Thread: Algebra of all continuous function

1. 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.

2. Originally Posted by Arczi1984
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.