[proofplan] The first two parts use a generic principle: the closure of a commutative subalgebra is commutative, and adjoining $1$ to a commutative subalgebra produces a commutative subalgebra. By maximality, both these enlargements must equal $C$ itself, giving closedness and unitality. Part (iii), spectral equality, requires more work: we already have $\sigma_C(x) \supseteq \sigma_A(x)$ from the **Spectrum in a Subalgebra** theorem. For the reverse, given $\lambda \notin \sigma_A(x)$ we set $z := (\lambda 1 - x)^{-1} \in A$; the goal is to show $z \in C$. The inverse of an element commuting with all of $C$ also commutes with all of $C$ (a small algebraic fact we prove explicitly). Hence $C \cup \{z\}$ generates a commutative subalgebra of $A$ properly containing $C$ unless $z \in C$, so by maximality $z \in C$, giving $\lambda \notin \sigma_C(x)$. [/proofplan] [step:Show $C$ is closed using maximality and continuity of multiplication] Let $\overline{C}$ denote the closure of $C$ in $A$. We claim $\overline{C}$ is also a commutative subalgebra of $A$. [claim:$\overline{C}$ is a commutative subalgebra of $A$] We verify the subalgebra axioms and commutativity. *Subalgebra*: Let $u, v \in \overline{C}$ and $\alpha, \beta \in \mathbb{C}$. Choose $u_n, v_n \in C$ with $u_n \to u$, $v_n \to v$. Then $\alpha u_n + \beta v_n \in C$ (since $C$ is a subalgebra), and $\alpha u_n + \beta v_n \to \alpha u + \beta v$ by continuity of vector-space operations. Hence $\alpha u + \beta v \in \overline{C}$. Similarly $u_n v_n \in C$ and $u_n v_n \to uv$ by joint continuity of multiplication in the Banach algebra $A$ (the bound $\|u_n v_n - uv\| \le \|u_n\| \|v_n - v\| + \|u_n - u\| \|v\|$ converges to $0$ since $(u_n)$ is bounded and both differences tend to $0$). Hence $uv \in \overline{C}$. *Commutativity*: With the same notation, $u_n v_n = v_n u_n$ in $C$. Passing to the limit (continuity of multiplication), $uv = vu$. [/claim] [proof] Self-contained: the subalgebra and commutativity axioms are verified by limit-passage, justified by joint continuity of multiplication, which holds in any Banach algebra by submultiplicativity $\|ab\| \le \|a\|\|b\|$. [/proof] Since $C \subseteq \overline{C}$ and $\overline{C}$ is a commutative subalgebra, the maximality of $C$ forces $\overline{C} = C$, so $C$ is closed. [/step] [step:Show $1 \in C$ using maximality] Consider $C' := C + \mathbb{C} \cdot 1 = \{u + \alpha 1 : u \in C, \alpha \in \mathbb{C}\}$. We claim $C'$ is a commutative subalgebra of $A$ containing $C$. Closure under linear combinations is immediate. For products: for $u + \alpha 1, v + \beta 1 \in C'$, \begin{align*} (u + \alpha 1)(v + \beta 1) = uv + \alpha v + \beta u + \alpha \beta 1 \in C + \mathbb{C} \cdot 1 = C' \end{align*} since $uv, \alpha v, \beta u \in C$. Commutativity: \begin{align*} (u + \alpha 1)(v + \beta 1) - (v + \beta 1)(u + \alpha 1) = uv - vu = 0 \end{align*} because $u, v \in C$ commute and $1$ commutes with everything in $A$. Each element of $C$ lies in $C'$ (take $\alpha = 0$ in the definition). By maximality of $C$, $C' = C$, so $1 \in C$. In particular $C$ is unital with $1_C = 1_A$. [/step] [step:Establish $\sigma_C(x) \supseteq \sigma_A(x)$ via the subalgebra inclusion] Apply [Spectrum in a Subalgebra](/theorems/2673) to the closed unital subalgebra $C \subseteq A$ (closedness from Step 1, unitality from Step 2 with $1_C = 1_A$). The hypotheses of theorem 2673 are: $A$ unital BA (given), $C$ closed unital subalgebra (verified). The conclusion gives $\sigma_C(x) \supseteq \sigma_A(x)$ for every $x \in C$. [/step] [step:Establish the reverse $\sigma_C(x) \subseteq \sigma_A(x)$ by showing inverses lie in $C$] Fix $x \in C$ and $\lambda \notin \sigma_A(x)$, so $\lambda 1 - x \in G(A)$. Set \begin{align*} z := (\lambda 1 - x)^{-1} \in A. \end{align*} We show $z \in C$ in two sub-steps. [claim:$z$ commutes with every $y \in C$] Fix $y \in C$. Since $x \in C$ and $C$ is commutative (and contains $1$), $\lambda 1 - x \in C$ and commutes with $y$: \begin{align*} y (\lambda 1 - x) = (\lambda 1 - x) y. \end{align*} Multiply both sides on the left and right by $z = (\lambda 1 - x)^{-1}$: \begin{align*} z y (\lambda 1 - x) z &= z (\lambda 1 - x) y z \\ z y \cdot 1 &= 1 \cdot y z \\ z y &= y z. \end{align*} Hence $z$ commutes with every $y \in C$. [/claim] [proof] The argument is the standard "inverse of a commuting element commutes": from $y a = ay$ with $a$ invertible, multiplying by $a^{-1}$ on both sides yields $a^{-1} y = y a^{-1}$. Self-contained above. [/proof] Now consider the subalgebra $C'' := \mathrm{alg}(C \cup \{z\})$ generated in $A$ by $C$ and $z$. Concretely, \begin{align*} C'' = \left\{ \sum_{k=0}^N u_k z^k : N \ge 0, \ u_0, \dots, u_N \in C \right\}, \end{align*} the set of polynomials in $z$ with coefficients in $C$. [claim:$C''$ is a commutative subalgebra of $A$] Closure under linear combinations and products of polynomials in $z$ with coefficients in $C$ follows from the polynomial algebra structure: the product of two such polynomials is again such a polynomial, since $C$ is a subalgebra (closed under products and sums of its own elements) and $z$ commutes with every $u \in C$ (by the previous claim). Commutativity: any two such polynomials $\sum u_k z^k$ and $\sum v_j z^j$ have product \begin{align*} \sum_{k,j} u_k v_j z^{k+j}, \end{align*} and reversing the order gives $\sum_{k,j} v_j u_k z^{k+j} = \sum_{k,j} u_k v_j z^{k+j}$ since $u_k v_j = v_j u_k$ in $C$. Hence $C''$ is commutative. [/claim] [proof] The polynomial-ring structure with central element $z$ (commuting with the coefficient ring $C$) is a standard construction; commutativity of products reduces to commutativity of $C$, since $z$ already commutes with all coefficients. [/proof] Since $C \subseteq C''$ (take $N = 0$) and $C''$ is a commutative subalgebra of $A$, maximality of $C$ forces $C'' = C$. In particular $z = z^1 \in C''= C$, so \begin{align*} (\lambda 1 - x)^{-1} \in C. \end{align*} This shows $\lambda 1 - x$ is invertible in $C$, i.e.\ $\lambda \notin \sigma_C(x)$. Contrapositively, $\sigma_C(x) \subseteq \sigma_A(x)$. Combined with Step 3, $\sigma_C(x) = \sigma_A(x)$. [guided] The structural pattern of this proof is **maximality forces equality with any natural enlargement**. Steps 1 and 2 instantiate this for closure and unit-adjunction. Step 4 instantiates it again, but with a less obvious enlargement: starting from an inverse $z = (\lambda 1 - x)^{-1}$ that lives in the larger algebra $A$, we ask whether $z \in C$. Maximality reduces this to the question: is the algebra generated by $C$ and $z$ commutative? If yes, then $C$ together with $z$ is a commutative subalgebra extending $C$, which by maximality must be $C$ itself, forcing $z \in C$. Why does $z$ commute with $C$? Because $\lambda 1 - x \in C$ does (since $C$ is commutative), and inverses of commuting elements commute. The algebraic identity $y a = ay \implies y a^{-1} = a^{-1} y$ is the engine: multiplying $y a = ay$ on the left and right by $a^{-1}$, \begin{align*} a^{-1} y a a^{-1} = a^{-1} a y a^{-1}, \quad \text{i.e.}, \quad a^{-1} y = y a^{-1}. \end{align*} The polynomial algebra $C[z]$ generated by $C$ together with the central element $z$ is then commutative: products commute because both the coefficients (in $C$) commute among themselves and with $z$, and powers of $z$ commute with each other. So $C[z]$ is a commutative subalgebra of $A$ containing $C$, and maximality slams the door: $C[z] = C$, $z \in C$. The conclusion $\sigma_C(x) = \sigma_A(x)$ is the key spectral fact: in a maximal commutative subalgebra, the spectrum is the same as in the ambient algebra. This is sharp — for non-maximal commutative subalgebras the inclusion $\sigma_C \supseteq \sigma_A$ can be strict, with extra "holes" appearing in $\sigma_C$ that are filled in $\sigma_A$. [/guided] [/step]