5 pagesInternational audienceLet $A$ be a unital $B_{0}$-algebra with an orthogonal basis, then every multiplicative linear functional on $A$ is continuous. This gives an answer to a problem posed by Z. Sawon and Z. Wronski

El Azhari, Mohammed

English

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Functional continuity of unital B_0-algebras with
orthogonal bases
Mohammed El Azhari
To cite this version:
Mohammed El Azhari. Functional continuity of unital B_0-algebras with orthogonal bases. Le
Matematiche, Università degli Studi di Catania, 2017, 72 (1), pp.97-102. ￿hal-01205148v4￿
Functional continuity of unital B0-algebras with orthogonal bases
M. El Azhari
Abstract. Let A be a unital B0-algebra with an orthogonal basis, then every
multiplicative linear functional on A is continuous. This gives an answer to a
problem posed by Z. Sawon and Z. Wronski.
2010 Mathematics Subject Classification: 46H40.
Keywords and phrases: B0-algebra, orthogonal basis, multiplicative linear func-
tional.
I. Preliminaries
A topological algebra is a complex associative algebra which is also a Hausdorff
topological vector space such that the multiplication is separately continuous. A
locally convex algebra is a topological algebra whose topology is determined by
a family of seminorms. A complete metrizable locally convex algebra is called a
B0-algebra. The topology of a B0-algebra A may be given by a countable family
(‖.‖i)i≥1 of seminorms such that ‖x‖i ≤ ‖x‖i+1 and ‖xy‖i ≤ ‖x‖i+1‖y‖i+1
for all i ≥ 1 and x, y ∈ A. A multiplicative linear functional on a complex
algebra A is an algebra homomorphism from A to the complex field. Let A be a
topological algebra. M∗(A) denotes the set of all nonzero multiplicative linear
functionals on A. M(A) denotes the set of all nonzero continuous multiplicative
linear functionals on A. A seminorm p on A is lower semicontinuous if the set
{x ∈ A : p(x) ≤ 1} is closed in A.
Let A be a topological algebra. A sequence (en)n≥1 in A is a basis if for each
x ∈ A there is a unique sequence (αn)n≥1 of complex numbers such that x =
Σ∞n=1αnen. Each linear functional e
∗
n : A→ C, e∗n(x) = αn, is called a coefficient
functional. If each e∗n is continuous, the basis (en)n≥1 is called a Schauder basis.
A basis (en)n≥1 is orthogonal if eiej = δijei where δij is the Kronecker symbol. If
(en)n≥1 is an orthogonal basis, then each e∗n is a multiplicative linear functional
on A. Let A be a topological algebra with an orthogonal basis (en)n≥1. If A
has a unity e, then e = Σ∞n=1en. Let (xk)k be a net in A converging to 0,
since the multiplication is separately continuous, enxk = e
∗
n(xk)en →k 0 and so
e∗n(xk)→k 0. Then each orthogonal basis in a topological algebra is a Schauder
basis. Let f be a multiplicative linear functional on A. If f(en0) 6= 0 for some
n0 ≥ 1, then f(x)f(en0) = f(xen0) = f(e∗n0(x)en0) = e∗n0(x)f(en0) for all x ∈ A
and therefore f = e∗n0 ∈M(A). This shows that M(A) = {e∗n : n ≥ 1}.
Here we consider unital B0-algebras with orthogonal bases. These algebras
were investigated in [3] where we can find examples of such algebras.
1
II. Results
Proposition II.1. Let (A, (‖.‖i)i≥1) be a unital B0-algebra with an orthogonal
basis (en)n≥1. Then there exists x ∈ A such that the sequence (e∗n(x))n≥1 is not
bounded.
Proof. Suppose that supn≥1 |e∗n(x)| < ∞ for all x ∈ A. For each x ∈ A, let
‖x‖ = supn≥1 |e∗n(x)|, ‖.‖ is a lower semicontinuous norm on A since (en)n≥1 is a
Schauder basis. Let τA be the topology on A determined by the family (‖.‖i)i≥1
of seminorms. We define a new topology τ on A described by the norm ‖.‖ and
the family (‖.‖i)i≥1 of seminorms. The topology τ is stronger than the topology
τA. By Garling’s completeness theorem [1, Theorem 1], (A, τ) is complete. The
topologies τA and τ are homeomorphic by the open mapping theorem. Then
there exist i0 ≥ 1 and M > 0 such that ‖x‖ ≤ M‖x‖i0 for all x ∈ A, hence
1 = |e∗n(en)| ≤ ‖en‖ ≤ M‖en‖i0 for all n ≥ 1. This contradicts the fact that
en →n 0.
Proposition II.2. Let A be a unital B0-algebra with an orthogonal basis
(en)n≥1. If x = Σ∞n=1tnen ∈ A such that tn ∈ R, tn ≤ tn+1 for n ≥ 1 and
tn →∞, then f(x) ∈ R for all f ∈M∗(A).
Proof. If f ∈ M(A), f = e∗n for some n ≥ 1, then f(x) = tn ∈ R. If f ∈
M∗(A) rM(A), then f(en) = 0 for all n ≥ 1. Suppose that f(x) /∈ R, f(x) =
α + iβ with β 6= 0. Since tn → ∞, there exists n0 ≥ 1 such that tn ≥ α + |β|
for all n ≥ n0. We define the sequence (sn)n≥1 by sn = tn0 for 1 ≤ n ≤ n0
and sn = tn for n ≥ n0 + 1. It is clear that y = Σ∞n=1snen ∈ A such that
sn ≥ α + |β|, sn ≤ sn+1 for n ≥ 1 and sn → ∞. Since f(en) = 0 for all
n ≥ 1, f(y) = f(x) = α + iβ. We have f(|β|−1y) = |β|−1α + i|β|−1β, then
f(|β|−1y − |β|−1αe) = i|β|−1β. Set z = |β|−1y − |β|−1αe = Σ∞n=1 sn−α|β| en. The
real sequence ( sn−α|β| )n≥1 is positive increasing and
sn−α
|β| → ∞, then z−1 =
Σ∞n=1
|β|
sn−αen ∈ A by [3, Theorem 0.1] and f(z−1) = −i|β|β−1, so f(z+z−1) = 0.
Set v = z + z−1 = Σ∞n=1(
sn−α
|β| +
|β|
sn−α )en and vn =
sn−α
|β| +
|β|
sn−α for all n ≥ 1.
Since the map g : [1,∞) → R, g(x) = x + 1x , is increasing and the sequence
( sn−α|β| )n≥1 ⊂ [1,∞) is increasing, it follows that (vn)n≥1 is a positive increasing
sequence and vn →∞. By [3, Theorem 0.1], v−1 = Σ∞n=1 1vn en ∈ A and therefore
f(e) = f(v)f(v−1) = 0. This contradicts the fact that f is nonzero.
The following two results are due to Sawon and Wronski [3], the proofs are
given for completeness.
Theorem II.3([3, Theorem 2.1]). Let A be a unital B0-algebra with an orthog-
onal basis (en)n≥1. If x = Σ∞n=1tnen ∈ A such that tn ∈ R, tn ≤ tn+1 for n ≥ 1
and tn →∞, then every multiplicative linear functional on A is continuous.
2
Proof. Suppose that M∗(A) rM(A) is nonempty. Let f ∈ M∗(A) rM(A),
then f(en) = 0 for all n ≥ 1. Put f(x) = α, then α ∈ R by Proposition II.2.
Since tn → ∞, there exists n0 ≥ 1 such that tn > α for n ≥ n0. Consider
y = Σ∞n=1λnen ∈ A such that λn = tn0 for 1 ≤ n ≤ n0 and λn = tn for
n ≥ n0 + 1. Since f(en) = 0 for all n ≥ 1, it follows that f(x) = f(y) = α.
We have y − αe = Σ∞n=1vnen ∈ A where vn = λn − α for all n ≥ 1. It is
clear that vn > 0, vn ≤ vn+1 for n ≥ 1 and vn → ∞. By [3, Theorem 0.1],
(y − αe)−1 = Σ∞n=1 1vn en ∈ A, so y − αe is invertible and f(y − αe) = 0, a
contradiction.
Proposition II.4([3, p.109]). Let (A, (‖.‖i)i≥1) be a unital B0-algebra with an
orthogonal basis (en)n≥1. Then the set N of all positive integers can be split
into two disjoint subsets N1 and N2 such that by putting A1 = span(en)n∈N1
and A2 = span(en)n∈N2 , we have
(1) A = A1 ⊕A2;
(2) if f is a multiplicative linear functional on A such that f /∈ M(A), then
f/A1 = 0.
Proof. By Proposition II.1, there is x = Σ∞n=1tnen ∈ A such that the sequence
(tn)n≥1 is not bounded. Then there exists a subsequence (tkn)n≥1 of (tn)n≥1
such that |tkn | ≥ n2 for all n ≥ 1. For each i ≥ 1, there is Mi > 0 such that
‖tnen‖i ≤ Mi for all n ≥ 1. Let i ≥ 1 and n ≥ 1, n2‖ekn‖i ≤ |tkn |‖ekn‖i =
‖tknekn‖i ≤Mi, then ‖ekn‖i ≤ n−2Mi. This implies that Σ∞n=1ekn is absolutely
convergent. Let A1 = span{ekn : n ≥ 1}, A1 is a unital B0-algebra with an
orthogonal basis (ekn)n≥1 and Σ
∞
n=1n
1
2 ekn ∈ A1 since n
1
2 ‖ekn‖i ≤ n−
3
2Mi for
all i ≥ 1 and n ≥ 1. Set N1 = {kn : n ≥ 1}, N2 = N rN1 and A2 = span{en :
n ∈ N2}. A2 is a B0-algebra with an orthogonal basis (en)n∈N2 and the unity
u2 = e − u1 where u1 = Σ∞n=1ekn is the unity of A1. Let x ∈ A, x = xe =
x(u1+u2) = xu1+xu2 ∈ A1+A2, then A = A1⊕A2. If f is a multiplicative linear
functional on A such that f /∈ M(A), f/A1 is a multiplicative linear functional
on A1 such that f/A1(ekn) = 0 for all n ≥ 1. Since Σ∞n=1n
1
2 ekn ∈ A1, f/A1 is
continuous on A1 by Theorem II.3 and therefore f/A1 = 0.
Sawon and Wronski [3, p.109] posed the following problem:
Problem. Let A be a unital B0-algebra with an orthogonal basis (en)n≥1. Does
there exist a maximal subalgebra A
′
1 = span{en : n ∈ N
′
1}(N
′
1 ⊂ N) of A for
which (1) and (2) hold?
Proposition II.5. Let A be a unital B0-algebra with an orthogonal basis
(en)n≥1. Then the following assertions are equivalent:
(i)A = span{en : n ∈ N} is a maximal subalgebra of itself for which (1) and
(2) hold;
(ii) every multiplicative linear functional on A is continuous.
3
Proof. (i) ⇒ (ii) : Let f be a multiplicative linear functional on A such that
f /∈M(A), f is zero on A by (i). Then every multiplicative linear functional on
A is continuous.
(ii) ⇒ (i) : It is clear that A satisfies (1). Let f be a multiplicative linear
functional on A such that f /∈M(A), then f is zero on A since f is continuous,
hence A satisfies (2).
Proposition II.6. Let (tn)n≥n0 be a complex sequence, the following assertions
are equivalent:
(i) Σ∞n=n0 |tn − tn+1| <∞;
(ii) there exists M > 0 such that |tq|+ Σq−1n=p|tn− tn+1| ≤M for all q > p ≥ n0.
Proof. (i)⇒ (ii) : Let  > 0, there exists n1 ≥ n0 such that Σ∞k=n|tk−tk+1| ≤ 
for every n ≥ n1. let m > n ≥ n1, |tn− tm| ≤ |tn− tn+1|+ ...+ |tm−1− tm| ≤ .
Then the sequence (tn)n≥n0 converges, so there is M0 > 0 such that |tn| ≤ M0
for all n ≥ n0. Let q > p ≥ n0, |tq|+ Σq−1n=p|tn − tn+1| ≤M0 + Σ∞n=n0 |tn − tn+1|.
(ii)⇒ (i) : Let p ≥ n0 and q = p+1, Σpn=n0 |tn−tn+1| ≤ |tq|+Σq−1n=n0 |tn−tn+1| ≤
M. Then the sequence (Σpn=n0 |tn−tn+1|)p≥n0 is positive increasing and bounded,
so it is convergent i.e. Σ∞n=n0 |tn − tn+1| <∞.
Proposition II.7. Let (A, (‖.‖i)i≥1) be a unital B0-algebra with an orthogonal
basis (en)n≥1. If (tn)n≥n0 is a complex sequence such that Σ
∞
n=n0 |tn−tn+1| <∞,
then Σ∞n=n0tnen ∈ A.
Proof. Let q > p ≥ n0, by using the equality tn = tq+Σq−1k=n(tk−tk+1) for every
p ≤ n < q, we obtain that Σq−1n=ptnen = tq(ep+ ...+ eq−1) + Σq−1k=p(tk− tk+1)(ep+
...+ek). Let i ≥ 1, ‖Σq−1n=ptnen‖i ≤ |tq|‖ep+...+eq−1‖i+Σq−1k=p|tk−tk+1|‖ep+...+
ek‖i ≤ (|tq|+ Σq−1k=p|tk− tk+1|) supp≤k≤q ‖ep+ ...+ ek‖i ≤M supp≤k≤q ‖ep+ ...+
ek‖i by Proposition II.6. Let  > 0, since e = Σ∞n=1en ∈ A, there is n1 ≥ n0 such
that ‖ep + ...+ ek‖i ≤ M−1 for n1 ≤ p ≤ k, hence supp≤k≤q ‖ep + ...+ ek‖i ≤
M−1 for n1 ≤ p < q. Then ‖Σq−1n=ptnen‖i ≤  for n1 ≤ p < q. This shows that
Σ∞n=n0tnen is convergent in A.
Theorem II.8. Let (A, (‖.‖i)i≥1) be a unital B0-algebra with an orthogonal
basis (en)n≥1. Then every multiplicative linear functional on A is continuous.
Proof. By Proposition II.7, x = Σ∞n=1
1
nen ∈ A. Let f ∈M∗(A) and α = f(x),
then f(αe − x) = 0. We have αe − x = Σ∞n=1(α − 1n )en = Σ∞n=1 αn−1n en. Let
N be the set of all positive integers. Put Iα = {n ∈ N : n 6= α−1}, Iα = N
if α−1 /∈ N and Iα = N r {α−1} if α−1 ∈ N. Let mα = inf{n ∈ N : |α|n −
1 > 0} and consider the complex sequence ( nαn−1 )n≥mα . Let n ≥ mα, nαn−1 −
n+1
α(n+1)−1 =
1
(αn−1)(α(n+1)−1) . We have |αn− 1| ≥ |α|n− 1 and |α(n+ 1)− 1| ≥
|α|(n + 1) − 1 = |α|n + |α| − 1 ≥ |α|n − 1. Let n ≥ mα, |α|n − 1 > 0, hence
1
|αn−1| ≤ 1|α|n−1 and 1|α(n+1)−1| ≤ 1|α|n−1 . Consequently | nαn−1 − n+1α(n+1)−1 | =
4
1
|αn−1||α(n+1)−1| ≤ 1(|α|n−1)2 . Then Σ∞n=mα nαn−1en ∈ A by Proposition II.7 and
therefore Σn∈Iα
n
αn−1en ∈ A. If α−1 /∈ N i.e. Iα = N, (αe− x)Σ∞n=1 nαn−1en = e,
then αe − x is invertible, a contradiction. If α−1 ∈ N, put nα = α−1, then
(αe − x)Σn∈Iα nαn−1en = Σn∈Iαen, hence Σn∈Iαen ∈ Ker(f) since αe − x ∈
Ker(f). If enα ∈ Ker(f), then e = enα + Σn∈Iαen ∈ Ker(f), a contradiction.
Finally f(enα) 6= 0 and therefore f = e∗nα .
Remark. Proposition II.5 and Theorem II.8 give an answer to Sawon and
Wronski’s problem.
Note. Let A be a unital B0-algebra with an orthogonal basis (en)n≥1 and let
δ be a translation invariant metric defining the topology of A. For each n ≥ 1,
define b(n) = Σ∞k=nek, this sequence converges to zero in A. For each n ≥ 1,
there exists pn ≥ 2 such that δ(b(k), 0) ≤ 1n2 for all k ≥ pn. We can assume that
(pn)n≥1 is a strictly increasing sequence in N. Then x = Σ∞n=1b(pn) converges
in A since δ(b(pn), 0) ≤ 1n2 for all n ≥ 1. Let k ≥ 1, we have e∗k(x) = 0 for
1 ≤ k < p1 and e∗k(x) = n for pn ≤ k < pn+1. Therefore (e∗k(x))k≥1 is an
increasing sequence of natural integers such that e∗k(x) → ∞. This shows that
the condition in [3, Theorem 2.1] is always satisfied and so Theorem II.8 can be
deduced from Theorem II.3.
References
[1] D. J. H. Garling, On topological sequence, Proc. Cambridge Philos. Soc.,
63 (1967), 997-1019.
[2] T. Husain, Orthogonal Schauder bases, Monographs and Textbooks in Pure
and Applied Mathematics, 143, Marcel Dekker, New York, 1991.
[3] Z. Sawon and Z. Wronski, Fre´chet algebras with orthogonal basis, Colloquium
Mathematicum, 48 (1984), 103-110.
[4] W. Zelazko, Metric generalizations of Banach algebras, Rozprawy Matem-
atyczne, 47, Warszawa, 1965.
Ecole Normale Supe´rieure
Avenue Oued Akreuch
Takaddoum, BP 5118, Rabat
Morocco
E-mail: mohammed.elazhari@yahoo.fr
5


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