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• 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
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 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? .

• 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
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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
• 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
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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 $a,b\in G$.

• 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
 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., $$=\int_G<\pi(x)a,b>$$ fotr all $a,b\in E.$ Best regardsShravan --- On Sat, 25/4/09, maslouhi mostafa wrote:From: maslouhi mostafa Subject: [harmonic] A questionTo: harmonicanalysis@yahoogroups.comDate: Saturday, 25 April, 2009, 9:28 PMDear  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$(a,b)=$ for all $a,b\in G$.Thanks in advance,Best regards,Mostafa MASLOUHI.