Я0 = /(Г'(у))= I «„/= I ».('/'(0)” n =

ROCZNIKÏ POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Séria 1: PRACE MATEMATYCZNE XXIII (1983)

S

W

(Warszawa)

J^-Algebras with cyclic bases

Abstract. In previous papers [3 ], [4 ] we studied the structure of topological algebras possessing (Schau.der) bases, where the bases satisfied various multiplicative conditions. Here we consider J*-algebras with Schauder bases of the form {z": n = 0, 1, We will call such a basis cyclic. Certain properties of such algebras are developed and a characterization of the algebra H (Q) of holomorphic functions on Q is given. We also consider the situation when the basis is unconditional.

1. We begin by giving three fundamental examples which will be useful in the sequel.

E

1.1. Let s t r be the space of functions / analytic in the disc Dr of c°

radius r > 0 such that the norm ||/||r = £

n= о n ! rn is finite. sé r is a Banach space in this nprm [1] and is also a Banach algebra with point wise multiplication. The sequence {z"}, where z is the identity function on Dr, is a cyclic basis in stfr.

E

1.2. For 1 < p < oo, Hp is a commutative Banach algebra with the product

i f *g){Q = ~

«( J f&-X)g{X)dX,

where f , g e H p, (e D , [7]. If z is the identity function on D, then an easy induction shows that z*"(Ç) = £n/nl, where z*n = z * z * . . . * z и-times. The sequence {z*"} is a basis for Hp which is clearly cyclic in our sense.

E

1.3. Let Q be a simply connected domain and let H(Q) be the algebra of holomorphic functions on Q in the compact-open topology. If Q

= C, then H(Q) = $ the algebra of entire functions and the sequence (z"|, z the identity function, is a basis in ê . If Q i§ a proper subset of C, let у = ф(О be a conformal map from Q onto D, the unit disc. For / eH(Q) we have

Я0 = /(Г'(у))= I «„/= I ».('/'(0)”

n = 0 n = 0

330 S. W a ts o n

and the series converge in the topology of H(Q). This series representation is unique and thus \фп] is a cyclic basis for H(Q).

2. Throughout this paper A will denote a complex ^-algebra (complete metrizable locally m-convex) with a cyclic basis n — 0, 1, We will denote by xk the kth coefficient functional (which is continuous [5]) as­

sociated with this basis. Also <r(x), {?(*), and M(A) will denote the spectrum of x, spectral radius of x, and maximal ideal space of A, respectively. For these and other definitions and results concerning bases and topological algebras we refer to [5] and [6].

We begin by noting that it is easy to show that, for x , y ,e A ,

у /

x = £ oc„zn, y = Y, Pnz\ then their product is the Cauchy product

n — 0 w = 0

oo oo

* У = X ( X * n - k P k ) z n- n = 0 f c = 0

Le m m a

2.1. Let A eC. Then, (a) z is non-invertible.

X

(b) z — Ae is .invertible if and only if £ (z/Я)" converges in A. In this case

n= 0

( z - X e Y 1 = - - f {z/A)n.

* n= 0

(c) I f Афс

(

), then \A\ ^

(

)..

■00 00

Proof, (a) If not and z_1 = £ a„z", then e — zz~1 = £ a„z"M con-

n= 0 n = 0

tradicting the uniqueness of the basis representation.

00 00 X'1

(b) If (z — Ae)~1 = ]T a«z"’ then e — (z — Ae) £ anzn = —kct0e + £ (a„ —

n = 0 n = 0 n = 0

— Aan+1)zn+1. Thus —Яа0 = 1 and x„ — Aocn+1 = 0 , n ^ 0, and since Я 0 by (a), we have x0 — —1/A and a„+1 =%JA, n ^ O . By induction we have

= —(1 /Я)"+1. Conversely, it is clear by direct multiplication that the inverse is given by this formula.

(c) For / eM(A) we have by (Ъ), f ((z — Ae)~x) = —| W ]. Hence / „=0 V я /

\f(z)/A\ < 1 for all / eM(/4) and thus g(z) = sup |/(z)| ^ |Я|.

f e M( A)

Part (c) of this lemma shows that if g(z) = r, then Dr ç u(z) Ç We make repeated use of the following lemma. For this we define (p: M(A)^>a(z) by f - > f ( z ) . This map is one-to-one, onto, and con­

tinuous [2].

Le m m a

2.2. Let V be a subset o f M(A) such that (p(V) has a limit point in

C. I f

A and f (x) = 0 for all f e V, then x = 0.

Proof. We have f (x) = £ x*(x)f(z)n and this power series converges n= 0

to zero on q>{V). Since this set has a limit point, by the identity theorem for power series x*(x) = 0, n ^ 0, and hence x = 0.

Th e o r e m

2.3. I f

> 0, then the topology of A is given by an increasing family o f norms.

Proof. If each compact subset of M(A) is finite, then M(A) being hemicompact [6] would be countable contradicting the fact that <p is a bijection of M(A) onto o(z) the latter being uncountable (Lemma 1.1 (c)). So let H be an infinite compact subset of M(A) and let {pk} be an increasing sequence of seminorms giving the topology of A. Now H is equicontinuous [6] so there exists M > 0, к > 0 such that |/ (x)| < Mpk{x) for all/

H,

A.

Thus if pk(x) = 0, then f (x) = 0 for a ll/ e H and hence, by Lemma 2.2, x = 0.

Therefore pk is a norm and it follows that pn is also a norm for all n ^ k.

By Lemma 1.1 (a) zero is always in the spectrum of z. We now show that if this spectrum contains points other than zero, then A is necessarily semisimple.

Th e o r e m 2.4. A is semisimple if and only if ^q(z)

> 0.

Proof. If

{

) = 0, then/ (z) = 0 for/

M(A). Since z generates A each /

M(A) is completely determined by its value at z so that there is only one such functional, namely x*. It follows that the radical of A, Rad {А) Ф (0}.

Conversely, if xeRad(v4), then /(x) = 0 for all f

M(A). N

since Dr ç a (z), r =

(

) > 0, it follows by Lemma 2.2 that x = 0.

3 .

In Example 1.3 we showed that

has a cyclic basis (here

will always denote a simply connected domain). We will prove a converse to this.

Le m m a

3.1. I f o(z) is open, then it is homeomorphic to M(A).

Proof. Define x on

by x(t)

x((p~

(r)), where x is the Gelfand transform of x and tp is as in Lemma 2.2. We have

1 OO 00

x(t) = x((p~l {t)) = X x*{x)zn((p~l (t))= X X* (x)

n = 0 n = 0

and since this series converges x is analytic (hence continuous) on cr(z), i.e., x

= xo<p_1 is continuous for each x. Since the topology of M(A) is the weak topology generated by the functions x it follows that tp~l is continuous.

Th e o r e m

3.2. An JF-algebra A is isomorphic to H(Q) if and only if A has a cyclic basis (z"} with o(z) open.

Proof. For each x e A define x as in Lemma 3.1. Let A be the algebra of Gelfand transforms of A and let A be the subalgebra of H(o{z)) consisting of the functions x and equipped with the compact-open topology. In view of

Lemma 3.1 it is clear that the map F: A -> A by x -> x is an algebraic and

332 S. W a ts o n

topological isomorphism onto A. Now, let G: A -> A be the Gelfand map and set D = F o G . We will show that D is the desired isomorphism onto H(o(z)). Since a (z) is open A is semisimple by Theorem 2.4. Thus G and hence D is one-to-one. If geH(a(z)), then by the functional calculus for 3F- algebras [6] there exists y e A with y ( f ) — g(z(f)) for / eM(A). Equivalently we have g(t) = ÿ(<p- 1 (0) for teo(z). Thus g —y and this shows that D is onto H(o(z)). Finally, it is well known that G is continuous. Thu*s D is - continuous and by the open mapping theorem D is also open.

Corollary

3.3. I f g (z) =

then A is isomorphic to the -algebra S o f . entire functions.

It is clear from Example 1.3 that H(Q) has different cyclic bases for different conformal maps of Q onto D. In general, for our algebras A we have

Theorem

3.4. Let (w"] be another cyclic basis for A. I f 0 <

(

) <

then w — A(z — ae)(r2e — az)~l, where \a\ < r = £>(z) and |Я| = ^(z)^(w). I f

(

)

= x , then w = A(z — ae), A, a e C , А Ф 0.

X)

Proof. Suppose w = X p„zn and consider ф:

(

)-+<

(

) by ф ^ ) n= 0

= /(w), where tf is the point of o(z) corresponding to the functional fe M ( A ) , i.e., tf = f(z). Clearly ф is one-to-one and onto. Now ф(tf ) =/(w)

X X

= X Pnf(z)n = X Pntf and this series converges on Int o(z), the interior of

n= 0 n= 0 '

(

(

), hence defines an analytic function there. Similarly ф~г is analytic on Int cr(w). From the Reimann Mapping Theorem ^(Int a{z)) and ф~х (Int o’(w)) are open hence ф(1пг <

(

>) = Int

from the Schwarz Lemma it can be deduced that ф, being a one-to-one analytic map from the disc of radius ^(z) onto the disc of radius g(w) is of the form

<И(г) = * r2 - at, ’

where a and A satisfy the conditions in the theorem, tf ea(z). Thus,

/ И = A “ = f[ A { z - a e ) { r 2e - a z ) ~ x~\.

r2- a f ( z ) The result now follows by Lemma 2.2.

The other part is proved similarly after we note that a one-to-one entire function onto C is of the form ф(r) = A (t-a), А ф 0.

4.

We consider some properties of A when the basis is unconditional [5]. So, let A be an .^-algebra with an unconditional cyclic basis (z"}.

Theorem 4.1.

Let A be as above. Then:

(a) о (z) is either open or compact.

(b) M(A) is homeomorphic to cr{z).

Proof, (a) If o(z) contains a boundary point h, then for every x e A , 00

£ x*(x)h" converges absolutely. (Because there exists f e M ( A ) with f(z )

n= 0

= h and since the series x = ]T x*(x)z" converges unconditionally in A the

x * = о

complex series f ( x ) = x* (x) / (z)" converges absolutely.) It follows that if

n= 0 x

d is such that \d\ = \b\, then the series £ x*(x)dn converges for all x e A and n= 0

so f(z) = d defines a multiplicative linear functional on A. By [6], Lemma 6.1 (a), d ea(z) and therefore the disc o{z) contains all its boundary points.

(b) If er(z) is open, then this is Lemma 3.1. If <r(z) is compact, then this is proved just as in Lemma 3.1 after we note that the series x(t) converges absolutely on the closed disc o(z) and so defines a continuous function on it.

Recall that a function algebra is a Banach algebra in which the spectral norm

is equivalent to the original norm.

Th e o r e m

4.2. I f A is a function algebra with an unconditional cyclic basis

\z n}, then A is isomorphic to stfr, r =

{

) [Example 1.1].

Proof. For x e A define x as in Lemma 3.1. Since the basis is uncondi­

tional each x(r) is an absolutely convergent series for tetr(z). Since er(z) is compact, g(z)e<r(z) [Lemma 1.1 (c)] and so for every x e A, xe^/r, r =

{

).

So, let R : A -> by x -> x. R is clearly one-to-one, linear, and multiplica-

X X

tive. Now, if f e.$/r, f { t ) ~ £ a„r", then £ |a„| g (z)" < oo and so v

x n = 0 n= 0

= £ xnzneA . Clearly y — f and hence R is also onto. Thus R is an

n= 0

algebraic isomorphism from A onto s4r and since sér is semisimple R is also a topological isomorphism [8].

References 1 2 * 4

[1 ] M. G. A rs o v e , P roper bases and linear homeomorphisms in spaces o f analytic functions, Math. Ann. 135 (1958), p. 235-243.

[2 ] R. M. B ro o k s , On the spectrum o f finitely-generated locally M -convex algebras, Studia Math. 29 (1968), p. 143 150.

T. H u sain . S. W a ts o n , Topological algebras with orthogonal Schauder bases, Pacific. J.

Math. 91 (1980), p. 3 3 9-347..

[4 ] —, —, Um-onditional orthogonal bases, Proc. Amer. Math. Soc. 79 (1980), p. 539-545.

334 S. W a ts o n

[5 ] J. T. M a rti, Introduction to the theory o f bases, Springer-Verlag, New York 1969.

[6 ] 1'. Л. M ich a e l, Locally m ulliplicatively-convex topological algebras, Amer. Math. Soc.

Mem. 11 (1952).

[7 ] N. M. W ig le y , A B an ach algebra structure f o r Hp, Canad. Math. Bull. 18 (1975), p. 597- 603.

[8 ] W. Z e la z k o , B an ach algebras, Elsevier, Amsterdam 1973.

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