It is common that in order to obtain a $C^*$-algebra from a $^*$-algebra $A$ one defines a norm on $A$ by $$\|x\|=\sup\{\|\pi(x)\|\,|\,\pi\ \text{is a }^*\text{-representation of }A\}.$$ However, I have seen in a lot of places that this is usually written as $$\|x\|=\sup\{\|\pi(x)\|\,|\,\pi\ \text{is a (cyclic) }^*\text{-representation of }A\},$$ implying that any for any $^*$-representation $\pi$ of $A$ and $\varepsilon>0$, one can find a cyclic $^*$-representation $\pi_0$ of $A$ such that $\|\pi(x)\|\leq\|\pi_0(x)\|+\varepsilon$. I want to know why this is true. I have been able to prove that it holds for unital $C^*$-algebras:
Let $\varepsilon>0$ and let $\pi\colon A\to B(H)$ be a representation. Let $P=\pi(1_A)$ and $H_1=P(H)$ (where $1_A$ denotes the identity of $A$). Note that for all $a\in A$ and $\eta\in H_1$, we have $P\pi(a)\eta=\pi(a)\eta$. Hence $\pi(a)\eta\in H_1$, so we can define a map $\pi_1\colon A\to B(H_1)$ by $$\pi_1(a)\eta=\pi(a)\eta,\quad a\in A,\ \eta\in H_1.$$ $\pi'$ is then well-defined. Note moreover that $\pi(a)=\pi(a)P=P\pi(a)$ for all $a\in A$. Then $$\pi(a)\eta=\pi(a)P\eta=\pi_1(a)P\eta.$$ Hence $\|\pi(a)\eta\|=\|\pi_1(a)P\eta\|\leq\|\pi'(a)\|\|\eta\|$, so $\|\pi(a)\|=\|\pi_1(a)\|$. Choose $\xi\in H_1$ with $\|\xi\|=1$ such that $\|\pi_1(x)\xi\|\geq\|\pi_1(x)\|-\varepsilon$. Put $H_0=\overline{\pi_1(A)\xi}\subseteq H_1$. Let $\eta\in H_0$ and $(y_n)_{n\geq 1}$ in $A$ such that $\varphi(y_n)\xi\to\eta$. Then $$\pi_1(a)\eta=\lim_{n\to\infty}\pi_1(a)\pi_1(y_n)\xi=\lim_{n\to\infty}\pi_1(ay_n)\xi\in\overline{\pi_1(A)\xi}= H_0$$ by continuity of $\pi_1(a)$ for all $a\in A$, so $\pi_1(A)H_0\subseteq H_0$. Define a map $\pi_0\colon A\to B(H_0)$ by $$\pi_0(a)\eta=\pi_1(a)\eta,\quad a\in A,\ \eta\in H_0.$$ Then $\pi_0$ is a well-defined representation, since $\pi_1(a)\eta\in H_0$ for all $a\in A$ and $\eta\in H_0$. Note that $\pi_0(x)\eta=\pi(x)\eta$ for all $\eta\in H_0$. Since $\xi=P\xi=\pi_1(1_A)\xi\in H_0$, we then have $\pi_1(A)\xi=\pi_0(A)\xi,$ so that $\pi_0$ is cyclic, and $$\|\pi_0(x)\|\geq\|\pi_0(x)\xi\|=\|\pi_1(x)\xi\|\geq\|\pi_1(x)\|-\varepsilon=\|\pi(x)\|-\varepsilon,$$ as wanted.
Does anybody know of another proof of the wanted result, or just of a way to generalize the above to non-unital $^*$-algebras?
I think it is a lot simpler than that.
Fix $x\in A$. Since the norm satisfies $\|x\|^2=\|x^*x\|$, it is enough to assume that $x$ is positive. Let $\pi:A\to B(H)$ be a $*$-representation with $\pi(x)\ne0$. Let $B\subset B(H)$ be the closure of $\pi(A)$.
Let $\varphi$ be a state of $B$ with $|\varphi(\pi(x))|=\|\pi(x)\|$. Now consider the GNS representation $\pi_\varphi$ corresponding to $\varphi$. This is a cyclic representation, and $$ \|\pi(x)\|=|\varphi(\pi(x))|=|\langle \pi_\varphi(\pi(x))\Omega_\varphi,\Omega_\varphi\rangle|\leq\|\pi_\varphi\circ\pi(x)\|. $$ The representation $\pi_\varphi\circ\pi:A\to B(H_\varphi)$ is cyclic. So this shows that the second norm in your question is at least as big as the first one. The other inequality is trivial.
Note: cyclicity can be guaranteed if one replaces $H_\varphi$ with $\overline{\pi_\varphi(\pi(A))H_\varphi}$.