1. ## Spectrum

Hi,

I've problem with following statement:

For $A=(0,1)$ (real Banach algebra) the functional
$f(x)=\int_0^1 x(t)dt$
satisfy
$f(x) \in \sigma(x)$

Any help or advices will be highly appreciated,
Best regars

2. Originally Posted by Arczi1984
Hi,

I've problem with following statement:

For $A=(0,1)$ (real Banach algebra) the functional
$f(x)=\int_0^1 x(t)dt$
satisfy
$f(x) \in \sigma(x)$
I don't know what is meant by the "real Banach algebra $A=(0,1)$". The unit interval (0,1) is not a Banach algebra. Do you mean C(0,1), the algebra of continuous functions on the (closed) unit interval, with the supremum norm? Or does it mean something else?

3. Hi,

Below is link to the paper where is that statement. Remark after proof of second Theorem.
I didn't see that mistake befor, of course it should be
$A=C(0,1)$.
And You are right it is algebra of all continuous functions on the unit interval.
I'm sorry for that mistake.

PS. Here is link to the original paper http://matwbn.icm.edu.pl/ksiazki/sm/sm29/sm29126.pdf

4. Originally Posted by Arczi1984
Hi,

I've problem with following statement:

For $A=(0,1)$ (real Banach algebra) the functional
$f(x)=\int_0^1 x(t)dt$
satisfy
$f(x) \in \sigma(x)$

Any help or advices will be highly appreciated,
Best regars
Here $A = C(0,1)$ the Banach algebra of all real valued bounded continuous functions on $(0,1)$. For $x \in A$ let $I = \int_0^1 x(t)dt$. Then one easily proves that $inf \{x(t) : t \in (0,1) \} \le I \le sup \{x(t) : t \in (0,1)\}$. Though the infimum and the supremum may not be attained, one can choose $m, M$ in the range of $x$ such that $inf \{x(t) : t \in (0,1)\} \le m \le I \le M \le sup \{x(t) : t \in (0,1)\}$. Now applying the Intermediate Value Theorem, one concludes that there is some $c \in (0,1)$ such that $x(c) = I = \int_0^1 x(t)dt$ is in the range of $x$. Consequently, $\frac{1}{x - I}$ is not a continuous function, in fact it is not defined at $c$. Thus $\int_0^1 x(t)dt \in \sigma(x)$. Since $x \in C(0,1)$ is chosen arbitrary, we see that the result is true for all $x \in C(0,1)$