In Lemma 4.1.4. Murphy is doing something strange that I don't seem to understand. for each $x\in H$ we find $v_n(x)$ such that it converges to $u(x)$. we wish to prove that is holds for EVERY $x$.
Murphy goes on to make this construction on $H^n$ so that in the end he can take $W$ (neighbourhood of $u$ in the strong topology) and show that there there is at least some $v_n(x)$ so it is indeed the sot closure. What I don't understand is why he uses this $x_1,...x_n$ what is he trying to say?
surely not that $W$ is a sot neighbourhood of $u$ if $|(t-u)x_j|<\epsilon$ for finitely many $x_j$?!
Can someone illuminate this argument, please?
Basic neighbourhoods are precisely those $W$. The argument in the first paragraph shows how to get $|(t-u)x|<\epsilon$, but for a single $x$. That's not enough for sot convergence, because if $$ W=\{t:\ \|(t-u)x_j\|<\epsilon,\ j=1,\ldots,n\}, $$ after you apply the first paragraph to $x_1$ and get a $v$, you have no reason for $\|(v-u)x_2\|<\epsilon$, etc.
The trick with moving things to $H^n$ is that now points are of the form $(x_1,\ldots,x_n)$. So now you can apply the first paragraph to $(x_1,\ldots,x_n)$, while the algebra now consists of operators of the form $$ \begin{bmatrix}v\\ &\ddots\\ && v\end{bmatrix}. $$ So the first paragraph, applied to $\varphi(A)$ on $H^n$, produces $v$ such that $$ \left\|\left(\begin{bmatrix}v\\ &\ddots\\ && v\end{bmatrix}-\begin{bmatrix}u\\ &\ddots\\ && u\end{bmatrix} \right) \begin{bmatrix} x_1\\ \vdots\\ x_n\end{bmatrix}\right\|<\epsilon. $$ This inequality is, by definition $$ \Big(\sum_{j=1}^n\|(v-u)x_j\|^2\Big)^{1/2}<\epsilon. $$ In particular, $$ \|(v-u)x_j\|<\epsilon,\qquad\qquad j=1,\ldots,n, $$ which says precisely that $v\in W$.