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The resolvent norm blows up at an eigenvalue

Example

Assume the setting of Non-invertible elliptic shifts form a discrete set in the self-adjoint case and let (λk,ek) be a weak eigenpair with ∥ek∥L2=1 (Symmetric elliptic weak eigenpairs). Then for every real λ∉{λj} ∥(L−λ)−1∥ ≥ 1∣λk−λ∣, because (L−λ)−1ek=ek/(λk−λ) has L2 norm 1/∣λk−λ∣; combined with the exact formula of the spectral-series corollary this gives ∥(L−λ)−1∥=1/dist⁡(λ,{λj}). Hence the resolvent norm is unbounded on every neighbourhood of an eigenvalue: at λ=λk−ε, for every sufficiently small ε>0, it is at least 1/ε. The Fredholm alternative is consistent with this: at λ=λk the homogeneous problem has the nonzero solution ek, and uniqueness and bounded invertibility both fail.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a bounded open set Ω⊆Rn; the symmetric divergence-form operator L with eigenvalues {λj} and orthonormal eigenbasis; a weak eigenpair (λk,ek) with ∥ek∥L2=1; and a real λ∉{λj}.

[F1]

Eigenpair data: ek∈H01(Ω) and a(ek,v)=λk(ek,v)L2 for all v, equivalently ek∈D(L) and Lek=λkek; the eigenbasis is orthonormal in L2 (Symmetric elliptic weak eigenpairs, Discrete spectrum of a symmetric elliptic Dirichlet operator).

[F2]

Real resolvent data: for real λ∉{λj}, the base-field operator L−λ is bijective with bounded inverse Rλ=(L−λ)−1:L2(Ω;K)→D(L), and ∥Rλ∥=1/dist⁡(λ,{λj}). In the real case this inverse complexifies to (L~−λ)−1=−(λ−L~)−1 and has the same norm; thus its complexification is the negative of the library resolvent of L~, with the same operator norm (Non-invertible elliptic shifts form a discrete set in the self-adjoint case, The complex L2 pairing on equivalence classes, Complex Lp classes and Euclidean test-function conventions, Complexification of a real-linear map, Complexification as C⊗RV with its canonical real-linear embedding, Resolvent and spectrum of an unbounded operator, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, The space Lp(μ) as the quotient by null functions).

Verification

technique · direct
1.1F1F2givenalgebra

Action on the eigenfunction. Since Lek=λkek by [F1], for real λ≠λk one has (L−λ)ek=(λk−λ)ek, and applying the inverse Rλ of [F2] (which exists because λ≠λk and λ∉{λj}) gives Rλek=1λk−λek. Taking L2 norms and using ∥ek∥L2=1, ∥Rλek∥L2=1/∣λk−λ∣.

2.1F2step 1.1givenalgebra

Lower bound for the operator norm. By definition of the operator norm, ∥Rλ∥≥∥Rλek∥L2/∥ek∥L2=1/∣λk−λ∣; combined with the exact formula ∥Rλ∥=1/dist⁡(λ,{λj}) of [F2] the lower bound for this fixed k is an equality exactly when ∣λk−λ∣=dist⁡(λ,{λj}), that is, when λk is a nearest eigenvalue.

3.1F1F2step 2.1given∎

Blow-up near an eigenvalue. Fix k and ε>0 such that λk−ε∉{λj} (possible for all sufficiently small ε because the eigenvalue set is discrete); then step 2.1 with λ=λk−ε gives ∥Rλk−ε∥≥1/ε, so the resolvent norm is unbounded on every neighbourhood of λk. At λ=λk itself no bounded inverse exists: ek is a nonzero homogeneous solution, so L−λk is not injective, in agreement with the criterion that L−λ is bijective with bounded inverse exactly for λ∉{λj}; uniqueness and bounded invertibility both fail at an eigenvalue.

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