strong limit-operator is multiplicative

Question

I have questions regarding the proof of the following theorem, there are some gaps in the proof which we discussed in lecture.

Theorem: Let $A,\;B$ two $C^*$-algebras and $\pi:A\odot B\to B(H)$ a non-degenerate $*$-representation ($A\odot B$ denotes the tensor product of $A$ and $B$ as a $*$-algebra). Then there exist unique non-degenerate $*$-representations $\pi_A:A\to B(H)$, $\pi_B:B\to B(H)$ such that $$\pi(a\otimes b)=\pi_A(a)\pi_B(b)=\pi_B(b)\pi_A(a)$$ for all $a\in A,\; b\in B$.

Proof (only the important passage): Let $(u_\lambda)_{\lambda}\subset A$ and $(v_\mu)_{\mu}\subset B$ be approximate units. Define for fixed $a\in A$, $b\in B$ $$\pi_A(a):=\lim\limits_{\mu} \pi(a\otimes v_\mu),\; \pi_B(b):=\lim\limits_{\lambda} \pi(u_\lambda\otimes b),$$ where the limits denote the strong operator topology-limit ( i.e. $\|\pi_A(a)h-\pi(a\otimes v_\mu)h\|_H\to 0$ for all $h\in H$).

My questions:

1. How to justify the uniqueness? The uniqueness comes from the uniqueness of strong operator topology-limits, is it correct?

2. How to prove that $\pi_A$ (for $\pi_B$ analogously) is multiplicative?

**Edit:**I'm also stuck to prove that the definition of $\pi_A(a)$ doesn't depend on the choice of the approximate unit $(v_\mu )\subseteq B$. (Or shall I open a new question for this?)

For this let $(f_k)\subseteq B$ be another approximate unit. We know that the strong operator topology -limit $\lim_k \pi (a\otimes f_k)$ exists. Set $y_a=\lim_k \pi (a\otimes f_k)$. We must prove $y_a=\pi_A(a)$ (or equivalently $\|(y_a-\pi_A(a) )x\|=0$ for all $x\in H$). But I'm stuck to prove this.

Answer 1

For the uniqueness, if $\pi(a\times b)=\pi_A'(a)\pi_B'(b)=\pi_B'(b)\pi_A'(a)$, then $$ \pi_A'(a)=\lim \pi_A'(a)\pi_B'(v_\mu)=\lim \pi(a\otimes v_\mu) =\lim \pi_A(a)\pi_B(v_\mu)=\pi_A(a), $$ where we are using that a representation maps an approximate unit to an approximate unit (no strong topology here).

The sot is multiplicative when the nets involved are bounded: if $\|x_k\|\leq \alpha$ for all $k$, then if $x_k\to x$ and $y_j\to y$ sot, $$ \|(x_ky_j-xy)h\|\leq\|x_k(y_j-y)h\|+\|(x_k-x)yh\|\leq\alpha\|(y_j-y)h\|+\|(x_k-x)yh\|\to0. $$

So, now using sot limits, $$ \pi_A(a_1a_2)=\lim \pi(a_1a_2\otimes v_mu)=\lim\pi(a_1\otimes v_\mu)\pi(a_2\otimes v_\mu)=\lim\pi(a_1\otimes v_\mu)\lim\pi(a_2\otimes v_\mu) =\pi_A(a_1)\pi_A(a_2), $$ where we are using that $\|\pi(a_1\otimes v_\mu)\|\leq\|a_1\otimes v_\mu\|\leq\|a_1\|$, and the aforementioned fact that the sot is multiplicative when the nets are bounded.