[proofplan] We realize $A$ concretely by applying the [universal representation theorem](/theorems/8566) to obtain a faithful unital *-representation $\pi_u:A\to\mathcal{L}(H_u)$ on a [Hilbert space](/page/Hilbert%20Space) $H_u$. Faithfulness lets us invoke the isometry theorem for faithful *-representations, so $\pi_u$ preserves the C*-norm. Its image is therefore a unital *-subalgebra isometric to the [Banach space](/page/Banach%20Space) $A$, and a direct completeness argument shows that this image is closed in the operator norm. [/proofplan]

[step:Apply the universal representation to obtain a faithful unital representation]Let $S(A)$ denote the state space of $A$, that is, the set of positive linear functionals $\phi:A\to\mathbb C$ satisfying $\phi(1_A)=1$. By the universal representation theorem for unital C*-algebras [citetheorem:8566], applied to the given unital C*-algebra $A$, there exist a complex Hilbert space $H_u$ and a faithful *-representation \begin{align*} \pi_u:A\to\mathcal{L}(H_u) \end{align*} defined as the direct sum of the GNS representations over $S(A)$. The cited theorem states precisely that this representation is faithful, hence injective, and is a *-representation into the C*-algebra $\mathcal{L}(H_u)$ of bounded linear operators on $H_u$. Since it is the universal representation of a unital C*-algebra, it is unital, so \begin{align*} \pi_u(1_A)=I_{H_u}, \end{align*} where $I_{H_u}$ denotes the identity operator on $H_u$.[/step]

[guided]The first task is to construct the Hilbert space on which $A$ will act. Let $S(A)$ denote the state space of $A$, that is, the set of positive linear functionals $\phi:A\to\mathbb C$ satisfying $\phi(1_A)=1$. We use the universal representation theorem [citetheorem:8566]. Its hypotheses are satisfied because the theorem applies to every unital C*-algebra, and $A$ is a unital C*-algebra by assumption. The theorem gives a complex Hilbert space $H_u$ and a map \begin{align*} \pi_u:A\to\mathcal{L}(H_u) \end{align*} which is a faithful *-representation. Here $\mathcal{L}(H_u)$ is the C*-algebra of bounded linear operators on $H_u$. Being a *-representation means that for all $a,b\in A$ and all $\lambda\in\mathbb C$, \begin{align*} \pi_u(a+\lambda b)=\pi_u(a)+\lambda\pi_u(b), \end{align*} \begin{align*} \pi_u(ab)=\pi_u(a)\pi_u(b), \end{align*} and \begin{align*} \pi_u(a^*)=\pi_u(a)^*. \end{align*} Faithful means that $\ker\pi_u=\{0\}$, so $\pi_u$ is injective. The representation is also unital in the unital universal representation, which gives \begin{align*} \pi_u(1_A)=I_{H_u}. \end{align*} This is the construction step: instead of trying to build operators directly from abstract elements of $A$, we let the universal representation assemble all state-induced GNS representations into one faithful action.[/guided]

[step:Use faithfulness to identify the norm on $A$ with the operator norm]The hypotheses of the isometry theorem for faithful representations are satisfied: $A$ is a C*-algebra, $\pi_u:A\to\mathcal{L}(H_u)$ is a *-representation, and $\pi_u$ is faithful. Therefore, by [citetheorem:8563], \begin{align*} \|\pi_u(a)\|_{\mathcal{L}(H_u)}=\|a\|_A \end{align*} for every $a\in A$.[/step]

[guided]At this point we have a faithful *-representation \begin{align*} \pi_u:A\to\mathcal{L}(H_u). \end{align*} The isometry theorem for faithful representations [citetheorem:8563] applies because its hypotheses are exactly that $A$ is a C*-algebra, that $H_u$ is a complex Hilbert space, that $\pi_u$ is a *-representation into $\mathcal{L}(H_u)$, and that $\pi_u$ is faithful. The first hypothesis is the theorem assumption, and the remaining hypotheses were obtained from the universal representation theorem. Therefore [citetheorem:8563] gives, for every $a\in A$, \begin{align*} \|\pi_u(a)\|_{\mathcal{L}(H_u)}=\|a\|_A. \end{align*} This is the norm-identification step: faithfulness prevents the representation from collapsing nonzero elements, and the C*-identity then forces the represented norm to agree with the original C*-norm.[/guided]

[step:Show that the image is a norm-closed unital star subalgebra]Define \begin{align*} B:=\pi_u(A)\subset \mathcal{L}(H_u). \end{align*} Because $\pi_u$ is linear and multiplicative, $B$ is a complex subalgebra of $\mathcal{L}(H_u)$. Because $\pi_u(a^*)=\pi_u(a)^*$ for every $a\in A$, the set $B$ is closed under the adjoint operation. Because $\pi_u(1_A)=I_{H_u}$, the subalgebra $B$ is unital and has the same unit as $\mathcal{L}(H_u)$. It remains to prove norm-closedness. Let $(T_n)_{n\in\mathbb N}$ be a sequence in $B$ converging in the operator norm $\|\cdot\|_{\mathcal{L}(H_u)}$ to some operator $T\in\mathcal{L}(H_u)$. For each $n\in\mathbb N$, choose $a_n\in A$ such that \begin{align*} T_n=\pi_u(a_n). \end{align*} For $m,n\in\mathbb N$, the isometry already proved gives \begin{align*} \|a_n-a_m\|_A=\|\pi_u(a_n-a_m)\|_{\mathcal{L}(H_u)}. \end{align*} By linearity of $\pi_u$, \begin{align*} \|\pi_u(a_n-a_m)\|_{\mathcal{L}(H_u)}=\|T_n-T_m\|_{\mathcal{L}(H_u)}. \end{align*} Since $(T_n)_{n\in\mathbb N}$ converges in $\mathcal{L}(H_u)$, it is Cauchy; hence $(a_n)_{n\in\mathbb N}$ is Cauchy in $A$. The space $A$ is complete because it is a C*-algebra, so there exists $a\in A$ such that \begin{align*} a_n\to a \end{align*} in the norm $\|\cdot\|_A$. Applying the isometry again, \begin{align*} \|\pi_u(a_n)-\pi_u(a)\|_{\mathcal{L}(H_u)}=\|a_n-a\|_A\to 0. \end{align*} Thus $T_n\to\pi_u(a)$ in $\mathcal{L}(H_u)$. Since limits in the normed space $\mathcal{L}(H_u)$ are unique and also $T_n\to T$, we have \begin{align*} T=\pi_u(a)\in B. \end{align*} Therefore $B$ is norm-closed in $\mathcal{L}(H_u)$.[/step]

[guided]Define the candidate concrete algebra by \begin{align*} B:=\pi_u(A)\subset \mathcal{L}(H_u). \end{align*} Because $\pi_u$ is linear and multiplicative, sums, scalar multiples, and products of elements of $B$ remain in $B$. Because $\pi_u(a^*)=\pi_u(a)^*$ for every $a\in A$, the set $B$ is closed under taking adjoints. Because $\pi_u(1_A)=I_{H_u}$, the identity operator of $\mathcal{L}(H_u)$ belongs to $B$, so $B$ is a unital *-subalgebra with the inherited unit. It remains to check the topological condition: $B$ must be closed in the operator norm. Let $(T_n)_{n\in\mathbb N}$ be a sequence in $B$ converging in $\mathcal{L}(H_u)$ to an operator $T\in\mathcal{L}(H_u)$ with respect to the norm $\|\cdot\|_{\mathcal{L}(H_u)}$. Since each $T_n$ lies in $B=\pi_u(A)$, for every $n\in\mathbb N$ choose $a_n\in A$ such that \begin{align*} T_n=\pi_u(a_n). \end{align*} For $m,n\in\mathbb N$, the isometry from the preceding step gives \begin{align*} \|a_n-a_m\|_A=\|\pi_u(a_n-a_m)\|_{\mathcal{L}(H_u)}. \end{align*} By linearity of $\pi_u$ and the identities $T_n=\pi_u(a_n)$ and $T_m=\pi_u(a_m)$, this becomes \begin{align*} \|a_n-a_m\|_A=\|T_n-T_m\|_{\mathcal{L}(H_u)}. \end{align*} Since $(T_n)_{n\in\mathbb N}$ converges in the normed space $\mathcal{L}(H_u)$, it is Cauchy. The displayed equality therefore implies that $(a_n)_{n\in\mathbb N}$ is Cauchy in $A$. Now we use completeness. A C*-algebra is a Banach *-algebra, so $A$ is complete for $\|\cdot\|_A$. Hence there exists $a\in A$ such that $a_n\to a$ in $\|\cdot\|_A$. Applying the same isometry again gives \begin{align*} \|\pi_u(a_n)-\pi_u(a)\|_{\mathcal{L}(H_u)}=\|a_n-a\|_A\to 0. \end{align*} Thus $T_n\to\pi_u(a)$ in $\mathcal{L}(H_u)$. But also $T_n\to T$ by assumption, and limits in a normed space are unique. Therefore \begin{align*} T=\pi_u(a)\in B. \end{align*} Every operator-norm limit of a sequence in $B$ lies in $B$, so $B$ is norm-closed in $\mathcal{L}(H_u)$.[/guided]

[step:Conclude that the representation is the required isometric star isomorphism]Define \begin{align*} \Phi:A\to B \end{align*} by \begin{align*} \Phi(a):=\pi_u(a) \end{align*} for every $a\in A$. The map $\Phi$ is surjective by the definition $B=\pi_u(A)$ and injective because $\pi_u$ is faithful. Since $\pi_u$ is linear, multiplicative, unital, and *-preserving, $\Phi$ is a unital *-isomorphism from $A$ onto $B$. The norm identity from the isometry step gives \begin{align*} \|\Phi(a)\|_{\mathcal{L}(H_u)}=\|a\|_A \end{align*} for every $a\in A$. Taking $H:=H_u$, we have realized $A$ as the norm-closed unital *-subalgebra $B\subset\mathcal{L}(H)$, isometrically and as a unital *-algebra.[/step]

[guided]Define the map \begin{align*} \Phi:A\to B \end{align*} by \begin{align*} \Phi(a):=\pi_u(a) \end{align*} for every $a\in A$. This map has codomain $B$ because $B$ was defined to be $\pi_u(A)$. It is surjective by that same definition: every element of $B$ has the form $\pi_u(a)=\Phi(a)$ for some $a\in A$. It is injective because $\pi_u$ is faithful, equivalently $\ker\pi_u=\{0\}$. The algebraic structure is inherited from the representation. Since $\pi_u$ is linear, multiplicative, unital, and *-preserving, the map $\Phi$ is a unital *-isomorphism from $A$ onto $B$. The norm identity already proved says that for every $a\in A$, \begin{align*} \|\Phi(a)\|_{\mathcal{L}(H_u)}=\|\pi_u(a)\|_{\mathcal{L}(H_u)}=\|a\|_A. \end{align*} Thus $\Phi$ is isometric. Finally set $H:=H_u$. The preceding step proved that $B\subset\mathcal{L}(H_u)=\mathcal{L}(H)$ is a norm-closed unital *-subalgebra, so $A$ is realized as such a concrete operator algebra through the isometric unital *-isomorphism $\Phi$.[/guided]