[proofplan] We verify the four parts in order. First, that the Gelfand map is well-defined and is a unital algebra homomorphism: the values $\hat{x}(\varphi) = \varphi(x)$ assemble into a continuous function on $\Phi_A$ by the very definition of the weak-$*$ topology, and homomorphism properties pass through evaluation at each $\varphi$. Statement (i) is the spectral-radius identity from theorem 2677 combined with $r_A(x) \le \|x\|$ (theorem 2672). Statement (ii) computes the spectrum in $C(\Phi_A)$ as the image of the function (a standard fact: a continuous function on a compact Hausdorff space is invertible iff it is non-vanishing, so its spectrum is its image), and matches it against the description $\sigma_A(x) = \hat{x}(\Phi_A)$ from theorem 2677(ii). Statement (iii) is the case $\lambda = 0$ of (ii). [/proofplan] [step:Verify the Gelfand map is well-defined and continuous as a map $A \to C(\Phi_A)$] Fix $x \in A$. The function $\hat{x} : \Phi_A \to \mathbb{C}$, $\varphi \mapsto \varphi(x)$, is the restriction to $\Phi_A \subseteq A^*$ of the canonical evaluation functional $\operatorname{ev}_x : A^* \to \mathbb{C}$, $\operatorname{ev}_x(f) = f(x)$. By definition, the weak-$*$ topology on $A^*$ is the coarsest topology making each $\operatorname{ev}_x$ continuous. Hence $\operatorname{ev}_x$ is continuous on $A^*$ for the weak-$*$ topology, and its restriction $\hat{x}$ to the weak-$*$-closed subset $\Phi_A$ is continuous. The character space $\Phi_A$ is compact Hausdorff: by [Characters are Continuous](/theorems/2675), each $\varphi \in \Phi_A$ has $\|\varphi\|_{A^*} \le 1$, so $\Phi_A$ sits inside the closed unit ball $B_{A^*}$. By the **Banach–Alaoglu Theorem**, $B_{A^*}$ is weak-$*$ compact. The set $\Phi_A$ is weak-$*$ closed in $B_{A^*}$ — it is the intersection of the closed sets $\{\varphi \in B_{A^*} : \varphi(xy) = \varphi(x)\varphi(y)\}$ over all $x, y \in A$ and $\{\varphi \in B_{A^*} : \varphi(1) = 1\}$, each cut out by the continuous evaluation functionals $\operatorname{ev}_x$. As a closed subset of a compact Hausdorff space, $\Phi_A$ is compact Hausdorff. Therefore $\hat{x} \in C(\Phi_A)$ for every $x \in A$, so the Gelfand map $\widehat{\;\;} : A \to C(\Phi_A)$ is well-defined. [/step] [step:Verify the Gelfand map is a unital algebra homomorphism] Let $x, y \in A$ and $\lambda \in \mathbb{C}$. For every $\varphi \in \Phi_A$: \begin{align*} \widehat{x + y}(\varphi) &= \varphi(x + y) = \varphi(x) + \varphi(y) = \hat{x}(\varphi) + \hat{y}(\varphi),\\ \widehat{\lambda x}(\varphi) &= \varphi(\lambda x) = \lambda \varphi(x) = \lambda \hat{x}(\varphi),\\ \widehat{xy}(\varphi) &= \varphi(xy) = \varphi(x)\varphi(y) = \hat{x}(\varphi)\hat{y}(\varphi) = (\hat{x} \cdot \hat{y})(\varphi),\\ \widehat{1_A}(\varphi) &= \varphi(1_A) = 1 = 1_{C(\Phi_A)}(\varphi). \end{align*} The first three identities use linearity ($\varphi$ is linear), multiplicativity ($\varphi$ is multiplicative), and the fourth uses that a character is unital ($\varphi(1) = 1$). Pointwise equality on $\Phi_A$ gives equality of continuous functions on $\Phi_A$. Therefore the Gelfand map is linear, multiplicative, and sends $1_A$ to the constant function $1$, hence is a unital algebra homomorphism. [/step] [step:Prove (i): $\|\hat{x}\|_\infty = r_A(x) \le \|x\|$] By the sup-norm definition on $C(\Phi_A)$ (where $\Phi_A$ is compact), \begin{align*} \|\hat{x}\|_\infty = \sup_{\varphi \in \Phi_A} |\hat{x}(\varphi)| = \sup_{\varphi \in \Phi_A} |\varphi(x)|. \end{align*} By [Spectrum via Characters](/theorems/2677) (iii), this last supremum equals $r_A(x)$. By the [Beurling–Gelfand Spectral Radius Formula](/theorems/2672), \begin{align*} r_A(x) = \lim_{n \to \infty} \|x^n\|^{1/n} \le \|x\|, \end{align*} where the inequality uses submultiplicativity $\|x^n\| \le \|x\|^n$ to bound $\|x^n\|^{1/n} \le \|x\|$. Combining, \begin{align*} \|\hat{x}\|_\infty = r_A(x) \le \|x\|. \end{align*} In particular, the Gelfand map is a bounded linear map with operator norm at most $1$, hence continuous. [/step] [step:Prove (ii): $\sigma_{C(\Phi_A)}(\hat{x}) = \sigma_A(x)$] We show both sides equal the image $\hat{x}(\Phi_A) = \{\varphi(x) : \varphi \in \Phi_A\}$. For the right-hand side, [Spectrum via Characters](/theorems/2677) (ii) gives \begin{align*} \sigma_A(x) = \{\varphi(x) : \varphi \in \Phi_A\} = \hat{x}(\Phi_A). \end{align*} For the left-hand side, fix $\lambda \in \mathbb{C}$. By definition, \begin{align*} \lambda \in \sigma_{C(\Phi_A)}(\hat{x}) \iff (\lambda \cdot 1 - \hat{x}) \notin G(C(\Phi_A)). \end{align*} The function $\lambda \cdot 1 - \hat{x} : \Phi_A \to \mathbb{C}$ is continuous on the compact Hausdorff space $\Phi_A$. A standard fact about $C(K)$ for $K$ compact Hausdorff: $f \in C(K)$ is invertible iff $f$ is nowhere zero, with inverse $1/f$ continuous on $K$ (continuity of $1/f$ uses that $f$ is bounded away from zero, which follows from compactness of $K$ when $f$ does not vanish). Therefore \begin{align*} \lambda \cdot 1 - \hat{x} \notin G(C(\Phi_A)) &\iff \exists\, \varphi \in \Phi_A : (\lambda \cdot 1 - \hat{x})(\varphi) = 0\\ &\iff \exists\, \varphi \in \Phi_A : \lambda - \varphi(x) = 0\\ &\iff \lambda \in \hat{x}(\Phi_A). \end{align*} Hence $\sigma_{C(\Phi_A)}(\hat{x}) = \hat{x}(\Phi_A) = \sigma_A(x)$. [guided] The strategy is to compute both spectra explicitly and observe they coincide. **Spectrum of $\hat{x}$ in $C(\Phi_A)$.** Recall that for a compact Hausdorff space $K$, the spectrum of $f \in C(K)$ is the image $f(K)$. Why? Because $f - \lambda \cdot 1$ is invertible in $C(K)$ iff it is nowhere zero on $K$ — and on a compact Hausdorff space, a nowhere-zero continuous function is bounded below in modulus (by compactness of $K$ and continuity of $|f - \lambda|$, $\inf_K |f - \lambda| > 0$), so $1/(f - \lambda)$ is continuous. Conversely, if $(f - \lambda)(\varphi_0) = 0$ then any function $g$ with $g(f - \lambda) = 1$ would have to satisfy $0 = 1$ at $\varphi_0$, impossible. So $\lambda \in \sigma(f) \iff \lambda \in f(K)$. Applying this with $f = \hat{x}$, $K = \Phi_A$: \begin{align*} \sigma_{C(\Phi_A)}(\hat{x}) = \hat{x}(\Phi_A) = \{\varphi(x) : \varphi \in \Phi_A\}. \end{align*} **Spectrum of $x$ in $A$.** [Theorem 2677](/theorems/2677) part (ii) gives \begin{align*} \sigma_A(x) = \{\varphi(x) : \varphi \in \Phi_A\}. \end{align*} **Conclusion.** Both sides have been described as $\{\varphi(x) : \varphi \in \Phi_A\}$, so they are equal. The point of the Gelfand representation is that it makes this equality structural: $x$ and $\hat{x}$ have the same spectrum because $\hat{x}$ literally records the values $\varphi(x)$ over all characters. [/guided] [/step] [step:Prove (iii) by specialising (ii) to $\lambda = 0$] By the standard characterisation of invertibility, \begin{align*} x \in G(A) &\iff 0 \notin \sigma_A(x),\\ \hat{x} \in G(C(\Phi_A)) &\iff 0 \notin \sigma_{C(\Phi_A)}(\hat{x}). \end{align*} By (ii), $\sigma_A(x) = \sigma_{C(\Phi_A)}(\hat{x})$, so $0 \notin \sigma_A(x) \iff 0 \notin \sigma_{C(\Phi_A)}(\hat{x})$. Therefore \begin{align*} x \in G(A) \iff \hat{x} \in G(C(\Phi_A)), \end{align*} proving (iii) and completing the proof. [/step]