ORTHOGONAL BASES IN A TOPOLOGICAL ALGEBRA ARE SCHAUDER BASES SUBBASH

In a topological algebra with separately continuous multiplication, the result quoted in the title is proved.


INTRODUCTION.
A topological algebra A is a linear associative algebra over complex scalars which is a Hausdorff topological vector space (TVS) in which multiplication is separately continuous, i.e., for each xeA, the operators Lx and Rz, Lzy xy, Rzy yx (yeA), are continuous.A basis (en) in A is Schauder (where , Otnen) are continuous (respectively b-Schauder) if the functionals en, en(x)= an x (respectively bounded i.e. map bounded sets to bounded sets).An orthogonal basis is a basis (en) satisfying en em nrn en for all n,m.
Recently S. E1-Helaly and T. Husain [1] showed that an orthogonal basis in A is Schauder if multiplication is jointly continuous (i.e.continuous as a bilinear map on AA).Now joint continuity is a very stringent requirement.In fact, abundance of examples have forced upon some other weaker modes of continuity in literature.Multiplication in A is hypocontinuous (respectively sequentially jointly continuous) if given a o-neighborhood U and a bounded set B, there is a o- neighborhood V such that BVCU, VBCU (respectively for sequences (Xn) (Yn) in A,x n ---, x,y n -, y imply Xny n xy).In a topological algebra, joint continuity gives hypocontinuity which in turn implies sequential joint continuity; and if A is barelled (respectively complete matrizable or m-convex), multiplication is hypocontinuous (respectively jointly continuous).We extend the above result of E1-Helaly and Husain in its final form by modifying their arguments, and also obtain its variant in a more general frame-work.
THEOREM.Let A be a Hausdorff TVS that is an algebra (1) If A is a topological algebra, then every orthogonal basis in A is Schauder.
(2) If multiplication in A is sequentially separately continuous (i.e. for a sequence (Xn) in A, Xn 0 implies Xny -O, yx n 0 for all y), then every orthogonal basis in A is b-Schauder.
PROOF.Let (en) be an orthogonal basis in A. Let ne N be fixed.Orthogonality applied to e(x)e n xe n for all x in A. Choose a balanced o- the expansion x en(X)e n implies that enx neighborhood U such that enU.Let r inf {d > 0" enedU}.Then r > 1.
(1) Let (xa) be a net in A such that lira xa 0. Hence lim xaen 0. Given an > 0, there is an ene(eU for all a > a o.As V is balanced, ]en(Xa) len o o such that en(xa)en x a (eU) for o > Hence by the definition of r, ler(xa) le>_r>l, and so en(x,)l < for all o_>a o.Thus * lim en (xa) O.
(2) Since a subset in a TVS is bounded iff each of its countable subset is bounded, it is sufficient * maps a bounded sequence (xk) to a bounded sequence.is bounded, and for all k, e(xk)enAU for some ,X A(n,U) > 0. Again by Hence (e(xk) n)l definition of r, e,* (z)l < for all .
(2) In a topological algebra, a basis which is not orthogonal need not be Schauder even if multiplication is sequentially jointly continuous.The algebra 11 of summable scalar sequences with weak topology a=a(ll, co) is a topological algebra in which multiplication (pointwise) is sequentially jointly continuous.
Now for any sequence k > 0, xk/r k --* O.By sequential separate continuity of multiplication, e n xk/rk O. r