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The Resolvent and the Spectral Radius

The resolvent (T − λI)⁻¹ is an analytic function on the resolvent set, and its analyticity controls exactly how big the spectrum can be through Gelfand's spectral-radius formula.

The resolvent as an analytic family

On the resolvent set ρ(T) define R(λ) = (T − λI)⁻¹, the resolvent. A two-line computation gives the resolvent identity R(λ) − R(μ) = (λ − μ)R(λ)R(μ). Dividing by λ − μ and letting μ → λ shows R is differentiable with R′(λ) = R(λ)², so the resolvent is an analytic operator-valued function on the open set ρ(T). This analyticity is the engine behind every theorem about the spectrum.

Resolvent identity, derived from a single algebraic trick:

  R(lambda) - R(mu)
    = R(lambda) [ (T - mu I) - (T - lambda I) ] R(mu)      (insert I = (T-muI)R(mu) on left,
                                                            I = R(lambda)(T-lambdaI) on right)
    = R(lambda) [ (lambda - mu) I ] R(mu)
    = (lambda - mu) R(lambda) R(mu).

So  [R(lambda) - R(mu)] / (lambda - mu) = R(lambda) R(mu).
Let mu -> lambda:  R'(lambda) = R(lambda)^2.
=> R is analytic on rho(T), with all the consequences of analyticity.
The resolvent identity, then differentiability of R falls right out of it.

The Neumann series and a radius bound

For |λ| > ‖T‖ we expand R(λ) as a geometric series in operators: −(1/λ)(I − T/λ)⁻¹ = −Σ_{n≥0} Tⁿ/λⁿ⁺¹. This converges in operator norm because ‖Tⁿ/λⁿ⁺¹‖ ≤ ‖T‖ⁿ/|λ|ⁿ⁺¹, a convergent geometric series. So every such λ is in ρ(T), confirming σ(T) ⊆ {|λ| ≤ ‖T‖} and giving the bound r(T) ≤ ‖T‖, where r(T) = sup{|λ| : λ ∈ σ(T)} is the spectral radius.

Gelfand's spectral-radius formula

The sharp statement is Gelfand's formula: r(T) = lim_{n→∞} ‖Tⁿ‖^{1/n}, and this limit always exists. The intuition: the Neumann series for R(λ) is a power series in 1/λ whose coefficients are Tⁿ, and a power series converges exactly outside its radius of convergence. Since R is analytic on all of ρ(T), the largest λ where it can fail — the boundary of σ(T) — is governed by the Cauchy–Hadamard root test applied to ‖Tⁿ‖.

  1. Sub-multiplicativity ‖T^{m+n}‖ ≤ ‖Tᵐ‖‖Tⁿ‖ makes log‖Tⁿ‖ subadditive, so lim ‖Tⁿ‖^{1/n} = inf_n ‖Tⁿ‖^{1/n} exists (Fekete's lemma).
  2. The series Σ Tⁿ/λⁿ⁺¹ converges for |λ| > lim‖Tⁿ‖^{1/n}, so r(T) ≤ lim‖Tⁿ‖^{1/n}.
  3. Analyticity of R on |λ| > r(T) forces the series to diverge for |λ| < lim‖Tⁿ‖^{1/n} only if equality fails; matching radii gives r(T) = lim‖Tⁿ‖^{1/n}.