[proofplan] The general Gelfand theory of commutative unital Banach algebras already gives that $\Gamma$ is a continuous unital algebra homomorphism into $C(\Phi_A)$, with $\Phi_A$ compact Hausdorff in the weak-$*$ topology and $\|\hat{x}\|_\infty = r_A(x)$ (the spectral radius). We must add three properties special to the $C^*$-setting: (a) $\Gamma$ respects the involution, i.e.\ $\widehat{x^*} = \overline{\hat{x}}$; (b) $\Gamma$ is isometric, which uses the spectral-radius-equals-norm result for normal elements (every element of a commutative algebra is normal); (c) $\Gamma$ is surjective, which follows from the Stone-Weierstrass theorem applied to the image $\hat{A} \subseteq C(\Phi_A)$ — a unital, conjugation-closed, point-separating, norm-closed subalgebra is the entire $C(\Phi_A)$. [/proofplan]

[step:Recall the general Gelfand theory of commutative unital Banach algebras] Since $A$ is a commutative unital Banach algebra (a $C^*$-algebra is a Banach algebra), the [Gelfand Representation Theorem](/theorems/2678) applies. Theorem 2678 requires $A$ commutative and unital Banach (verified). Its conclusions, which we use, are: (a) The character space $\Phi_A$, equipped with the weak-$*$ topology induced from $A^*$, is a non-empty compact Hausdorff space. (b) The Gelfand map $\Gamma: A \to C(\Phi_A)$, $x \mapsto \hat{x}$, $\hat{x}(\varphi) := \varphi(x)$, is well-defined (each $\hat{x}$ is continuous on $\Phi_A$), linear, multiplicative, unital ($\hat{1} = 1_{C(\Phi_A)}$, the constant function $1$), and continuous with $\|\Gamma\| \leq 1$. (c) For every $x \in A$, $\|\hat{x}\|_\infty = r_A(x) \leq \|x\|$, where $r_A(x)$ is the spectral radius. These are the building blocks. We add three additional properties below: $*$-preservation, isometry, and surjectivity. [/step]

[step:Verify $\Gamma$ is a $*$-homomorphism using that characters are star-preserving] Fix $x \in A$. By the previous theorem **Characters Are Star-Homomorphisms**, since $A$ is a unital $C^*$-algebra and every $\varphi \in \Phi_A$ is a character on $A$, we have \begin{align*} \varphi(x^*) = \overline{\varphi(x)} \qquad \text{for all } \varphi \in \Phi_A. \end{align*} Translate this into a statement about Gelfand transforms. The involution on $C(\Phi_A)$ is pointwise complex conjugation: for $f \in C(\Phi_A)$, $f^*(\varphi) = \overline{f(\varphi)}$. Then for every $\varphi \in \Phi_A$, \begin{align*} \widehat{x^*}(\varphi) = \varphi(x^*) = \overline{\varphi(x)} = \overline{\hat{x}(\varphi)} = (\hat{x})^*(\varphi). \end{align*} The two continuous functions $\widehat{x^*}$ and $(\hat{x})^*$ agree on every $\varphi \in \Phi_A$, hence are equal in $C(\Phi_A)$: \begin{align*} \widehat{x^*} = (\hat{x})^*, \qquad \text{i.e.\ } \Gamma(x^*) = \Gamma(x)^*. \end{align*} Thus $\Gamma$ is a $*$-homomorphism. [/step]

[step:Prove $\Gamma$ is isometric using $r_A(x) = \|x\|$ for normal elements] Fix $x \in A$. Since $A$ is commutative, every element commutes with its adjoint: $x x^* = x^* x$. Hence $x$ is normal. By [Spectral Radius Equals Norm for Normal Elements](/theorems/2686) applied to the unital $C^*$-algebra $A$ and the normal element $x$ (verified hypotheses), we have \begin{align*} r_A(x) = \|x\|. \end{align*} Combining with Step 1(c), $\|\hat{x}\|_\infty = r_A(x) = \|x\|$. Hence $\Gamma$ is isometric: \begin{align*} \|\Gamma(x)\|_\infty = \|x\| \qquad \text{for all } x \in A. \end{align*} [/step]

[step:Prove $\Gamma$ is surjective via the Stone-Weierstrass theorem]Let $\hat{A} := \Gamma(A) = \{\hat{x} : x \in A\} \subseteq C(\Phi_A)$. We show $\hat{A} = C(\Phi_A)$. The Stone-Weierstrass theorem (complex version) states: if $K$ is a compact Hausdorff space and $\mathcal{B} \subseteq C(K)$ is a sub-$*$-algebra (closed under addition, scalar multiplication, multiplication, and complex conjugation) that is unital (contains the constant function $1$) and separates points (for every $\varphi \neq \psi$ in $K$ there exists $f \in \mathcal{B}$ with $f(\varphi) \neq f(\psi)$), then $\mathcal{B}$ is dense in $C(K)$ in the supremum norm. We verify each hypothesis for $\mathcal{B} = \hat{A}$ and $K = \Phi_A$: *Sub-algebra and unitality.* $\Gamma$ is a unital algebra homomorphism (Step 1(b)), so $\hat{A}$ is the image of an algebra homomorphism, hence closed under addition, scalar multiplication, and multiplication; and $\hat{1} = 1 \in \hat{A}$. *Conjugation-closed.* For $\hat{x} \in \hat{A}$, the complex conjugate is $\overline{\hat{x}} = \hat{x}^* = \widehat{x^*}$ by Step 2; since $x^* \in A$, $\widehat{x^*} \in \hat{A}$. *Separation of points.* Suppose $\varphi, \psi \in \Phi_A$ are distinct characters. Then there exists $x \in A$ with $\varphi(x) \neq \psi(x)$ (otherwise $\varphi = \psi$ as functionals on $A$). For this $x$, $\hat{x}(\varphi) = \varphi(x) \neq \psi(x) = \hat{x}(\psi)$, so $\hat{x}$ separates $\varphi$ and $\psi$. By Stone-Weierstrass, $\hat{A}$ is dense in $C(\Phi_A)$. *Closure.* By Step 3, $\Gamma$ is isometric. Since $A$ is complete (a $C^*$-algebra is in particular a Banach space) and $\Gamma$ is an isometric linear map, $\hat{A} = \Gamma(A)$ is complete in the sup norm, hence closed in $C(\Phi_A)$. A dense and closed subset of $C(\Phi_A)$ is all of $C(\Phi_A)$. Hence $\hat{A} = C(\Phi_A)$, i.e.\ $\Gamma$ is surjective.[/step]

[guided]We need to show $\Gamma: A \to C(\Phi_A)$ is **surjective**. The strategy is to identify $\hat{A} = \Gamma(A)$ as a subset of $C(\Phi_A)$ and apply the Stone-Weierstrass theorem to conclude $\hat{A}$ is **dense**, and then use isometry of $\Gamma$ to conclude $\hat{A}$ is **closed** — together giving $\hat{A} = C(\Phi_A)$. **The Stone-Weierstrass theorem (complex version).** Let $K$ be compact Hausdorff. A subset $\mathcal{B} \subseteq C(K)$ is dense in $(C(K), \|\cdot\|_\infty)$ provided: 1. $\mathcal{B}$ is a $*$-subalgebra (closed under $+$, scalar $\cdot$, pointwise product, and complex conjugation), 2. $\mathcal{B}$ is unital (contains the constant function $1$), 3. $\mathcal{B}$ separates points of $K$. We verify these for $\mathcal{B} = \hat{A}$ and $K = \Phi_A$. **Why $\hat{A}$ is a subalgebra.** $\Gamma$ is a unital algebra homomorphism (this is general Gelfand theory, theorem 2678): $\Gamma(x + y) = \Gamma(x) + \Gamma(y)$, $\Gamma(xy) = \Gamma(x)\Gamma(y)$, $\Gamma(1_A) = 1_{C(\Phi_A)}$. So the image is a unital subalgebra. **Why $\hat{A}$ is conjugation-closed.** Here is where Step 2 enters: $\widehat{x^*} = \overline{\hat{x}}$. This says: the complex conjugate of any element of $\hat{A}$ is again in $\hat{A}$ (it's the Gelfand transform of $x^*$). *Without the previous theorem* (**Characters Are Star-Homomorphisms**), this property would not be available; the Gelfand theory of *general* commutative Banach algebras gives a homomorphism $\Gamma$, but not a $*$-homomorphism. **Why $\hat{A}$ separates points of $\Phi_A$.** Two distinct characters $\varphi, \psi$ are *by definition* distinct as linear functionals on $A$, so there exists $x \in A$ with $\varphi(x) \neq \psi(x)$. The function $\hat{x}: \Phi_A \to \mathbb{C}$, $\varphi \mapsto \varphi(x)$, separates them: $\hat{x}(\varphi) \neq \hat{x}(\psi)$. This is **automatic** for the Gelfand topology — the weak-$*$ topology on $\Phi_A$ is exactly the topology making all the $\hat{x}$ continuous, so distinct points are distinguished by some $\hat{x}$. **Conclusion of Stone-Weierstrass.** $\hat{A}$ is dense in $C(\Phi_A)$. **Why $\hat{A}$ is closed.** Step 3 gives $\|\Gamma(x)\|_\infty = \|x\|$ for all $x \in A$. So $\Gamma: A \to \hat{A}$ is an isometric bijection. Since $A$ is complete (Banach), the image $\hat{A}$ is complete, hence closed in $C(\Phi_A)$ (a complete subset of a metric space is closed). **Density + closedness = surjectivity.** A dense closed subset of any topological space is the whole space. Hence $\hat{A} = C(\Phi_A)$. **Where the $C^*$-axiom enters.** The whole argument breaks for general commutative Banach algebras. The two crucial $C^*$-specific inputs are: - *$*$-preservation* (Step 2), needed to show $\hat{A}$ is conjugation-closed. Stone-Weierstrass requires conjugation-closure to handle complex-valued functions; without it we would need to work over $\mathbb{R}$ and miss the imaginary parts. - *Isometry* (Step 3), needed to show $\hat{A}$ is closed. In a general Banach algebra, $\|\hat{x}\|_\infty = r_A(x)$ can be strictly less than $\|x\|$ (e.g.\ a non-zero nilpotent has $r_A = 0$ but positive norm), so $\Gamma$ would be strictly contractive and the image would not be closed. It is precisely the convergence of these two $C^*$-features — both flowing from the $C^*$-identity $\|x^*x\| = \|x\|^2$ — that makes $\Gamma$ an isometric $*$-isomorphism, and hence makes $A$ literally the algebra of continuous functions on its character space.[/guided]

[step:Conclude that $\Gamma$ is an isometric $*$-isomorphism with $K = \Phi_A$] Combining Steps 1-4: $\Gamma$ is a unital algebra homomorphism (Step 1), a $*$-homomorphism (Step 2), isometric (Step 3), and surjective (Step 4). An isometric linear map is automatically injective. Hence $\Gamma: A \to C(\Phi_A)$ is an isometric $*$-isomorphism. The space $K = \Phi_A$ is compact Hausdorff (Step 1(a)). This completes the proof. [/step]