Let $A$ be a C$^*$-algebra. Let $\mathrm{Irr}(A)=\{[\pi]: \pi \text{ is an irreducible representation of } A\}$ and $\rho\in [\pi]$ if there is an unitary operator $V:H_\pi \to H_\rho$ such that $V\pi(a)=\rho(a)V$ for all $a\in A$.
Why is the representation $$\bigoplus_{[\pi]\in \mathrm{Irr}(A)}\pi:A\to L(\bigoplus_{[\pi]}H_\pi)$$ of A faithful? Here is $\bigoplus_{[\pi]\in \mathrm{Irr}(A)} \pi(a)(\sum_\pi h_\pi)=\sum_\pi \pi(a)h_\pi$.
We proved in lecture: Let $A$ be a C$^*$-Algebra. For all $a\in A$ is $$\| a\|=\sup_{[\pi]\in Irr(A)}\|\pi(a)\|.$$ I think this should give us that the direct sum is an isometry, i.e. faithful. But is $\sup_{[\pi]\in \mathrm{Irr}(A)}\|\pi(a)\|=\|\bigoplus_{[\pi]\in \mathrm{Irr}(A)}\pi(a)\|_{op}$? $\|\bigoplus_{[\pi]\in \mathrm{Irr}(A)}\pi(a)\|_{op}$ means the operatornorm of $\bigoplus_{[\pi]\in \mathrm{Irr}(A)}\pi(a)$.
Yes, it is the same norm. And you don't have to work hard to know it is an isometry: it is enough to know that $\bigoplus\pi$ is one-to-one, and then it is automatically an isometry.
That $\bigoplus\pi$ is one-to-one follows from the equality $\|a\|=\sup_{[\pi]}\|\pi(a)\|$: if $\bigoplus\pi(a)=0$, then $\pi(a)=0$ for all irrep $\pi$, and so $\|a\|=0$.