A multiplicative Banach-Stone theorem
Kristopher Lee
Abstract
The Banach-Stone theorem states that any surjective, linear mapping T between spaces of
continuous functions that satisfies
kT (f ) − T (g)k = kf − gk,
where k · k denotes the uniform norm, is a weighted composition operator. We study a multiplicative analogue, and demonstrate that a surjective mapping T , not necessarily linear, between
algebras of continuous functions with
kT (f )T (g)k = kf gk
must be a composition operator in modulus.
1
Introduction
As stated by Weyl [11, p.144], when studying a mathematical object it is worthwhile to characterize
mappings that preserve any structural relations. The collection C(X) of complex-valued continuous
functions on a compact Hausdorff space X has many algebraic and topological structures; in particular, when equipped with the uniform norm k · k, the set C(X) is a normed vector space. It is then
advantageous to analyze the mappings that leave this structure undisturbed, and the celebrated
Banach-Stone theorem [1, 10] states that any surjective linear isometry between spaces of continuous functions must be a weighted composition operator. More precisely, if T : C(X) → C(Y ) is
surjective, linear, and
kT (f ) − T (g)k = kf − gk
holds for all f, g ∈ C(X), then T (1) is unimodular on Y and there exists a homeomorphism
ψ : Y → X such that
T (f ) = T (1) · (f ◦ ψ).
This classic result has since been generalized to mappings between certain linear subspaces, see [2,
Chapter 2] for a thorough discussion.
One striking aspect of the Banach-Stone theorem is that is reveals how the vector space structure
of C(X) intertwines with the other structures. For example, the mapping f 7→ T (1)T (f ) is an
algebra isomorphism, and thus the multiplicative structure of C(X) is preserved, even though the
assumptions have no clear connection to point-wise multiplication of functions. A natural question
to ask now is if the multiplicative structure of C(X) itself determines any of the other structures.
Recently, there has been much work regarding mappings that leave multiplicative proprieties
invariant under their action. For example, Molnár [9] demonstrated that given a first-countable,
compact Hausdorff space X and a surjective, not necessarily linear, mapping T : C(X) → C(X) that
1
satisfies σ(T (f )T (g)) = σ(f g), where σ(·) denotes the spectrum, then T (1)2 = 1 and there exists
a homeomorphism ψ : X → X such that T (f ) = T (1) · (f ◦ ψ), i.e. T is a weighted composition
operator. This work has attracted much attention, as its conclusion is reminiscent to that of
Banach-Stone, but the linearity of T was not included as an assumption. In light of this, Molnár’s
result has inspired a wave of research into similar problems, and it is now considered the inaugural
spectral preserver problem. See [3] and the references therein for a general survey on such problems.
Molnár’s assumption of σ(T (f )T (g)) = σ(f g) implies that
kT (f )T (g)k = kf gk,
(1.1)
and such mappings have come to be known as norm multiplicative. Analyzing such mappings is
the first step in some spectral preserver problems; however, these mappings are of interest in their
own right. The condition (1.1) is similar to the assumption of isometry, with subtraction replaced
by multiplication, and thus characterizing such mappings can be seen as a multiplicative analogue
to the Banach-Stone theorem. It is worth noting that norm multiplicative mappings need not be
linear (for example, T (f ) = f ), and any linear T satisfying (1.1) is automatically an isometry, and
thus Banach-Stone would apply. Consequently, norm multiplicative mappings are not assumed to
be linear.
Mappings T : A → B between subalgebras A ⊂ C(X) and B ⊂ C(Y ) that satisfy (1.1) have
been studied for a variety of settings, such as Lipschitz algebras [5, Section 3], uniform algebras
[7, Theorem 1], and real function algebras [8, Proposition 3.1]. In each case, it is shown that such
mappings are composition operators in modulus. This is to say that
|T (f )| = |f ◦ ψ|,
(1.2)
where ψ : Ch(B) → Ch(A) is a homeomorphism between the Choquet boundaries, i.e. the points
x ∈ X such that the point-evaluation mapping ϕx : A → C defined by ϕx (f ) = f (x) is an extreme
point of the unit ball of the dual space of A.
As the same conclusion is made, it is natural to consider that there may be a general theorem
that envelops the previous work; however the arguments used each situation relies on the fact that
x ∈ Ch(A) if and only if given an open neighborhood U of x, there exists an f ∈ A such that
kf k = |f (x)| = 1 and |f | < 1 on X \ U . For a general subalgebra A, this characterization may
fail to be true (see [4] for such an example), and so the usual techniques cannot be used in a wider
setting.
In this work, we give a novel approach for characterizing norm multiplicative mappings between
general subalgebras that serves as an all encompassing technique. In particular, the conclusion that
(1.1) implies (1.2) is strengthened, as the homeomorphism ψ is constructed on the larger Shilov
boundary, i.e. points x ∈ X such that given an open neighborhood U of x, there exists an f such
that kf k = 1 and |f | < 1 on X \ U , and this is done using topological divisors of zero, which are
functions f ∈ A such that there exists a normalized sequence {hn } ⊂ A such that kf hn k → 0. We
begin in Section 2 with the basic notation and results that will be needed throughout, and then in
Section 3 we prove the following:
Main Theorem. Let X and Y be compact Hausdorff spaces, and let A ⊂ C(X) and B ⊂ C(Y )
be complex subalgebras that contain the constants and separate points. If T : A → B is a surjective
mapping, not necessarily linear, such that
kT (f )T (g)k = kf gk
holds for all f, g ∈ A, then there exists a homeomorphism ψ : ∂B → ∂A between the Shilov boundaries such that
|T (f )(y)| = |f (ψ(y))|
for all f ∈ A and all y ∈ ∂B.
2
Notation and Preliminary Results
Throughout this section, X denotes a compact Hausdorff space, C(X) denotes the collection of
complex-valued continuous functions on X, and A ⊂ C(X) is a complex subalgebra that contains
the constants and separates points, i.e. αf + βg ∈ A and f g ∈ A for all α, β ∈ C and f, g ∈ A,
1 ∈ A, and given distinct x, y ∈ X, there exists an f ∈ A such that f (x) 6= f (y). For x ∈ X, define
Ix (A) = {f ∈ A : f (x) = 0}.
As A separates points and 1 ∈ A, then Ix (A) ⊂ Iy (A) implies that x = y.
Given f ∈ A, the maximizing set of f is the non-empty, compact set
M (f ) = {x ∈ X : |f (x)| = kf k}.
The unit sphere of A is the set S(A) = {f ∈ A : kf k = 1}. A non-empty closed subset B ⊂ X is a
closed boundary for A if M (f ) ∩ B 6= ∅ for all f ∈ S(A). As A separates points, it must be that
the intersection of all closed boundaries for A is again a closed boundary for A [6, Theorem 3.3.2],
and this boundary is known as the Shilov boundary, which is denoted by ∂A. A point x belongs
to ∂A if and only if given an open set U of x, there exists an f ∈ S(A) such that M (f ) ⊂ U [6,
Corollary 3.3.4].
For f1 , . . . , fn ∈ A, define


n
X

d(f1 , . . . , fn ) = inf
kfj hk : h ∈ S(A) .


j=1
A non-empty subset J ⊂ A consists of joint topological divisors of zero (which will henceforth
be abbreviated to JTDZ) if given any finite collection f1 , . . . , fn ∈ J , then d(f1 , . . . , fn ) = 0.
Equivalently, J ⊂ A consists
Pn of JTDZ if and only if given f1 , . . . , fn ∈ J , there exists a sequence
{hn } ⊂ S(A) such that j=1 kfj hn k → 0.
There is a deep connection between subsets J that consist of JTDZ and the Shilov boundary.
For any x ∈ ∂A, the set Ix (A) consists of JTDZ [6, Theorem 3.4.10], and conversely, any set of
JTDZ is contained in an Ix (A), for some x ∈ ∂A.
Lemma 2.1. Let J ⊂ A consist of JTDZ. Then there exists an x ∈ ∂A such that J ⊂ Ix (A).
Proof. For each f ∈ J , denote the zero set of f by Z(f ) = {x ∈ X : f (x) = 0}. We will prove that
the family of closed subsets {Z(f ) ∩ ∂A : f ∈ J } has the finite intersection property, which yields
the desired result.
P
Indeed, let f1 , . . . , fn ∈ J and set g = nj=1 |fj |. Suppose that g is non-zero on ∂A, then there
exists an ε > 0 such that ε < g(x) for all x ∈ ∂A. Since fj ∈ J for all 1 ≤ j ≤ n, it must be that
P
d(f1 , . . . , fn ) = 0, hence there exists an h ∈ S(A) such that nj=1 kfj hk < ε. Furthermore, as ∂A
is a boundary for A, there exists an x0 ∈ ∂A such |h(x0 )| = 1; however,
ε < g(x0 ) =
n
X
|fj (x0 )| =
j=1
n
X
|fj (x0 )h(x0 )| ≤
j=1
n
X
kfj hk < ε,
j=1
which is aTcontradiction. Therefore, there exists a z ∈ ∂A such that 0 = g(z) =
thus z ∈ nj=1 Z(fj ) ∩ ∂A.
3
Pn
j=1 |fj (z)|,
and
Proof of Main Theorem
In this section, we prove the following:
Main Theorem. Let X and Y be compact Hausdorff spaces, and let A ⊂ C(X) and B ⊂ C(Y )
be complex subalgebras that contain the constants and separate points. If T : A → B is a surjective
mapping, not necessarily linear, such that
kT (f )T (g)k = kf gk
(3.1)
holds for all f, g ∈ A, then there exists a homeomorphism ψ : ∂B → ∂A between the Shilov boundaries such that
|T (f )(y)| = |f (ψ(y))|
for all f ∈ A and all y ∈ ∂B.
The Main Theorem shall be proven via a sequence of lemmas, and its hypotheses shall be
assumed throughout, i.e. T : A → B denotes a surjective mapping between complex subalgebras
that contain the constants and separates points that satisfies (3.1). Note that (3.1) implies that
kT (f )k = kf k
for all f ∈ A, which implies S(A) = T −1 [S(B)] and T [S(A)] = S(B).
Lemma 3.1. Let f1 , . . . , fn ∈ A. Then d(f1 , . . . , fn ) = d(T (f1 ), . . . , T (fn )).
Proof. Let h ∈ S(A), then T (h) ∈ S(B). By (3.1),


n
X

d(T (f1 ), . . . , T (fn )) = inf
kT (fj )kk : k ∈ S(B)


j=1
≤
n
X
j=1
kT (fj )T (h)k =
n
X
kfj hk.
j=1
As h was chosen arbitrarily, it must be that d(T (f1 ), . . . , T (fn )) ≤ d(f1 , . . . , fn ). A similar argument
yields the reverse inequality.
Lemma 3.2. Let J ⊂ A and D ⊂ B consist of JTDZ. Then both T [J ] and T −1 [D] consist of
JTDZ.
Proof. Let g1 , . . . , gn ∈ T [J ], then there exist f1 , . . . , fn ∈ J such that T (fj ) = gj for 1 ≤ j ≤ n.
Lemma 3.1 implies that
0 = d(f1 , . . . , fn ) = d(T (f1 ), . . . , T (fn )) = d(g1 , . . . , gn ).
Since g1 , . . . , gn were chosen arbitrarily, it must be that T [J ] consists of JTDZ. The fact that
T −1 [D] consits of JTDZ is proven similarly.
Lemma 3.3. Let y ∈ ∂B. Then there exists a unique x ∈ ∂A such that T −1 [Iy (B)] = Ix (A).
Proof. As Iy (B) consists of JTDZ, Lemma 3.2 implies that T −1 [Iy (B)] consists of JTDZ. Consequently, Lemma 2.1 guarantees the existence of an x ∈ ∂A such that T −1 [Iy (B)] ⊂ Ix (A).
Now, the surjectivity of T yields that
Iy (B) = T [T −1 [Iy (B)]] ⊂ T [Ix (A)]
Appealing to Lemmas 2.1 and 3.2 again implies that there exists a z ∈ ∂B such that Iy (B) ⊂
T [Ix (A)] ⊂ Iz (B). Since B separates points and contains 1, it must be that y = z and thus
T [Ix (A)] = Iy (B). Therefore,
Ix (A) ⊂ T −1 [T [Ix (A)]] = T −1 [Iy (B)].
The uniqueness of such an x follows from the fact that if Iw (A) = T −1 [Iy (B)] = Ix (A), then
w = x.
Define the mapping ψ : ∂B → ∂A by
T −1 [Iy (B)] = Iψ(y) (A).
(3.2)
Note that this is well-defined by the previous lemma.
Lemma 3.4. The mapping defined by (3.2) is bijective.
Proof. Let x ∈ ∂A, then there must exist a y ∈ ∂B such that T [Ix (A)] ⊂ Iy (B). Therefore
Ix (A) ⊂ T −1 [T [Ix (A)]] ⊂ T −1 [Iy (B)] = Iψ(y) (A),
and since A separates points and 1 ∈ A, we have that ψ(y) = x. Consequently, ψ is surjective.
Now, let y, z ∈ ∂B be such that ψ(y) = ψ(z). This implies that
T −1 [Iy (B)] = Iψ(y) (A) = Iψ(z) (A) = T −1 [Iz (B)],
and the surjectivity of T yields that
Iy (B) = T [T −1 [Iψ(y) (A)]] = T [T −1 [Iψ(z) (A)]] = Iz (B)
Therefore y = z, hence ψ is injective.
We now demonstrate that T is a composition operator in modulus.
Lemma 3.5. Let f ∈ A and let y ∈ ∂B. Then |T (f )(y)| = |f (ψ(y))|.
Proof. Set α = f (ψ(y)) and β = T (f )(y). As f − α ∈ Iψ(y) (A), then (3.2) yields that T (f −
α) ∈ Iy (B). Moreover, T (f ) − β ∈ Iy (B) and since Iy (B) consists of JTDZ, it must be that
d(T (f − α), T (f ) − β) = 0. Thus there exists a sequence {kn } ⊂ S(B) such that kT (f − α)kn k +
k[T (f ) − β]kn k → 0, which yields that both {kT (f − α)kn k} and {k[T (f ) − β]kn k} converge to zero.
Let hn ∈ A be such that T (hn ) = kn , then hn ∈ S(A). Furthermore, (3.1) yields
kf hn k − |α| = kf hn k − kαhn k ≤ k[f − α]hn k = kT (f − α)kn k,
and thus kf hn k → |α|. Additionally,
kT (f )kn k − |β| = kT (f )kn k − kβkn k ≤ k[T (f ) − β]kn k,
which implies that kT (f )kn k → |β|.
By (3.1), kf hn k = kT (f )T (hn )k = kT (f )kn k for all n ∈ N. Therefore, by the uniqueness of
limits of sequences for metric spaces, it must be that
|T (f )(y)| = |β| = |α| = |f (ψ(y))|.
We now complete the proof of the Main Theorem.
Lemma 3.6. The mapping defined by (3.2) is a homeomorphism.
Proof. Since ψ is a bijection between a compact space and a Hausdorff space, it is only to show
that ψ is continuous. Indeed, let U ⊂ ∂A be an open set and let y0 ∈ ψ −1 [U ]. Since A separates
points and contains the complex constant functions, there exist functions f1 , . . . , fn ∈ A such that
ψ(y0 ) ∈
n
\
{x ∈ ∂A : |fj (x)| < 1} ⊂ U
j=1
(cf. [6, Proposition 2.2.14]). As |T (f )(y)| = |f (ψ(y))| for all y ∈ ∂B, it follows that
y0 ∈
n
\
{y ∈ ∂B : |T (fj )(y)| < 1} ⊂ ψ −1 [U ].
j=1
Therefore ψ −1 [U ] is open, hence ψ is continuous.
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