[proofplan] We prove both directions. First, the kernel of a representation is closed because $*$-homomorphisms between $C^*$-algebras are contractive, hence continuous, and it is a two-sided ideal by linearity and multiplicativity. Conversely, from a closed two-sided ideal $I$ we form the quotient $C^*$-algebra $A/I$, represent it faithfully on a [Hilbert space](/page/Hilbert%20Space), and compose that faithful representation with the quotient map. The kernel computation then reduces exactly to the identity $\ker q=I$ for the quotient map $q:A\to A/I$. [/proofplan]

[step:Show that the kernel of a representation is a closed two-sided ideal]Assume that $H$ is a complex Hilbert space and that \begin{align*} \pi:A\to B(H) \end{align*} is a $*$-representation. Since $B(H)$ is a $C^*$-algebra and $\pi$ is a $*$-homomorphism, [Automatic Continuity of Star Homomorphisms][citetheorem:8547] gives \begin{align*} \|\pi(a)\|_{B(H)}\leq \|a\|_A \end{align*} for every $a\in A$. Hence $\pi$ is continuous. The singleton $\{0\}$ is closed in the norm topology of $B(H)$, so \begin{align*} \ker \pi=\pi^{-1}(\{0\}) \end{align*} is closed in $A$ by continuity of $\pi$. We next verify the algebraic ideal properties. Since $\pi$ is linear, $\ker\pi$ is a linear subspace of $A$. Let $x\in\ker\pi$ and let $a\in A$. Multiplicativity of $\pi$ gives \begin{align*} \pi(ax)=\pi(a)\pi(x)=\pi(a)0=0. \end{align*} Thus $ax\in\ker\pi$. Similarly, \begin{align*} \pi(xa)=\pi(x)\pi(a)=0\pi(a)=0, \end{align*} so $xa\in\ker\pi$. Therefore $\ker\pi$ is a closed two-sided ideal of $A$.[/step]

[guided]Assume that $H$ is a complex Hilbert space and that \begin{align*} \pi:A\to B(H) \end{align*} is a $*$-representation. We need to prove three things about $\ker\pi$: it is closed, it is a linear subspace, and it is stable under multiplication on both sides by arbitrary elements of $A$. The closedness is the only topological point. The algebra $B(H)$ is a $C^*$-algebra with its operator norm, and $\pi$ is a $*$-homomorphism between $C^*$-algebras. Therefore [Automatic Continuity of Star Homomorphisms][citetheorem:8547] applies and gives the contractive estimate \begin{align*} \|\pi(a)\|_{B(H)}\leq \|a\|_A \end{align*} for every $a\in A$. In particular, $\pi$ is continuous as a map from the normed space $A$ to the normed space $B(H)$. Now $\{0\}$ is closed in $B(H)$ because normed spaces are Hausdorff. Since the preimage of a [closed set](/page/Closed%20Set) under a continuous map is closed, \begin{align*} \ker \pi=\pi^{-1}(\{0\}) \end{align*} is closed in $A$. The remaining part uses only the homomorphism properties. Since $\pi$ is linear, if $x,y\in\ker\pi$ and $\lambda,\mu\in\mathbb C$, then \begin{align*} \pi(\lambda x+\mu y)=\lambda\pi(x)+\mu\pi(y)=0, \end{align*} so $\lambda x+\mu y\in\ker\pi$. Thus $\ker\pi$ is a linear subspace. Finally, let $x\in\ker\pi$ and let $a\in A$. Multiplicativity gives \begin{align*} \pi(ax)=\pi(a)\pi(x)=\pi(a)0=0, \end{align*} and hence $ax\in\ker\pi$. The same computation on the other side gives \begin{align*} \pi(xa)=\pi(x)\pi(a)=0\pi(a)=0, \end{align*} so $xa\in\ker\pi$. Therefore $\ker\pi$ is a closed two-sided ideal of $A$.[/guided]

[step:Represent the quotient by the ideal faithfully] Assume conversely that $I\subset A$ is a closed two-sided ideal. By [Quotient Norm Theorem For C Star Algebras][citetheorem:8568], the quotient [vector space](/page/Vector%20Space) $A/I$, equipped with the quotient norm, quotient multiplication, and involution \begin{align*} (a+I)^*=a^*+I, \end{align*} is a $C^*$-algebra. Let \begin{align*} q:A\to A/I \end{align*} be the quotient map defined by $q(a)=a+I$. This map is a surjective $*$-homomorphism and satisfies \begin{align*} \ker q=I. \end{align*} If $I=A$, take $H=\{0\}$ and let \begin{align*} \pi:A\to B(H) \end{align*} be the zero $*$-representation. Then $\ker\pi=A=I$. It remains to treat the case $I\neq A$. Then $A/I$ is a nonzero $C^*$-algebra. By the Gelfand-Naimark representation theorem for $C^*$-algebras, obtained in the unital case from [Abstract Gelfand-Naimark Theorem][citetheorem:8567] and in the nonunital case by applying the unital theorem to the $C^*$-unitization and restricting to the original algebra, there exist a complex Hilbert space $H$ and a faithful $*$-representation \begin{align*} \rho:A/I\to B(H). \end{align*} Faithfulness means \begin{align*} \ker\rho=\{0_{A/I}\}. \end{align*} [/step]

[step:Compose the quotient map with the faithful representation and compute the kernel] Define \begin{align*} \pi:A\to B(H) \end{align*} by \begin{align*} \pi(a)=\rho(q(a)) \end{align*} for every $a\in A$. Since $q$ and $\rho$ are $*$-homomorphisms, their composition $\pi=\rho\circ q$ is a $*$-representation. We compute its kernel. For $a\in A$, \begin{align*} a\in\ker\pi \end{align*} if and only if \begin{align*} \rho(q(a))=0. \end{align*} Since $\rho$ is faithful, this is equivalent to \begin{align*} q(a)=0_{A/I}. \end{align*} By the definition of the quotient map, this is equivalent to \begin{align*} a\in I. \end{align*} Therefore \begin{align*} \ker\pi=I. \end{align*} Together with the first direction, this proves that the closed two-sided ideals of $A$ are exactly the kernels of $*$-representations of $A$. [/step]