Loading ...
Sorry, an error occurred while loading the content.

A question

Expand Messages
  • Fa Ha
    In the book A Course in Functional Analysi by J. B. Conway, 2nd ed., in the proof of Theorem 3.8, chapter VII, pages 175, 176, he has claimed that the set B
    Message 1 of 6 , Dec 12, 2008
    • 0 Attachment
      In the book " A Course in Functional Analysi" by J. B. Conway, 2nd ed., in the proof of Theorem 3.8, chapter VII, pages 175, 176, he has claimed that the set B is" finite" which seems unusual. Could anyone explain it? .

    • Andrew B. Frigyik
      Hi Fa Ha, What do you mean unusual? Here is how to see it: If the function f is in F (script F) then the range of the function is in the closed unit disk D of
      Message 2 of 6 , Dec 12, 2008
      • 0 Attachment
        Hi Fa Ha,

        What do you mean unusual?

        Here is how to see it: If the function f is in F (script F) then the
        range of the function is in the closed unit disk D of C, the set of
        complex numbers (I assumed C(X) means complex valued functions). We
        can cover the set D with finitely many epsilon/6-sized neighborhoods,
        all centered in D, since D is compact in the standard topology of C.
        Let's call the collection of centers E. Now B sits inside of the
        n-fold Cartesian product of E which is again finite.

        I Hope this helps.

        Andrew

        On Fri, Dec 12, 2008 at 12:13 PM, Fa Ha <f.ha88@...> wrote:
        > In the book " A Course in Functional Analysi" by J. B. Conway, 2nd ed., in
        > the proof of Theorem 3.8, chapter VII, pages 175, 176, he has claimed that
        > the set B is" finite" which seems unusual. Could anyone explain it? .



        --
        "Az ember egy okos állat, amelyik úgy viselkedik, mint egy hülye."
        "Man is a clever animal who behaves like an imbecile."

        Albert Schweitzer
      • maslouhi mostafa
        Dear members, I don t see how to prove the following: Let $G$ be a compact group and $( pi, E)$ a finite linear representation of $G$. We consider a a
        Message 3 of 6 , Apr 25, 2009
        • 0 Attachment
          Dear  members,

          I don't see how to prove the following:

          Let $G$ be a compact group and  $(\pi, E)$ a finite linear representation of $G$. We consider a  a hermitian form where $<,>$ on $E$ and set
          $(a,b)=\int_G <\pi_x(a),\pi_x(b)> dx$, $a,b\in G$,  where $dx$ is a  Haar measure on $G$.

          The question is: Show that there exists an invertible  operator  $A:E\to E$ such that
          $(a,b)=<A(a),A(b)> $ for all $a,b\in G$.

          Thanks in advance,

          Best regards,

          Mostafa MASLOUHI.

        • shravan kumar
          Dear Mostafa Maslouhi   Define the operator $$A:E rightarrow E$$ as $$A(a)= int_G pi(x)a dx$$ where the integral has to be interpreted in the weak sense
          Message 4 of 6 , Apr 27, 2009
          • 0 Attachment
            Dear Mostafa Maslouhi
             
            Define the operator $$A:E\rightarrow E$$ as $$A(a)=\int_G\pi(x)a dx$$ where the integral has to be interpreted in the weak sense i.e., $$<A(a),b>=\int_G<\pi(x)a,b>$$ fotr all $a,b\in E.$
             
            Best regards
            Shravan

            --- On Sat, 25/4/09, maslouhi mostafa <maslouhi_mostafa@...> wrote:

            From: maslouhi mostafa <maslouhi_mostafa@yahoo..fr>
            Subject: [harmonic] A question
            To: harmonicanalysis@yahoogroups.com
            Date: Saturday, 25 April, 2009, 9:28 PM

            Dear  members,

            I don't see how to prove the following:

            Let $G$ be a compact group and  $(\pi, E)$ a finite linear representation of $G$. We consider a  a hermitian form where $<,>$ on $E$ and set
            $(a,b)=\int_ G <\pi_x(a),\pi_ x(b)> dx$, $a,b\in G$,  where $dx$ is a  Haar measure on $G$.

            The question is: Show that there exists an invertible  operator  $A:E\to E$ such that
            $(a,b)=<A(a),A(b)> $ for all $a,b\in G$.

            Thanks in advance,

            Best regards,

            Mostafa MASLOUHI.



            Now surf faster and smarter ! Check out the new Firefox 3 - Yahoo! Edition * Click here!
          Your message has been successfully submitted and would be delivered to recipients shortly.