Let $H$ be a Hilbert space and $A$ a C*-subalgebra of $B(H)$, and $1_H\in A$. Show that the unitaries of $A$ are strongly dense in the unitaries of $\overline{A}^{sot}$.
Suppose $U(A)$ be unitaries of $A$. To prove $\overline{U(A)}^s \subset U(\overline{A}^s)$, let $x\in \overline{U(A)}^s$. I need to show $x\in U(\overline {A}^s)$, in other words, $x$ is unitary. It's possible, because there is the net $\{u_i\}$ of unitaries of $A$ such that $u_i\to x$ (sot). Using these two points;
1- involution is sot continuous on normal operators.
2- sot is jointly continuous on bounded subsets of $B(H)$.
we have $x$ is unitary.
But It's may possible that $x$ is not unitary, and also normal, for instance, let $\{e_n\}$ is a basis for $H$. Define unitary operator $u_n$ such that $u_ne_i=e_{i+1}$ for $i=1,...,n-1$, u_n(e_n) = e_1, and $u_n(e_i) = e_i$ for $i>n$. we have $x_n\to x$(sot), when $xe_i=e_{i+1}$ for $i\in \Bbb N$.
I'm confused, how should prove $\overline{U(A)}^s \subset U(\overline{A}^s)$. Please help me. Thanks.
As you noticed, you cannot prove that $\overline{U(A)}^s\subset U(\overline{A}^s)$, because it is not true.
What the assertion "the unitaries of $A$ are strongly dense in the unitaries of $\overline{A}^s$" means is exactly that: you obviously have that $U(A)\subset U(\overline{A}^s)$. Here "density" means that every unitary in $U(\overline{A}^s)$ is a sot limit of unitaries in $U(A)$.
Here is the argument in Davidson's book: given $u\in U(\overline{A}^s)$, there exists $a\in\overline{A}^s$, selfadjoint, with $u=e^{ia}$. Assuming that you already know that the selfadjoints of $\overline{A}^s$ are limits of selfadjoints in $A$, you have $a=\lim a_j$ with $a_j\in A_{\rm sa}$. Then $$ u=\lim e^{ia_j}. $$