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 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 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. - Dear Mostafa MaslouhiDefine 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 regardsShravan
--- On

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

From: maslouhi mostafa <maslouhi_mostafa@yahoo..fr>

Subject: [harmonic] A question

To: harmonicanalysis@yahoogroups.com

Date: 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)=<A(a),A(b)> $ for all $a,b\in G$.

Thanks in advance,

Best regards,

Mostafa MASLOUHI.

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