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I would like someone to help me check if I understand this proof and for this reason I would like to give the proof here in my own words. The statement I am proving is this:

Let $A$ be a unital $C^\ast$-algebra and $a \in A$ normal. Then for all $f \in C(\sigma (a))$: $$ f(\sigma (a)) = \sigma (f(a))$$

Ok here is the proof (as I understand it):

Let $\tau \in \Omega (A)$ be a character and $p: \sigma (a) \to \mathbb C$ a polynomial. Then $p(\tau(a)) = \tau (p(a))$ since $\tau$ is linear and multiplicative. Since $\tau$ is continuous and since every $f \in C(\sigma (a))$ is the (uniform) limit of a sequence of polynomials $p_n$ we therefore have

$$ f(\tau(a)) = \lim p_n (\tau (a)) = \lim \tau (p_n (a)) = \tau (\lim p_n (a)) = \tau (f(a))$$

Again from a previous theorem we have that $\sigma (a) = \{\tau (a) \mid \tau \in \Omega (A)\}$. Putting these things together we therefore have:

$$ \sigma (f(a)) = \{\tau (f(a)) \mid \tau \in \Omega (A) \} = \{f(\tau (a)) \mid \tau \in \Omega (A) \} = f(\sigma (a))$$

Is this correct reasoning?

2 Answers2

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The argument looks fine to me. But you need to restrict $A$ to be commutative.

Martin Argerami
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  • Oh right, because $\sigma(a) = {\tau(a)\mid \tau \in \Omega (A)}$ only for commutative Banach algebras! But the statement of the theorem is as it should be so my proof is wrong. I will correct my proof and make it a new question. Thank you for your help! –  Nov 07 '14 at 05:17
  • The theorem works for arbitray $ A $. What you do is to work on the C $^$-subalgebra generated by $ a $, which is commutative (that's what requires normality). And you take advantage of the fact that for C $^$-algebras the spectrum does not depend on the algebra. – Martin Argerami Nov 07 '14 at 06:28
  • I tried to fix my proof and posted it here in this thread. Would you have a look and tell me if it is correct now? –  Nov 08 '14 at 02:25
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Let $A$ be a unital $C^\ast$ algebra and let $a\in A$ be normal. Then for $f \in C(\Omega(A))$ we have

$$ f(\sigma(a)) = \sigma(f(a))$$

Proof:

We let $B$ be the algebra generated by $1$ and $a$. Then $B$ is commutative and unital. Hence the Gelfand representation $\varphi: B \to C(\Omega(B))$ is an isometric isomorphism. Also, $\sigma_B(a) = \sigma_A(a)$.

Let $\tau \in \Omega(B)$ and $p: B \to B$ be a polynomial (identify the complex coefficients $c$ with $c \cdot 1 \in B$). Then $\tau \circ p = p \circ \tau$ because characters are linear and multiplicative.

By Stone Weierstrass, $f \in C(\Omega(B))$ is a limit of a sequence of polynomials $p_n$. Then

$$ \begin{align} f(\sigma_A(a)) &= f(\sigma_B(a)) \\ &= \{f \circ \tau (a) \mid \tau \in \Omega (B)\}\\ &=\{\lim_n p_n\circ \tau (a) \mid \tau \in \Omega (B)\}\\ &=\{\lim_n \tau \circ p_n (a) \mid \tau \in \Omega (B)\}\\ &=\{\tau \circ f (a) \mid \tau \in \Omega (B)\}\\ &=\sigma_B(f(a)) = \sigma_A(f(a)) \end{align}$$

where $\sigma_B(f(a)) = \sigma_A(f(a))$ holds because $f$ is in the closure of the algebra of polynomials which is the algebra $B$, that is, $f(a)$ is in fact in $B$.

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    Yes, this is fine. A minor point is that $B$ is the C$^*$-algebra generated by $1$ and $a$ (and not just the algebra). And one thing that you need to have clear in my mind is how to use the Gelfand transform to argue that $\sigma(x)={\tau(x):\ \tau\in \Omega(B)}$ for all $x\in B$. – Martin Argerami Nov 08 '14 at 04:23
  • @MartinArgerami Thank you very much for checking my proof! I don't think $\sigma (x) = {\tau(x)\mid \tau \in \Omega (B)}$ is related to the Gelfand transform: the proof of it on page 14 in Murphy uses that a character corresponds to a maximal ideal. At this point in the book the Gelfand transform has not been introduced. –  Nov 08 '14 at 04:38
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    Fair enough. I haven't read that carefully in a while. – Martin Argerami Nov 08 '14 at 04:47