FRICHET ALGEBRAS GENERATED BY CERTAIN OF THEIR ELEMENTS

We consider F-algci)ras A that are generated by elements of the form z, (z- Ale)- 1, (z ANe)- 1, where is the identity. If A has no topclogical divisors of zero we show that A is isomorphic to H(), where f is finitely connected region. We also study F-algebras in which


INTRODUCTION
The algebra H(f) of holomorphic functions on a region l], with the compact-open topology has been characterized among F-algebras in several ways.Rudin [1] proved that a uniform F- algebra which satisfies a form of the maximum modulus principle is an algebra of holomorphic functions.Birtel [2] showed that under certain conditions a singly-generated F-algebra is the algebra of entire functions, and under other conditions it is shown in [3] that such an algebra is the algebra of holomorphic functions on a simply connected domain.Meyers [4] characterized H(f) by using the property that bounded sets are relatively compact.Carpenter [5] used the existence of derivations to characterize H(fl).Arens [6] gave conditions on a singly-rationally- generated F-algebra which ensure that the algebra is the direct sum of its radical and an algebra of holomorphic functions on a region.Brooks [7] showed that the same conclusion holds for locally m-convex algebras in which a certain boundary is empty.In [8] and [9] conditions are given on an F-algebra generated by a finite number of elements that guarantee that the algebra is an algebra of holomorphic functions.These and other F-algebra characterizations of H(I]) are in terms of well known properties of H(I]), such as Liouville's theorem [2], the maximum modulus principle [1], Montel's theorem [4], the Cauchy estimate [6], Taylor's theorem [9], the existence of certain derivations [5], and others.
In this paper we consider F-algebras A that are generated by certain of their elements, namely Z, (Z-,IC)-1, (Z-,N)-1, where e is the identity of A and Aa, ,A N (:.Under one additional assumption, that A has no topological divisors of zero, we show that A is isomorphic to H(f), where 9t is a finitely connected region.This can be viewed as a characterization of H(fl) in terms of Runge's ~-1 .2theorem.We also consider F-algebras in which {e,z,.. is a basis for A.
An F-algebra is a complete metrizable locally m-convex algebra over C. (We consider only commutative F-algebras.)The topology of an F-algebra A is given by an increasing sequence of seminorms {p,,:nGN} and A is isomorphic to an inverse limit lim(A,,,Tr )of Banach algebras as follows" A,, is the completion of A/ker(p,) in the induced norm, and for n _< m the nap rr,,,,,:A,,,A,, is the natural homomorphisxn.The maximal ideal space consists of all non-zero continuous multiplicative linear functionals on A endowed with the weak topology generated by the Gelfand transforms 2:.Ag(A)-C, where '(f)= f(x).The algebra of all Gelfand transforms is A {':x A} equipped with the compact-open topology.
The Gelfand map F:AA is continuous and is a bijection if A is semisimple.The quotient map r,:A--A/ker(p,)induces a homeomorphism of the maximal ideal space 3t(A,) onto a compact subset M, of dtt(A).For n < m we have M,, C M,, and MI(A)= U M,,. (f(Wl),f(w2),..,f(WN))f .AI(A) }.
An element z in a Banach algebra B is a topological divisor of zero if the map Tz:BzB is not an isomorphism.In an F-algebra A, z is a topological divisor of zero if for each sequence {p,:n } of seminorms defining the topology of A there exists a k such that rt,(z is a topological divisor of zero in the Banach algebra A k (See Michael [10]).When we say that an algebra has no topological divisors of zero, we will mean, of course, that it has no nonzero ones.

A CHARACTERIZATION OF RUNGE ALGEBRAS
If f C C is a finitely connected domain, and H(t) is endowed with the compact-open topology, then by Runge's Theorem, H(fl) is generated by rational functions with poles in C\fl.In fact, if A, ...,AN are points chosen one from each component of C\, then the rational functions with poles at .kx, ...,v suffice to generate H(gt) [11, page 200].It is also well-known that H() has no topological divisors of zero.We will show that these two properties characterize H(f) as an F-algebra.
An F-algebra is a Runge algebra if there is an element z G A and complex numbers ,--.,AN such that A is generated by {z, w,...,WN} where w=(z-e)-1, If A zs a Runge algebra wztlt no topological divisors of zero, then a(z) is a finitely connected domazn.
PROOF.If c(z) were to contain a boundary point c, then z-he would be a topological divisor of zero [10,Proposition 11.8].Thus a(z)is open.Now suppose a(z)is the disjoint union of nonempty open sets S and T, and let Xs and XT be their characteristic functions.
These are analytic on a(z), so by the functional calculus [12] we can find x,y A such that Xs o " and Xr o S; hence z3ot 0. Since A has no topological divisors of zero, it follows that A is semisimple and so xy 0. But then x and y would be proper zero-divisors, a contra- diction.Thus a(z) is connected, and so it is a domain.Let 9:.A(A)--r(z, wl,...,WN) be the spectrum map and let be the map taking (f(z),f(w,),...,f(wN)) to f(z).Clearly /, is a continuous bijection, and hence so is 0 o.For t a(z)let ft -'(t), and for x A define ":a(z)-C to be the map taking to (t) (ft) ft(z), as indicated in the commutative diagram below.
( ( Let A= {':z A}, equipped with the compact-open topology.We first show that A c_ H(a(z)).
Let x A, and let P be a sequence of polynomials in N+ variables such that Pk(z,w,...,Wg) converges to z.
First note that since A has no topological divisors of zero, for each n 5 there exists m > n such that M, C int M, (see Arens [13]).So, without loss of generality (by replacing the original sequence of seminorms by an equivalent one) we may assume that M, C int M,+ C M,+ (n 1,2,...).Now qo01M,, is a homeomorphism onto its image (because 1M, is, and is a bijection).Thus it follows that o(M,)C intcp0(M,+a)C 0(M,+a), n 1,2,..., and a(z)= LI0(M,)= Uint 0(M,).Now let S be a compact subset of a(z).By the preceding paragraph, there exists n such that S C int0(M,), so that '(S) C M,. Thus '(S) is a compact (and hence equicontinuous) subset of 31(A).Now P, z, so the continuity of the Gelfand map implies that.P,in A, i.e., uniformly on compact subsets of .Ab(A).Thus for > 0 and sufficiently large k,]Pt.(ft)-'2(ft)< for ftGg'(S), which is the same as]r.(t)-"t)]< for teS.
This shows that each ' is the limit of rational functions whose poles lie outside a(z); the convergence is uniform on compact subsets of or(z), and hence ' is analytic on a(z).It follows that A C_ H(a(z)).Now if h H(a(z)), then by the functional calculus for F-algebras there exists y A such that (f)= h('i(f))for fG al(A).Therefore h='ff and so A= H(a(z)).This last equality shows that the polynomials in t,(t-la)-, ,(t-N) -(i.e. the rational functions with poles at A,, ...,AN) are dense in H(a(z)).Runge's Theorem [11, page 200] implies that a(z)is a finitely connected domain.
TIIEOREM 2.2.The algebra A "ts a Rung algebra wzth no topological divisors of zero if and only ff A "ts somorphic to H() for some finitely connected domain .
PROOF.That H(f) has the indicated properties is discussed in the first paragraph of this section.Conversely, suppose A has no topological divisors of zero and is generated by elements (z-,e)-', ,(Z--Ne)-1, where A ,$N C. The map o:-Ag(A)a(z) defined in the proof of Proposition 2.1 is a homeomorphism, because the topology on .At,(A) is the weak topology generated by A, and 5 o ff ='' is continuous for each A. Let G:AA be the map G()= . .We claim that A F_ , G____, is an isomorphism.Since G(Y) " o -1, G is an isomorphism, and since A is semisimple, F is a bijection.Thus G o F is a continuous bijection onto the F-algebra A, hence is open by the open mapping theorem, and so A A. But, as in Proposition 2.1, A H(a(z)) and a(z)is finitely connected.This completes the proof.
If each of the seminorms p, of an F-algebra A satisfies p(z) p(z) for all z A, then A is a uniform algebra.For such an algebra, p(z)= sup {l(f)l f M,} for all Thus the Gelfand map F is a homeomorphism of A onto a complete subalgebra A of A derivation on an F-algebra A is a linear transformation D:AA satisfying D(z9) zD(y)+ D(x)9.Carpenter [5] used the existence of a derivation on a uniform F- algebra to characterize H(f).His result provides one of the equivalences in the following theorem.
THEOREM 2.3.Let A be a uniform Rung algebra which is generated by the elements z,(z $e) -, (z $ge) -, where , ,A N C.. The following are equivalent conditions on ,4.(d) A ts isomorphic to g() for a finitely connected domain .
PROOF.Clearly (d)implies (a), (b), and (c).That (a)implies (d) follows from Theorem 2.2, and that (c)implies (d) follows from Carpenter [5].We prove that (b) implies (d).Let w, ("-A,e)-l, z= 1,...,N.Observe first that a(z, wl,...,wN)is homeomorphic to a(z), and a,(Z, Wl,...,WN) to a,(z), via the map z/, defined in the proof of Proposition 2.1.We show that {a,(Z, Wl,...,WN) is a /,'-covering sequence for a(z,w,...,wu) by showing that {a,(z)} is a k- covering sequence for a(z) (k-covering means that the sequence covers every compact subset).But a(z)is the increasing union of the a,(z) and is open, so a(z)= [.J int a,( z ).Thus {inta,(z)} is an open cover for any compact subset It" of a(z), so It" C_ a,,(z)for some n e N.This proves that {a,(Z, Wl,...,WN) is a k-covering for a(z, wa,...,WN) and so by [7, Theorem .al the map :J(A)--,a(z,w,...,WN)is a homeomorphism.Thus _(A)is homeomorphic to a(z) by the map o:ff(z).This, together with the fact that A is a uniform algebra, allows us to consider A (and hence A) as an algebra of continuous functions on a(z) with the compact-open topology.Under this identification, for (5 a(z) we have '(t) '(l(t)) [ffl(t)](Z) l, and similarly moreover, these functions generate A. By Runge's theorem [11, page 200], A (and hence A) is isomorphic to H() for a finitely connected domain f.

BASES GENERATED BY z AND z -1
In this section we assume that A is an F-algebra with a basis generated by z and z-; that is, for each z A, there is a unique sequence {c,,: n E ?'} of scalars such that z--Z CnZn where the series converges independently in the positive and negative directions of summation.
(In [14] algebras with bases of the form {z":n N} are studied.)We fit .th spectrum of z for such algebras.
LEMMA a.1.Let A be an F-algebra with a basis {z n Z}, and let r= l/p(z-), R= p(z), where p denotes the spectral radius.If p(z)p(z-) 1, let S {A C r < IAI < R}, and if p(z)#(z-) 1, let S {A C" al R}.Then S C a(z) G -.
TIiEOREM 3.2.Let A be an F-algebra wth a baszs {z:n Z}.Then A is semsimple and its topology s gzven by an increasing sequence of norms.
PROOF.Suppose x e Rad(A), x= E_%a,z".Then f(x)= E_a,,f(z)"=O for all f (A).Thus the Laurcnt series E_%a,t" converges to 0 for a(z), and since a(z) h a limit point (Lemma 3.1), o,, 0 for all n.Thus x 0 and A is semisimple.
Let K be an infinite compact subset of (A).(Such a set must exist, for otherwise (A) being hemicompact [10, page 22] would be countable, contradicting Lemma 3.1 and the fact that is a bijection.)Since K is cquicontinuous there exists c > 0 and a seminorm p such that [f(x)[cp(x) for all f K, xA.Thus p(x)=O implies f(x)=O for all fK, and since T(K) has a limit point in C, an argument as in the preceding paragraph shows that x 0. This means that p is a norm, and hence so is p, for n k.THEOREM 3.3.Let A be a Banach algebra with a basis homeomorphic to fl {A G C" 1/p(z-) <_ [A[ _< p(z)}.
Then J(A is PROOF.Since A is a Banach algebra, .Jig(A) is homeomorphic to a(z, z-l), which in turn is homeomorphic to a(z) via the projection map rl:a(z,z-X)--a(z)which takes (f(z),f(z-')) to f(z).(r, is one-to-one because each fE Ag(A)is completely determined by its value at z.) Since a(z) is compact, it is equal to by Lemma 3.1.
An example of a Banach algebra with a basis of the type being considered is t'(l).We now construct another example.
If f(t)= _a,t is analytic on the annulus Ann(0;r,R)={tC r< It <R}, and if r<s<R, let Ilfllo= ]-oola,ls"-The limits lim Ilfll Ill]l, and lim ]]fll I]flln both exist.Let A(r,R) be the space of such functions + R for which the norm Ilfllmax {llfll.Ilfll} is finite.By an argument as in [15], A(r,n) is a Banach algebra with basis {z":n e l}, where z is the function z(t)= t.Abel's theorem shows that A(r, R) consists of those analytic functions f(t) ,_%oa,t" on Ann(0;r,R) for which (3.2) We show that this example is typical of Banach algebras with bases of the type under consideration.THEOREM 3.4.Let A be a Banach algebra with an unconditional basis {z n" n Z}.If p(z)--II z and p(z-')= z-' II, th, A is isomorphic to A(r,R), where r 1/p(z-1) and R (z).
(a) A has no topological divisors of zero.(b) The spectrum a(z) is an open subset oft.(c) A has a derivation D such that D(z)= e.