Resolvents & Operator Spectrum

An eigenvalue detects failure of injectivity, but infinite-dimensional inversion can fail even with no eigenvector: the range may be non-dense, not closed, or dense but not all of the space. The spectrum collects every scalar where bounded invertibility fails.

Resolvent and spectrum #

Let $T\in\mathcal B(X)$ on a complex Banach space. The resolvent set is

$$\rho(T)=\{\lambda\in\mathbb C:T-\lambda I \text{ is bijective with bounded inverse}\}.$$

The spectrum is $\sigma(T)=\mathbb C\setminus\rho(T)$, and $R(\lambda,T)=(T-\lambda I)^{-1}$ is the resolvent. The bounded-inverse condition is redundant once bijectivity and the Banach hypotheses are known, but keeping it in the definition makes the stability meaning explicit.

For $|\lambda|>\|T\|$,

$$T-\lambda I=-\lambda\left(I-\frac{T}{\lambda}\right)$$

is invertible by the Neumann series. Therefore

$$\sigma(T)\subseteq\{\lambda:|\lambda|\le\|T\|\}.$$

The spectrum of a bounded operator on a nonzero complex Banach space is nonempty and compact.

Why the resolvent is analytic #

The first resolvent identity is

$$R(\lambda,T)-R(\mu,T) =(\lambda-\mu)R(\lambda,T)R(\mu,T).$$

Near $\lambda_0\in\rho(T)$,

$$R(\lambda,T) =R(\lambda_0,T) \sum_{n=0}^{\infty} \left[(\lambda-\lambda_0)R(\lambda_0,T)\right]^n$$

with the sign adjusted to the chosen $T-\lambda I$ convention; direct factorization confirms it. This norm-convergent power series makes $\rho(T)$ open and the resolvent analytic. It also shows that

$$\|R(\lambda,T)\|\ge \frac{1}{\mathop{\mathrm{dist}}(\lambda,\sigma(T))}.$$

For normal Hilbert-space operators equality holds; for nonnormal operators the resolvent can be much larger, producing pseudospectral sensitivity far from the spectrum.

Spectral failure modes #

A common disjoint classification for $T-\lambda I$ is:

• point spectrum: not injective; $\lambda$ is an eigenvalue;
• continuous spectrum: injective with dense range, but not surjective;
• residual spectrum: injective with non-dense range.

Terminology varies slightly between texts, so the injective/range statement is more reliable than the label. Approximate eigenvectors $\|x_n\|=1$ with $\|(T-\lambda I)x_n\|\to0$ explain how stable inversion can fail without an actual eigenvector.

Spectral radius #

The spectral radius

$$r(T)=\sup\{|\lambda|:\lambda\in\sigma(T)\}$$

satisfies the Gelfand formula

$$r(T)=\lim_{n\to\infty}\|T^n\|^{1/n} =\inf_{n\ge1}\|T^n\|^{1/n}.$$

Always $r(T)\le\|T\|$, and strict inequality is possible. Thus $r(T)<1$ does not imply $\|T\|<1$, although it does imply that sufficiently large powers decay geometrically after choosing a rate above $r(T)$.

Examples #

• For scalar multiplication $T=\alpha I$, the spectrum is $\{\alpha\}$.
• For multiplication $M_g$ on a standard $L^p$ space, the spectrum is the essential range of $g$.
• The Volterra operator on $C[0,1]$ has spectrum $\{0\}$ but is not the zero operator: it is quasinilpotent rather than nilpotent.
• A compact operator has a particularly discrete nonzero spectrum; only zero can be an accumulation point.

Self-adjoint and normal structure #

For bounded self-adjoint $T$ on a Hilbert space,

$$\sigma(T)\subset\mathbb R, \qquad r(T)=\|T\|,$$

and the spectral extremes control the quadratic form. For normal $T$, $\|T\|=r(T)$ and the resolvent norm equals reciprocal distance to the spectrum. In general the spectrum is contained in the closure of the numerical range

$$W(T)=\{\langle Tx,x\rangle:\|x\|=1\};$$

the closure cannot be omitted for an arbitrary bounded operator.

Spectral checklist #

• Work over a complex Banach space, or explicitly complexify a real problem.
• Fix whether the resolvent uses $T-\lambda I$ or $\lambda I-T$ before using identities.
• Distinguish eigenvalues from the full spectrum.
• Inspect the range and inverse norm, not only the kernel.
• For nonnormal operators, use resolvent or pseudospectral information when perturbation sensitivity matters.