[proofplan] For each $a\in A$, contractivity of the GNS representations gives a uniform bound on the family $(\pi_\phi(a))_{\phi\in S}$, so the componentwise direct sum operator on $H_S$ is bounded. The algebra and adjoint identities are then checked componentwise, proving that $\pi_S$ is a $*$-representation. Finally, if $\pi_S(a)=0$, then every GNS representation annihilates $a$; evaluating on the cyclic vectors gives $\phi(a^*a)=0$ for all $\phi\in S$, and the separating hypothesis forces $a^*a=0$, hence $a=0$. [/proofplan]

[step:Use contractivity to define the direct sum operator]Fix $a\in A$. For each $\phi\in S$, the GNS representation \begin{align*} \pi_\phi:A\to\mathcal{L}(H_\phi) \end{align*} is a $*$-homomorphism between $C^*$-algebras, so [citetheorem:8547] gives \begin{align*} \|\pi_\phi(a)\|_{\mathcal{L}(H_\phi)}\leq \|a\|_A. \end{align*} Thus the family $(\pi_\phi(a))_{\phi\in S}$ is uniformly bounded. Let $\eta=(\eta_\phi)_{\phi\in S}\in H_S$. By definition of the Hilbert direct sum, \begin{align*} \sum_{\phi\in S}\|\eta_\phi\|_{H_\phi}^2<\infty, \end{align*} where the sum over $S$ is understood as the supremum of finite partial sums. For every finite subset $F\subset S$, \begin{align*} \sum_{\phi\in F}\|\pi_\phi(a)\eta_\phi\|_{H_\phi}^2 \leq \sum_{\phi\in F}\|\pi_\phi(a)\|_{\mathcal{L}(H_\phi)}^2\|\eta_\phi\|_{H_\phi}^2. \end{align*} Using the uniform bound above, \begin{align*} \sum_{\phi\in F}\|\pi_\phi(a)\eta_\phi\|_{H_\phi}^2 \leq \|a\|_A^2\sum_{\phi\in F}\|\eta_\phi\|_{H_\phi}^2. \end{align*} Taking the supremum over all finite $F\subset S$ gives \begin{align*} \sum_{\phi\in S}\|\pi_\phi(a)\eta_\phi\|_{H_\phi}^2 \leq \|a\|_A^2\sum_{\phi\in S}\|\eta_\phi\|_{H_\phi}^2. \end{align*} Therefore $(\pi_\phi(a)\eta_\phi)_{\phi\in S}\in H_S$, and the map \begin{align*} \pi_S(a):H_S&\to H_S \end{align*} \begin{align*} (\eta_\phi)_{\phi\in S}&\mapsto(\pi_\phi(a)\eta_\phi)_{\phi\in S} \end{align*} is bounded with \begin{align*} \|\pi_S(a)\|_{\mathcal{L}(H_S)}\leq \|a\|_A. \end{align*}[/step]

[guided]Fix $a\in A$. The only possible issue in defining $\pi_S(a)$ is boundedness on the Hilbert direct sum, especially when $S$ is infinite or uncountable. We address this by proving a uniform operator norm estimate for all components. For each $\phi\in S$, the GNS representation \begin{align*} \pi_\phi:A\to\mathcal{L}(H_\phi) \end{align*} is a $*$-homomorphism from the $C^*$-algebra $A$ into the $C^*$-algebra $\mathcal{L}(H_\phi)$. Hence the hypotheses of [citetheorem:8547] are satisfied, and that theorem gives \begin{align*} \|\pi_\phi(a)\|_{\mathcal{L}(H_\phi)}\leq \|a\|_A. \end{align*} The important point is that the right-hand side does not depend on $\phi$. Now let $\eta=(\eta_\phi)_{\phi\in S}\in H_S$. The Hilbert direct sum condition means \begin{align*} \sum_{\phi\in S}\|\eta_\phi\|_{H_\phi}^2<\infty, \end{align*} where for an arbitrary index set $S$ this sum is the supremum of sums over finite subsets of $S$. To prove that $(\pi_\phi(a)\eta_\phi)_{\phi\in S}$ also belongs to $H_S$, it is therefore enough to bound every finite partial sum uniformly. Let $F\subset S$ be finite. For each $\phi\in F$, the definition of the operator norm gives \begin{align*} \|\pi_\phi(a)\eta_\phi\|_{H_\phi} \leq \|\pi_\phi(a)\|_{\mathcal{L}(H_\phi)}\|\eta_\phi\|_{H_\phi}. \end{align*} Squaring and summing over $\phi\in F$ gives \begin{align*} \sum_{\phi\in F}\|\pi_\phi(a)\eta_\phi\|_{H_\phi}^2 \leq \sum_{\phi\in F}\|\pi_\phi(a)\|_{\mathcal{L}(H_\phi)}^2\|\eta_\phi\|_{H_\phi}^2. \end{align*} Using the uniform contractive estimate, \begin{align*} \sum_{\phi\in F}\|\pi_\phi(a)\eta_\phi\|_{H_\phi}^2 \leq \|a\|_A^2\sum_{\phi\in F}\|\eta_\phi\|_{H_\phi}^2. \end{align*} Taking the supremum over all finite subsets $F\subset S$ yields \begin{align*} \sum_{\phi\in S}\|\pi_\phi(a)\eta_\phi\|_{H_\phi}^2 \leq \|a\|_A^2\sum_{\phi\in S}\|\eta_\phi\|_{H_\phi}^2. \end{align*} Thus $(\pi_\phi(a)\eta_\phi)_{\phi\in S}\in H_S$, and the assignment \begin{align*} \pi_S(a):H_S&\to H_S \end{align*} \begin{align*} (\eta_\phi)_{\phi\in S}&\mapsto(\pi_\phi(a)\eta_\phi)_{\phi\in S} \end{align*} defines a [bounded linear operator](/page/Bounded%20Linear%20Operator) satisfying \begin{align*} \|\pi_S(a)\|_{\mathcal{L}(H_S)}\leq \|a\|_A. \end{align*}[/guided]

[step:Check the representation identities componentwise] Let $a,b\in A$, let $\lambda\in\mathbb C$, and let $\eta=(\eta_\phi)_{\phi\in S}\in H_S$. Since each $\pi_\phi$ is linear, \begin{align*} \pi_S(a+\lambda b)\eta = (\pi_\phi(a+\lambda b)\eta_\phi)_{\phi\in S} = (\pi_\phi(a)\eta_\phi+\lambda\pi_\phi(b)\eta_\phi)_{\phi\in S}. \end{align*} Hence \begin{align*} \pi_S(a+\lambda b)=\pi_S(a)+\lambda\pi_S(b). \end{align*} Since each $\pi_\phi$ is multiplicative, \begin{align*} \pi_S(ab)\eta = (\pi_\phi(ab)\eta_\phi)_{\phi\in S} = (\pi_\phi(a)\pi_\phi(b)\eta_\phi)_{\phi\in S} = \pi_S(a)\pi_S(b)\eta. \end{align*} Thus \begin{align*} \pi_S(ab)=\pi_S(a)\pi_S(b). \end{align*} It remains to verify preservation of adjoints. Let $\eta=(\eta_\phi)_{\phi\in S}$ and $\zeta=(\zeta_\phi)_{\phi\in S}$ be elements of $H_S$. The Hilbert direct sum [inner product](/page/Inner%20Product) is \begin{align*} (\eta,\zeta)_{H_S} = \sum_{\phi\in S}(\eta_\phi,\zeta_\phi)_{H_\phi}. \end{align*} Using the adjoint identity for each $\pi_\phi$, \begin{align*} (\pi_S(a)\eta,\zeta)_{H_S} = \sum_{\phi\in S}(\pi_\phi(a)\eta_\phi,\zeta_\phi)_{H_\phi}. \end{align*} Therefore \begin{align*} (\pi_S(a)\eta,\zeta)_{H_S} = \sum_{\phi\in S}(\eta_\phi,\pi_\phi(a^*)\zeta_\phi)_{H_\phi} = (\eta,\pi_S(a^*)\zeta)_{H_S}. \end{align*} Thus $\pi_S(a)^*=\pi_S(a^*)$. Therefore $\pi_S:A\to\mathcal{L}(H_S)$ is a $*$-representation. [/step]

[step:Use the cyclic GNS vectors to convert kernel membership into state evaluations] Let $a\in\ker\pi_S$. Then $\pi_S(a)=0$, so for every $\phi\in S$ the component operator satisfies \begin{align*} \pi_\phi(a)=0. \end{align*} Indeed, if $\eta\in H_\phi$, the vector in $H_S$ with $\eta$ in the $\phi$-component and $0$ in all other components is sent by $\pi_S(a)$ to the vector with $\pi_\phi(a)\eta$ in the $\phi$-component and $0$ in all other components. For each $\phi\in S$, the GNS vector $\xi_\phi\in H_\phi$ satisfies \begin{align*} \phi(x)=(\pi_\phi(x)\xi_\phi,\xi_\phi)_{H_\phi} \end{align*} for every $x\in A$. Applying this with $x=a^*a$ gives \begin{align*} \phi(a^*a) = (\pi_\phi(a^*a)\xi_\phi,\xi_\phi)_{H_\phi}. \end{align*} Since $\pi_\phi$ is a $*$-representation, \begin{align*} \phi(a^*a) = (\pi_\phi(a)^*\pi_\phi(a)\xi_\phi,\xi_\phi)_{H_\phi}. \end{align*} By the [Hilbert space](/page/Hilbert%20Space) adjoint identity, \begin{align*} \phi(a^*a) = (\pi_\phi(a)\xi_\phi,\pi_\phi(a)\xi_\phi)_{H_\phi} = \|\pi_\phi(a)\xi_\phi\|_{H_\phi}^2. \end{align*} Because $\pi_\phi(a)=0$, this gives \begin{align*} \phi(a^*a)=0 \end{align*} for every $\phi\in S$. [/step]

[step:Apply separation and the $C^*$-identity to prove faithfulness] Suppose, toward proving injectivity, that $a\in\ker\pi_S$. The preceding step shows that \begin{align*} \phi(a^*a)=0 \end{align*} for every $\phi\in S$. If $a^*a\neq 0$, then the separating property of $S$ applied to the nonzero element $a^*a\in A$ would give a state $\phi\in S$ such that \begin{align*} \phi(a^*a)\neq 0, \end{align*} contradicting the displayed equality for all $\phi\in S$. Hence \begin{align*} a^*a=0. \end{align*} Using the $C^*$-identity in $A$, \begin{align*} \|a\|_A^2=\|a^*a\|_A=0. \end{align*} Therefore $a=0$. Thus $\ker\pi_S=\{0\}$, so $\pi_S$ is faithful. This completes the proof. [/step]