Operators on a Banach Algebra

Introduction

A Banach algebra $A$ is first of all a Banach space, so the natural operators on it are the bounded linear maps $A \to A$, and the norm makes the space $B(A)$ of them complete. The same object carries a product, and the product singles out, among all bounded operators, those built from the algebra itself: the one-sided multiplications, the derivations generated by the elements, and the multipliers that commute with the product. This article lays down that operator layer. It fixes the bounded operators and their algebra, shows that the one-sided multiplications are bounded and form the two regular representations, defines the multipliers and the derivations with the identities that govern them, and records the automatic continuity a complete norm imposes on a multiplicative functional and on a derivation.

The article assumes the Banach algebra, its submultiplicative norm, its unit group, its spectrum and its spectral radius from Topological Algebras and Banach Algebras; the one-sided multiplications, the regular representation and the double centraliser statement from Left and Right Multiplication in a Ring; the ideals, the unit group and the annihilators from Rings; and the bounded linear maps, the operator norm and the completeness of the operator space from Normed and Banach Spaces. The involution on the elements and the adjoint on the operators are the subjects of the later groups of this category and are not used. The multiplier algebra and its strict topology are Multipliers of a Banach Algebra, later in this group; the operators of a $\mathrm{C}^*$-algebra are Operators on a $\mathrm{C}^*$-Algebra; the derivation and the automorphism group of an algebra are named in Automorphisms and Derivations of Algebras, and only the definitions recorded here are used. No form, no measure and no Fourier theory occurs.

Throughout, $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$; $A$ is a Banach algebra over $\mathbb{K}$ with submultiplicative norm $\lVert\cdot\rVert$, unital with unit $1$ exactly where a statement names $1$; $B(A)$ is the algebra of bounded $\mathbb{K}$-linear operators $A \to A$ with the operator norm, $\operatorname{End}(A)$ the algebra of all $\mathbb{K}$-linear maps, $A^\times$ the unit group and $Z(A)$ the centre. For $a \in A$ the one-sided multiplications are $L_a(x) = ax$ and $R_a(x) = xa$, and their product is the two-sided operator $L_aR_b = R_bL_a$, $x \mapsto axb$.

The Bounded Operators

Definition. For $T \in \operatorname{End}(A)$ the operator norm is

$$ \lVert T\rVert = \sup\{\lVert Tx\rVert : x \in A,\ \lVert x\rVert \leq 1\} \in [0,+\infty] , $$

and $T$ is bounded when $\lVert T\rVert < +\infty$. The bounded operators form the subspace $B(A)$ of $\operatorname{End}(A)$.

Proposition (boundedness is continuity). For $T \in \operatorname{End}(A)$ the following are equivalent: $T$ is bounded; $T$ is continuous; $T$ is continuous at $0$; there is $C \geq 0$ with $\lVert Tx\rVert \leq C\lVert x\rVert$ for all $x$. Consequently $B(A)$ is exactly the algebra of continuous $\mathbb{K}$-linear operators on $A$, and $\lVert T\rVert$ is the least such $C$.

Proof. If $T$ is bounded with norm $M$ then $\lVert Tx - Ty\rVert \leq M\lVert x - y\rVert$, so $T$ is Lipschitz and continuous. Conversely a continuous $T$ is continuous at $0$, so there is a neighbourhood of $0$ on which $\lVert Tx\rVert \leq 1$; by the definition of the topology of a normed space this neighbourhood contains a ball $B(0,r)$, whence $\lVert Tx\rVert \leq 1$ whenever $\lVert x\rVert \leq r$ and $\lVert Tx\rVert \leq r^{-1}\lVert x\rVert$ for all $x$ by scaling. The last two conditions are recastings of the first, and the least constant is the supremum in the definition. $\square$

Theorem ($B(A)$ is a Banach algebra). $B(A)$ is a unital algebra under the pointwise linear structure, the composition of operators and the identity $\mathrm{id}$, and

$$ \lVert ST\rVert \leq \lVert S\rVert\,\lVert T\rVert , \qquad \lVert \mathrm{id}\rVert = 1 \ (A \neq 0), $$

so the operator norm is a submultiplicative norm on $B(A)$; $B(A)$ is complete for it, hence a unital Banach algebra, and the multiplication of $B(A)$ is continuous.

Proof. Composition is bilinear and associative, and $\mathrm{id}$ is its unit. Submultiplicativity is $\lVert STx\rVert \leq \lVert S\rVert\lVert Tx\rVert \leq \lVert S\rVert\lVert T\rVert\lVert x\rVert$. For completeness, let $(T_n)$ be Cauchy in $B(A)$; then $(T_nx)$ is Cauchy in the Banach space $A$ for every $x$, so $Tx = \lim_n T_nx$ defines a linear $T$, and the uniform bound $\lVert T_n\rVert \leq M$ transfers to $\lVert Tx\rVert \leq M\lVert x\rVert$, so $T$ is bounded, and $\lVert T - T_n\rVert \leq \sup_{m \geq n}\lVert T_m - T_n\rVert \to 0$. Continuity of the product is submultiplicativity. $\square$

Corollary (the closed graph and the open map). A linear map $T : A \to A$ is bounded if and only if its graph is closed in $A \times A$; a bounded bijection $T \in B(A)$ has bounded inverse, and its inverse is again in $B(A)$.

Proof. A bounded map has closed graph because $x_n \to x$ and $Tx_n \to y$ give $Tx_n \to Tx = y$ by continuity. Conversely a linear map with closed graph is bounded by the closed graph theorem of Normed and Banach Spaces, whose hypotheses are the completeness of $A$. A bounded bijection is continuous, surjective and injective, so its inverse is bounded by the open mapping theorem. $\square$

Remark (the unit group of the operator algebra). An operator $T \in B(A)$ is invertible in the algebra $B(A)$ exactly when it is a bijection, and then $T^{-1} \in B(A)$ by the corollary; the unit group of $B(A)$ is therefore the group of bounded bijections of $A$, and the spectrum of $T$ in $B(A)$ is the set of $\lambda$ for which $T - \lambda\,\mathrm{id}$ is not a bijection.

The One-Sided Multiplications

Definition. For $a \in A$ the left multiplication and the right multiplication by $a$ are

$$ L_a : A \to A, \quad L_a(x) = ax , \qquad\qquad R_a : A \to A, \quad R_a(x) = xa . $$

Proposition (they are bounded, with the norm of the element). For every $a$ the operators $L_a$ and $R_a$ are bounded and linear, with

$$ \lVert L_a\rVert \leq \lVert a\rVert , \qquad \lVert R_a\rVert \leq \lVert a\rVert , $$

and when $A$ is unital both inequalities are equalities, $\lVert L_a\rVert = \lVert R_a\rVert = \lVert a\rVert$.

Proof. $\lVert L_ax\rVert = \lVert ax\rVert \leq \lVert a\rVert\lVert x\rVert$ by submultiplicativity, so $L_a$ is bounded with norm at most $\lVert a\rVert$, and likewise on the right. If $A$ has a unit then $L_a(1) = a$ and $R_a(1) = a$, so $\lVert L_a\rVert \geq \lVert a\rVert$ and $\lVert R_a\rVert \geq \lVert a\rVert$ on the unit vector $1/\lVert 1\rVert$. $\square$

Proposition (composition and commutation). For all $a,b \in A$,

$$ L_{ab} = L_aL_b , \qquad R_{ab} = R_bR_a , \qquad L_aR_b = R_bL_a . $$

Hence $a \mapsto L_a$ is a unital algebra homomorphism and $a \mapsto R_a$ a unital anti-homomorphism, and the two families commute.

Proof. $L_a(L_bx) = a(bx) = (ab)x = L_{ab}x$; $R_a(R_bx) = (xb)a = x(ba) = R_{ba}x$; and $L_aR_bx = a(xb) = (ax)b = R_bL_ax$ by associativity. $\square$

Corollary (the regular representations). The maps

$$ \lambda : A \to B(A), \quad a \mapsto L_a , \qquad\qquad \rho : A \to B(A), \quad a \mapsto R_a , $$

are a bounded unital algebra homomorphism and a bounded unital algebra anti-homomorphism, with $\lVert\lambda\rVert \leq 1$ and $\lVert\rho\rVert \leq 1$; when $A$ is unital both are isometric, and both are injective, since $L_a = 0$ or $R_a = 0$ forces $a = L_a(1) = 0$ or $a = R_a(1) = 0$.

Proof. The homomorphism properties are the composition laws, and the norm bound is the proposition. Injectivity on a unital algebra is evaluation at $1$, and isometry then follows from the norm equalities. $\square$

Proposition (invertibility and the units). For $a \in A$ the operators $L_a$ and $R_a$ are invertible in $B(A)$ if and only if $a \in A^\times$, and then $L_a^{-1} = L_{a^{-1}}$ and $R_a^{-1} = R_{a^{-1}}$. If $L_a$ or $R_a$ is a bijection of $A$ then $a$ is a unit.

Proof. If $a$ is a unit then $L_aL_{a^{-1}} = L_{aa^{-1}} = L_1 = \mathrm{id}$ and $L_{a^{-1}}L_a = \mathrm{id}$, so $L_a$ is invertible with the stated inverse, and likewise on the right. Conversely if $L_a$ is surjective then $a = L_a(1)$ lies in $LA = A$, so $a$ has a right inverse $b$ with $ab = 1$; if $L_a$ is injective then $ax = 0$ forces $x = 0$, so $a$ is not a left zero divisor, and then $ba = 1$ from $(ba - 1)a = a(ba) - a = a - a = 0$; hence $a$ is a unit. The right-handed argument is the mirror image. $\square$

The Multipliers

Definition. A left multiplier is a bounded operator $T \in B(A)$ with $T(ab) = T(a)b$ for all $a,b$; a right multiplier is a bounded $S$ with $S(ab) = aS(b)$; a double centraliser is a pair $(T,S)$ with $T$ a left multiplier, $S$ a right multiplier, and

$$ a\,T(b) = S(a)\,b \qquad \text{for all } a,b \in A . $$

Proposition (the multiplications are multipliers). For every $a$ the operator $L_a$ is a left multiplier and $R_a$ a right multiplier, and $(L_c,R_c)$ is a double centraliser for every $c$; moreover $L_a$ is a left multiplier and $R_b$ a right multiplier for every pair $a,b$, with the centralising relation $a\,L_c(b) = R_c(a)\,b$.

Proof. $L_c(ab) = c(ab) = (ca)b = L_c(a)b$ and $R_c(ab) = (ab)c = a(bc) = aR_c(b)$; the centralising relation is $a(cb) = (ac)b$. $\square$

Proposition (the unital case). If $A$ is unital then every left multiplier is a left multiplication and every right multiplier is a right multiplication, $T = L_{T(1)}$ and $S = R_{S(1)}$, the centralising relation forces $T(1) = S(1)$, and the double centralisers are exactly the pairs $(L_c,R_c)$ with $c \in A$. Hence the map $a \mapsto (L_a,R_a)$ is a bijection of $A$ onto the double centralisers, and it is an algebra isomorphism for the algebra structure fixed in Multipliers of a Banach Algebra.

Proof. A left multiplier satisfies $T(b) = T(1b) = T(1)b = L_{T(1)}(b)$, and a right multiplier $S(b) = S(b1) = bS(1) = R_{S(1)}(b)$. Setting $a = b = 1$ in $aT(b) = S(a)b$ gives $T(1) = S(1) = c$, so the pair is $(L_c,R_c)$; conversely every such pair is a double centraliser by the previous proposition, and $c \mapsto (L_c,R_c)$ is injective because $L_c(1) = c$. $\square$

Remark (the boundary of the section). On a non-unital algebra the multipliers are genuinely more than the multiplications, and they carry a Banach-algebra structure of their own, the multiplier algebra $M(A)$, whose product is the composition of double centralisers and whose topology of interest is the strict topology; both are the subject of Multipliers of a Banach Algebra, later in this group. The present article fixes only the operator-theoretic definition and the unital identification, and it uses no strict topology.

The Derivations

Definition. A derivation of $A$ is a linear map $D : A \to A$ with the Leibniz rule

$$ D(ab) = D(a)\,b + a\,D(b) \qquad \text{for all } a,b \in A . $$

For $x \in A$ the inner derivation determined by $x$ is $\operatorname{ad}_x = \operatorname{ad}(x) : a \mapsto [x,a] = xa - ax$.

Proposition (inner derivations are bounded derivations). For every $x$ the map $\operatorname{ad}_x$ is a derivation, it is bounded, and

$$ \lVert \operatorname{ad}_x\rVert \leq 2\lVert x\rVert , $$

with equality not asserted; moreover $\operatorname{ad}_x = L_x - R_x$.

Proof. $\operatorname{ad}_x(ab) = x(ab) - (ab)x = (xa - ax)b + a(xb - bx) = \operatorname{ad}_x(a)b + a\operatorname{ad}_x(b)$ by associativity, and $\operatorname{ad}_x = L_x - R_x$ is bounded with norm at most $\lVert L_x\rVert + \lVert R_x\rVert \leq 2\lVert x\rVert$. $\square$

Proposition (the Leibniz calculus). A derivation $D$ satisfies $D(1) = 0$ when $A$ is unital, $D(a^n) = \sum_{i=0}^{n-1} a^i\,D(a)\,a^{n-1-i}$ for $n \geq 1$, and $D(a^{-1}) = -a^{-1}D(a)a^{-1}$ for $a \in A^\times$. The inner derivations satisfy

$$ \operatorname{ad}_{[x,y]} = [\operatorname{ad}_x,\operatorname{ad}_y] , \qquad \operatorname{ad}_x(1) = 0 , $$

so $x \mapsto \operatorname{ad}_x$ is a homomorphism of the commutator algebra of $A$ into the commutator algebra of $B(A)$.

Proof. $D(1) = D(1 \cdot 1) = D(1) + D(1)$. The power rule is induction: $D(a^{n+1}) = D(a^n)a + a^nD(a) = \sum_{i=0}^{n-1} a^iD(a)a^{n-i} + a^nD(a)$, which is the stated sum for $n+1$. For the inverse, $0 = D(aa^{-1}) = D(a)a^{-1} + aD(a^{-1})$ gives $D(a^{-1}) = -a^{-1}D(a)a^{-1}$. Finally $\operatorname{ad}_{[x,y]} = \operatorname{ad}(xy - yx) = [L_{xy} - R_{xy}] - [L_{yx} - R_{yx}] = \operatorname{ad}_x\operatorname{ad}_y - \operatorname{ad}_y\operatorname{ad}_x$ by the composition laws. $\square$

Proposition (the space of derivations). The bounded derivations of $A$ form a closed linear subspace $\operatorname{Der}(A)$ of $B(A)$, and a Lie subalgebra under the commutator bracket $[D_1,D_2] = D_1D_2 - D_2D_1$; the inner derivations form a linear subspace $\operatorname{Inn}(A)$ and a Lie ideal of $\operatorname{Der}(A)$.

Proof. If $D_1,D_2$ are derivations then so is $D_1 - D_2$ and so is $[D_1,D_2]$: $[D_1,D_2](ab) = D_1(D_2(ab)) - D_2(D_1(ab))$, expanded, gives $[D_1,D_2](a)b + a[D_1,D_2](b)$. If $D_n \to D$ in $B(A)$ and each $D_n$ is a derivation, then $D(ab) = \lim D_n(ab) = \lim(D_n(a)b + aD_n(b)) = D(a)b + aD(b)$, so the derivation property is closed; $\operatorname{Der}(A)$ is closed in the Banach space $B(A)$. The inner derivations are in $\operatorname{Der}(A)$ and are an ideal by $\operatorname{ad}_{[x,y]} = [\operatorname{ad}_x,\operatorname{ad}_y]$ and $[D,\operatorname{ad}_x] = \operatorname{ad}_{D(x)}$. $\square$

Theorem (the exponential of a bounded derivation). Let $D \in \operatorname{Der}(A) \cap B(A)$ be a bounded derivation. Then the series $\exp D = \sum_{n \geq 0} D^n/n!$ converges in the Banach algebra $B(A)$, its sum is a bounded algebra automorphism of $A$, and

$$ (\exp D)^{-1} = \exp(-D) , \qquad \exp D = \sum_{n\geq0}\frac{D^n}{n!} , \qquad \exp(D)(ab) = (\exp D\,a)(\exp D\,b) . $$

Hence the exponential carries the bounded derivations into the automorphism group $\operatorname{Aut}(A)$.

Proof. The series converges absolutely because $\lVert D^n\rVert \leq \lVert D\rVert^n$ and $\sum\lVert D\rVert^n/n!$ is finite; $B(A)$ is complete. That $\exp D$ is multiplicative is the Leibniz rule reorganized: $D^n(ab) = \sum_{i=0}^{n} \binom{n}{i} D^i(a)D^{n-i}(b)$, which is the binomial theorem for the derivation, and summing against $1/n!$ gives $(\exp D)(ab) = (\exp Da)(\exp Db)$. The identity $\exp(D)\exp(-D) = \mathrm{id}$ is the Cauchy product and $\exp(0) = \mathrm{id}$. $\square$

Example (the matrix algebra). On $A = M_n(\mathbb{K})$ every derivation is inner, $D = \operatorname{ad}_X$ for some $X$, and the exponential of $\operatorname{ad}_X$ is the inner automorphism $Y \mapsto e^{X}Ye^{-X}$, whose inverse is conjugation by $e^{-X}$; the derivation $\operatorname{ad}_X$ is the operator $Y \mapsto XY - YX$ of commutator with $X$.

Automatic Continuity

Completeness constrains the unital multiplicative maps and the derivations. The device is the separating space of a linear map.

Definition. For a linear map $T : A \to A$ the separating space is

$$ \mathfrak{S}(T) = \{y \in A : \exists (x_n) \subseteq A,\ x_n \to 0 \text{ and } Tx_n \to y\} . $$

Proposition (the separating space is closed). $\mathfrak{S}(T)$ is a linear subspace of $A$, closed in $A$, and $T$ is bounded if and only if $\mathfrak{S}(T) = \{0\}$.

Proof. If $y_n \in \mathfrak{S}(T)$ and $y_n \to y$, pick $x_n \to 0$ with $Tx_n \to y_n$, and a diagonal subsequence $x_{n_k} \to 0$ with $Tx_{n_k} \to y$; hence $y \in \mathfrak{S}(T)$ and the space is closed. If $T$ is bounded then $x_n \to 0$ gives $Tx_n \to 0$, so $\mathfrak{S}(T) = \{0\}$. If $\mathfrak{S}(T) = \{0\}$ then $T$ is bounded: otherwise there are $x_n$ with $\lVert x_n\rVert = 1$ and $\lVert Tx_n\rVert \to \infty$, and $x_n/\lVert Tx_n\rVert \to 0$ while $T(x_n/\lVert Tx_n\rVert)$ has norm $1$, whose limit points lie in $\mathfrak{S}(T)$ and are nonzero. $\square$

Theorem (characters are continuous). Let $A$ be a unital Banach algebra and let $\chi : A \to \mathbb{K}$ be a nonzero multiplicative linear functional, $\chi(ab) = \chi(a)\chi(b)$ and $\chi(1) = 1$. Then $\chi$ is bounded, with $\lVert\chi\rVert = 1$, and

$$ \lvert \chi(a)\rvert \leq \lVert a\rVert \qquad \text{for all } a \in A . $$

Proof. Suppose $\lvert\chi(a)\rvert > \lVert a\rVert$ for some $a$. Then $u = a/\chi(a)$ has $\lVert u\rVert < 1$, so $1 - u$ is invertible, by the Neumann series of Topological Algebras and Banach Algebras; applying $\chi$ gives $\chi(1 - u) = 1 - 1 = 0$, and a multiplicative functional sends a unit to a nonzero scalar, since $\chi(v)\chi(v^{-1}) = 1$; contradiction. Hence $\lvert\chi(a)\rvert \leq \lVert a\rVert$, and $\chi(1) = 1$ makes the norm exactly $1$. $\square$

Corollary (the character space). The set $\operatorname{Max}(A)$ of nonzero multiplicative functionals of a unital commutative Banach algebra $A$ is contained in the unit sphere of the dual $A'$ and is closed in the weak-$*$ topology; it is the Gelfand spectrum of $A$, and the evaluation $a \mapsto \hat a$, $\hat a(\chi) = \chi(a)$, is the Gelfand transform, the subject of The Gelfand Transform as an Operator later in this group.

Proof. Boundedness is the theorem, so $\operatorname{Max}(A) \subseteq A'$; the relations $\chi(ab) = \chi(a)\chi(b)$ and $\chi(1) = 1$ are closed conditions under pointwise convergence, so a weak-$*$ limit of characters is a character, and $\operatorname{Max}(A)$ is closed. $\square$

Theorem (automatic continuity of derivations, Johnson–Sinclair; standard). Every derivation of a semisimple Banach algebra is continuous, hence bounded, and every derivation of a $\mathrm{C}^*$-algebra is continuous.

Proof (sketch). For a derivation $D$ the separating space $\mathfrak{S}(D)$ is a two-sided ideal, because $D$ satisfies the Leibniz rule: if $x_n \to 0$ and $Dx_n \to y$ then for fixed $a$ one has $ax_n \to 0$ and $D(ax_n) = D(a)x_n + aDx_n \to ay$, and $x_na \to 0$ with $D(x_na) = D(x_n)a + x_nD(a) \to ya$; so $\mathfrak{S}(D)$ absorbs multiplication on both sides. For a semisimple algebra the radical vanishes, and one shows $\mathfrak{S}(D)$ is nilpotent, hence zero; for a $\mathrm{C}^*$-algebra the same argument with the closed ideal structure gives $\mathfrak{S}(D) = \{0\}$. The details, due to Johnson and Sinclair, are standard and are quoted here. $\square$

Remark (what is not automatic). The continuity of characters does not extend to every homomorphism. A unital algebra homomorphism $\varphi : A \to B$ between Banach algebras need not be continuous when $B$ is only a Banach algebra, even a commutative one: Dales constructed discontinuous homomorphisms from certain $\mathrm{C}^*$-algebras into commutative Banach algebras. The positive statements are the character theorem above, the continuity of derivations, and the contractivity of a ${}^*$-homomorphism of $\mathrm{C}^*$-algebras recorded in Operators on a $\mathrm{C}^*$-Algebra; the general homomorphism problem is named here and deferred, not used.

Remark (the separating space of a homomorphism). When $\varphi$ is an algebra homomorphism, $\mathfrak{S}(\varphi)$ is a two-sided ideal, by the same computation as for a derivation, and $\varphi$ is continuous exactly when that ideal is zero; this is the reason the automatic-continuity proofs above turn on the absence of nilpotent ideals.

Summary

The operator layer of a Banach algebra $A$ is the Banach algebra $B(A)$ of bounded linear operators with the operator norm, in which boundedness, continuity and the closed graph coincide and the invertible elements are the bounded bijections. The one-sided multiplications $L_a(x) = ax$ and $R_a(x) = xa$ are bounded with norm at most $\lVert a\rVert$, equal to it on a unital algebra; they satisfy $L_{ab} = L_aL_b$, $R_{ab} = R_bR_a$ and $L_aR_b = R_bL_a$, so $\lambda : a \mapsto L_a$ is a bounded unital homomorphism and $\rho : a \mapsto R_a$ a bounded unital anti-homomorphism, isometric and injective on a unital algebra, and $L_a$ or $R_a$ is invertible exactly for a unit $a$. A left multiplier is a bounded $T$ with $T(ab) = T(a)b$, a right multiplier satisfies $S(ab) = aS(b)$, and a double centraliser is a pair with $aT(b) = S(a)b$; on a unital algebra the multipliers are exactly the multiplications, $T = L_{T(1)}$ and $S = R_{S(1)}$, and the double centralisers are the pairs $(L_c,R_c)$, while the general multiplier algebra and the strict topology are Multipliers of a Banach Algebra. A derivation satisfies the Leibniz rule, the inner derivation $\operatorname{ad}_x = L_x - R_x$ is bounded with norm at most $2\lVert x\rVert$, the bounded derivations form a closed Lie subalgebra of $B(A)$ containing the inner derivations as an ideal, and a bounded derivation $D$ has an exponential $\exp D$ that is an algebra automorphism with inverse $\exp(-D)$. Finally completeness forces continuity: a nonzero multiplicative functional of a unital Banach algebra is bounded with norm one, and every derivation of a semisimple Banach algebra is continuous, while a general algebra homomorphism of Banach algebras need not be.

Summary of Notation

Symbol	Meaning
$A$, $\lVert\cdot\rVert$, $A^\times$, $Z(A)$	Banach algebra, its norm, its units, its centre
$B(A)$, $\operatorname{End}(A)$	Bounded and all linear operators $A \to A$
$\lVert T\rVert = \sup_{\lVert x\rVert\leq1}\lVert Tx\rVert$	The operator norm
$L_a(x) = ax$, $R_a(x) = xa$	The one-sided multiplications, bounded
$L_{ab} = L_aL_b$, $R_{ab} = R_bR_a$, $L_aR_b = R_bL_a$	Composition and commutation
$\lambda : a \mapsto L_a$, $\rho : a \mapsto R_a$	The regular representations, bounded and isometric on a unital algebra
$T(ab) = T(a)b$, $S(ab) = aS(b)$	Left and right multiplier
$aT(b) = S(a)b$	The double centraliser relation
$\operatorname{ad}_x(a) = [x,a] = xa - ax$	The inner derivation
$\operatorname{Der}(A)$, $\operatorname{Inn}(A)$	Bounded derivations, inner derivations
$\exp D = \sum_n D^n/n!$	The exponential of a bounded derivation, an automorphism
$\mathfrak{S}(T) = \{y : x_n\to0,\ Tx_n\to y\}$	The separating space, closed
$\operatorname{Max}(A)$, $\hat a$	Characters of a commutative Banach algebra, the Gelfand transform