Commutativity of Banach Algebras with Derivations and Local Identities

Abstract

In this paper we prove a transfer principle for Banach algebras and use it to show that, for local identities of product type, the algebraic hypotheses of the commutativity theorems of Yood type are superfluous. Such theorems assume an algebraic identity only on a small part of the algebra and only with an exponent that may change from one point to the next. Completeness then restores both globality and uniformity. Every proof in this line opens with the same Baire category argument, and that argument uses nothing about the map to which the identity is applied. The transfer principle isolates that argument, and it holds for an arbitrary countable family of continuous maps of several variables subject to two elementary polynomial conditions. Its proof also supplies a selection step that has been used in the literature without proof. Combined with a linearisation at the identity element, the principle shows that continuity and additivity alone force the map to agree with the identity map modulo the prescribed closed subspace and force every commutator to lie in that subspace. For a closed two-sided ideal the result is best possible. The local hypothesis is then equivalent to the commutativity of the quotient algebra together with the requirement that the map induce the identity map on it. The theorems of Yood, of Ali and Khan and of Ashraf and Wani are recovered without their primeness, non-degeneracy and auxiliary hypotheses. The analogues for skew, b-generalised, Jordan and higher derivations follow at the same time, and no derivation can satisfy the hypothesis modulo a proper closed ideal. Identities of power, commutator, Jordan and multiplicative type are treated by the same method, and a unital $C^{*}$-algebra satisfying the hypothesis is commutative.

1. Introduction

Commutativity theorems for rings begin with Herstein [1,2]: a ring without non-zero nil ideals is commutative as soon as some fixed integer $n>1$ satisfies ${\left(xy\right)}^n=x^n{y}^n$ identically, a result presented together with its ring-theoretic background in [3], which is refined in many directions by Bell [4] and Abu-Khuzam [5] and surveyed in [6]. A second strand imposes the identity not on the elements of the ring but on the values of a derivation: Posner [7], Daif and Bell [8], Brešar [9], Vukman [10], Mayne [11], Bell and Martindale [12], Herstein [13], Ashraf and Rehman [14] and Ali, Khan, Khan and Muthana [15] are representative. A third strand, analytic in flavor, goes back to Singer and Wermer [16], whose theorem and the automatic continuity questions surrounding it were developed by Johnson [17], Johnson and Sinclair [18], Sinclair [19] and Thomas [20], and are presented systematically in the monograph of Dales [21]. Within this third strand stand the papers of Yood [22,23,24], which rest on the observation that for a Banach algebra the hypothesis may be weakened twice over: the identity need hold only for x and y in a pair of non-empty open sets, and the exponent may vary from point to point. Completeness restores both globality and uniformity through the Baire category theorem. Yood’s prototype [23] (Theorem 1) states that if $\mathcal{A}$ is a unital Banach algebra, M a closed subspace, and if each $x,y$ in two non-empty open sets admit an integer $n=n(x,y)>1$ with ${\left(xy\right)}^n-x^n{y}^n\in M$ or ${\left(xy\right)}^n-y^n{x}^n\in M$, then $[\mathcal{A},\mathcal{A}]\subseteq M$; with $M=Z\left(\mathcal{A}\right)$ this gives commutativity for semiprime $\mathcal{A}$.

The two later strands were joined by Ali and Khan [25], who replaced ${\left(xy\right)}^n$ by $d\left({\left(xy\right)}^n\right)$ for a continuous derivation d of a unital prime Banach algebra, and by Ashraf and Wani [26], who did the same for generalised derivations under the additional hypothesis $d\left(Z\right(\mathcal{A}\left)\right)\ne 0$; skew, b-generalised, Jordan and higher derivations have been treated since. All of these proofs open with the same two or three pages: a Baire category argument passing from the local hypothesis with a variable exponent to a global identity with a fixed one. That argument uses no multiplicative property of the map. It uses only that the map is continuous, and that substituting $u+tw$ for one of its variables produces an $\mathcal{A}$-valued polynomial in $t\in \mathbb{R}$ whose leading coefficient is the value of the map at w.

The purpose of this paper is to isolate that topological content once and for all, and then to determine what, if anything, the algebra contributes. The first task is carried out in Theorem 1, a transfer principle valid for an arbitrary countable family $\{{\Phi }_i^{\nu }:{\mathcal{A}}^n\to \mathcal{A}\}$ of continuous maps subject to the two polynomial conditions just described: if for $x_r$ ranging over non-meagre sets $G_r\subseteq \mathcal{A}$ some level $\nu$ and some index i satisfy ${\Phi }_i^{\nu }(x_1,\dots ,x_n)\in M$, then a single level $\rho$ serves at every point of ${\mathcal{A}}^n$. One theorem thus disposes of an arbitrary number of variables, a countable family of exponents, finitely many alternatives holding pointwise, and local sets that are merely of the second category rather than open, and it does so with no multiplicative hypothesis on the maps involved. The polynomial condition imposed on the family is not claimed to be necessary; it is a sufficient condition tailored to the extraction of a coefficient, and it is satisfied by every identity that has occurred in this literature. The transfer principle rests on Lemma 5; applied at a single level, that lemma also yields Lemma 6, which passes from “at each point one of finitely many alternatives holds” to “one alternative holds at every point”. This step is used in the existing literature without proof, and it is a genuine assertion, resting on the polynomial condition through Lemma 5.

The second task is settled by combining the transfer principle with a linearisation at the identity element. For identities of product type the algebraic hypotheses turn out to be entirely superfluous: if $\mathcal{A}$ is unital, M is a closed subspace, $F:\mathcal{A}\to \mathcal{A}$ is any continuous additive map, and each $x\in G_1$, $y\in G_2$ admit an integer $m=m(x,y)>1$ with $F\left({\left(xy\right)}^m\right)-x^m{y}^m\in M$ or $F\left({\left(xy\right)}^m\right)-y^m{x}^m\in M$, then $F\left(x\right)-x\in M$ for all x and $[\mathcal{A},\mathcal{A}]\subseteq M$. When M is a closed two-sided ideal this is best possible, and the resulting equivalence, Theorem 6, is the sharpest statement we obtain: the local hypothesis holds if and only if $\mathcal{A}/M$ is commutative and F induces the identity map on it. A pointwise power identity on a pair of small sets is in this sense no more than a disguised description of the quotient. The theorems of Yood, of Ali and Khan and of Ashraf and Wani follow without their primeness, non-degeneracy and auxiliary hypotheses, and the analogues for skew, b-generalised, Jordan and higher derivations follow simultaneously. The membership $F\left(x\right)-x\in M$, which is the first of the two conclusions of Theorem 5, also carries a rigidity phenomenon: a map with $F\left(e\right)\in M$ cannot satisfy the hypothesis at all modulo a proper closed ideal, so no derivation does.

The pattern applies beyond the theorems proved here: the analytic hypotheses upgrade a local identity with a variable exponent to a global identity with a fixed exponent; the identity element then linearises that identity; and only afterwards, if at all, does the algebraic nature of the map enter. For product-type identities the third stage is empty. For power-type and commutator-type identities it is not, and the ring theory needed there is the subject of a separate work in preparation, which takes the results of Section 4 as its input; the present paper is self-contained, and none of its results depends on that work.

Section 2 fixes the algebraic notation and the elementary lemmas on which every extraction of a coefficient rests. Section 3 proves the transfer principle, Section 4 the linearisation and the theorems on identities of product type, Section 5 their consequences for derivations and related maps, and Section 6 treats identities in which the right-hand side is built from F itself. Section 7 collects the conclusions of the paper.

2. Ring-Theoretic and Algebraic Preliminaries

Throughout, R denotes an associative ring and we write $[x,y]=xy-yx,x\circ y=xy+yx.$ The centre of R is $Z\left(R\right)=\{a\in R:[a,x]=0\mathrm{for}\mathrm{all}x\in R\}$. Recall that R is semiprime if $aRa=\left(0\right)$ implies $a=0$, equivalently if R has no non-zero nilpotent ideals, and prime if $aRb=\left(0\right)$ implies $a=0$ or $b=0$.

Lemma 1. 

Let R be a semiprime ring with $[x,y]\in Z\left(R\right)$ for all $x,y\in R$. Then R is commutative.

Proof. 

Let $x,y,z\in R$. From the derivation identity for the commutator,

$$[xy,z]=x[y,z]+[x,z]y,$$

and since $[y,z]$ and $[x,z]$ are central this may be rewritten as

$$[xy,z]=[y,z]x+[x,z]y.$$

(1)

The left-hand side of (1) is again a commutator, hence central, so it commutes with x. Computing the commutator of the right-hand side with x and using centrality of $[y,z]$ and $[x,z]$ once more, we get

$$0=\left[[y,z]x+[x,z]y,x\right]=[y,z][x,x]+[x,z][y,x]=[x,z][y,x].$$

Thus $[x,z][y,x]=0$ for all $x,y,z\in R$. Specialising $z=y$ gives $[x,y][y,x]=0$, that is,

$${[x,y]}^2=0\mathrm{for}\mathrm{all}x,y\in R.$$

(2)

Now fix $x,y$ and put $a=[x,y]$. Since a is central, $aRa=a^{2}R=\left(0\right)$ by (2), and semiprimeness gives $a=0$. □

The following example shows that the semiprimeness cannot be omitted from Lemma 1.

Example 1. 

Let $N\subseteq M_3\left(\mathbb{C}\right)$ be the algebra of strictly upper triangular matrices, so that $N^3=\left(0\right)$, and put

$$\mathcal{A}=\mathbb{C}e\oplus N\subseteq M_3\left(\mathbb{C}\right),$$

a unital finite-dimensional, hence Banach, algebra. Since $e_{12}{e}_{23}=e_{13}$ and $e_{23}{e}_{12}=0$,

$$Z\left(\mathcal{A}\right)=\mathbb{C}e\oplus \mathbb{C}{e}_{13},[\mathcal{A},\mathcal{A}]=[N,N]=\mathbb{C}{e}_{13}\subseteq Z\left(\mathcal{A}\right),$$

so every commutator of $\mathcal{A}$ is central while $\mathcal{A}$ is not commutative. Here $\mathcal{A}$ is not semiprime, N being a non-zero nilpotent ideal.

2.1. Derivations and Their Relatives

We now recall the definitions of the classes of maps that will occur later. Throughout, $\alpha$ denotes an automorphism of R.

Definition 1.

An additive map $d:R\to R$ is

(a)

a derivation if $d\left(xy\right)=d\left(x\right)y+xd\left(y\right)$ for all $x,y\in R$;

(b)

an α-derivation, or skew derivation, if $d\left(xy\right)=d\left(x\right)y+\alpha \left(x\right)d\left(y\right)$ for all $x,y\in R$;

(c)

a Jordan derivation if $d\left(x^2\right)=d\left(x\right)x+xd\left(x\right)$ for all $x\in R$.

An additive map $F:R\to R$ is a generalised derivation associated with a derivation d, in the sense of [27,28], if

$$F\left(xy\right)=F\left(x\right)y+xd\left(y\right)\mathit{for}\mathit{all}x,y\in R;$$

a b-generalised derivation associated with d and an element b, in the sense of [29], if $F\left(xy\right)=F\left(x\right)y+bxd\left(y\right)$ for all $x,y\in R$; and a b-generalised skew derivation associated with an element b and with a map d that is an α-derivation of R in the sense of (b) above, α being the fixed automorphism of R occurring there, if

$$F\left(xy\right)=F\left(x\right)y+b\alpha \left(x\right)d\left(y\right)\mathit{for}\mathit{all}x,y\in R.$$

(3)

Here b is taken either in R itself or, when R is prime, in its Martindale quotient ring [30]. A left multiplier is an additive map with $F\left(xy\right)=F\left(x\right)y$, and a higher derivation of R is a sequence $D={\left(d_n\right)}_{n\ge 0}$ of additive maps with $d_0={\mathrm{id}}_R$ and

$$d_n\left(xy\right)=\sum _{i+j=n}{d}_i\left(x\right)d_j\left(y\right)\mathit{for}\mathit{all}x,y\in R,n\ge 1;$$

equivalently [31], D is a homomorphism of R into the ring of formal power series over R that lifts the identity map. Finally, given automorphisms $\alpha ,\beta$ of R, an additive map $d:R\to R$ is an $(\alpha ,\beta )$-derivation if $d\left(xy\right)=d\left(x\right)\alpha \left(y\right)+\beta \left(x\right)d\left(y\right)$ for all $x,y\in R$; it is a Jordan triple derivation if $d\left(xyx\right)=d\left(x\right)yx+xd\left(y\right)x+xyd\left(x\right)$ for all $x,y\in R$; and it is a Lie derivation if $d\left(\right[x,y\left]\right)=\left[d\right(x),y]+[x,d(y\left)\right]$ for all $x,y\in R$. An elementary operator of R is a map of the form $x\mapsto {\sum }_{i=1}^q{a}_{i}x{b}_i$ with $a_1,\dots ,a_q,b_1,\dots ,b_q\in R$ and $q\in \mathbb{N}$.

Every map recorded in Definition 1 is additive, and additivity together with continuity is all that the theorems of Section 4, Section 5 and Section 6 require of the map to which they are applied; Theorem 5, which governs those applications, is stated for an arbitrary continuous additive map $F:\mathcal{A}\to \mathcal{A}$. In particular, in the Banach-algebra applications of Section 5 the element b of a b-generalised derivation or of a b-generalised skew derivation may be taken in $\mathcal{A}$ itself, and no appeal to the Martindale quotient ring is needed there.

When the ring has an identity, all of the maps just defined take a common shape: each of them is a left multiplication plus a fixed element times a derivation.

Proposition 1

(Hvala [27]; Lee [32]). Let R be a unital ring with identity e and let F be a b-generalised skew derivation of R associated with an α-derivation d and an element b. Then $d\left(e\right)=0$ and

$$F\left(x\right)=px+bd\left(x\right)(x\in R),\mathit{where}p:=F\left(e\right).$$

(4)

Conversely, given any p, any b and any α-derivation d of R, formula (4) defines a b-generalised skew derivation of R associated with d and b.

Lemma 2.

Let R be unital. Then $d\left(e\right)=0$ for every derivation, every α-derivation and every Jordan derivation d of R; and $d_n\left(e\right)=0$ for every $n\ge 1$ if $D=\left(d_n\right)$ is a higher derivation of R.

Proof. 

For a derivation and for an $\alpha$-derivation the assertion is contained in Proposition 1. If d is a Jordan derivation, then putting $x=e$ in $d\left(x^2\right)=d\left(x\right)x+xd\left(x\right)$ gives $d\left(e\right)=2d\left(e\right)$, and hence $d\left(e\right)=0$.

Let now $D=\left(d_n\right)$ be a higher derivation and suppose, inductively, that $d_i\left(e\right)=0$ for $1\le i<n$; for $n=1$ this hypothesis is empty. Evaluating the defining identity at $x=y=e$ yields

$$d_n\left(e\right)=\sum _{i+j=n}{d}_i\left(e\right)d_j\left(e\right).$$

Every term with $0<i<n$ vanishes by the inductive hypothesis, while the two extreme terms, those with $i=0$ and with $j=0$, each contribute $d_n\left(e\right)$, because $d_0={\mathrm{id}}_R$. The identity therefore reduces to $d_n\left(e\right)=2d_n\left(e\right)$, and so $d_n\left(e\right)=0$. □

2.2. Two Elementary Lemmas on Polynomials

The two lemmas below are used at every extraction of a coefficient. The first is Vandermonde interpolation, in the form employed by Yood [23], and the second is the non-commutative binomial expansion.

Lemma 3

(Yood [23]). Let X be a real vector space, $M\subseteq X$ a linear subspace, $b_0,\dots ,b_N\in X$, and let

$$p\left(t\right)={\displaystyle \sum _{j=0}^N}{b}_j{t}^j(t\in \mathbb{R}).$$

If $p\left(t\right)\in M$ for infinitely many $t\in \mathbb{R}$, then $b_j\in M$ for every $j=0,\dots ,N$.

For the second lemma, let $\Lambda$ be a commutative unital ring and let A be an associative, not necessarily commutative, $\Lambda$-algebra; in every application below $\Lambda =\mathbb{R}$ and $A=\mathcal{A}$.

Lemma 4.

Let $a,b\in A$, let $t\in \Lambda$ and let $\rho \ge 1$ be an integer. Then

$${(a+tb)}^{\rho }={\displaystyle \sum _{j=0}^{\rho }}{t}^j{c}_j(a,b),$$

(5)

where $c_j(a,b)$ denotes the sum of the $\left({\displaystyle \genfrac{}{}{0pt}{}{\rho }{j}}\right)$ products of length ρ in which b occurs j times and a occurs $\rho -j$ times. In particular

$$c_0(a,b)=a^{\rho },c_1(a,b)={\displaystyle \sum _{j=0}^{\rho -1}}{a}^{j}b{a}^{\rho -1-j},c_{\rho }(a,b)=b^{\rho }.$$

(6)

If A is unital and $a=e$, then $c_j(e,b)=\left({\displaystyle \genfrac{}{}{0pt}{}{\rho }{j}}\right)b^j$, so that (5) becomes the binomial expansion

$${(e+tb)}^{\rho }={\displaystyle \sum _{j=0}^{\rho }}\left({\displaystyle \genfrac{}{}{0pt}{}{\rho }{j}}\right)t^j{b}^j,$$

(7)

whose coefficient of $t^1$ is $\rho b$.

Proof. 

Regard ${(a+tb)}^{\rho }$ as a product of $\rho$ factors, each equal to $a+tb$, and expand it by distributivity: one chooses from every factor either a or $tb$, and the order of the factors is retained. As t lies in $\Lambda$ and is therefore central, each of the $2^{\rho }$ terms obtained in this way has the form $t^{j}w$, where j is the number of factors from which $tb$ was taken and w is the corresponding product of j copies of b and $\rho -j$ copies of a, in the order in which they occur. Collecting the terms according to the value of j gives (5), the coefficient of $t^j$ being the sum of the $\left({\displaystyle \genfrac{}{}{0pt}{}{\rho }{j}}\right)$ such words.

Of these words, the only one containing no b is $a^{\rho }$ and the only one consisting entirely of b’s is $b^{\rho }$; and a word containing exactly one b is determined by the place at which that b stands, so it is $a^{j}b{a}^{\rho -1-j}$ for some $0\le j\le \rho -1$. This is (6). If finally $a=e$, then every word with j occurrences of b collapses to $b^j$, and there are $\left({\displaystyle \genfrac{}{}{0pt}{}{\rho }{j}}\right)$ of them; this gives (7), whose coefficient of $t^1$ is $\left({\displaystyle \genfrac{}{}{0pt}{}{\rho }{1}}\right)b=\rho b$. □

The identity of product type studied in Section 4 has the two sides $x^{\rho }{y}^{\rho }$ and $y^{\rho }{x}^{\rho }$, and it is handled by substituting $e+tx$ for the first variable and then for the second. The four maps occurring in the next corollary are exactly the four expressions produced by those two substitutions: ${\left((e+tx)y\right)}^{\rho }$ and ${(e+tx)}^{\rho }{y}^{\rho }$ come from the first substitution, ${\left(y(e+tx)\right)}^{\rho }$ and $y^{\rho }{(e+tx)}^{\rho }$ from the second. Their coefficients of $t^1$, computed once here, are all that the argument uses.

Corollary 1.

Let A be unital, let $x,y\in A$, let $\rho >1$ be an integer and set

$$Q=Q_{\rho }(x,y):={\displaystyle \sum _{j=1}^{\rho -1}}{y}^{j}x{y}^{\rho -j}.$$

(8)

Then each of the four maps

$$t\mapsto {\left((e+tx)y\right)}^{\rho },t\mapsto {\left(y(e+tx)\right)}^{\rho },t\mapsto {(e+tx)}^{\rho }{y}^{\rho },t\mapsto y^{\rho }{(e+tx)}^{\rho }$$

is a polynomial in $t\in \Lambda$ of degree at most ρ with constant term $y^{\rho }$, and their coefficients of $t^1$ are, respectively,

$$x{y}^{\rho }+Q,y^{\rho }x+Q,\rho x{y}^{\rho },\rho y^{\rho }x.$$

(9)

Proof. 

Each of the four maps is a polynomial in t of degree at most $\rho$ by Lemma 4, and each takes the value $y^{\rho }$ at $t=0$; only the coefficients of $t^1$ require attention.

Consider the first map. Since $(e+tx)y=y+txy$, we may apply (6) with $a=y$ and $b=xy$, and the coefficient in question is

$${\displaystyle \sum _{j=0}^{\rho -1}}{y}^j\left(xy\right)y^{\rho -1-j}=x{y}^{\rho }+{\displaystyle \sum _{j=1}^{\rho -1}}{y}^{j}x{y}^{\rho -j}=x{y}^{\rho }+Q,$$

where the term $j=0$, which is the only one that begins with x, has been separated from the rest. The second map is treated in the same way, this time with $a=y$ and $b=yx$; here it is the term $j=\rho -1$ that must be separated, and after the change of index $k=j+1$ the remaining terms again make up Q:

$${\displaystyle \sum _{j=0}^{\rho -1}}{y}^j\left(yx\right)y^{\rho -1-j}={\displaystyle \sum _{j=0}^{\rho -1}}{y}^{j+1}x{y}^{\rho -1-j}=y^{\rho }x+{\displaystyle \sum _{k=1}^{\rho -1}}{y}^{k}x{y}^{\rho -k}=y^{\rho }x+Q.$$

The last two maps are simpler, since $y^{\rho }$ is a fixed element there: multiplying the binomial expansion (7) by $y^{\rho }$ on the right, and then on the left, and reading off the coefficient of $t^1$, which is $\rho x$, gives $\rho x{y}^{\rho }$ and $\rho y^{\rho }x$ respectively. □

3. The Transfer Principle

In this section $\mathcal{A}$ is a complex Banach algebra and $M\subseteq \mathcal{A}$ a closed linear subspace; no identity element is required.

A subset of a topological space is meagre if it is a countable union of nowhere dense sets and non-meagre otherwise, and it is comeagre if its complement is meagre; for these notions and for the Baire category theorem we refer to [33,34]. By the Baire category theorem a Banach space is non-meagre in itself, and so is each of its non-empty open subsets. We use constantly, and without further comment, the fact that a non-meagre set meets every comeagre set, since otherwise it would be contained in a meagre set. For $n\ge 1$ we regard ${\mathcal{A}}^n$ as a Banach space with the maximum norm.

Recall further that $Z\left(\mathcal{A}\right)$ is closed in $\mathcal{A}$, that $\mathbb{C}e\subseteq Z\left(\mathcal{A}\right)$ when $\mathcal{A}$ is unital, and that the Jacobson radical $rad\mathcal{A}$ is a closed two-sided ideal [35]; $\mathcal{A}$ is semisimple if $rad\mathcal{A}=\left(0\right)$, and semisimple algebras are semiprime.

A continuous additive map $F:\mathcal{A}\to \mathcal{A}$ is $\mathbb{R}$-linear, additivity giving $\mathbb{Q}$-homogeneity and continuity extending it to $\mathbb{R}$. All that is required of F below is that it may be applied coefficientwise to an $\mathcal{A}$-valued polynomial in a real variable, that is,

$$F\left(\sum _j{t}_j{a}_j\right)=\sum _j{t}_{j}F\left(a_j\right)(t_j\in \mathbb{R},a_j\in \mathcal{A});$$

(10)

complex linearity is nowhere assumed and nowhere used.

Definition 2.

Let $n\ge 1$, let $\mathcal{N}$ be a non-empty countable set, whose elements we call levels, and let $k\in \mathbb{N}$. A family

$$\Phi =\left\{{\Phi }_i^{\nu }:{\mathcal{A}}^n\to \mathcal{A}:\nu \in \mathcal{N},1\le i\le k\right\}$$

of maps is admissible if every ${\Phi }_i^{\nu }$ is continuous and if for every $\nu \in \mathcal{N}$ there are integers $d_{\nu }^{\left(1\right)},\dots ,d_{\nu }^{\left(n\right)}\ge 1$ with the following property, to which we refer as(A_r): for every $r\in \{1,\dots ,n\}$, every $i\le k$, all $a_1,\dots ,a_n\in \mathcal{A}$ and all $u,w\in \mathcal{A}$, the map

$$t⟼{\Phi }_i^{\nu }\left(a_1,\dots ,a_{r-1},u+tw,a_{r+1},\dots ,a_n\right)$$

is an $\mathcal{A}$-valued polynomial in $t\in \mathbb{R}$ of degree at most $d_{\nu }^{\left(r\right)}$, whose coefficient of $t^{d_{\nu }^{\left(r\right)}}$ equals ${\Phi }_i^{\nu }(a_1,\dots ,a_{r-1},w,a_{r+1},\dots ,a_n)$.

For an admissible family $\Phi$ and a level $\nu$ we abbreviate

$$C_{\nu }:=\left\{x\in {\mathcal{A}}^n:{\Phi }_i^{\nu }\left(x\right)\in M\mathrm{for}\mathrm{at}\mathrm{least}\mathrm{one}i\le k\right\}={\displaystyle \bigcup _{i=1}^k}{\left({\Phi }_i^{\nu }\right)}^{-1}\left(M\right).$$

(11)

Since each ${\Phi }_i^{\nu }$ is continuous and M is closed, each set ${\left({\Phi }_i^{\nu }\right)}^{-1}\left(M\right)$ is closed, and a finite union of closed sets is closed; hence

$$C_{\nu }\mathrm{is}\mathrm{a}\mathrm{closed}\mathrm{subset}\mathrm{of}{\mathcal{A}}^n(\nu \in \mathcal{N}).$$

(12)

Lemma 5.

Let $\mathcal{A}$ be a normed algebra, $M\subseteq \mathcal{A}$ a closed linear subspace, and let ${\Psi }_1,\dots ,{\Psi }_k:{\mathcal{A}}^n\to \mathcal{A}$ be continuous maps satisfying the conditions (A₁), …,(A_n) of Definition 2 at a single level, with degrees $d^{\left(1\right)},\dots ,d^{\left(n\right)}$. Put

$$C=\left\{x\in {\mathcal{A}}^n:{\Psi }_i\left(x\right)\in M\mathit{for}\mathit{at}\mathit{least}\mathit{one}i\le k\right\}.$$

If C has non-empty interior, then $C={\mathcal{A}}^n$.

Proof. 

Since C has an interior point, there are non-empty open sets $U_1,\dots ,U_n$ in $\mathcal{A}$ with $U_1\times \cdots \times U_n\subseteq C$; fix $u_r\in U_r$ for each r. For $0\le r\le n$ let $\mathrm{P}\left(r\right)$ be the assertion that ${\Psi }_i(x_1,\dots ,x_n)\in M$ for some $i\le k$ whenever $x_1,\dots ,x_r\in \mathcal{A}$ are arbitrary and $x_{r+1}\in U_{r+1},\dots ,x_n\in U_n$. Then $\mathrm{P}\left(0\right)$ is the displayed inclusion and $\mathrm{P}\left(n\right)$ is the equality $C={\mathcal{A}}^n$, so it suffices to pass from $\mathrm{P}(r-1)$ to $\mathrm{P}\left(r\right)$ for $1\le r\le n$.

Assume $\mathrm{P}(r-1)$ and fix $x_1,\dots ,x_{r-1}\in \mathcal{A}$, $x_{r+1}\in U_{r+1},\dots ,x_n\in U_n$ and an arbitrary $x_r\in \mathcal{A}$. As $U_r$ is open and contains $u_r$, there is $\delta >0$ with $u_r+t{x}_r\in U_r$ for $\left|t\right|<\delta$, so $\mathrm{P}(r-1)$ provides, for each such t, an index $i\le k$ with ${\Psi }_i(x_1,\dots ,u_r+t{x}_r,\dots ,x_n)\in M$. The indices are finite in number while t runs through an interval, so one index $i_0$ serves for infinitely many t. By (A_r) the map $t\mapsto {\Psi }_{i_0}(x_1,\dots ,u_r+t{x}_r,\dots ,x_n)$ is a polynomial of degree at most $d^{\left(r\right)}$ whose coefficient of $t^{d^{\left(r\right)}}$ is ${\Psi }_{i_0}(x_1,\dots ,x_n)$, and it takes values in M for infinitely many t; Lemma 3 puts all its coefficients in M, that one among them. As $x_r$ was arbitrary, this is $\mathrm{P}\left(r\right)$. □

Lemma 6.

Let $\mathcal{A}$ be a Banach algebra, $M\subseteq \mathcal{A}$ a closed linear subspace, and let ${\Psi }_1,\dots ,{\Psi }_k:{\mathcal{A}}^n\to \mathcal{A}$ be continuous maps satisfying (A₁), …,(A_n) at a single level. If for every $x\in {\mathcal{A}}^n$ there is an index i with ${\Psi }_i\left(x\right)\in M$, then there is a single index $i_0$ with

$${\Psi }_{i_0}\left(x\right)\in M\mathit{for}\mathit{all}x\in {\mathcal{A}}^n.$$

Proof. 

For $1\le i\le k$ put

$$C_i={\Psi }_i^{-1}\left(M\right)=\left\{x\in {\mathcal{A}}^n:{\Psi }_i\left(x\right)\in M\right\}.$$

Each ${\Psi }_i$ is continuous and M is closed in $\mathcal{A}$, so every $C_i$ is closed in ${\mathcal{A}}^n$. The hypothesis asserts that each $x\in {\mathcal{A}}^n$ satisfies ${\Psi }_i\left(x\right)\in M$ for at least one i, which is to say

$${\mathcal{A}}^n={\displaystyle \bigcup _{i=1}^k}{C}_i.$$

(13)

Suppose that every $C_i$ had empty interior. A closed set with empty interior is nowhere dense, so (13) would exhibit ${\mathcal{A}}^n$ as a union of finitely many, and in particular countably many, nowhere dense sets; that is, ${\mathcal{A}}^n$ would be meagre in itself. Since $\mathcal{A}$ is a Banach space, so is ${\mathcal{A}}^n$ under the maximum norm, and ${\mathcal{A}}^n$ is therefore a complete metric space; the Baire category theorem forbids it to be meagre in itself. Hence

$$int{C}_{i_0}\ne \varnothing \mathrm{for}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{index}{i}_0\le k.$$

Fix such an $i_0$ and apply Lemma 5 to the single map ${\Psi }_{i_0}$, that is, with $k=1$. Its hypotheses are met: ${\Psi }_{i_0}$ is continuous and satisfies (A₁), …, (A_n) at the level in question, with the same degrees $d^{\left(1\right)},\dots ,d^{\left(n\right)}$, and the set which the lemma calls C is here $\{x\in {\mathcal{A}}^n:{\Psi }_{i_0}\left(x\right)\in M\}=C_{i_0}$, which has just been shown to have non-empty interior. The lemma yields $C_{i_0}={\mathcal{A}}^n$, that is, ${\Psi }_{i_0}\left(x\right)\in M$ for every $x\in {\mathcal{A}}^n$. □

Proposition 2.

Let $\mathcal{A}$ be a Banach algebra, $M\subseteq \mathcal{A}$ a closed linear subspace and $\Phi =\left\{{\Phi }_i^{\nu }\right\}$ an admissible family of n variables with countable level set $\mathcal{N}$ and k alternatives. Fix an index $r\in \{1,\dots ,n\}$ and elements $a_s\in \mathcal{A}$ for $s\ne r$, and for $y\in \mathcal{A}$ write

$${\psi }_i^{\nu }\left(y\right)={\Phi }_i^{\nu }\left(a_1,\dots ,a_{r-1},y,a_{r+1},\dots ,a_n\right)(\nu \in \mathcal{N},i\le k),$$

so that y occupies the r-th slot and the remaining slots are filled by the fixed elements $a_s$. Let $G\subseteq \mathcal{A}$ be non-meagre and assume that every $y\in G$ admits a level $\nu \in \mathcal{N}$ and an index $i\le k$ with ${\psi }_i^{\nu }\left(y\right)\in M$. Then there is a single level $\rho \in \mathcal{N}$ such that every $y\in \mathcal{A}$ admits an index $i\le k$ with ${\psi }_i^{\rho }\left(y\right)\in M$.

Proof. 

For $\nu \in \mathcal{N}$ put

$$E_{\nu }={\displaystyle \bigcup _{i=1}^k}{\left({\psi }_i^{\nu }\right)}^{-1}\left(M\right)\subseteq \mathcal{A}.$$

Each ${\psi }_i^{\nu }$ is continuous, so $E_{\nu }$ is closed and $V_{\nu }:=\mathcal{A}\setminus E_{\nu }$ is open. Suppose, for contradiction, that every $V_{\nu }$ were dense in $\mathcal{A}$. Then ${\bigcap }_{\nu \in \mathcal{N}}{V}_{\nu }$, a countable intersection of dense open sets, is comeagre, and therefore meets the non-meagre set G, say at a point $y_0$. But $y_0\in V_{\nu }$ for every $\nu$ means that no level and no alternative works at $y_0$, contradicting the hypothesis. Hence some $V_{\rho }$ fails to be dense; its closure is then a proper closed subset of $\mathcal{A}$, and the non-empty open set $\Gamma :=\mathcal{A}\setminus \overline{V_{\rho }}$ is contained in $E_{\rho }$.

Thus $E_{\rho }$ has non-empty interior. The maps ${\psi }_1^{\rho },\dots ,{\psi }_k^{\rho }$ are continuous and satisfy the one-slot polynomial condition (A_r) of Definition 2, with degree $d_{\rho }^{\left(r\right)}$, so Lemma 5, applied in one variable, gives $E_{\rho }=\mathcal{A}$. □

Theorem 1.

Let $\mathcal{A}$ be a Banach algebra, $M\subseteq \mathcal{A}$ a closed linear subspace, and let $\Phi =\{{\Phi }_i^{\nu }:{\mathcal{A}}^n\to \mathcal{A}\}$ be an admissible family with countable level set $\mathcal{N}$ and k alternatives. Let $G_1,\dots ,G_n\subseteq \mathcal{A}$ be non-meagre sets and assume the local hypothesis

$$\forall x_1\in G_1,\dots ,\forall x_n\in G_n\exists \nu \in \mathcal{N},\exists i\le k:{\Phi }_i^{\nu }(x_1,\dots ,x_n)\in M.$$

(14)

Then there is a single level $\rho \in \mathcal{N}$ such that

$$\forall x_1,\dots ,x_n\in \mathcal{A}\exists i\le k:{\Phi }_i^{\rho }(x_1,\dots ,x_n)\in M.$$

(15)

Proof. 

The hypothesis (14) restricts the r-th variable to the set $G_r$, for each of the n slots, and it lets the level $\nu$ and the index i depend on the point. We first remove the restrictions on the variables, retaining that dependence, and then remove the dependence of the level on the point.

For the first part, let $\mathrm{S}\left(r\right)$, where $0\le r\le n$, denote the statement that for all $x_1,\dots ,x_r\in \mathcal{A}$ and all $x_{r+1}\in G_{r+1},\dots ,x_n\in G_n$ there are $\nu \in \mathcal{N}$ and $i\le k$ with ${\Phi }_i^{\nu }(x_1,\dots ,x_n)\in M$. In $\mathrm{S}\left(r\right)$ the first r variables are thus unrestricted, while the last $n-r$ still range over the prescribed sets. At one end, $\mathrm{S}\left(0\right)$ restricts every variable and is therefore the hypothesis (14) itself; at the other, $\mathrm{S}\left(n\right)$ restricts none of them and is the assertion (16) below. The passage from $\mathrm{S}(r-1)$ to $\mathrm{S}\left(r\right)$ releases the r-th variable from $G_r$ to the whole of $\mathcal{A}$, and it is this passage that we carry out, for an arbitrary r with $1\le r\le n$.

Assume, then, that $\mathrm{S}(r-1)$ holds, and let $x_1,\dots ,x_{r-1}\in \mathcal{A}$ and $x_{r+1}\in G_{r+1},\dots ,x_n\in G_n$ be arbitrary but fixed. To keep the r-th slot in view we abbreviate

$${\Theta }_i^{\nu }\left(y\right)={\Phi }_i^{\nu }(x_1,\dots ,x_{r-1},y,x_{r+1},\dots ,x_n)(y\in \mathcal{A}).$$

By $\mathrm{S}(r-1)$, every $y\in G_r$ satisfies ${\Theta }_i^{\nu }\left(y\right)\in M$ for some $\nu \in \mathcal{N}$ and some $i\le k$. Since $G_r$ is non-meagre, Proposition 2 applies in the r-th slot and yields one level $\rho \in \mathcal{N}$, which may of course depend on $x_1,\dots ,x_{r-1},x_{r+1},\dots ,x_n$, such that

$${\Theta }_i^{\rho }\left(y\right)\in M\mathrm{for}\mathrm{some}i\le k,\mathrm{for}\mathrm{every}y\in \mathcal{A}.$$

Every $y\in \mathcal{A}$ thus admits a level and an index, and the remaining variables were arbitrary; this is $\mathrm{S}\left(r\right)$. Carrying out the passage for $r=1,\dots ,n$ in turn, we reach $\mathrm{S}\left(n\right)$:

$$\forall x\in {\mathcal{A}}^n\exists \nu \in \mathcal{N},\exists i\le k:{\Phi }_i^{\nu }\left(x\right)\in M.$$

(16)

The level supplied by (16) still varies from point to point, and it is this dependence that must now be removed. For $\nu \in \mathcal{N}$ recall the set introduced in (11),

$$C_{\nu }={\displaystyle \bigcup _{i=1}^k}{\left({\Phi }_i^{\nu }\right)}^{-1}\left(M\right)=\left\{x\in {\mathcal{A}}^n:{\Phi }_i^{\nu }\left(x\right)\in M\mathrm{for}\mathrm{at}\mathrm{least}\mathrm{one}i\le k\right\},$$

which consists of the points at which some alternative holds at the level $\nu$. In this notation (16) states that every point of ${\mathcal{A}}^n$ lies in at least one of these sets:

$${\mathcal{A}}^n=\bigcup _{\nu \in \mathcal{N}}{C}_{\nu }.$$

(17)

What we must produce is a single level $\rho$ for which $C_{\rho }$ is already the whole of ${\mathcal{A}}^n$.

Suppose, to the contrary of what is wanted, that $int{C}_{\nu }=\varnothing$ for every $\nu \in \mathcal{N}$. Each $C_{\nu }$ is closed by (12), so that $\overline{C_{\nu }}=C_{\nu }$ and hence $int\overline{C_{\nu }}=int{C}_{\nu }=\varnothing$; that is, every $C_{\nu }$ is nowhere dense. As $\mathcal{N}$ is countable, (17) would then exhibit ${\mathcal{A}}^n$ as a countable union of nowhere dense sets, so that ${\mathcal{A}}^n$ would be meagre in itself. This is impossible: $\mathcal{A}$ is a Banach space, hence ${\mathcal{A}}^n$ is complete in the maximum norm, and by the Baire category theorem a complete metric space is non-meagre in itself. Consequently

$$int{C}_{\rho }\ne \varnothing \mathrm{for}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{level}\rho \in \mathcal{N}.$$

Fix such a $\rho$ and apply Lemma 5 to the maps ${\Phi }_1^{\rho },\dots ,{\Phi }_k^{\rho }$. They are continuous; they satisfy (A₁), …, (A_n) at the single level $\rho$, with degrees $d_{\rho }^{\left(1\right)},\dots ,d_{\rho }^{\left(n\right)}$; and the set which that lemma associates with them is precisely $C_{\rho }$, whose interior has just been shown to be non-empty. The lemma gives $C_{\rho }={\mathcal{A}}^n$, which says that for every $x\in {\mathcal{A}}^n$ there is an $i\le k$ with ${\Phi }_i^{\rho }\left(x\right)\in M$. □

4. Local Identities and Linearisation at the Unit

Throughout this section $\mathcal{A}$ is a unital complex Banach algebra with identity e, $M\subseteq \mathcal{A}$ is a closed linear subspace and $F:\mathcal{A}\to \mathcal{A}$ is a continuous additive map, so that (10) is available.

Theorem 2.

Let $\mathcal{A}$ be a unital complex Banach algebra, M a closed linear subspace and $F:\mathcal{A}\to \mathcal{A}$ a continuous additive map. Suppose there is a fixed integer $\rho >1$ such that for all $x,y\in \mathcal{A}$

$$F\left({\left(xy\right)}^{\rho }\right)-x^{\rho }{y}^{\rho }\in M\mathit{or}F\left({\left(xy\right)}^{\rho }\right)-y^{\rho }{x}^{\rho }\in M.$$

(18)

Then:

(i)

$F\left(x\right)-x\in M$ for every $x\in \mathcal{A}$;

(ii)

for all $x,y\in \mathcal{A}$, either ${\left(xy\right)}^{\rho }-x^{\rho }{y}^{\rho }\in M$ or ${\left(xy\right)}^{\rho }-y^{\rho }{x}^{\rho }\in M$;

(iii)

$[x,y]\in M$ for all $x,y\in \mathcal{A}$.

Proof. 

(i) Put $y=e$ in (18). Since $e^{\rho }=e$, both alternatives reduce to the single relation

$$F\left(x^{\rho }\right)-x^{\rho }\in M(x\in \mathcal{A}),$$

(19)

so that no selection between them is needed here. Replace x by $e+tx$ in (19), where $t\in \mathbb{R}$. Expanding by (7), and applying F coefficientwise as its $\mathbb{R}$-linearity permits,

$$F\left({(e+tx)}^{\rho }\right)-{(e+tx)}^{\rho }={\displaystyle \sum _{j=0}^{\rho }}\left({\displaystyle \genfrac{}{}{0pt}{}{\rho }{j}}\right)t^j\left(F\left(x^j\right)-x^j\right).$$

The left-hand side lies in M for every real t, by (19); so Lemma 3 places every coefficient on the right in M. The coefficient of $t^1$ is $\rho \left(F\right(x)-x)$, and division by the non-zero scalar $\rho$ gives $F\left(x\right)-x\in M$.

(ii) Let $x,y\in \mathcal{A}$. Applying (i) to the element ${\left(xy\right)}^{\rho }$ gives $F\left({\left(xy\right)}^{\rho }\right)-{\left(xy\right)}^{\rho }\in M$. Subtracting this from whichever of the two memberships in (18) holds for the pair $(x,y)$ removes F altogether and leaves (ii).

(iii) The argument is Yood’s [23], applied here to the identity (ii) in place of the hypothesis. Fix $x,y\in \mathcal{A}$. We substitute $e+tx$ for the first variable of (ii), then for the second, compare the coefficients of $t^1$ in the two resulting relations, and subtract; the term $Q=Q_{\rho }(x,y)$ of (8) occurs in both comparisons and cancels.

Apply (ii) first to the pair $(e+tx,y)$. For each $t\in \mathbb{R}$ at least one of

$${\left((e+tx)y\right)}^{\rho }-{(e+tx)}^{\rho }{y}^{\rho }\in M,{\left((e+tx)y\right)}^{\rho }-y^{\rho }{(e+tx)}^{\rho }\in M$$

holds; as there are two alternatives and infinitely many values of t, one alternative holds for infinitely many t. Both expressions are polynomials in t, and by Corollary 1 their coefficients of $t^1$ are

$$\left(x{y}^{\rho }+Q\right)-\rho x{y}^{\rho }=(1-\rho )x{y}^{\rho }+Q\mathrm{and}\left(x{y}^{\rho }+Q\right)-\rho y^{\rho }x$$

respectively. Lemma 3, applied to the alternative that recurs, therefore yields

$$(1-\rho )x{y}^{\rho }+Q\in M\mathrm{or}x{y}^{\rho }+Q-\rho y^{\rho }x\in M.$$

(20)

Apply (ii) next to the pair $(y,e+tx)$. The two expressions are now ${\left(y(e+tx)\right)}^{\rho }-y^{\rho }{(e+tx)}^{\rho }$ and ${\left(y(e+tx)\right)}^{\rho }-{(e+tx)}^{\rho }{y}^{\rho }$, whose coefficients of $t^1$ are, again by Corollary 1,

$$\left(y^{\rho }x+Q\right)-\rho y^{\rho }x=(1-\rho )y^{\rho }x+Q\mathrm{and}\left(y^{\rho }x+Q\right)-\rho x{y}^{\rho },$$

so that the same reasoning gives

$$(1-\rho )y^{\rho }x+Q\in M\mathrm{or}{y}^{\rho }x+Q-\rho x{y}^{\rho }\in M.$$

(21)

For our pair $(x,y)$ one of the four combinations of an alternative in (20) with an alternative in (21) must hold. In each combination Q occurs in both memberships and cancels when one is subtracted from the other, and M is a linear subspace, so the difference lies in M:

$$\begin{array}{cc}\hfill \left((1-\rho )x{y}^{\rho }+Q\right)-\left((1-\rho )y^{\rho }x+Q\right)& =(1-\rho )[x,y^{\rho }],\hfill \\ \hfill \left((1-\rho )x{y}^{\rho }+Q\right)-\left(y^{\rho }x+Q-\rho x{y}^{\rho }\right)& =[x,y^{\rho }],\hfill \\ \hfill \left(x{y}^{\rho }+Q-\rho y^{\rho }x\right)-\left((1-\rho )y^{\rho }x+Q\right)& =[x,y^{\rho }],\hfill \\ \hfill \left(x{y}^{\rho }+Q-\rho y^{\rho }x\right)-\left(y^{\rho }x+Q-\rho x{y}^{\rho }\right)& =(1+\rho )[x,y^{\rho }].\hfill \end{array}$$

Since $\rho >1$, the real scalars $1-\rho$ and $1+\rho$ are non-zero; this is the one place where the hypothesis $\rho >1$ is used. Dividing by them where they occur, we obtain in all four cases

$$[x,y^{\rho }]\in M(x,y\in \mathcal{A}).$$

(22)

It remains to lower the exponent in (22). Replace y by $e+ty$ there: by (7),

$$\left[x,{(e+ty)}^{\rho }\right]={\displaystyle \sum _{j=0}^{\rho }}\left({\displaystyle \genfrac{}{}{0pt}{}{\rho }{j}}\right)t^j[x,y^j],$$

a polynomial in t with values in M for every $t\in \mathbb{R}$, whose coefficient of $t^1$ is $\rho [x,y]$. Lemma 3 now gives $[x,y]\in M$. □

Theorem 3.

Let $\mathcal{A}$ be a unital complex Banach algebra, M a closed linear subspace and $F:\mathcal{A}\to \mathcal{A}$ continuous and additive. If there is a fixed integer $\rho >1$ such that for every $x\in \mathcal{A}$

$$F\left(x^{\rho }\right)-x^{\rho }\in M\mathit{or}F\left(x^{\rho }\right)+x^{\rho }\in M,$$

then there is $\epsilon \in \{1,-1\}$ with $F\left(x\right)-\epsilon x\in M$ for all $x\in \mathcal{A}$.

Proof. 

Put

$${\Psi }_1\left(x\right)=F\left(x^{\rho }\right)-x^{\rho },{\Psi }_2\left(x\right)=F\left(x^{\rho }\right)+x^{\rho }(x\in \mathcal{A}),$$

so that the hypothesis reads: every $x\in \mathcal{A}$ satisfies ${\Psi }_1\left(x\right)\in M$ or ${\Psi }_2\left(x\right)\in M$. Both maps are continuous, and both satisfy the one-variable condition (A₁) with degree $\rho$; for by Lemma 4 the map $t\mapsto {(u+tw)}^{\rho }$ is a polynomial of degree $\rho$ with leading coefficient $w^{\rho }$, and F, being $\mathbb{R}$-linear, may be applied to it coefficientwise, so that the coefficient of $t^{\rho }$ in $t\mapsto {\Psi }_{1,2}(u+tw)$ is $F\left(w^{\rho }\right)\mp w^{\rho }={\Psi }_{1,2}\left(w\right)$. By Lemma 6, with $n=1$ and $k=2$, one of the two maps takes all of $\mathcal{A}$ into M: there is $\epsilon \in \{1,-1\}$, namely $\epsilon =1$ if that map is ${\Psi }_1$ and $\epsilon =-1$ if it is ${\Psi }_2$, such that

$$F\left(x^{\rho }\right)-\epsilon x^{\rho }\in M(x\in \mathcal{A}).$$

(23)

Replace x by $e+tx$ in (23), where $t\in \mathbb{R}$. Expanding by (7) and applying F coefficientwise,

$$F\left({(e+tx)}^{\rho }\right)-\epsilon {(e+tx)}^{\rho }={\displaystyle \sum _{j=0}^{\rho }}\left({\displaystyle \genfrac{}{}{0pt}{}{\rho }{j}}\right)t^j\left(F\left(x^j\right)-\epsilon x^j\right),$$

a polynomial in t whose values lie in M for every real t. Lemma 3 places each of its coefficients in M; that of $t^1$ is $\rho \left(F\left(x\right)-\epsilon x\right)$, and division by the non-zero scalar $\rho$ gives $F\left(x\right)-\epsilon x\in M$. □

Theorem 4

(Commutator and Jordan type). Let $\mathcal{A}$, M and F be as in Theorem 3 and suppose there are fixed integers $r,s>1$ such that for all $x,y\in \mathcal{A}$

$$F\left([x^r,y^s]\right)-[x^r,y^s]\in M\mathit{or}F\left([x^r,y^s]\right)+[x^r,y^s]\in M.$$

Then there is $\epsilon \in \{1,-1\}$ with

$$F\left([x,y]\right)-\epsilon [x,y]\in M(x,y\in \mathcal{A}).$$

If the commutator is replaced by the Jordan product in the hypothesis, so that $F(x^r\circ y^s)-x^r\circ y^s\in M$ or $F(x^r\circ y^s)+x^r\circ y^s\in M$ for all $x,y$, then in the same way there is $\epsilon \in \{1,-1\}$ with $F(x\circ y)-\epsilon (x\circ y)\in M$ for all $x,y\in \mathcal{A}$.

Proof. 

Put

$${\Psi }_1(x,y)=F\left([x^r,y^s]\right)-[x^r,y^s],{\Psi }_2(x,y)=F\left([x^r,y^s]\right)+[x^r,y^s],$$

so that the hypothesis reads: every pair $(x,y)$ satisfies ${\Psi }_1(x,y)\in M$ or ${\Psi }_2(x,y)\in M$. Both maps are continuous, and they satisfy (A₁) and (A₂) with degrees r and s. Indeed, by Lemma 4 the map $t\mapsto {(u+tw)}^r$ is a polynomial of degree r with leading coefficient $w^r$, so that $t\mapsto [{(u+tw)}^r,y^s]$ has $[w^r,y^s]$ as its coefficient of $t^r$; applying the $\mathbb{R}$-linear map F coefficientwise, the coefficient of $t^r$ in $t\mapsto {\Psi }_{1,2}(u+tw,y)$ is $F\left([w^r,y^s]\right)\mp [w^r,y^s]={\Psi }_{1,2}(w,y)$, which is (A₁); and (A₂) follows in the same way from the second variable. Lemma 6, with $n=2$ and $k=2$, therefore singles out one of the two maps: there is $\epsilon \in \{1,-1\}$ with

$$F\left([x^r,y^s]\right)-\epsilon [x^r,y^s]\in M(x,y\in \mathcal{A}).$$

(24)

The exponents are now lowered one variable at a time. Replace x by $e+tx$ in (24), where $t\in \mathbb{R}$. By (7),

$$\left[{(e+tx)}^r,y^s\right]={\displaystyle \sum _{j=0}^r}\left({\displaystyle \genfrac{}{}{0pt}{}{r}{j}}\right)t^j[x^j,y^s],$$

so that the left-hand side of (24) becomes, F being $\mathbb{R}$-linear,

$${\displaystyle \sum _{j=0}^r}\left({\displaystyle \genfrac{}{}{0pt}{}{r}{j}}\right)t^j\left(F\left([x^j,y^s]\right)-\epsilon [x^j,y^s]\right),$$

a polynomial in t with values in M for every real t. Its coefficient of $t^1$ is $r\left(F\left([x,y^s]\right)-\epsilon [x,y^s]\right)$, so Lemma 3 and division by r give

$$F\left([x,y^s]\right)-\epsilon [x,y^s]\in M(x,y\in \mathcal{A}).$$

Replacing y by $e+ty$ in this relation and repeating the computation in the second variable, the coefficient of $t^1$ is now $s\left(F\left(\right[x,y\left]\right)-\epsilon [x,y]\right)$, and division by s yields $F\left(\right[x,y\left]\right)-\epsilon [x,y]\in M$ for all $x,y\in \mathcal{A}$.

For the Jordan product the argument is word for word the same, because ${(e+tx)}^r\circ y^s={\sum }_j\left({\displaystyle \genfrac{}{}{0pt}{}{r}{j}}\right)t^j\left(x^j\circ y^s\right)$, and likewise in the second variable. □

The theorems that follow appeal to the transfer principle in two variables, in which form we record it here.

Corollary 2.

Let $\mathcal{A}$ be a Banach algebra, $M\subseteq \mathcal{A}$ a closed linear subspace and $\{{\Phi }_i^{\nu }:\mathcal{A}\times \mathcal{A}\to \mathcal{A}\}$ an admissible family of two variables with countable level set $\mathcal{N}$ and k alternatives. If $G_1,G_2\subseteq \mathcal{A}$ are non-meagre and for each $x\in G_1$ and $y\in G_2$ there are $\nu \in \mathcal{N}$ and $i\le k$ with ${\Phi }_i^{\nu }(x,y)\in M$, then there is $\rho \in \mathcal{N}$ such that for all $x,y\in \mathcal{A}$ some $i\le k$ satisfies ${\Phi }_i^{\rho }(x,y)\in M$.

Proof. 

This is Theorem 1 for $n=2$, with $G_1$ and $G_2$ as the two non-meagre sets. □

Theorem 5.

Let $\mathcal{A}$ be a unital complex Banach algebra, M a closed linear subspace of $\mathcal{A}$, $F:\mathcal{A}\to \mathcal{A}$ a continuous additive map, and $G_1,G_2$ non-meagre subsets of $\mathcal{A}$. Assume that for all $x\in G_1$ and $y\in G_2$ there is an integer $m=m(x,y)>1$ with

$$F\left({\left(xy\right)}^m\right)-x^m{y}^m\in M\mathit{or}F\left({\left(xy\right)}^m\right)-y^m{x}^m\in M.$$

(25)

Then

(i)

$F\left(x\right)-x\in M$ for all $x\in \mathcal{A}$;

(ii)

$[x,y]\in M$ for all $x,y\in \mathcal{A}$.

If moreover $M=Z\left(\mathcal{A}\right)$ and $\mathcal{A}$ is semiprime, then $\mathcal{A}$ is commutative.

Proof. 

Consider the two-variable family

$${\Phi }_1^m(x,y)=F\left({\left(xy\right)}^m\right)-x^m{y}^m,{\Phi }_2^m(x,y)=F\left({\left(xy\right)}^m\right)-y^m{x}^m(m\ge 2),$$

(26)

whose level set $\mathcal{N}=\{m\in \mathbb{N}:m>1\}$ is countable and which has $k=2$ alternatives. We claim that it is admissible, with $d_m^{\left(1\right)}=d_m^{\left(2\right)}=m$. By Lemma 4 the map $t\mapsto {\left((u+tw)y\right)}^m={(uy+twy)}^m$ is a polynomial of degree m whose leading coefficient is ${\left(wy\right)}^m$, and $t\mapsto {(u+tw)}^m{y}^m$ is one whose leading coefficient is $w^m{y}^m$; since F is $\mathbb{R}$-linear it may be applied to such polynomials coefficientwise. Hence the coefficient of $t^m$ in $t\mapsto {\Phi }_1^m(u+tw,y)$ is

$$F\left({\left(wy\right)}^m\right)-w^m{y}^m={\Phi }_1^m(w,y),$$

which is (A₁); the same computation in the second variable gives (A₂), and ${\Phi }_2^m$ is treated identically.

The hypothesis (25) is exactly the local hypothesis (14) for this family, with the non-meagre sets $G_1$ and $G_2$. Corollary 2 therefore supplies a single level $\rho >1$ such that, for all $x,y\in \mathcal{A}$,

$$F\left({\left(xy\right)}^{\rho }\right)-x^{\rho }{y}^{\rho }\in M\mathrm{or}F\left({\left(xy\right)}^{\rho }\right)-y^{\rho }{x}^{\rho }\in M,$$

which is the hypothesis (18) of Theorem 2. That theorem gives (i) and (ii).

Finally, let $M=Z\left(\mathcal{A}\right)$, a closed linear subspace of $\mathcal{A}$, and let $\mathcal{A}$ be semiprime. Then (ii) reads $[\mathcal{A},\mathcal{A}]\subseteq Z\left(\mathcal{A}\right)$, so every commutator of $\mathcal{A}$ is central, and Lemma 1, applied to $\mathcal{A}$ as a ring, gives the commutativity of $\mathcal{A}$. □

When M is a closed two-sided ideal, the local hypothesis admits a description that involves neither the sets $G_1,G_2$ nor the exponent.

Theorem 6. 

Let $\mathcal{A}$ be a unital complex Banach algebra, M a closed two-sided ideal of $\mathcal{A}$, and $F:\mathcal{A}\to \mathcal{A}$ a continuous additive map. The following are equivalent.

(i)

There are non-meagre sets $G_1,G_2\subseteq \mathcal{A}$ such that for each $x\in G_1$ and $y\in G_2$ there is an integer $m=m(x,y)>1$ with $F\left({\left(xy\right)}^m\right)-x^m{y}^m\in M$ or $F\left({\left(xy\right)}^m\right)-y^m{x}^m\in M$.

(ii)

$F\left({\left(xy\right)}^m\right)-x^m{y}^m\in M$ and $F\left({\left(xy\right)}^m\right)-y^m{x}^m\in M$ for all $x,y\in \mathcal{A}$ and all integers $m\ge 1$.

(iii)

$[\mathcal{A},\mathcal{A}]\subseteq M$ and $F\left(x\right)-x\in M$ for all $x\in \mathcal{A}$; equivalently, the quotient Banach algebra $\mathcal{A}/M$ is commutative and F induces on it the identity map.

Proof. 

(i)⇒(iii). This is Theorem 5, whose conclusions (ii) and (i) are the two inclusions in (iii). As to the reformulation, suppose $F\left(x\right)-x\in M$ for every x. If $x\in M$ then $F\left(x\right)=x+\left(F\left(x\right)-x\right)\in M$, so $F\left(M\right)\subseteq M$ and F induces an additive map $\overline{F}$ on $\mathcal{A}/M$; and $\overline{F}={\mathrm{id}}_{\mathcal{A}/M}$ says precisely that $F-{\mathrm{id}}_{\mathcal{A}}$ maps $\mathcal{A}$ into M. Likewise $\mathcal{A}/M$ is commutative if and only if $[\mathcal{A},\mathcal{A}]\subseteq M$.

(iii)⇒(ii). Since M is a closed two-sided ideal, $\mathcal{A}/M$ is a Banach algebra and the quotient map $\pi :\mathcal{A}\to \mathcal{A}/M$ is a continuous homomorphism with $ker\pi =M$; see [35]. Let $x,y\in \mathcal{A}$ and $m\ge 1$. Because $F-{\mathrm{id}}_{\mathcal{A}}$ maps $\mathcal{A}$ into M we may replace $F\left({\left(xy\right)}^m\right)$ by ${\left(xy\right)}^m$ under $\pi$, and because $\mathcal{A}/M$ is commutative the images of x and y commute; hence

$$\pi \left(F\left({\left(xy\right)}^m\right)-x^m{y}^m\right)=\pi \left({\left(xy\right)}^m\right)-\pi {\left(x\right)}^m\pi {\left(y\right)}^m={\left(\pi \left(x\right)\pi \left(y\right)\right)}^m-\pi {\left(x\right)}^m\pi {\left(y\right)}^m=0.$$

Thus $F\left({\left(xy\right)}^m\right)-x^m{y}^m\in ker\pi =M$, and the same computation with the factors interchanged gives $F\left({\left(xy\right)}^m\right)-y^m{x}^m\in M$. Nothing in this computation excludes the value $m=1$, so (iii) yields the case $m=1$ of (ii) as well, although that case is excluded from the local hypothesis (i); this is why (ii) may be stated for every $m\ge 1$.

(ii)⇒(i). Take $G_1=G_2=\mathcal{A}$ and $m=2$; the whole of $\mathcal{A}$ is non-meagre in itself by the Baire category theorem. □

Corollary 3. 

Let $\mathcal{A}$ be a unital complex Banach algebra and F a continuous additive map satisfying the hypothesis of Theorem 5 with $M=rad\mathcal{A}$. Then $\mathcal{A}/rad\mathcal{A}$ is commutative and F induces the identity map on it. If $\mathcal{A}$ is semisimple, then $F={\mathrm{id}}_{\mathcal{A}}$ and $\mathcal{A}$ is commutative.

Proof. 

The Jacobson radical is a closed two-sided ideal, so Theorem 6 applies and its statement (iii) holds. If $\mathcal{A}$ is semisimple, that is $rad\mathcal{A}=\left(0\right)$, then (iii) reads $[\mathcal{A},\mathcal{A}]=\left(0\right)$ and $F-{\mathrm{id}}_{\mathcal{A}}=0$. □

Corollary 4. 

Let $\mathcal{A}$ be a unital $C^{*}$-algebra and $F:\mathcal{A}\to \mathcal{A}$ a continuous additive map for which there are non-meagre sets $G_1,G_2\subseteq \mathcal{A}$ such that for each $x\in G_1$ and $y\in G_2$ some integer $m>1$ satisfies

$$F\left({\left(xy\right)}^m\right)=x^m{y}^m\mathit{or}F\left({\left(xy\right)}^m\right)=y^m{x}^m.$$

Then $F={\mathrm{id}}_{\mathcal{A}}$, the algebra $\mathcal{A}$ is commutative, and $\mathcal{A}$ is isometrically ∗-isomorphic to $C(\Omega )$ for some compact Hausdorff space Ω.

Proof. 

The zero subspace is closed, and the two displayed equalities say exactly that the corresponding differences lie in $M=\left(0\right)$. Theorem 5 therefore applies and gives $F\left(x\right)-x=0$ for all x and $[x,y]=0$ for all $x,y$. A commutative unital $C^{*}$-algebra is isometrically ∗-isomorphic to $C(\Omega )$, with $\Omega$ its maximal ideal space, by the commutative Gelfand–Naimark theorem [36]. □

Theorem 6 does not cover the case $M=Z\left(\mathcal{A}\right)$, since the centre is not, in general, an ideal. For semiprime algebras the following takes its place.

Corollary 5. 

Let $\mathcal{A}$ be a unital semiprime complex Banach algebra and $F:\mathcal{A}\to \mathcal{A}$ a continuous additive map. Then $\mathcal{A}$ is commutative if and only if there are non-meagre sets $G_1,G_2\subseteq \mathcal{A}$ such that for each $x\in G_1$ and $y\in G_2$ some integer $m=m(x,y)>1$ satisfies

$$F\left({\left(xy\right)}^m\right)-x^m{y}^m\in Z\left(\mathcal{A}\right)\mathit{or}F\left({\left(xy\right)}^m\right)-y^m{x}^m\in Z\left(\mathcal{A}\right).$$

Proof. 

If such sets exist, $\mathcal{A}$ is commutative by the last statement of Theorem 5. Conversely, let $\mathcal{A}$ be commutative. Then $Z\left(\mathcal{A}\right)=\mathcal{A}$, so both memberships hold trivially for every F, every $m>1$ and every pair $(x,y)$, and one may take $G_1=G_2=\mathcal{A}$, which are non-meagre by the Baire category theorem. □

5. Consequences for Derivations and Related Maps

In this section $\mathcal{A}$ is a unital complex Banach algebra. Each of the maps recorded in Definition 1 is additive, so Theorem 5 applies to it as soon as it is continuous.

The first corollary is the case $F={\mathrm{id}}_{\mathcal{A}}$, in which conclusion (i) of Theorem 5 is vacuous; open sets may be replaced by non-meagre ones because a non-empty open subset of a Banach space is non-meagre.

Corollary 6 

(Yood [23] (Theorem 1)). Let $\mathcal{A}$ be a unital complex Banach algebra and M a closed linear subspace of $\mathcal{A}$. Suppose there are non-empty open sets $G_1,G_2\subseteq \mathcal{A}$ such that for each $x\in G_1$ and $y\in G_2$ there is an integer $n=n(x,y)>1$ with

$${\left(xy\right)}^n-x^n{y}^n\in M\mathit{or}{\left(xy\right)}^n-y^n{x}^n\in M.$$

Then $[x,y]\in M$ for all $x,y\in \mathcal{A}$. Moreover, the sets $G_1$ and $G_2$ need only be non-meagre.

Proof. 

The identity map ${\mathrm{id}}_{\mathcal{A}}$ is continuous and additive, and with $F={\mathrm{id}}_{\mathcal{A}}$ the two memberships in the hypothesis are precisely the two memberships (25) of Theorem 5. Non-empty open subsets of a Banach space are non-meagre by the Baire category theorem, so that theorem applies with $G_1$ and $G_2$ as they stand, and its conclusion (ii) is the assertion $[x,y]\in M$ for all $x,y\in \mathcal{A}$; its conclusion (i) reads ${\mathrm{id}}_{\mathcal{A}}\left(x\right)-x=0\in M$ and says nothing. Since the only property of $G_1$ and $G_2$ used is that they are non-meagre, the hypothesis may be imposed on non-meagre sets from the outset. □

The next corollary is Theorem 5 with $M=Z\left(\mathcal{A}\right)$, taken first with $F=d$ and then with $F=f$.

Corollary 7 

(Ali–Khan [25] (Theorem 1.2), Ashraf–Wani [26] (Theorem 3.1)). Let $\mathcal{A}$ be a unital semiprime complex Banach algebra and let F be either a continuous derivation d of $\mathcal{A}$ or a continuous generalised derivation f of $\mathcal{A}$ associated with a derivation d. Suppose there are non-meagre sets $G_1,G_2\subseteq \mathcal{A}$ such that for each $x\in G_1$ and $y\in G_2$ there is an integer $m=m(x,y)>1$ with

$$F\left({\left(xy\right)}^m\right)-x^m{y}^m\in Z\left(\mathcal{A}\right)\mathit{or}F\left({\left(xy\right)}^m\right)-y^m{x}^m\in Z\left(\mathcal{A}\right).$$

Then $\mathcal{A}$ is commutative. The conclusion uses neither the primeness of $\mathcal{A}$, nor $d\ne 0$, nor $f\ne 0$, nor the hypothesis $d\left(Z\right(\mathcal{A}\left)\right)\ne 0$ of [26] (Theorem 3.1), and the open sets of [25,26] may be replaced by non-meagre ones.

Proof. 

A derivation and a generalised derivation of $\mathcal{A}$ are additive by Definition 1, so a continuous one is a continuous additive map and is admitted as the map F of Theorem 5. The centre $Z\left(\mathcal{A}\right)$ is a closed linear subspace of $\mathcal{A}$, and with $M=Z\left(\mathcal{A}\right)$ the displayed hypothesis is the hypothesis (25) of that theorem. Its last assertion, $\mathcal{A}$ being semiprime, gives the commutativity of $\mathcal{A}$; the argument is the same for $F=d$ and for $F=f$. No property of F beyond continuity and additivity, and no property of $G_1,G_2$ beyond non-meagreness, enters at any point, whence the hypotheses listed in the statement are not needed and non-empty open sets, being non-meagre, are covered as a special case. □

The hypothesis $d\left(Z\right(\mathcal{A}\left)\right)\ne 0$ is a severe restriction on $\mathcal{A}$ rather than a technical convenience. A derivation of a unital algebra annihilates e by Lemma 2, so $d\left(Z\right(\mathcal{A}\left)\right)\ne 0$ forces $Z\left(\mathcal{A}\right)$ to be strictly larger than $\mathbb{C}e$; by [37] (Corollary 2.4.5) no unital primitive Banach algebra has that property, and in particular neither $M_n\left(\mathbb{C}\right)$ nor $B\left(H\right)$ does.

Since Theorem 5 constrains F only through continuity and additivity, the following list is immediate.

Corollary 8. 

Let $\mathcal{A}$ be a unital semiprime complex Banach algebra and let $F:\mathcal{A}\to \mathcal{A}$ be any continuous additive map satisfying the hypothesis of Theorem 5 with $M=Z\left(\mathcal{A}\right)$. Then $\mathcal{A}$ is commutative. In particular this applies to every continuous map recorded in Definition 1, to the components $d_n$$(n\ge 1)$ of a continuous higher derivation, to right multipliers as well as left ones, and to arbitrary bounded linear operators on $\mathcal{A}$.

Proof. 

Every map named is additive: for the maps recorded in Definition 1 additivity is part of the definition, a right multiplier is additive for the same reason as a left one, and a bounded linear operator is additive by definition. A continuous map from the list is therefore a continuous additive map $F:\mathcal{A}\to \mathcal{A}$, and $Z\left(\mathcal{A}\right)$ is a closed linear subspace of $\mathcal{A}$; Theorem 5 applies with $M=Z\left(\mathcal{A}\right)$ and its last assertion, $\mathcal{A}$ being semiprime, gives the commutativity of $\mathcal{A}$. □

Continuity is required in each of these cases whenever it is not automatic, since Theorem 5, which governs them, applies to continuous additive maps. For some of the maps it is automatic. Every derivation of a semisimple Banach algebra is continuous by the theorem of Johnson and Sinclair [18], and higher derivations are continuous under the hypotheses of Loy [38] and Jewell [39]. For the remaining maps continuity is part of the hypothesis of the corollary rather than a consequence of it.

For a map that annihilates the identity, the conclusion $F\left(x\right)-x\in M$ of Theorem 5 is not extra information but an obstruction.

Corollary 9. 

Let $\mathcal{A}\ne \left(0\right)$ be a unital complex Banach algebra, let M be a proper closed two-sided ideal of $\mathcal{A}$, and let $F:\mathcal{A}\to \mathcal{A}$ be a continuous additive map with $F\left(e\right)\in M$. Then there are no non-meagre sets $G_1,G_2\subseteq \mathcal{A}$ satisfying the hypothesis (25) of Theorem 5. In particular, no such sets exist when F is a continuous derivation, α-derivation or Jordan derivation of $\mathcal{A}$, or the component $d_n$$(n\ge 1)$ of a continuous higher derivation of $\mathcal{A}$; and M may be taken to be $rad\mathcal{A}$.

Proof. 

Suppose such sets $G_1,G_2$ existed. Theorem 5(i) would give $F\left(x\right)-x\in M$ for every $x\in \mathcal{A}$, and in particular

$$F\left(e\right)-e\in M.$$

Since $F\left(e\right)\in M$ and M is a linear subspace, $e\in M$; but a two-sided ideal containing the identity is the whole algebra, so $M=\mathcal{A}$, contradicting the properness of M. For each of the maps listed, $F\left(e\right)=0\in M$ by Lemma 2. Finally, in a non-zero unital Banach algebra the identity is not quasi-regular, so $e\notin rad\mathcal{A}$ and $rad\mathcal{A}$ is a proper closed two-sided ideal. □

6. Identities of Multiplicative Type

The right-hand side of the local identity may also be built from F itself. The transfer principle and the selection lemma apply to such identities unchanged; what the linearisation then yields is a partial structural result, the case $F\left(e\right)=e$ remaining open.

Proposition 3. 

Let $\mathcal{A}$ be a unital complex Banach algebra, M a closed linear subspace, $F:\mathcal{A}\to \mathcal{A}$ continuous and additive, and $G_1,G_2$ non-meagre subsets of $\mathcal{A}$. Suppose that for all $x\in G_1$ and $y\in G_2$ there is an integer $m=m(x,y)>1$ with

$$F\left({\left(xy\right)}^m\right)-F\left(x^m\right)F\left(y^m\right)\in M\mathit{or}F\left({\left(xy\right)}^m\right)-F\left(y^m\right)F\left(x^m\right)\in M,$$

and put $q=F\left(e\right)$. Then

$$q-q^2\in M,F\left(x\right)(e-q)\in M\mathit{and}(e-q)F\left(x\right)\in M(x\in \mathcal{A}).$$

In particular, if $F\left(e\right)=0$ then $F\left(\mathcal{A}\right)\subseteq M$; so $F\left(\mathcal{A}\right)\subseteq Z\left(\mathcal{A}\right)$ when $M=Z\left(\mathcal{A}\right)$, and $F=0$ when $M=\left(0\right)$.

Proof. 

Put

$${\Phi }_1^m(x,y)=F\left({\left(xy\right)}^m\right)-F\left(x^m\right)F\left(y^m\right),{\Phi }_2^m(x,y)=F\left({\left(xy\right)}^m\right)-F\left(y^m\right)F\left(x^m\right)$$

for $m\ge 2$. This family is admissible with $d_m^{\left(1\right)}=d_m^{\left(2\right)}=m$: by Lemma 4 and the $\mathbb{R}$-linearity of F, the map $t\mapsto F\left({(u+tw)}^m\right)$ is a polynomial of degree at most m with $F\left(w^m\right)$ as its coefficient of $t^m$, and multiplication by the fixed element $F\left(y^m\right)$ is linear, so the coefficient of $t^m$ in $t\mapsto {\Phi }_1^m(u+tw,y)$ is $F\left({\left(wy\right)}^m\right)-F\left(w^m\right)F\left(y^m\right)={\Phi }_1^m(w,y)$; this is (A₁), and (A₂) follows in the same way.

Corollary 2 now yields a fixed level $\rho >1$ at which, for every pair $(x,y)$, one of the two alternatives holds. The maps ${\Phi }_1^{\rho }$ and ${\Phi }_2^{\rho }$ satisfy (A₁) and (A₂) at that single level, so Lemma 6 makes the alternative uniform. Assume first that it is the first one:

$$F\left({\left(xy\right)}^{\rho }\right)-F\left(x^{\rho }\right)F\left(y^{\rho }\right)\in M(x,y\in \mathcal{A}).$$

(27)

Setting $x=y=e$ in (27) gives $q-q^2\in M$. Setting only $y=e$ gives

$$F\left(x^{\rho }\right)-F\left(x^{\rho }\right)q=F\left(x^{\rho }\right)(e-q)\in M(x\in \mathcal{A}).$$

Replace x by $e+tx$ in this relation, with $t\in \mathbb{R}$. By (7), and because both F and right multiplication by $e-q$ are $\mathbb{R}$-linear,

$$F\left({(e+tx)}^{\rho }\right)(e-q)={\displaystyle \sum _{j=0}^{\rho }}\left({\displaystyle \genfrac{}{}{0pt}{}{\rho }{j}}\right)t^{j}F\left(x^j\right)(e-q),$$

a polynomial in t with values in M; its coefficient of $t^1$ is $\rho F\left(x\right)(e-q)$, so Lemma 3 and division by $\rho$ give $F\left(x\right)(e-q)\in M$. Setting $x=e$ in (27) gives $(e-q)F\left(y^{\rho }\right)\in M$, and the same substitution in the second variable gives $(e-q)F\left(y\right)\in M$ for all $y\in \mathcal{A}$.

If instead the uniform alternative is the second one, the two substitutions above interchange left and right and produce the same pair of conclusions. Finally, if $q=0$ then $F\left(x\right)(e-q)=F\left(x\right)$, so that $F\left(\mathcal{A}\right)\subseteq M$; the two special cases follow on taking $M=Z\left(\mathcal{A}\right)$ and $M=\left(0\right)$. □

When M is a closed ideal and $q-e\in M$, the three conclusions of Proposition 3 hold automatically: the memberships $F\left(x\right)(e-q)\in M$ and $(e-q)F\left(x\right)\in M$ follow from $e-q\in M$ because M absorbs multiplication by elements of $\mathcal{A}$, and $q-q^2=q(e-q)\in M$ for the same reason. For a general closed subspace M this argument is not available, since $a\in M$ does not imply $a^2\in M$ there. In either case it is presently unclear whether the hypothesis imposes any further restriction on F in this case.

Problem 1. 

Determine the continuous additive maps F with $F\left(e\right)-e\in M$ satisfying the hypothesis of Proposition 3.

7. Conclusions

Theorem 1, the transfer principle, is the analytic core of the paper. It takes an identity assumed only on a non-meagre set, and only at a level that varies from point to point, and returns it on all of $\mathcal{A}$ at a single level. Nothing multiplicative about the maps is used; an arbitrary countable family of continuous maps of several variables satisfying the two polynomial conditions of Definition 2 suffices. Lemma 6 supplies the passage from a pointwise alternative to a uniform one.

For product-type identities this and the linearisation at e settle the matter. A continuous additive map obeying the local hypothesis differs from the identity by a map into M, and $[\mathcal{A},\mathcal{A}]\subseteq M$ (Theorem 5); when M is a closed two-sided ideal the hypothesis becomes an equivalent description of $\mathcal{A}/M$ (Theorem 6). Corollaries 6 and 7 recover the theorems of Yood, of Ali and Khan and of Ashraf and Wani without their primeness and non-degeneracy hypotheses, and Corollary 8 extends the conclusion to skew, $(\alpha ,\beta )$-, b-generalised, Jordan, Jordan triple and Lie derivations, to the components of higher derivations, and to multipliers, elementary operators and bounded linear operators. Corollary 9 runs the other way: a continuous additive map annihilating e, a derivation for instance, cannot satisfy the hypothesis modulo a proper closed two-sided ideal. Corollaries 3 and 4 specialise to the Jacobson radical and to unital $C^{*}$-algebras, where $M=\left(0\right)$ makes $\mathcal{A}$ commutative and hence isometrically ∗-isomorphic to an algebra of continuous functions.

Multiplicative identities are not settled. Proposition 3 determines F only up to $q=F\left(e\right)$, idempotent modulo M, and the case $q-e\in M$ is Problem 1. Theorems 3 and 4 carry power-type and commutator-type identities as far as the analytic argument goes; what remains is ring-theoretic and is taken up in a separate work in preparation that begins from Section 4.