Operators on a Banach Space

Introduction

On a Banach space the bounded operators again form a unital Banach algebra, but the adjoint is not available unless the space carries an involution, and the spectral theory is consequently the theory of a single element of a Banach algebra rather than the theory of an operator with a conjugation. Two facts survive: the spectrum of a bounded operator is a nonempty compact subset of the plane, and it is controlled by the resolvent, which is analytic on its complement and whose behaviour at infinity yields the spectral radius formula. From the resolvent one builds the holomorphic functional calculus, in which every function holomorphic on a neighbourhood of the spectrum may be evaluated at the operator; the calculus is the Banach-algebra substitute for the spectral theorem and the reason a Banach-algebraic spectral theory exists at all.

This article fixes the algebra $B(X)$ of bounded operators, the spectrum and the resolvent, the spectral radius, the holomorphic functional calculus and its spectral mapping theorem, the module structure of $X$ over $B(X)$ and the invariant subspace problem. The normed-space background is Normed and Banach Spaces, and the Banach-algebra framework with the Gelfand theory is Topological Algebras and Banach Algebras; the Riemann-integral form of the calculus used here is the Riesz–Dunford calculus of that article, and the analytic function theory is Complex Analysis (Part II). The Hilbert-space specialisation, where the adjoint and the spectral theorem exist, is Bounded Operators on a Hilbert Space and Self-Adjoint Operators and the Spectral Theorem; the von Neumann algebra generated by an operator on a Hilbert space is Operator Algebras (Part II). Nothing here needs an inner product.

Throughout, $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, $X$ and $Y$ are Banach spaces over $\mathbb{K}$ with norms written $\|\cdot\|$, and $B(X)=B(X,X)$ is the bounded operators $X\to X$ with the operator norm. For $T\in B(X)$ the resolvent set is $\rho(T)$, its complement $\sigma(T)$ is the spectrum, the resolvent is $R(\lambda,T)=(\lambda I-T)^{-1}$, and the spectral radius is $r(T)=\sup_{\lambda\in\sigma(T)}|\lambda|$. An operator $T$ is invertible when it is bijective with bounded inverse.

The Algebra of Bounded Operators

Definition. For Banach spaces $X,Y$ the set $B(X,Y)$ of bounded linear maps carries the operator norm $\|T\|=\sup_{\|x\|\le1}\|Tx\|$, and $B(X)=B(X,X)$ carries composition as multiplication.

Proposition (unital Banach algebra). $B(X)$ is a unital associative algebra over $\mathbb{K}$ with unit $I$, complete for the operator norm, which is submultiplicative, $\|ST\|\le\|S\|\,\|T\|$ and $\|I\|=1$ when $X\neq0$. The product is jointly continuous in the norm topology, and the invertible operators form a group open in $B(X)$ on which inversion is continuous.

Proof. Completeness is the standard argument of Normed and Banach Spaces: a norm-Cauchy sequence is pointwise Cauchy, its limit is linear and bounded, and the norm converges. Multiplicativity of the norm is immediate. The openness of the invertible group is the Neumann series: if $\|S\|<1$ then $I-S$ is invertible with inverse $\sum_{n\ge0}S^n$, convergent in norm.

Remark (when the algebra is a $C^*$-algebra). The involution and the $C^*$-identity require a conjugation on $X$, hence an inner product in the complex case; the whole difference between this article and the Hilbert-space operator theory is the absence of the adjoint. The one exception that survives without an inner product is the transpose: for $T\in B(X)$ the dual map $T'\in B(X')$ is defined and is isometric and multiplicative, and it is the closest Banach-space substitute for the adjoint.

The Spectrum

Definition. The resolvent set of $T\in B(X)$ is $\rho(T)=\{\lambda\in\mathbb{K}:\lambda I-T\ \text{is invertible}\}$, the spectrum is $\sigma(T)=\mathbb{K}\setminus\rho(T)$, and the spectral radius is $r(T)=\sup_{\lambda\in\sigma(T)}|\lambda|$.

Theorem (the spectrum is nonempty and compact). For $T\in B(X)$ over $\mathbb{C}$ the spectrum $\sigma(T)$ is a nonempty compact subset of the closed disc $\{|\lambda|\le\|T\|\}$; over $\mathbb{R}$ the spectrum may be empty, but the complexification has nonempty spectrum. Consequently $r(T)\le\|T\|$.

Proof. The resolvent $\lambda\mapsto(\lambda I-T)^{-1}$ is analytic on $\rho(T)$ and satisfies $R(\lambda,T)=\sum_{n\ge0}\lambda^{-n-1}T^n$ for $|\lambda|>\|T\|$, so $\rho(T)$ contains the exterior of the disc and is open; $\sigma(T)$ is closed and bounded, hence compact. Were $\sigma(T)$ empty the resolvent would be entire and, by the expansion at infinity, bounded and tending to $0$, hence identically zero, which is impossible; so the spectrum is nonempty.

Proposition (the resolvent identity and analyticity). For $\lambda,\mu\in\rho(T)$,

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

and the resolvent is an analytic $B(X)$-valued function on $\rho(T)$ with derivative $-R(\lambda,T)^2$.

Proof. $(\lambda I-T)-(\mu I-T)=(\lambda-\mu)I$, and multiplying by the two resolvents on the left and right gives the identity; analyticity is the local expansion $R(\lambda+h,T)=(I+hR(\lambda,T))^{-1}R(\lambda,T)=\sum_{n\ge0}(-h)^nR(\lambda,T)^{n+1}$ for $|h|<\|R(\lambda,T)\|^{-1}$.

Theorem (spectral radius formula). For every $T\in B(X)$ over $\mathbb{C}$,

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

Proof. The convergence of the series $\sum_n\lambda^{-n-1}T^n$ at $|\lambda|=r(T)$ is obstructed exactly at the radius of convergence of the power series, which is the reciprocal of the growth rate $\limsup\|T^n\|^{1/n}$; the analytic continuation of the resolvent beyond the disc forces this radius to equal $r(T)$, and the standard submultiplicative argument makes the limsup a limit and identifies it with the infimum.

The Resolvent and the Holomorphic Functional Calculus

Definition. Let $\Omega\subseteq\mathbb{C}$ be open with $\sigma(T)\subseteq\Omega$ and let $\Gamma$ be a finite cycle in $\Omega\setminus\sigma(T)$ winding once positively around each point of $\sigma(T)$. For $f$ holomorphic on $\Omega$ the holomorphic functional calculus is

$$ f(T)=\frac{1}{2\pi i}\oint_\Gamma f(\lambda)\,R(\lambda,T)\,d\lambda , $$

the integral converging in the norm of $B(X)$.

Theorem (the calculus). The assignment $f\mapsto f(T)$ is a unital algebra homomorphism from the algebra of functions holomorphic on a neighbourhood of $\sigma(T)$ into $B(X)$, it is continuous, it sends the constant $1$ to $I$ and the identity function to $T$, and it depends only on the germ of $f$ near $\sigma(T)$. If $f(z)=\sum_{n\ge0}a_nz^n$ has radius of convergence exceeding $r(T)$, then $f(T)=\sum_{n\ge0}a_nT^n$.

Proof. The homomorphism property is the residue computation: the product $f(T)g(T)$ is a double contour integral over independent cycles, the resolvent identity converts the integrand to a difference of simple terms, and Cauchy's theorem leaves the single integral of $fg$; dependence on the germ is the deformation of the contour inside $\Omega\setminus\sigma(T)$; the power-series statement is the evaluation of the contour integral by the expansion of the resolvent at infinity.

Theorem (spectral mapping). With the calculus as above, for every $f$ holomorphic on $\Omega$,

$$ \sigma(f(T))=f(\sigma(T)). $$

Proof. If $\mu\notin f(\sigma(T))$ then $g(z)=(f(z)-\mu)^{-1}$ is holomorphic on $\Omega$ and $g(T)(f(T)-\mu I)=I$ by the homomorphism property, so $\mu\in\rho(f(T))$; conversely if $\mu=f(\lambda_0)$ for some $\lambda_0\in\sigma(T)$, then $f(T)-\mu I=(z-\lambda_0)h(T)$ for a holomorphic $h$ with $h(\lambda_0)\neq0$ after factoring, and the first factor is not invertible, so the product is not invertible.

Remark. The point spectrum, the residual spectrum and the continuous spectrum partition $\sigma(T)$ according to whether $\lambda I-T$ fails injectivity, fails density of the range, or has dense range and is injective but not surjective; the decomposition is empty of content in finite dimension, where the spectrum is the set of eigenvalues. The holomorphic calculus refines none of them, but it does preserve them under $f$ in the form of the spectral mapping theorem above.

The Topologies and the Module Structure

Proposition (the operator topologies). The norm topology of $\|\cdot\|$, the strong operator topology of pointwise convergence and the weak operator topology of convergence of $\langle Tx,\varphi\rangle$ against $\varphi\in X'$ are successively coarser; the strong and weak topologies agree with their Hilbert-space names when $X$ is a Hilbert space. The unit ball is compact in the weak operator topology by Banach–Alaoglu, metrisable there when $X$ is separable.

Proof. The weak topology is by definition the topology generated by the functionals $T\mapsto\varphi(Tx)$ for $x\in X$, $\varphi\in X'$; the strong topology is generated by the seminorms $T\mapsto\|Tx\|$; the norm topology is the metric topology, which is finer than either. Compactness of the ball is the Banach–Alaoglu theorem applied to the product of the discs $\{Tx\}$.

Proposition (the module structure). $X$ is a Banach left module over the unital Banach algebra $B(X)$, the action being $(T,x)\mapsto Tx$, and the action is continuous; a closed subspace $M\subseteq X$ is the same thing as a closed invariant subspace of some operator, and the invariant subspace problem asks whether every bounded operator on a separable infinite-dimensional complex Banach space has a nontrivial closed invariant subspace.

Proof. The module axioms are those of the action of maps on a set, the continuity is the bound $\|Tx\|\le\|T\|\|x\|$, and the correspondence between invariant subspaces and submodules is the definition of invariance. The problem is stated as a question, not solved here.

Example (the shift and the Volterra operator). On $\ell^p$, $1\le p<\infty$, the unilateral shift $Se_n=e_{n+1}$ is an isometry with $\sigma(S)=\{|\lambda|\le1\}$; the Volterra operator $Vf(x)=\int_0^xf$ on $C[0,1]$ or $L^p[0,1]$ is compact and quasinilpotent, $\sigma(V)=\{0\}$ and $\|V^n\|=1/n!$, so $r(V)=0$; and on $C(K)$ the multiplication operator $M_gf=gf$ by a continuous $g$ has spectrum the range $g(K)$, realising the spectral radius formula as the sup norm of $g$.

Summary

For a Banach space $X$ the bounded operators form a unital Banach algebra $B(X)$ under composition and the operator norm, with the invertible operators an open group; the adjoint exists only through the dual map $T'$, and the algebra is a $C^*$-algebra only when $X$ is a Hilbert space. The spectrum of $T\in B(X)$ over $\mathbb{C}$ is a nonempty compact subset of the disc of radius $\|T\|$, the resolvent is analytic on its complement and satisfies the resolvent identity, and the spectral radius satisfies $r(T)=\lim\|T^n\|^{1/n}=\inf\|T^n\|^{1/n}$. The holomorphic functional calculus $f(T)=\frac{1}{2\pi i}\oint_\Gamma f(\lambda)R(\lambda,T)\,d\lambda$ is a continuous unital homomorphism from the functions holomorphic near the spectrum into $B(X)$, and it satisfies the spectral mapping theorem $\sigma(f(T))=f(\sigma(T))$. The space $X$ is a Banach module over $B(X)$, and the invariant subspace problem is the question whether every bounded operator on a separable infinite-dimensional complex Banach space has a proper closed invariant subspace; the shift, the Volterra operator and the multiplication operators are the standard examples.

Summary of Notation

Symbol	Meaning
$B(X)$	the unital Banach algebra of bounded operators on $X$
$\rho(T)$, $\sigma(T)$	resolvent set and spectrum
$R(\lambda,T)=(\lambda I-T)^{-1}$	the resolvent
$r(T)=\lim\\|T^n\\|^{1/n}$	the spectral radius
$f(T)=\frac{1}{2\pi i}\oint_\Gamma f(\lambda)R(\lambda,T)\,d\lambda$	the holomorphic functional calculus
$\sigma(f(T))=f(\sigma(T))$	the spectral mapping theorem
$T'\in B(X')$	the dual (transpose) operator
invariant subspace	a closed $T$-invariant $M\subseteq X$
$r(V)=0$	quasinilpotent operator, for example the Volterra operator