[proofplan] The proof is pointwise on the character space. First we verify that each transformed function is a bounded continuous scalar-valued function on $\Delta(A)$, using weak-* continuity of evaluations and the standard spectral-radius estimate for characters. Then we check linearity and multiplicativity by evaluating at an arbitrary character. Finally, under the involutive hypothesis, we compute the transform of $a^*$ pointwise and identify it with the pointwise involution of $\Gamma(a)$. [/proofplan]

[step:Show that each transformed element is a bounded continuous function]Fix $a\in A$. By definition, \begin{align*} \Gamma(a):\Delta(A)\to\mathbb C \end{align*} is the map given by $\tau\mapsto \tau(a)$. The weak-* topology on $\Delta(A)\subseteq A^*$ is the topology of pointwise convergence on $A$. Hence the evaluation map \begin{align*} \operatorname{ev}_a:A^*\to\mathbb C \end{align*} defined by $\operatorname{ev}_a(\phi)=\phi(a)$ is weak-* continuous, and $\Gamma(a)=\operatorname{ev}_a|_{\Delta(A)}$ is continuous. We use the standard character estimate for complex Banach algebras: every character $\tau$ on $A$ is continuous and satisfies \begin{align*} |\tau(x)|\le r_A(x)\le \|x\|_A \end{align*} for every $x\in A$, where $r_A(x)$ denotes the spectral radius of $x$ in $A$ or, in the nonunital case, in the unitization of $A$. Applying this estimate to $x=a$ gives \begin{align*} |\Gamma(a)(\tau)|=|\tau(a)|\le \|a\|_A \end{align*} for every $\tau\in\Delta(A)$. Taking the supremum over $\Delta(A)$ gives \begin{align*} \|\Gamma(a)\|_\infty=\sup_{\tau\in\Delta(A)}|\Gamma(a)(\tau)|\le \|a\|_A. \end{align*} Thus $\Gamma(a)\in C_b(\Delta(A))$, and the same inequality proves contractivity.[/step]

[guided]Fix $a\in A$. The Gelfand transform of $a$ is the scalar-valued function \begin{align*} \Gamma(a):\Delta(A)\to\mathbb C \end{align*} defined by $\Gamma(a)(\tau)=\tau(a)$ for each character $\tau\in\Delta(A)$. To prove that this is an element of $C_b(\Delta(A))$, we need two facts: continuity and boundedness. First consider continuity. The space $\Delta(A)$ is equipped with the weak-* topology inherited from $A^*$. By definition of the weak-* topology, for each fixed element $a\in A$, the evaluation map \begin{align*} \operatorname{ev}_a:A^*\to\mathbb C \end{align*} given by $\operatorname{ev}_a(\phi)=\phi(a)$ is continuous. Since $\Gamma(a)$ is exactly the restriction of $\operatorname{ev}_a$ to $\Delta(A)$, the restriction is continuous on $\Delta(A)$. Now consider boundedness. We use the standard character estimate for complex Banach algebras: every character $\tau$ on $A$ is continuous and satisfies \begin{align*} |\tau(x)|\le r_A(x)\le \|x\|_A \end{align*} for every $x\in A$, where $r_A(x)$ is the spectral radius of $x$ in $A$ or, if $A$ is nonunital, in the unitization of $A$. Applying this estimate with $x=a$ gives, for every $\tau\in\Delta(A)$, \begin{align*} |\Gamma(a)(\tau)|=|\tau(a)|\le \|a\|_A. \end{align*} Because the right-hand side is independent of $\tau$, the supremum over all characters is bounded by the same number: \begin{align*} \|\Gamma(a)\|_\infty=\sup_{\tau\in\Delta(A)}|\Gamma(a)(\tau)|\le \|a\|_A. \end{align*} This proves both that $\Gamma(a)$ is bounded and that the map $\Gamma:A\to C_b(\Delta(A))$ is contractive.[/guided]

[step:Verify linearity pointwise on characters] Let $a,b\in A$ and let $\lambda\in\mathbb C$. For every $\tau\in\Delta(A)$, the character $\tau:A\to\mathbb C$ is linear, so \begin{align*} \Gamma(a+b)(\tau)=\tau(a+b)=\tau(a)+\tau(b)=\Gamma(a)(\tau)+\Gamma(b)(\tau). \end{align*} Since this holds for every $\tau\in\Delta(A)$, we have \begin{align*} \Gamma(a+b)=\Gamma(a)+\Gamma(b). \end{align*} Similarly, \begin{align*} \Gamma(\lambda a)(\tau)=\tau(\lambda a)=\lambda\tau(a)=\lambda\Gamma(a)(\tau) \end{align*} for every $\tau\in\Delta(A)$, hence \begin{align*} \Gamma(\lambda a)=\lambda\Gamma(a). \end{align*} Therefore $\Gamma$ is complex-linear. [/step]

[step:Verify multiplicativity pointwise on characters] Let $a,b\in A$. For every $\tau\in\Delta(A)$, the character $\tau:A\to\mathbb C$ is multiplicative, so \begin{align*} \Gamma(ab)(\tau)=\tau(ab)=\tau(a)\tau(b)=\Gamma(a)(\tau)\Gamma(b)(\tau). \end{align*} The multiplication on $C_b(\Delta(A))$ is pointwise multiplication. Therefore the preceding identity for all $\tau\in\Delta(A)$ gives \begin{align*} \Gamma(ab)=\Gamma(a)\Gamma(b). \end{align*} Combining this with complex-linearity, $\Gamma$ is an algebra homomorphism. Together with the estimate from the first step, it is a contractive algebra homomorphism. [/step]

[step:Identify the transform of the involution with pointwise conjugation] Assume now that $A$ is involutive and that every character $\tau\in\Delta(A)$ satisfies $\tau(x^*)=\overline{\tau(x)}$ for every $x\in A$. Define the involution on $C_b(\Delta(A))$ by \begin{align*} f^*(\tau)=\overline{f(\tau)} \end{align*} for every $f\in C_b(\Delta(A))$ and every $\tau\in\Delta(A)$. Let $a\in A$. For every $\tau\in\Delta(A)$, the assumed $*$-preservation of $\tau$ gives \begin{align*} \Gamma(a^*)(\tau)=\tau(a^*)=\overline{\tau(a)}=\overline{\Gamma(a)(\tau)}=\Gamma(a)^*(\tau). \end{align*} Since the equality holds at every $\tau\in\Delta(A)$, we have \begin{align*} \Gamma(a^*)=\Gamma(a)^*. \end{align*} Thus $\Gamma$ preserves the involution. Since the previous steps already proved that $\Gamma$ is an algebra homomorphism, $\Gamma$ is a $*$-homomorphism under the stated additional hypotheses. [/step]