The Spectral Radius Formula: Gelfand's r(T) = lim ‖Tⁿ‖^{1/n}

Precise statement

Let A be a Banach algebra over ℂ with unit e, and let T ∈ A. The spectrum is σ(T) = { λ ∈ ℂ : T − λe is not invertible in A }, a nonempty compact subset of ℂ. Define the spectral radius r(T) = sup{ |λ| : λ ∈ σ(T) }. Gelfand's theorem (1941) asserts:

The limit limₙ→∞ ‖Tⁿ‖^{1/n} exists, and r(T) = limₙ→∞ ‖Tⁿ‖^{1/n} = infₙ≥1 ‖Tⁿ‖^{1/n}.

Three claims are packed in here. (i) The spectrum is nonempty and compact, so the sup is attained and finite — indeed r(T) ≤ ‖T‖. (ii) The sequence ‖Tⁿ‖^{1/n} converges. (iii) The two apparently unrelated numbers — the geometric radius of σ(T) and the analytic growth rate of the powers — coincide. The infimum characterization is a bonus: the limit never exceeds any individual term, so r(T) ≤ ‖Tⁿ‖^{1/n} for every n, giving computable upper bounds. For bounded operators on a Banach space X, take A = B(X).

The picture: growth rate is geometric decay of the resolvent

Fix T and consider the resolvent R(λ) = (T − λe)⁻¹, defined for λ ∉ σ(T). For |λ| large it has the convergent Neumann/Laurent expansion R(λ) = −∑ₙ≥0 λ⁻ⁿ⁻¹ Tⁿ, a power series in the variable 1/λ. The picture to hold in your head: R is an analytic function of λ on the complement of σ(T), and this Laurent series converges precisely outside the smallest disk containing σ(T) — a disk of radius r(T).

So the growth of ‖Tⁿ‖ controls where the series converges, and analyticity controls the same thing from the spectral side. If the powers grow like ρⁿ, the series ∑ λ⁻ⁿ⁻¹Tⁿ converges for |λ| > ρ; conversely the series must diverge somewhere on |λ| = r(T) because the resolvent has a singularity there (the spectrum touches that circle). Matching these two descriptions of the same radius of convergence forces limsup ‖Tⁿ‖^{1/n} = r(T). The formula is Cauchy–Hadamard's root test, read through the lens of operator-valued analyticity.

Key idea of the proof (the mechanism)

The proof squeezes r(T) between the limsup and liminf of ‖Tⁿ‖^{1/n}, forcing convergence and equality. Two inequalities do the work.

r(T) ≤ liminf. By the spectral mapping theorem for polynomials, λ ∈ σ(T) ⟹ λⁿ ∈ σ(Tⁿ), so |λ|ⁿ ≤ r(Tⁿ) ≤ ‖Tⁿ‖. Taking nth roots, |λ| ≤ ‖Tⁿ‖^{1/n} for all n, hence r(T) ≤ infₙ ‖Tⁿ‖^{1/n} ≤ liminf ‖Tⁿ‖^{1/n}.

limsup ≤ r(T). This is the deep half, using that A is a complex Banach algebra. The map λ ↦ R(λ) is A-valued analytic on ℂ∖σ(T), and → 0 as λ → ∞. Apply a continuous linear functional φ: then φ(R(λ)) is a scalar analytic function whose Laurent series ∑ φ(Tⁿ) λ⁻ⁿ⁻¹ converges for |λ| > r(T) (analyticity gives no singularities there). By Cauchy–Hadamard the scalar coefficients satisfy limsup |φ(Tⁿ)|^{1/n} ≤ r(T) for every φ. The uniform boundedness principle upgrades this to the operator norm: limsup ‖Tⁿ‖^{1/n} ≤ r(T). Chaining the two inequalities: r(T) ≤ liminf ≤ limsup ≤ r(T). ∎

Worked example: a nilpotent, a shift, and a Volterra operator

Nilpotent. Let T be the 2×2 Jordan block [[0,1],[0,0]]. Then ‖T‖ = 1 but T² = 0, so ‖Tⁿ‖^{1/n} = 0 for n ≥ 2 and r(T) = 0. Indeed σ(T) = {0}. Here r(T) = 0 ≪ ‖T‖ = 1: the norm badly overestimates the spectral radius, and Gelfand's limit corrects it.

Weighted shift / quasinilpotent. The Volterra operator (Vf)(x) = ∫₀ˣ f(t) dt on L²[0,1] satisfies ‖Vⁿ‖ ~ 1/n! (its iterated kernel is (x−t)ⁿ⁻¹/(n−1)!). Then ‖Vⁿ‖^{1/n} → 0 by Stirling, so r(V) = 0 and σ(V) = {0} — yet V ≠ 0 and is not nilpotent. This is a genuinely quasinilpotent operator, invisible to the spectrum but detected by the norms.

Normaloid. For a normal (in particular self-adjoint) operator on a Hilbert space, the C*-identity gives ‖T²‖ = ‖T‖², so ‖T^{2ᵏ}‖ = ‖T‖^{2ᵏ} and r(T) = ‖T‖ exactly. The formula collapses to the norm.

Why the hypotheses matter

ℂ is essential. Over ℝ the spectrum can be empty. Take rotation by 90° on ℝ², T = [[0,−1],[1,0]]. It has no real eigenvalues, so its real spectrum is empty and sup∅ = 0 would give a false r(T), while ‖Tⁿ‖^{1/n} = 1 for all n. The proof of limsup ≤ r(T) needs the resolvent to be analytic on a punctured neighborhood of ∞ in ℂ and Liouville's theorem to force σ(T) ≠ ∅ — both are ℂ-only phenomena. This is exactly why Banach-algebra spectral theory is done over ℂ.

Completeness is essential. The Neumann series ∑ λ⁻ⁿ⁻¹Tⁿ must converge in A to produce the inverse (T − λe)⁻¹; that convergence of an absolutely summable series is precisely completeness. In an incomplete normed algebra the resolvent set can shrink and σ(T) need not be closed or nonempty.

Unitality is a convenience — one adjoins a unit if absent, and the spectrum is unchanged up to {0}. Connections: the formula underlies the Gelfand–Mazur theorem and the whole Gelfand representation of commutative Banach algebras.

Applications and significance

Gelfand's formula is the load-bearing lemma of Banach-algebra theory. It makes the spectral radius subadditive and submultiplicative on commuting elements: if ST = TS then r(S+T) ≤ r(S)+r(T) and r(ST) ≤ r(S)r(T), because the limit turns ‖(ST)ⁿ‖ = ‖SⁿTⁿ‖ inequalities into the clean statements. Without the formula the spectral radius is only defined geometrically and these bounds are opaque.

Concretely it powers: (i) iterative numerical analysis — a stationary iteration xₖ₊₁ = Txₖ + b converges for every start iff r(T) < 1, and the asymptotic convergence rate is exactly r(T); (ii) stability of dynamical systems — discrete linear systems are asymptotically stable iff r(T) < 1 (the spectral-radius / Lyapunov criterion); (iii) Perron–Frobenius theory, where r(T) is the dominant eigenvalue governing Markov chains and population growth; (iv) the proof that in a C*-algebra the norm is determined by the algebraic structure via ‖T‖² = ‖T*T‖ = r(T*T), the rigidity at the heart of Gelfand–Naimark.

The three quantities Gelfand's formula ties together, and how each is computed.

Quantity	Definition	How you compute it
Spectral radius r(T)	sup{ |λ| : λ ∈ σ(T) }	Locate the spectrum; take the farthest point from 0
Gelfand limit	limₙ→∞ ‖Tⁿ‖^{1/n}	Grow powers of T, take nth roots, pass to the limit
Radius of convergence	1 / limsupₙ ‖Tⁿ‖^{1/n}	Cauchy–Hadamard applied to ∑ λ⁻ⁿ⁻¹Tⁿ
Norm bound	r(T) ≤ ‖T‖ always; equality iff normaloid	Compare with ‖Tⁿ‖^{1/n} ≤ ‖T‖
Self-adjoint case	r(T) = ‖T‖	C*-identity ‖T²‖ = ‖T‖² forces equality