On decomposition of a commutative p-normed algebra into a direct sum of ideals
Wiesław Żelazko (1963)
Colloquium Mathematicae
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Wiesław Żelazko (1963)
Colloquium Mathematicae
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W. Żelazko (1960)
Studia Mathematica
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C. J. Read (2005)
Studia Mathematica
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It is a long standing open problem whether there is any infinite-dimensional commutative Banach algebra without nontrivial closed ideals. This is in some sense the Banach algebraists' counterpart to the invariant subspace problem for Banach spaces. We do not here solve this famous problem, but solve a related problem, that of finding (necessarily commutative) infinite-dimensional normed algebras which do not even have nontrivial closed subalgebras. Our examples are incomplete normed...
Béla Bollobás (1974)
Studia Mathematica
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Gustavo Corach, Fernando Suárez (1987)
Studia Mathematica
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W. Żelazko (1963)
Studia Mathematica
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Kent Merryfield, Saleem Watson (1991)
Colloquium Mathematicae
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M.ª Isabel Garrido, Javier Gómez Gil, Jesús Angel Jaramillo (1992)
Extracta Mathematicae
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Suppose that A is an algebra of continuous real functions defined on a topological space X. We shall be concerned here with the problem as to whether every nonzero algebra homomorphism φ: A → R is given by evaluation at some point of X, in the sense that there exists some a in X such that φ(f) = f(a) for every f in A. The problem goes back to the work of Michael [19], motivated by the question of automatic continuity of homomorphisms in a symmetric *-algebra. More recently, the problem...
John Lindberg (1971)
Studia Mathematica
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Vladimir G. Pestov (1992)
Atti della Accademia Nazionale dei Lincei. Classe di Scienze Fisiche, Matematiche e Naturali. Rendiconti Lincei. Matematica e Applicazioni
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We show that a free graded commutative Banach algebra over a (purely odd) Banach space is a Banach-Grassmann algebra in the sense of Jadczyk and Pilch if and only if is infinite-dimensional. Thus, a large amount of new examples of separable Banach-Grassmann algebras arise in addition to the only one example previously known due to A. Rogers.