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I want to prove the next theorem:

If $\pi: A \rightarrow B$ is a star homomorphism, meaning it's an algebra homomorphism which also satisfies: $\pi(x^*)=(\pi(x))^*$, where $A$ is an involutive Banach Algebra, and $B$ a C*-algebra, then $$||\pi(x)|| \leq ||x||$$

I am given a hint to prove, which I am not sure how to prove it.

The hint says to prove that: $$\sigma_{B_{I}} (\pi(y)) \subset \sigma_{A_{I}} (y); \forall y \in A$$

where $A_I=A\oplus \mathbb{C}$

Here's what I tried thusfar, I want to prove the converse of this inculsion: Take $\lambda \in (\sigma_{A_I}(y))^c$, thus for $\lambda e -y=b$ is invertible, i.e there exists $b^{-1}$, now $$\pi(b^{-1})=(\pi(b))^{-1}=(\pi(b))^* = \pi(b^*)= \bar{\lambda} e_B - \pi(y)^*$$, and I want to show that $\lambda e_B -\pi(y)$ is invertible in $B$, but I don't see how $\pi(b)^{-1}=\bar{\lambda} e_B - \pi(y^*)$ is an inverse of $\pi(b)=\lambda e_B -\pi(y)$.

Anyone can enlighten my eyes?


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up vote 3 down vote accepted

$\pi(b)^{-1}=\pi(b)^*$ is incorrect unless $\pi(b)$ is unitary, which it usually is not. You otherwise had a good start. The point is the previous equation, $\pi(b^{-1})=\pi(b)^{-1}$. Thus $\pi(b)=\lambda e_B -\pi(y)$ is invertible, which means that $\lambda$ is not in the spectrum of $\pi(y)$ as desired. (Technically this is abusing notation using $\pi$ denote the unital extension of $\pi$.) Briefly, unital ring homomorphisms send invertible elements to invertible elements, and therefore algebra homomorphisms shrink spectra.

Once you have that containment, you will want to use the fact that the norm dominates the spectral radius in $A$, and the norm equals the spectral radius for positive elements of $B$. (I am guessing that part of your definition of involutive Banach algebra is that $\|a^*\|=\|a\|$ for all $a\in A$.)

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Note that $e_B=\pi(e_A)=\pi(b b^{-1})=\pi(b)\pi(b^{-1})$. Thus, $\pi(b^{-1})=\pi(b)^{-1}$.

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