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Math Help - Algebra of all continuous function

  1. #1
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    Algebra of all continuous function

    This is the last problem.

    If A is the algebra of all continuous functions definied on the unit disc, holomorfic in its interior, then the mapping
    T:A \rightarrow A \oplus A
    given by the formula
    x(t) \rightarrow [x(2t),0]
    is a multiplicative mapping satisfying neither
    \sigma(Tx) \in \sigma(x)
    nor
    \sigma(x) \in \sigma(Tx).

    Once more many thanks for any help or advices.
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  2. #2
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    Quote Originally Posted by Arczi1984 View Post
    This is the last problem.

    If A is the algebra of all continuous functions definied on the unit disc, holomorfic in its interior, then the mapping
    T:A \rightarrow A \oplus A
    given by the formula
    x(t) \rightarrow [x(2t),0]
    is a multiplicative mapping satisfying neither
    \sigma(Tx)\: {\color{red}\subseteq}\: \sigma(x)
    nor
    \sigma(x)\: {\color{red}\subseteq}\: \sigma(Tx).
    (Notice that you should be using the subset symbol here, not the set membership symbol.)

    Once more many thanks for any help or advices.
    I am very hesitant to suggest that mathematicians as eminent as Kahane and Żelazko would make a mistake in their paper. But it looks as though the formula x(t) \rightarrow [x(2t),0] should read x(t) \rightarrow [x(t/2),0]. The reason is that if |t|>1/2 then x(2t) will not be defined (because x(t) is only defined for |t|\leqslant1).

    In this algebra (as in any algebra of functions), the spectrum of a function in the algebra is the set of values of the function. For the function x(t) = t, for example, the spectrum of x is the whole unit disc. But the function Tx given by  t\mapsto [x(t/2),0] only takes the values \{t\in\mathbb{C}:|t|\leqslant1/2\}. On the other hand, if x is an invertible function then 0 will not be in its spectrum, but 0 is in the spectrum of Tx.
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