The Operator Algebra of a Banach Space

Introduction

A Banach space $X$ carries no product, but the bounded linear operators on it do: composition is an associative multiplication, the identity is the identity operator, and the operator norm is submultiplicative and complete. The result is the operator algebra $B(X)$, the canonical example of a Banach algebra and the source of every example of a noncommutative one. Its subalgebras, closed in the norm, are themselves Banach algebras, and the most important of them are generated by a single operator or defined by a commutation condition. This article develops $B(X)$ as a Banach algebra: its completeness and continuity of composition, its norm-closed subalgebras and the algebra generated by a set, the commutant of a subalgebra, the unit group with its openness and the spectrum and resolvent of an operator, the closed two-sided ideals with the compact operators as the basic case, and the finite-dimensional and classical examples.

The article assumes the Banach space, its norm, its completeness, the bounded linear maps, the operator norm and the completeness of the operator space from Normed and Banach Spaces; the Banach algebra, its submultiplicative norm, its unit group, its spectrum, its spectral radius and the Neumann series from Topological Algebras and Banach Algebras; and the operator-layer conventions, the regular representation and the bounded operators of an algebra from Operators on a Banach Algebra, the first article of this group. The involution on the elements and the adjoint on the operators are the later groups of this category and are not used; the algebra of bounded operators of a Hilbert space with its adjoint is The Adjoint of a Bounded Operator, and the von Neumann bicommutant theorem is Involutive Operator Algebras and the Commutant and Operator Algebras. The compact operators are named at the boundary; their theory, the ideals of $B(X)$ and the Calkin algebra are the subjects of that literature and are only pointed at. No form, no measure and no Fourier theory occurs.

Throughout, $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$; $X$ is a Banach space over $\mathbb{K}$ with norm $\lVert\cdot\rVert$, $X \neq 0$; $B(X)$ is the unital Banach algebra of bounded linear operators $X \to X$ with the operator norm $\lVert T\rVert = \sup_{\lVert x\rVert \leq 1}\lVert Tx\rVert$, and $X'$ is the dual space. For a subset $\mathcal{S} \subseteq B(X)$ the commutant is $\mathcal{S}' = \{T \in B(X) : TS = ST \text{ for all } S \in \mathcal{S}\}$.

The Bounded Operators of a Banach Space

Definition. For $T : X \to X$ linear, $T$ is bounded when $\lVert T\rVert = \sup\{\lVert Tx\rVert : \lVert x\rVert \leq 1\} < +\infty$, and $B(X)$ is the set of bounded linear operators on $X$.

Theorem ($B(X)$ is a unital Banach algebra). $B(X)$ is a unital associative algebra under pointwise linear structure, composition and the identity $\mathrm{id}$, the operator norm is submultiplicative,

$$ \lVert ST\rVert \leq \lVert S\rVert\,\lVert T\rVert , \qquad \lVert \mathrm{id}\rVert = 1 , $$

and $B(X)$ is complete for it. The multiplication $(S,T) \mapsto ST$ is continuous, and a linear $T$ is bounded exactly when it is continuous.

Proof. Composition is associative and bilinear, with unit $\mathrm{id}$. Submultiplicativity is $\lVert STx\rVert \leq \lVert S\rVert\lVert Tx\rVert \leq \lVert S\rVert\lVert T\rVert\lVert x\rVert$. Completeness and the equivalence of boundedness with continuity are those of the operator space of Normed and Banach Spaces: a Cauchy sequence of operators converges uniformly on the unit ball, its limit is linear and bounded, and continuity of $T$ is the existence of the bound $\lVert Tx\rVert \leq \lVert T\rVert\lVert x\rVert$. $\square$

Proposition (the isometric embedding of the scalars). The map $\mathbb{K} \to B(X)$, $\lambda \mapsto \lambda\,\mathrm{id}$, is an isometric unital algebra homomorphism, so $B(X)$ contains a copy of $\mathbb{K}$ as its centre when $X \neq 0$; the centre of $B(X)$ is exactly $\mathbb{K}\,\mathrm{id}$ whenever $X$ has dimension at least two.

Proof. Isometry: $\lVert\lambda\,\mathrm{id}\rVert = \lvert\lambda\rvert$. An operator commuting with every bounded operator is a scalar multiple of the identity, by evaluating on vectors; the proof is standard and is quoted. $\square$

Remark (the two-sided reading). Reading $B(X)$ as the operator algebra of the Banach space $X$ generalizes Operators on a Banach Algebra: if $X = A$ is a Banach algebra, then $B(X)$ is the algebra $B(A)$ of that article, and the one-sided multiplications $L_a$ and $R_a$ are particular elements of it. The present article treats $B(X)$ for an arbitrary Banach space, so it uses no product on $X$, and the multiplications of the algebra case are named only in the examples.

Closed Subalgebras and the Generated Algebra

Definition. A subalgebra of $B(X)$ is a linear subspace closed under composition; it is closed when it is closed in the norm topology. A Banach subalgebra is a closed subalgebra, which is then a Banach algebra for the restricted norm. The closed subalgebra generated by a set $\mathcal{S} \subseteq B(X)$ is the smallest closed subalgebra containing $\mathcal{S}$ and $\mathrm{id}$, namely the norm-closure of the algebra of noncommutative polynomials in the elements of $\mathcal{S}$.

Proposition (a closed subalgebra is a Banach algebra). Every closed subalgebra of $B(X)$ is a Banach algebra for the restricted norm and the restricted product, and its unit, when it contains $\mathrm{id}$, is $\mathrm{id}$. The closure of a subalgebra is a subalgebra, and the intersection of closed subalgebras is a closed subalgebra, so the closed subalgebras form a complete lattice.

Proof. A closed subspace of a Banach space is complete, and the product of a subalgebra stays in the subalgebra; the restricted norm is submultiplicative. The closure of a subalgebra is closed under products because composition is continuous on $B(X) \times B(X)$, and the intersection of closed subalgebras is closed and closed under products. $\square$

Proposition (the algebra generated by one operator). Let $T \in B(X)$. The closed subalgebra generated by $T$ and $\mathrm{id}$ is

$$ \mathbb{K}[T]^{\mathrm{cl}} = \overline{\{\,p(T) : p \in \mathbb{K}[x]\,\}} , $$

the closure of the algebra of polynomials in $T$; it is a unital commutative Banach algebra, and the map $p \mapsto p(T)$ is the unital algebra homomorphism $\mathbb{K}[x] \to B(X)$ of evaluation at $T$, whose continuity is the boundedness of $p(T)$.

Proof. The algebra of polynomials in $T$ is commutative because $T$ commutes with itself, and its closure is commutative because the commutator is continuous and vanishes on a dense subalgebra. The map $p \mapsto p(T)$ is a unital homomorphism by the powers of $T$, and it is bounded on the closed algebra by definition of that algebra. $\square$

Corollary (the kernel and the spectrum). The kernel of $p \mapsto p(T)$ is the ideal of polynomials vanishing on the spectrum, $\ker = \{\,p : p|_{\sigma(T)} = 0\,\}$, under the spectral mapping theorem; the algebra generated by $T$ is isomorphic to a quotient of the polynomial algebra, and it is a field only when the spectrum is a single point.

Proof. The spectral mapping theorem for polynomials, $p(\sigma(T)) = \sigma(p(T))$, is from Topological Algebras and Banach Algebras; $p(T) = 0$ forces $p$ to vanish on $\sigma(T)$, and conversely a polynomial vanishing on $\sigma(T)$ lies in the kernel by the theorem. $\square$

Remark (the boundary of the section). A closed subalgebra need not be closed under the adjoint or the involution, and the involution-closed subalgebras of the bounded operators of a Hilbert space are the C-algebras and the von Neumann algebras of The Adjoint of a Bounded Operator and Operator Algebras*; the present article constructs only the algebraic and norm-closed subalgebras, and the involutive layer is the later groups of this category.

The Commutant

Definition. For $\mathcal{S} \subseteq B(X)$ the commutant is $\mathcal{S}' = \{T \in B(X) : TS = ST \text{ for all } S \in \mathcal{S}\}$; the bicommutant is $\mathcal{S}'' = (\mathcal{S}')'$.

Proposition (the commutant is a closed subalgebra). For every $\mathcal{S}$, the commutant $\mathcal{S}'$ is a unital closed subalgebra of $B(X)$ containing $\mathrm{id}$; the assignment is order-reversing, $\mathcal{S} \subseteq \mathcal{T}$ implies $\mathcal{T}' \subseteq \mathcal{S}'$, and $\mathcal{S} \subseteq \mathcal{S}''$.

Proof. $\mathcal{S}'$ is a linear subspace closed under composition, because $T_1,T_2$ commuting with every $S$ implies $T_1T_2$ and $T_1 - T_2$ commute; it contains $\mathrm{id}$. It is closed: if $T_n \to T$ and $T_nS = ST_n$ for all $n$ and all $S$, then $TS = \lim T_nS = \lim ST_n = ST$ by continuity of the product. The order reversal and $\mathcal{S} \subseteq \mathcal{S}''$ are the definition read twice. $\square$

Corollary (the centraliser of a representation). If $\mathcal{A}$ is a closed unital subalgebra of $B(X)$, then $\mathcal{A}'$ is a closed unital subalgebra, every element of $\mathcal{A}'$ commutes with $\mathcal{A}$, and $\mathcal{A} \subseteq \mathcal{A}''$.

Proof. The proposition with $\mathcal{S} = \mathcal{A}$. $\square$

Remark (the bicommutant). The equality $\mathcal{A}'' = \mathcal{A}$ — the double commutant theorem — fails for a general closed subalgebra of $B(X)$ and holds for the weakly closed involutive subalgebras of the bounded operators of a Hilbert space, by von Neumann's theorem; that theorem is Involutive Operator Algebras and the Commutant, later in this category, and the von Neumann algebra setting is Operator Algebras. The present article states only the proposition above and does not use the bicommutant equality.

Invertibility, the Unit Group and the Spectrum

Definition. An operator $T \in B(X)$ is invertible when there is $S \in B(X)$ with $ST = TS = \mathrm{id}$; the group of invertible operators is $B(X)^{\times}$, the unit group of $B(X)$. The spectrum of $T$ is

$$ \sigma(T) = \{\lambda \in \mathbb{K} : T - \lambda\,\mathrm{id} \notin B(X)^{\times}\} , $$

and the resolvent set is its complement; the spectral radius is $r(T) = \sup\{\lvert\lambda\rvert : \lambda \in \sigma(T)\}$.

Theorem (the unit group is open and inversion is continuous). The unit group $B(X)^{\times}$ is open in $B(X)$, inversion $T \mapsto T^{-1}$ is continuous on it, and $B(X)^{\times}$ is a topological group for the subspace topology. If $\lVert T - \mathrm{id}\rVert < 1$ then $T$ is invertible with

$$ T^{-1} = \sum_{n\geq0}(\mathrm{id} - T)^n . $$

Proof. The Neumann series $\sum(\mathrm{id}-T)^n$ converges absolutely when $\lVert \mathrm{id} - T\rVert < 1$, by submultiplicativity and completeness, and its sum is an inverse. For openness at a general invertible $T$, write $S = T + H = T(\mathrm{id} + T^{-1}H)$, invertible when $\lVert T^{-1}H\rVert < 1$ by the series, so a small ball about $T$ lies in $B(X)^\times$. Continuity of inversion follows from $(T+H)^{-1} - T^{-1} = -T^{-1}H(T+H)^{-1}$ and the local boundedness of the inverse. $\square$

Theorem (the spectrum is compact and nonempty). For every $T$ the spectrum $\sigma(T)$ is a nonempty compact subset of $\mathbb{K}$ contained in the disk of radius $\lVert T\rVert$, and the spectral radius satisfies

$$ r(T) = \lim_{n\to\infty}\lVert T^n\rVert^{1/n} = \inf_n\lVert T^n\rVert^{1/n} . $$

Proof. Nonemptiness and the spectral radius formula are the Gelfand–Mazur and Beurling–Gelfand theorems of Topological Algebras and Banach Algebras; compactness follows because $\sigma(T) \subseteq \{\lvert\lambda\rvert \leq \lVert T\rVert\}$ is closed, the complement being the preimage of the open unit group under the continuous map $\lambda \mapsto T - \lambda\,\mathrm{id}$. $\square$

Corollary (the finite-dimensional case). If $X$ is finite-dimensional of dimension $n$ then $B(X)$ is isomorphic to $M_n(\mathbb{K})$ as a unital Banach algebra, every element is invertible exactly when its determinant is nonzero, the spectrum is the set of eigenvalues, and $\sigma(T)$ has at most $n$ points.

Proof. Choosing a basis identifies $B(X)$ with $M_n(\mathbb{K})$, and the operator norm is equivalent to the entrywise norm, so the algebra and its topology agree; invertibility is the nonvanishing of the determinant and the spectrum is the set of eigenvalues of the matrix of $T$. $\square$

Closed Ideals and the Compact Operators

Definition. A two-sided ideal of $B(X)$ is a linear subspace $\mathcal{I}$ with $\mathcal{I}B(X) \subseteq \mathcal{I}$ and $B(X)\mathcal{I} \subseteq \mathcal{I}$; it is closed when closed in the norm topology. An operator $K \in B(X)$ is compact when the image of the unit ball of $X$ has compact closure in $X$; the compact operators form the set $K(X)$.

Proposition (the closed ideals). A closed two-sided ideal of $B(X)$ is a closed subalgebra and a Banach algebra without identity unless it is all of $B(X)$; the intersection of closed ideals is closed, and the closure of an ideal is an ideal.

Proof. An ideal is closed under products and is a subspace, hence a subalgebra, and it is closed by hypothesis, hence complete. That a proper closed ideal does not contain $\mathrm{id}$ is the definition of a proper ideal, and an ideal containing $\mathrm{id}$ is all of $B(X)$. The closure of an ideal is an ideal by the continuity of the product, and an intersection of closed ideals is closed and absorbs multiplication. $\square$

Proposition (the compact operators). The compact operators form a closed two-sided ideal $K(X)$ of $B(X)$; $K(X) = B(X)$ when $X$ is finite-dimensional, and $K(X)$ is a proper closed ideal when $X$ is infinite-dimensional. The quotient $B(X)/K(X)$, the Calkin algebra, is a Banach algebra.

Proof. The sum of compact operators is compact, the composite of a bounded with a compact operator is compact, and a norm-limit of compact operators is compact; these are the standard facts of the compact operators, quoted from the functional-analysis literature. The quotient of a Banach algebra by a closed two-sided ideal is a Banach algebra by Ideals and Quotients of Algebras, and the ideal is proper in infinite dimension because the identity is not compact. $\square$

Remark (the boundary). The compact operators, the Fredholm operators and the Calkin algebra are named because they are the basic closed ideals of $B(X)$; their theory, the approximation property and the structure of the closed ideals of $B(X)$ are not developed here. On a Hilbert space the compact operators and the Schatten classes are Operator Algebras.

Examples

Example (the finite-dimensional algebra). For $X = \mathbb{K}^n$ the operator algebra is $M_n(\mathbb{K})$, a unital Banach algebra under any submultiplicative norm, all of whose norms are equivalent; every subspace is closed, so every subalgebra is closed, and $K(X) = M_n(\mathbb{K})$.

Example (the space $\ell^p$ and the shift). For $X = \ell^p$, $1 \leq p < \infty$, the shift $S(x_0,x_1,\ldots) = (0,x_0,x_1,\ldots)$ is an isometry, hence bounded with norm $1$, and the closed subalgebra generated by $S$ is commutative and is studied through the disk algebra; the backward shift $S^*$ on $\ell^p$ has the same norm. The algebra $B(\ell^p)$ is the model of a noncommutative operator algebra with a rich ideal structure.

Example (a multiplication operator). For $X = C(K)$ with $K$ compact Hausdorff and the sup norm, the map $f \mapsto M_f$ with $M_f(g) = fg$ is an isometric unital algebra homomorphism of $C(K)$ onto a closed commutative subalgebra of $B(C(K))$, with $\lVert M_f\rVert = \lVert f\rVert_\infty$; its commutant is the set of operators intertwining the point evaluations and is the algebra of all continuous functions acting as multipliers in the classical cases.

Example (a non-closed subalgebra). Let $X = \ell^2$ and let $\mathcal{P}$ be the algebra of finite-rank operators. It is a subalgebra of $B(X)$, but it is not closed: its closure is $K(X)$, so the closed subalgebra generated by $\mathcal{P}$ is the compact operators. This shows the closure in the definition of the generated algebra is not optional.

Summary

The bounded operators on a Banach space $X$ form the unital Banach algebra $B(X)$ under composition and the operator norm, in which boundedness, continuity and the closed graph coincide and the invertible elements are the bounded bijections. A closed subalgebra of $B(X)$ is a Banach algebra, the closure of a subalgebra is a subalgebra, and the closed subalgebra generated by a set is the norm-closure of the algebra of noncommutative polynomials in it; the closed subalgebra generated by a single operator $T$ is the closure of the polynomials $\mathbb{K}[T]$, a commutative Banach algebra whose kernel is the polynomials vanishing on $\sigma(T)$. The commutant $\mathcal{S}'$ of any set is a unital closed subalgebra, the assignment is order-reversing, and $\mathcal{S} \subseteq \mathcal{S}''$; the bicommutant equality fails for a general closed subalgebra and holds for the weakly closed involutive subalgebras of a Hilbert space, by von Neumann's theorem in Involutive Operator Algebras and the Commutant. The unit group $B(X)^\times$ is open and inversion is continuous, the spectrum $\sigma(T)$ is a nonempty compact subset of the disk of radius $\lVert T\rVert$ and $r(T) = \lim\lVert T^n\rVert^{1/n}$, and in finite dimension $B(X)$ is $M_n(\mathbb{K})$. The compact operators form a closed two-sided ideal $K(X)$, proper in infinite dimension, with Calkin algebra $B(X)/K(X)$, and the closed subalgebras and the closed ideals are the structures this article fixes for the involutive operator theory of the later groups.

Summary of Notation

Symbol	Meaning
$X$, $\lVert\cdot\rVert$, $X'$	Banach space, its norm, its dual
$B(X)$	Unital Banach algebra of bounded linear operators, operator norm
$\lVert T\rVert = \sup_{\lVert x\rVert\leq1}\lVert Tx\rVert$	The operator norm
$\mathcal{S}'$, $\mathcal{S}''$	Commutant and bicommutant of a subset
$\overline{\mathbb{K}[T]}$	Closed unital subalgebra generated by $T$
$B(X)^\times$	Unit group, open; inversion continuous
$\sigma(T)$, $r(T)$	Spectrum, nonempty compact; spectral radius $\lim\lVert T^n\rVert^{1/n}$
$T - \lambda\,\mathrm{id}$	The operator whose non-invertibility defines the spectrum
$\mathcal{I}$ closed ideal	Closed two-sided ideal, a Banach algebra
$K(X)$	Compact operators, a proper closed ideal in infinite dimension
$B(X)/K(X)$	The Calkin algebra