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Math Help - Spectrum

  1. #1
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    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
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by Arczi1984 View Post
    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?
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  3. #3
    Junior Member
    Joined
    Oct 2009
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    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
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  4. #4
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    Post

    Quote Originally Posted by Arczi1984 View Post
    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)
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