[proofplan] For (1) and (2): we localize to a commutative $C^*$-subalgebra. The closed $*$-subalgebra $A(x)$ generated by $x$ and the unit is commutative (because $x$ is Hermitian or unitary, hence normal). By [Spectrum in a Subalgebra](/theorems/2673) we have $\sigma_A(x) \subseteq \sigma_{A(x)}(x)$, and by [Spectrum via Characters](/theorems/2677) the spectrum in the commutative algebra $A(x)$ is $\{\varphi(x) : \varphi \in \Phi_{A(x)}\}$. The previous theorem (**Characters Are Star-Homomorphisms**) restricts each $\varphi(x)$ to $\mathbb{R}$ (resp.\ $S^1$). For (3): we first handle the Hermitian case using that $\sigma_B(x) \subset \mathbb{R}$ has empty interior in $\mathbb{C}$, hence $\sigma_B(x) = \partial \sigma_B(x)$, then apply the boundary-inclusion principle from [Spectrum in a Subalgebra](/theorems/2673). The general normal case reduces to the Hermitian case via the auxiliary Hermitian element $\eta_\lambda := (\bar\lambda \cdot 1 - x^*)(\lambda \cdot 1 - x)$, whose invertibility encodes that of $\lambda \cdot 1 - x$ when $x$ is normal. [/proofplan] [step:Define the closed $*$-subalgebra generated by $x$ and verify it is commutative] Let $x \in A$ be normal: $x x^* = x^* x$. Define \begin{align*} A(x) := \overline{\{p(x, x^*) : p \in \mathbb{C}\langle X, Y\rangle\}}^{\|\cdot\|}, \end{align*} the norm-closure in $A$ of the set of non-commutative polynomials in $x$ and $x^*$ (with complex coefficients) evaluated at $(x, x^*)$, including the constant polynomials. By construction, $A(x)$ is closed under addition, scalar multiplication, multiplication, the involution (since $p(x,x^*)^* = \tilde{p}(x, x^*)$ where $\tilde p$ is obtained by swapping $X \leftrightarrow Y$ and conjugating coefficients), and contains the unit $1 \in A$. The norm closure preserves all algebraic and isometric properties (limits of self-adjoint elements are self-adjoint by continuity of the involution in a $C^*$-algebra). Thus $A(x)$ is a unital $C^*$-subalgebra of $A$. The polynomial subalgebra $\{p(x, x^*) : p\}$ is commutative because $x$ and $x^*$ commute (normality). By continuity of multiplication in a Banach algebra, the closure $A(x)$ is also commutative. [/step] [step:Prove (1) and (2) via $\sigma_A(x) \subseteq \sigma_{A(x)}(x) = \{\varphi(x) : \varphi \in \Phi_{A(x)}\}$] Suppose $x$ is Hermitian (so in particular normal); the unitary case is handled symmetrically below. By [Spectrum in a Subalgebra](/theorems/2673), if $B \subseteq A$ is a unital subalgebra with the same unit and $y \in B$, then $\sigma_A(y) \subseteq \sigma_B(y)$. Theorem 2673 requires $B$ to be a unital subalgebra of $A$ with the same unit; we verified in Step 1 that $A(x)$ is such a subalgebra and $x \in A(x)$. Hence \begin{align*} \sigma_A(x) \subseteq \sigma_{A(x)}(x). \end{align*} Now $A(x)$ is a commutative unital Banach algebra (Step 1). By [Spectrum via Characters](/theorems/2677) applied to the commutative unital Banach algebra $A(x)$ and the element $x$, \begin{align*} \sigma_{A(x)}(x) = \{\varphi(x) : \varphi \in \Phi_{A(x)}\}, \end{align*} where $\Phi_{A(x)}$ denotes the set of characters on $A(x)$. Theorem 2677 requires $A(x)$ commutative unital Banach (verified) and $x \in A(x)$ (verified). By **Characters Are Star-Homomorphisms** applied to the unital $C^*$-algebra $A(x)$ and the Hermitian element $x \in A(x)$ — note that $x$ is Hermitian in $A$ and the involution on $A(x)$ is the restriction of the involution on $A$, so $x$ is Hermitian in $A(x)$ — every character $\varphi \in \Phi_{A(x)}$ satisfies $\varphi(x) \in \mathbb{R}$. Combining the two inclusions: \begin{align*} \sigma_A(x) \subseteq \sigma_{A(x)}(x) = \{\varphi(x) : \varphi \in \Phi_{A(x)}\} \subseteq \mathbb{R}. \end{align*} This proves (1). The unitary case (2) is identical, except that $x$ unitary implies $x$ is normal ($x^*x = 1 = xx^*$), so $A(x)$ is a commutative unital $C^*$-subalgebra, and the previous theorem now gives $\varphi(x) \in S^1$ for every character $\varphi \in \Phi_{A(x)}$. The same chain of inclusions yields $\sigma_A(x) \subseteq S^1$. [/step] [step:Prove (3) for Hermitian $x$ via the boundary-inclusion principle] Now assume $B$ is a unital $C^*$-subalgebra of $A$ and $x \in B$ is Hermitian. We must show $\sigma_A(x) = \sigma_B(x)$. The inclusion $\sigma_A(x) \subseteq \sigma_B(x)$ holds by [Spectrum in a Subalgebra](/theorems/2673) (verified hypotheses: $B$ unital subalgebra of $A$ with the same unit, and $x \in B$). For the reverse inclusion $\sigma_B(x) \subseteq \sigma_A(x)$, we use the following two facts: *Fact A.* By part (1) applied in $B$, $\sigma_B(x) \subseteq \mathbb{R}$. (Here we apply (1) to the unital $C^*$-algebra $B$ and Hermitian $x \in B$.) *Fact B (boundary-inclusion principle, from* [Spectrum in a Subalgebra](/theorems/2673)*).* For any unital subalgebra $B \subseteq A$ with the same unit and $y \in B$, \begin{align*} \partial \sigma_B(y) \subseteq \sigma_A(y) \subseteq \sigma_B(y), \end{align*} where $\partial$ denotes the topological boundary in $\mathbb{C}$. We claim $\sigma_B(x) = \partial \sigma_B(x)$ in $\mathbb{C}$. To see this, observe that $\sigma_B(x) \subseteq \mathbb{R}$ (Fact A). A subset of $\mathbb{R}$ has empty interior in $\mathbb{C}$ (any open ball in $\mathbb{C}$ around a real point contains non-real points). Hence the interior of $\sigma_B(x)$ in $\mathbb{C}$ is empty, so $\sigma_B(x)$ equals its boundary in $\mathbb{C}$: \begin{align*} \sigma_B(x) = \overline{\sigma_B(x)} \setminus \operatorname{int}(\sigma_B(x)) = \overline{\sigma_B(x)} = \partial \sigma_B(x), \end{align*} using also that $\sigma_B(x)$ is closed in $\mathbb{C}$ (the spectrum is always compact, hence closed). Combining with Fact B: \begin{align*} \sigma_B(x) = \partial \sigma_B(x) \subseteq \sigma_A(x) \subseteq \sigma_B(x), \end{align*} which forces equality $\sigma_A(x) = \sigma_B(x)$ for Hermitian $x$. [/step] [step:Prove (3) for normal $x$ by reducing $(\lambda - x)$-invertibility to $(\bar\lambda - x^*)(\lambda - x)$-invertibility] Now let $B$ be a unital $C^*$-subalgebra of $A$ and $x \in B$ normal. Fix $\lambda \in \mathbb{C}$. Define the auxiliary element \begin{align*} \eta_\lambda := (\bar\lambda \cdot 1 - x^*)(\lambda \cdot 1 - x) \in B. \end{align*} We verify $\eta_\lambda$ is Hermitian: \begin{align*} \eta_\lambda^* = (\lambda \cdot 1 - x)^* (\bar\lambda \cdot 1 - x^*)^* = (\bar\lambda \cdot 1 - x^*)(\lambda \cdot 1 - x) = \eta_\lambda, \end{align*} using $((\bar\lambda \cdot 1 - x^*))^* = \lambda \cdot 1 - x$ (the involution is conjugate-linear and squares to the identity), and that $\bar\lambda \cdot 1 - x^*$ commutes with $\lambda \cdot 1 - x$ because $x$ and $x^*$ commute (normality). Indeed, $(\bar\lambda - x^*)(\lambda - x) = \lambda \bar\lambda - \bar\lambda x - \lambda x^* + x^*x$ and similarly $(\lambda - x)(\bar\lambda - x^*) = \lambda \bar\lambda - \lambda x^* - \bar\lambda x + x x^*$; these agree iff $x^* x = x x^*$, which is the normality hypothesis. We now claim: \begin{align*} \lambda \cdot 1 - x \in G(A) \iff \eta_\lambda \in G(A), \tag{$\star$} \end{align*} where $G(A)$ denotes the group of invertible elements of $A$, and similarly with $A$ replaced by $B$. $(\Rightarrow)$ If $\lambda - x \in G(A)$, then so is its adjoint $\bar\lambda - x^* = (\lambda - x)^*$ (because invertibility is preserved by the involution: $(y^{-1})^* = (y^*)^{-1}$). The product of two invertibles is invertible, so $\eta_\lambda = (\bar\lambda - x^*)(\lambda - x) \in G(A)$. $(\Leftarrow)$ Suppose $\eta_\lambda \in G(A)$, with inverse $\eta_\lambda^{-1} \in A$. Then \begin{align*} \eta_\lambda^{-1} \cdot (\bar\lambda - x^*) \cdot (\lambda - x) = 1. \end{align*} This exhibits $u := \eta_\lambda^{-1}(\bar\lambda - x^*)$ as a left inverse of $\lambda - x$. Symmetrically, since $x$ is normal, the elements $\lambda - x$ and $\bar\lambda - x^*$ commute, so $\eta_\lambda = (\bar\lambda - x^*)(\lambda - x) = (\lambda - x)(\bar\lambda - x^*)$, and the same $\eta_\lambda^{-1}$ gives \begin{align*} (\lambda - x) \cdot (\bar\lambda - x^*) \cdot \eta_\lambda^{-1} = 1, \end{align*} exhibiting $v := (\bar\lambda - x^*)\eta_\lambda^{-1}$ as a right inverse of $\lambda - x$. By the standard Banach algebra fact that having a left and a right inverse forces the two to coincide and to be the (two-sided) inverse, $\lambda - x \in G(A)$ with inverse $u = v$. The same argument applies verbatim with $A$ replaced by $B$ (since $x, x^* \in B$, $\eta_\lambda \in B$, and inverses computed within $B$ are also inverses in $A$). Now apply $(\star)$ in both algebras and the Hermitian case (Step 3) to $\eta_\lambda$ (verified hypothesis: $\eta_\lambda \in B$ is Hermitian; $B$ is a unital $C^*$-subalgebra of $A$): \begin{align*} \eta_\lambda \in G(A) \iff \eta_\lambda \in G(B), \tag{Step 3 applied to $\eta_\lambda$} \end{align*} where the equivalence comes from $\sigma_A(\eta_\lambda) = \sigma_B(\eta_\lambda)$, i.e.\ $0 \in \sigma_A(\eta_\lambda) \iff 0 \in \sigma_B(\eta_\lambda)$, i.e.\ $\eta_\lambda$ is non-invertible in $A$ iff non-invertible in $B$. (The case $\lambda = 0$ corresponds to $\eta_0 = x^* x$ being non-invertible.) Combining the equivalence above with $(\star)$ in both algebras: \begin{align*} \lambda - x \in G(A) \iff \eta_\lambda \in G(A) \iff \eta_\lambda \in G(B) \iff \lambda - x \in G(B). \end{align*} Equivalently, $\lambda \in \rho_A(x) \iff \lambda \in \rho_B(x)$, where $\rho$ denotes the resolvent set. Taking complements in $\mathbb{C}$, $\sigma_A(x) = \sigma_B(x)$. [guided] The strategy is to bootstrap from the Hermitian case (Step 3) to the general normal case using a Hermitian auxiliary. We need a way to encode the invertibility of $\lambda - x$ (which is a *non*-Hermitian element in general, even for normal $x$) in terms of the invertibility of a *Hermitian* element. The natural candidate is $(\lambda - x)^*(\lambda - x) = (\bar\lambda - x^*)(\lambda - x) =: \eta_\lambda$. **Why is $\eta_\lambda$ Hermitian?** Compute $\eta_\lambda^* = (\lambda - x)^*((\bar\lambda - x^*)^*)^* = (\lambda - x)^* (\bar\lambda - x^*) \cdot \, ?$ — we need this product to swap. Here normality enters: $x$ and $x^*$ commute, so $\lambda - x$ and $\bar\lambda - x^*$ commute, so the product reorders, and $\eta_\lambda^* = (\bar\lambda - x^*)(\lambda - x) = \eta_\lambda$. **Without normality the argument breaks here**: $(\lambda - x)^*(\lambda - x)$ is always Hermitian, but $\eta_\lambda$ as written might not equal its adjoint. The form $(\lambda - x)^*(\lambda - x)$ is automatic; what we need is that this particular Hermitian element $(\bar\lambda - x^*)(\lambda - x)$ also equals $(\lambda - x)(\bar\lambda - x^*)$ (so we can build a two-sided inverse), and that is exactly normality. **Why does invertibility of $\eta_\lambda$ encode invertibility of $\lambda - x$?** $(\Rightarrow)$ Direction follows directly from preservation of invertibility under products: if $\lambda - x$ has an inverse, so does its adjoint $\bar\lambda - x^* = (\lambda - x)^*$ (involution preserves the unit group: $(y^{-1})^* = (y^*)^{-1}$), and the product of invertibles is invertible. $(\Leftarrow)$ This is the substance. From $\eta_\lambda^{-1} \cdot \eta_\lambda = 1$ we get $\eta_\lambda^{-1}(\bar\lambda - x^*) \cdot (\lambda - x) = 1$, so $\lambda - x$ has a *left* inverse $u := \eta_\lambda^{-1}(\bar\lambda - x^*)$. By the commutativity granted by normality, we can also write $\eta_\lambda = (\lambda - x)(\bar\lambda - x^*)$, so $(\lambda - x) \cdot (\bar\lambda - x^*)\eta_\lambda^{-1} = 1$, giving a *right* inverse $v$. In a unital ring, having left and right inverses forces them to coincide and gives a two-sided inverse: $u = u \cdot 1 = u(\lambda - x)v = 1 \cdot v = v$. Hence $\lambda - x$ is invertible. **Why does this argument work in $A$ and $B$ simultaneously?** All the elements $\lambda - x$, $\bar\lambda - x^*$, $\eta_\lambda$ live in $B$ (since $B$ is closed under involution and $x \in B$). The inverses, when they exist, may a priori live in $A$ or $B$. The Hermitian case (Step 3) tells us *invertibility of $\eta_\lambda$ in $A$ is equivalent to invertibility in $B$* — that is the spectral permanence we just proved for Hermitian elements. Combined with $(\star)$ in both algebras, this transfers to invertibility of $\lambda - x$. **The chain.** We get \begin{align*} \lambda - x \in G(A) \overset{(\star)}{\iff} \eta_\lambda \in G(A) \overset{\text{Step 3}}{\iff} \eta_\lambda \in G(B) \overset{(\star)}{\iff} \lambda - x \in G(B), \end{align*} and taking complements in $\mathbb{C}$ gives $\sigma_A(x) = \sigma_B(x)$. **Where Hermiticity of $\eta_\lambda$ is used.** Step 3 only applies to Hermitian elements; that is the entire point of constructing $\eta_\lambda$. If $x$ failed to be normal, we would not be able to form a Hermitian $\eta_\lambda$ as a polynomial in $\lambda - x$ and $(\lambda - x)^*$ that captures invertibility — and indeed spectral permanence can fail for non-normal elements in general subalgebras (it is a theorem about *normal* elements). [/guided] [/step]