Let $H$ be a Hilbert space and $M\subseteq B(H)$ a $*$-subalgebra containing $\Bbb1$. The following Lemma is used in the proof of the bicommutant theorem :
For every $u$ in the bicommutant $M''$ and $x\in H$ there is a sequence $a_n\in M$ with $a_n(x)\to u(x)$.
Proof: Define $V:=\overline{Mx}$, this is a subspace of $H$ invariant under $M$, since $M$ is self-adjoint $V^\perp$ is also invariant (since for $w\in V^\perp$ one has $(Aw,v)=(w,A^*v)=0$ for all $v\in V$). As such the projection $p$ onto $V$ is in $M'$ so one finds that $u$ commutes with $p$. Since $x\in V$ (as $\Bbb 1\in M$) one has: $u(x)=u(px)=pu(x)$ and $u(x)\in V$. By definition of $V$ there must then be a sequence $a_n\in M$ with $a_n(x)\to u(x)$.
The question:
Can one also achieve $a_n^*(u)\to u^*(u)$? Equivalently if $u$ is self-adjoint can $a_n$ also be assumed to be self-adjoint?
I know that the answer is affirmative, but could not modify the above elementary proof to get the result.