The picture below comes from Murphy's book.
My Questions: (1) In the sixth line of the picture, why $u(x)\in K$? Note that from $pu=up$, we only know that $K$ reduces $u$.
(2)I think the proof can end once we obtain that $u(x)=\lim_{n\to\infty}v_n(x)$ at the end of the first paragraph, because $v_n$ strongly converges to $u$ is equivalent to $u(x)=\lim_{n\to\infty}v_n(x)$ for every $x\in H$. Are the paragraphs below showing this in detail?
(3)Does the word "dense" in the lemma mean $A^{''}\subset \bar{A}$ rather than $A^{''}=\bar{A}$?
Update: I found just now that if $C$ is a *-subsubset of $B(H)$, then its commutant $C':=\{a\in A|ca=ac \forall c\in C\}$ is weakly closed, so there is no ambiguity of the meaning of the word "dense".
1: Since $x\in K$ (this is where one uses that $A$ is unital), you have $px=x$. Thus $$ ux=upx=pux\in K. $$
2: The fact that you find the sequence $v_n$ does not imply that $u$ is in the sot-closure. Being in the sot-closure means that there is a sequence $\{w_n\}$ such that for every $y$, $w_ny\to uy$. What Murphy is doing is applying the first paragraph to $n$-tuples: that's because the sot topology has a basis of neighbourhoods, around $u$, of the form $$ V_{x_1,\ldots,x_m}=\{t\in B(H):\ |(t-u)x_j|<1\} $$ indexed by all the $n$-tuples $x_1,\ldots,x_m$.