Bhatt, Subbash J.; Deheri, G. M. Orthogonal bases in a topological algebra are Schauder bases. (English) Zbl 0763.46031 Int. J. Math. Math. Sci. 15, No. 1, 203-204 (1992). Let \(A\) be a Hausdorff topological vector space over \(\mathbb{C}\) which is also an associative algebra in which each left-multiplication operator \(L_ x(y):=xy\) and each right-multiplication operator \(R_ x(y):=yx\) is continuous. A sequence \((e_ n)_{n\in\mathbb{N}}\subset A\) is a basis if each \(x\in A\) determines a unique sequence \((\alpha_ n)_{n\in\mathbb{N}}\) of scalars such that \(x=\lim_ N \sum_{n=1}^ N \alpha_ n e_ n\). If each coefficient functional \(e_ n^*(x):=\alpha_ n\) is continuous, \((e_ n)_{n\in\mathbb{N}}\) is a Schauder basis, and if \(e_ n e_ m=\delta_{nm}e_ m\) for all \(n,m\in\mathbb{N}\), then it is an orthogonal basis. These being the relevant definitions, the authors prove the assertion comprising the title of their paper. The converse assertion is disproved with a counterexample. If it is only required that each \(L_ x\) and each \(R_ x\) be sequentially continuous, then the functionals \(e_ n^*\) associated with an orthogonal basis are each bounded, i.e., each maps every bounded subset of \(A\) into a bounded subset of \(\mathbb{C}\). Here \(B\subset A\) is called bounded if every neighborhood of 0 contains a non-zero scalar multiple of \(B\). Reviewer: R.Burckel (Manhattan) Cited in 1 Document MSC: 46H05 General theory of topological algebras 46A35 Summability and bases in topological vector spaces Keywords:left-multiplication; Schauder basis; orthogonal basis PDFBibTeX XMLCite \textit{S. J. Bhatt} and \textit{G. M. Deheri}, Int. J. Math. Math. Sci. 15, No. 1, 203--204 (1992; Zbl 0763.46031) Full Text: DOI EuDML