ON LOCALLY PSEUDOCONVEX SQUARE ALGEBRAS

Let A be an algebra over the field of complex numbers with a (Hausdorff) topology given by a family Q = {qλ|λ ∈ Λ} of square preserving rλ-homogeneous seminorms (rλ ∈ (0, 1]). We shall show that (A, T (Q)) is a locally m-convex algebra. Furthermore we shall show that A is commutative. Introduction. Let A be a locally pseudoconvex algebra over the field of complex numbers. Let Q = {qλ|λ ∈ Λ} be a family of rλhomogeneous seminorms defining a Hausdorff topology on A. For each λ ∈ Λ the number rλ ∈ (0, 1] is fixed. By rλ-homogeniousity we mean that qλ(αx) = |α|λqλ(x) for all x ∈ A and α ∈ C. We shall say that the seminorm qλ is submultiplicative if qλ(xy) ≤ qλ(x)qλ(y) for all x and y in A. If every qλ ∈ Q is submultiplicative, then (A, T (Q)) is called a locally m-pseudoconvex algebra. If each qλ ∈ Q is square preserving in other words if qλ(x) = qλ(x) for all x ∈ A and λ ∈ Λ we shall say that (A, T (Q)) is a square algebra. Note that locally pseudoconvex algebras include as a special case better known locally convex algebras. Namely for locally convex algebras we have rλ = 1 for every λ ∈ Λ. For properties of commutative locally convex square algebras see [1], [4] or [14]. Commutative locally pseudoconvex square and star algebras have been studied in [2]. It is known that a commutative locally convex square algebra is automatically locally m-convex. See [1] and [5] and [16]. It was claimed in [2] that the corresponding result is valid also for locally pseudoconvex algebras. In this paper we shall show that indeed this claim is true and even the assumption of commutativity is superfluous. Main results. If ‖ ‖ is a r-homogeneous submultiplicative norm on a complex associatve algebra A, then (A, ‖ ‖) is called a locally bounded algebra (more precisely a r-normed algebra). See [18]. First we shall prove a locally bounded version of Theorem of [8] and Corollary 16.8 of [7]. See also [3], [12], [13] and [16].


Introduction.
Let A be a locally pseudoconvex algebra over the field of complex numbers.Let Q = {q λ |λ ∈ Λ} be a family of r λhomogeneous seminorms defining a Hausdorff topology on A. For each λ ∈ Λ the number r λ ∈ (0, 1] is fixed.By r λ -homogeniousity we mean that q λ (αx) = |α| r λ q λ (x) for all x ∈ A and α ∈ C. We shall say that the seminorm q λ is submultiplicative if q λ (xy) ≤ q λ (x)q λ (y) for all x and y in A. If every q λ ∈ Q is submultiplicative, then (A, T (Q)) is called a locally m-pseudoconvex algebra.If each q λ ∈ Q is square preserving in other words if q λ (x 2 ) = q λ (x) 2 for all x ∈ A and λ ∈ Λ we shall say that (A, T (Q)) is a square algebra.Note that locally pseudoconvex algebras include as a special case better known locally convex algebras.Namely for locally convex algebras we have r λ = 1 for every λ ∈ Λ.For properties of commutative locally convex square algebras see [1], [4] or [14].Commutative locally pseudoconvex square and star algebras have been studied in [2].It is known that a commutative locally convex square algebra is automatically locally m-convex.See [1] and [5] and [16].It was claimed in [2] that the corresponding result is valid also for locally pseudoconvex algebras.In this paper we shall show that indeed this claim is true and even the assumption of commutativity is superfluous.

A. Jorma
Lemma 1. Suppose that (A, ) is a r-Banach algebra for which there is a constant K > 0 such that Then A is commutative.
Proof: Let x be a given element of A. Furthermore, let B be a maximal commutative subalgebra of A including the element x.Then also (B, ) is a r-Banach algebra.By Theorems 3.3, 4.4 and 4.8 of [18]  By Theorem 1 of [11] it follows that A/ Rad A is commutative.(Rad A stands for the Jacobson radical of A).But it follows from (1) that Rad A = {0} and thus we can see that A is commutative.
Note that the topological dual of A was not used in [11] in proving the commutativity of A/ Rad A.
We shall now prove the generalization of the results of [5] and [6].For a seminorm q on an algebra A denote by N q = ker q = {x ∈ A|q(x) = 0}.Theorem 1.Let q be a r-homogeneous square preserving seminorm on a (complex associative) algebra A. Then q is submultiplicative, N q is an ideal of A, q 1 r is a seminorm on A, and the quotient algebra A/N q is commutative. 2for all x and y in A. Thus, if x and y ∈ A then So we have q(x • y) ≤ 2  4 r (q(x) + q(y)) 2 for all x and y in A. Let x, y ∈ A and > 0 be arbitrary.Denote by α = q(x) + and β = q(y) + .Then . This shows that q(x • y) ≤ 8  4 r q(x)q(y) for all x and y in A. If we now define p = 8 4 r q, then p is a r-homogeneous seminorm on A satisfying p(x • y) ≤ p(x)p(y) for all x and y in A and p(x) 2 ≤ 8 4 r p(x 2 ) for all x in A. For x, y ∈ A denote [x, y] = xy − yx.As in the proof of Proposition 1 of [9] it can be shown that there is some constant K for which p([x, y]) 2 ≤ Kp(x) 2 p(y) 2 for all x and y in A.
x, y ∈ A, we can see that there is some constant R such that p(xy) ≤ Rp(x)p(y) for all x and y in A. Thus there is some constant M for which q(xy) ≤ Mq(x)q(y) for all x and y in A. It follows from this inequality that N q is an ideal of A. Write B := A/N q and let q denote the induced r-homogeneous norm on B. Then .:= M q is a submultiplicative r-homogeneous norm on B satisfying x 2 ≤ M x 2 for all x ∈ A. Applying Lemma 1 to the completion of (B, ), we can see that B is commutative.To prove that q 1 r is a norm on B let ṡq be the spectral norm of (B, q) i.e. ṡq (x) = lim n→∞ n q(x n ) 1 n .By Theorem 3.3 of [18] ṡq satisfies the triangle inequality and on the otherhand we have q(x) = ṡq (x) for all x ∈ B (since q is square preserving).By Corollary 3.5 of [18] q 1 r is a usual (1-homogeneous) norm on B. Thus q 1 r is a seminorm on A. See also [2,Lemma 9].Note also that q 1 r is submultiplicative.See [1] or [5].
Corollary 1. Suppose that (A, T (Q)) is a square algebra.Then (A, T (Q)) is a locally m-convex commutative algebra.
Proof: It follows from Theorem 1 that each quotient algebra A/N λ is commutative (N λ = ker q λ ).This implies that q λ (xy − yx) = 0 for all λ ∈ Λ and since we assumed that T (Q) is a Hausdorff topology this implies that A is commutative.By Theorem 1 for each λ ∈ Λ, q 1 r λ is a usual 1-homogeneous submultiplicative seminorm on A. Thus T (Q) is equivalent with a locally m-convex topology T (P) where P = {q 1 r λ |λ ∈ Λ}.

2 r
we have s B (y) r = lim n→∞ y n 1 n for all y ∈ B. (Here s B stands for the spectral radius of y in B.) It follows from (1) that there is some constant M := M (K) > 0 such that y ≤ Ms B (y) r for all y ∈ B. Since B is a maximal commutative subalgebra of A we have s B (y) = s A (y) for all y ∈ B. (See for ex.[17, p. 46].)Since the above mentioned x is in B we can see that we have x 1 r ≤ M 1 r s A (x) and since x was chosen arbitrarily we can see that this same holds for all x ∈ A. But now we have s A (xy) ≤ xy 1 s A (x)s A (y) for all x and y ∈ A.