Confused on some parts in the proof of $p \in A'' \implies (A')_p=(A_p)'$

Question

I was reading C-star Algebra and operator theory by Murphy and I stuck at the proof of the lemma $4.1.6$. I am attaching below:

In the proof of the Lemma $4.1.6$ Murphy want to show that if $p \in A'' \implies (A')_p=(A_p)'.$ Now I understand the part that $(A')_p\subseteq(A_p)'$. But I am facing with the converse. Let me point them out.

$(1)$ If $v \in A$ then $v_pu=uv_p$, so $(pvp)_p\tilde{u}_p=\tilde{u}_p(pvp)p$. I understand that if $v \in A$ then $v_pu=uv_p$, but how does this implies $(pvp)_p\tilde{u}_p=\tilde{u}_p(pvp)p$?
$(2)$ And again, how does $(pvp)_p\tilde{u}_p=\tilde{u}_p(pvp)p$ implies $pvp\tilde{u}=\tilde{u}pvp$?
$(3)$ And lastly how does $pvp\tilde{u}=\tilde{u}pvp$ implies $v\tilde{u}=\tilde{u}v$?

Please help to understand me the aforementioned things. Thanks for your time.

Answer 1

1. Since $\tilde u_p=u$ and $(pvp)_p=v_p$, $$\tag1 (pvp)_p\tilde u_p=v_pu=uv_p=\tilde u_p(pvp)_p. $$

2. Because $\tilde u\in pB(H)p$, $$ \tilde u_p=p\tilde up=p\tilde u=\tilde u p. $$ The only difference between $pvp\,\tilde u$ and $(pvp)_p\,\tilde u_p$ is that the first operator acts on $H$ (but its $0$ on $(1-p)H$ while the second one acts on $pH$. So the equality $(1)$, seen in $H$ instead of $pH$, is $$ pvp\,\tilde u=\tilde u\,pvp. $$

3. Since $p\in A'$, $$ v\tilde u=vp\tilde u=pvp\tilde u=\tilde upvp=\tilde upv=\tilde u v. $$