An extension of representation

Question

Let $A,~B$ be two C*-algebras, if $A$ is an ideal in $B$, then do we have that any representation of $A$ can extend to a representation of $B$?

Answer 1

Yes.

Let $\pi:A\to B(H)$ is a representation. We can assume that $\overline{\pi(A)H}=H$ (otherwise, we replace $H$ with $\overline{\pi(A)H}$).

Now you can define $\tilde\pi:B\to B(H)$ on the dense subspace $\pi(A)H$ by $$ \tilde\pi(b)\,\pi(a)\xi=\pi(ba)\xi. $$ This is well-defined: if $\pi(a)\xi=\pi(a')\xi$, then $\pi(a-a')\xi=0$, and \begin{align} \|\pi(b(a-a'))\xi\|^2&=\langle \pi(b(a-a'))\xi,\pi(b(a-a'))\xi\rangle =\langle \pi((a-a')^*b^*)\pi(b(a-a'))\xi,\xi\rangle\\ &=\langle \pi((a-a')^*b^*b(a-a'))\xi,\xi\rangle\leq\|b\|^2\, \langle \pi((a-a')^*(a-a'))\xi,\xi\rangle\\ &=\|b\|^2\,\|\pi(a-a')\xi\|^2=0. \end{align} So $\pi(ba)\xi=\pi(ba')\xi$.

The same estimates above show that $\tilde\pi$ is bounded.

It is elementary to check that $\tilde\pi$ is a representation.