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The answer is no: $B^*B$ need not be positive, even if the algebra is unital and commutative and $B$ is self-adjoint. Consider the $\mathbb C$-algebra $\mathbb C[X]$ of polynomials in one indeterminate with coefficients in $\mathbb C$. We equip $\mathbb C[X]$ with involution $p \mapsto \overline{p}$, that is, complex conjugation of the coefficients. This ...

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Yes, that is one way to look at it. Here's another: even before the Gelfand transform, we already know from Lemma 1.2.4 that $\sigma(a)$ is compact. Here it is only proven for unital Banach algebras, but that is simply because Murphy only defines the spectrum of an element in a non-unital algebra on page 13, at the very end of section 1.2. (Think about it: ...

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If $g$ is an element of order $p$ (this always exists by Cauchy's Theorem), let $$Q=\frac1p\,\sum_{j=0}^{p-1} g^j.$$ Then $\text{tr}(Q)=1/p$ and $$Q^*=\frac1p\,\left(\sum_{j=0}^{p-1} g^j\right)^*=\frac1p\,\sum_{j=0}^{p-1} g^{-j} =\frac1p\,\sum_{j=0}^{p-1} g^{p-j}=\frac1p\,\sum_{k=1}^{p} g^k=\frac1p\,\sum_{k=0}^{p-1} g^k=Q.$$ Note that ...

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Note that $a-\lambda$ commutes with $be^{i\lambda}$. In any ring, if $x$ and $y$ commute and the product is invertible, then each is invertible. Proof: suppose that $zxy=xyz=I$. Then $zyx=zxy-I$, so $x$ has a left inverse. From $xyz=I$ we know that $x$ has a right inverse. Then $x$ is invertible.

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Of course. Because $C_r(\mathcal A,G)$ is the C$^*$-algebra generated--in the right environment--by $\mathcal A$ and $G$. You take a dense subset $\mathcal A_0$ of $\mathcal A$, and then all sums $$\sum_{j=1}^m a_jg_j,$$ are dense, where $a_j\in\mathcal A_0$ and $g_j\in G$. As you can see, you don't need $G$ finite: countable suffices.

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I think that one doesn't work. In fact, if $A$ is any separable infinite-dimensional C$^*$-algebra, then it has a faithful state $\varphi$. If we do GNS for $\varphi$, then $\pi_\varphi$ is a cyclic representation. And it is infinite-dimensional, because $A\subset H_\varphi$ (due to the fact that $\varphi$ is faithful).

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If $p_n(x)=p_n(y)$ for all $n$, then $f(x)=f(y)$, which we assumed was false. So $p_n(x)\neq p_n(y)$ for some $n$. Since $p_n$ is a projection, it can only take the values $0$ and $1$. If $p_n(x)=1$ and $p_n(y)=0$, we're done. The only other possibility is $p_n(x)=0$ and $p_n(y)=1$, in which case you can just consider the projection $1-p_n$.

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