Codex @codex  0

A algebra character $\phi$ is nonzero, linear and multiplicative, with $\phi (1)=1$. Since $\phi (a)\in {\sigma }_A(a)$, it satisfies $|\phi (a)|\le \Vert a\Vert$ and is automatically continuous. If $h=h^{\ast }$ and $t\in \mathbb{R}$, then $(e^{ith}{)}^{\ast }=e^{-ith}$, so $e^{ith}$ is a Unitary element of a C-star algebra and has norm one. HenceUsing both signs of $t$ forces $\mathrm{Im}\phi (h)=0$. Write $x=h+ik$ with $h=(x+x^{\ast })/2$ and $k=(x-x^{\ast })/(2i)$ both self-adjoint C-star elements. Their algebra character values are real, and thereforeThis is the conjugation in the original PDF; the converted TeX loses the bar.

For the final unheaded request, let $B=C^{\ast }(1,x)\subseteq A$. Normality makes this a commutative unital C-star subalgebra. We first justify spectral permanence, so that we do not silently use the possibly larger algebra spectrum in a mere non-star subalgebra. A self-adjoint C-star element $h$ in a C-star algebra has real algebra spectrum: $e^{ith}$ has norm one, and $\lambda \in \sigma (h)$ implies $e^{it\lambda }\in \sigma (e^{ith})$ for each real $t$. The latter assertion follows from the factorization $e^{ith}-e^{it\lambda }1=(h-\lambda 1)g_t(h)$ with commuting factors. An invertible product of commuting factors would make $h-\lambda 1$ invertible. Both signs of $t$ then force $\lambda \in \mathbb{R}$.

A compact subset of the real line has connected planar complement. Apply Question 1 to the self-adjoint C-star element $h\in B$: all of ${\rho }_A(h)$ is its unbounded component and is retained in $B$. Thus ${\sigma }_A(h)={\sigma }_B(h)$. If $a\in B$ is invertible in $A$, both $a^{\ast }a$ and $a{a}^{\ast }$ are self-adjoint C-star elements and invertible in $A$, hence invertible in $B$. The elements $(a^{\ast }a{)}^{-1}{a}^{\ast }$ and $a^{\ast }(a{a}^{\ast }{)}^{-1}$ are respectively a left and a right inverse in $B$, so coincide and give its inverse. This proves spectral permanence for C-star algebras.

In a commutative unital complex Banach algebra, algebra spectra are the ranges of the characters. Indeed a noninvertible $b-\lambda 1$ lies in a maximal ideal. Such an ideal is closed, since its proper closure cannot contain an element sufficiently near $1$. The quotient is a complex Banach division algebra; nonemptiness of the algebra spectrum in that quotient makes each element scalar, producing an algebra character with value $\lambda$ at $b$. Conversely, an algebra character value cannot correspond to an invertible difference.

Apply this to $B$. The preceding involution result gives $\phi (x^{\ast })=\overline{\phi (x)}$, and spectral permanence applies also to $p(x,x^{\ast })$. ConsequentlyThus the second argument is conjugated as in the original PDF, not a second copy of $\lambda$. This is the spectral mapping rule for a star polynomial in one normal element.