8.1 Spectrum and resolvent of bounded linear operators

Bounded linear operators on Hilbert spaces have a spectrum and resolvent set. The spectrum includes eigenvalues, continuous spectrum, and residual spectrum. Understanding these concepts is crucial for analyzing operator properties and solving related problems.

The resolvent set contains complex numbers where the operator minus that value times the identity is invertible. The spectrum is the complement of this set. These ideas help determine operator invertibility and behavior in various mathematical contexts.

Spectrum and Resolvent of Bounded Linear Operators

Spectrum and resolvent set definition

• Let $H$ be a Hilbert space and $T: H \to H$ be a bounded linear operator maps elements of $H$ to itself
• The resolvent set of $T$, denoted by $\rho(T)$, is the set of all complex numbers $\lambda$ such that:
  • $T - \lambda I$ is invertible has an inverse that is also a bounded linear operator
  • $(T - \lambda I)^{-1}$ is bounded its norm is finite
  • $(T - \lambda I)^{-1}$ is defined on all of $H$ maps every element of $H$ to an element of $H$
• The spectrum of $T$, denoted by $\sigma(T)$, is the complement of the resolvent set in the complex plane $\mathbb{C}$, i.e., $\sigma(T) = \mathbb{C} \setminus \rho(T)$ contains all complex numbers not in the resolvent set
• The spectrum can be further classified into:
  • Point spectrum: $\lambda \in \mathbb{C}$ such that $T - \lambda I$ is not injective there exist nonzero $x \in H$ such that $(T - \lambda I)x = 0$ (eigenvalues)
  • Continuous spectrum: $\lambda \in \mathbb{C}$ such that $T - \lambda I$ is injective, has dense range its image is dense in $H$, but is not surjective does not map onto all of $H$
  • Residual spectrum: $\lambda \in \mathbb{C}$ such that $T - \lambda I$ is injective, but its range is not dense in $H$ there exist $y \in H$ not in the closure of the image of $T - \lambda I$

Computation of spectrum examples

• Identity operator: $T = I$ maps each element to itself
  • $\sigma(I) = \{1\}$ (point spectrum) only eigenvalue is 1
  • $\rho(I) = \mathbb{C} \setminus \{1\}$ all complex numbers except 1
• Multiplication operator: $(Mf)(x) = x \cdot f(x)$ on $L^2([0, 1])$ multiplies each function by the independent variable $x$
  • $\sigma(M) = [0, 1]$ (continuous spectrum) no eigenvalues, but range is dense
  • $\rho(M) = \mathbb{C} \setminus [0, 1]$ all complex numbers outside $[0, 1]$
• Left shift operator: $(Lx)_n = x_{n+1}$ on $\ell^2(\mathbb{N})$ shifts sequence elements to the left
  • $\sigma(L) = \{\lambda \in \mathbb{C}: |\lambda| \leq 1\}$ unit disk in complex plane
    • $\{0\}$ is the point spectrum only eigenvalue is 0
    • $\{\lambda \in \mathbb{C}: 0 < |\lambda| < 1\}$ is the residual spectrum open unit disk excluding 0
    • $\{\lambda \in \mathbb{C}: |\lambda| = 1\}$ is the continuous spectrum unit circle
  • $\rho(L) = \{\lambda \in \mathbb{C}: |\lambda| > 1\}$ outside the closed unit disk

Properties of spectrum and resolvent

• The resolvent set $\rho(T)$ is an open subset of $\mathbb{C}$ any open disk centered at a point in $\rho(T)$ is contained in $\rho(T)$
  • For $\lambda_0 \in \rho(T)$, the resolvent operator $R(\lambda_0, T) = (T - \lambda_0 I)^{-1}$ is bounded by the inverse mapping theorem
  • For $\lambda$ close to $\lambda_0$, the resolvent operator can be expressed as a convergent series: $R(\lambda, T) = \sum_{n=0}^{\infty} (\lambda_0 - \lambda)^n R(\lambda_0, T)^{n+1}$ Neumann series expansion
  • This series converges for $|\lambda - \lambda_0| < \|R(\lambda_0, T)\|^{-1}$, proving that $\rho(T)$ is open contains an open disk around each of its points
• The spectrum $\sigma(T)$ is a nonempty, compact subset of $\mathbb{C}$ closed and bounded
  • $\sigma(T)$ is nonempty: if $\sigma(T) = \emptyset$, then $\rho(T) = \mathbb{C}$, implying that $T - \lambda I$ is invertible for all $\lambda \in \mathbb{C}$, which is impossible for bounded operators on infinite-dimensional spaces by the Fredholm alternative
  • $\sigma(T)$ is bounded: $\|T - \lambda I\| \geq |\lambda| - \|T\|$, so for $|\lambda| > \|T\|$, $T - \lambda I$ is invertible by the Neumann series, and thus $\lambda \in \rho(T)$
  • $\sigma(T)$ is closed: as the complement of the open set $\rho(T)$ in $\mathbb{C}$

Spectrum vs operator invertibility

• $T$ is invertible if and only if $0 \in \rho(T)$ zero is in the resolvent set
  • If $T$ is invertible, then $T - 0 \cdot I = T$ is invertible, so $0 \in \rho(T)$ by definition of the resolvent set
  • If $0 \in \rho(T)$, then $T = T - 0 \cdot I$ is invertible since $0$ is in the resolvent set
• $T$ is not invertible if and only if $0 \in \sigma(T)$ zero is in the spectrum
  • If $T$ is not invertible, then $T - 0 \cdot I = T$ is not invertible, so $0 \in \sigma(T)$ by definition of the spectrum
  • If $0 \in \sigma(T)$, then $T = T - 0 \cdot I$ is not invertible since $0$ is in the spectrum
• The spectral radius of $T$ is defined as $r(T) = \sup\{|\lambda|: \lambda \in \sigma(T)\}$ the supremum of the moduli of elements in the spectrum
  • The spectral radius formula states that $r(T) = \lim_{n \to \infty} \|T^n\|^{1/n}$ can be computed as a limit involving the operator norm of powers of $T$
  • If $r(T) < 1$, then $\sum_{n=0}^{\infty} T^n$ converges, and $I - T$ is invertible by the geometric series formula