Is a $*$-representation $\pi:A\to B(H)$ non-degenerate iff $\overline{\pi(A) B(H)} = B(H)$?

Question

Let $A$ be a $C^*$-algebra and $\pi:A\to B(H)$ a $*$-representation on a Hilbert space $H$ (that means a $*$-homomorphism). Is it true that $\pi$ is non-degenerate iff $\overline{\pi(A) B(H)} = B(H)$ where closure is taken with respect to the operator norm?

I call $\pi$ non-degenerate iff one of the following equivalent properties is true:

1. $\overline{\pi(A)H}=H$
2. $\overline{span(\pi(A)H)}=H$
3. $\bigcap_{T\in A} \ker \pi(T) = \{0\}$
4. $\pi(A)x\ne\{0\}$ for all $0\ne x\in H$
5. $\pi(A)T\ne\{0\}$ for all $0\ne T\in B(H)$
6. For all bounded approximate units $(u_\lambda)_\lambda$ of $A$ and $x\in H$ we have $\pi(u_\lambda) x\to x$

I was able to show the 6 properties above to be equivalent, but I don't know how to conclude from them that $\overline{\pi(A)B(H)}=B(H)$.

Answer 1

It's not true. The ideal of compact operators $A = K(H)$ is a proper norm closed ideal, and a C$^*$ subalgebra of $B(H)$. So let $A = K(H)$ and $\pi$ the identity representation of $A$ on $H$. This representation is non-degenerate, for example by property (4).