟩ = 0, then by Phase 5, we may assume u > 0 a.e. and v > 0 a.e. (or multiply by -1). But then ⟨u, v⟩ = ∫ u(x)v(x) dx > 0, contradicting orthogonality.
Phase 6: The top eigenvalue has multiplicity 1 (is simple).
Phase 7: Strict spectral gap #
For any eigenvalue μ ≠ lam₀, we have |μ| < lam₀.
Case 1: μ ≥ 0. Then μ < lam₀ by simplicity (if μ = lam₀, the eigenspace is 1-dimensional by Phase 6, so the eigenvector would be in the same eigenspace).
Case 2: μ < 0. Suppose |μ| = lam₀, i.e., μ = -lam₀. Let g be an eigenvector for -lam₀. By self-adjointness, ⟨e₀, g⟩ = 0. Apply the absolute value trick: |Tg| = |-lam₀g| = lam₀|g|. Phase 1 gives |Tg| ≤ T|g|, so lam₀|g| ≤ T|g|. The Rayleigh quotient bound gives ⟨|g|, T|g|⟩ ≤ lam₀ ‖g‖². Together: T|g| = lam₀|g|, so |g| is in the lam₀-eigenspace. By simplicity (Phase 6), |g| = c·e₀ for some c > 0. So |g| > 0 a.e. Phase 5 forces g to have constant sign. But ⟨e₀, g⟩ = 0 with e₀ > 0 and |g| > 0 contradicts constant sign.
Phase 7: All eigenvalues other than lam₀ satisfy |μ| < lam₀.
Spectral decomposition of the Rayleigh quotient #
Assembly: Jentzsch's theorem #
Combine all phases to prove the full theorem.
A positivity-improving compact self-adjoint operator has a strictly
positive eigenvalue. If all eigenvalues were ≤ 0, then ⟨f, Tf⟩ ≤ 0 for
every f by the spectral decomposition, contradicting ⟨|b j|, T|b j|⟩ > 0
which holds because T is positivity-improving.
Jentzsch's theorem (proved).
For a compact, self-adjoint, positivity-improving operator on L²(ℝ^n) with eigenbasis indexed by a type with ≥ 2 elements:
- The top eigenvalue lam₀ > 0 is simple.
- All other eigenvalues satisfy |λ| < lam₀.