A separating set which is not cyclic

Question

Let $H=L^2[0,1]$ , $T_g$ be the multiplication operator on $H$, i.e. $f\to fg$ . Let $A$ be the set of the $T_g$ as $g$ runs through the set of polynomials with complex coefficients. Let $h$ be te characteristic function of $[\frac{1}{2}, 1]$. Show that $h$ is separating for $A$, but is not cyclic for $A'$. Please help me.

Answer 1

If $T_gh=0$, then $g=0$ on $[1/2,1]$; any polynomial that is zero on an interval is equally zero, so $g=0$. That is, $h$ is separating.

The commutant of $A$ agrees with the commutant of the weak closure of $A$. This is the set of multiplication operators by all essentially bounded functions (i.e., $L^\infty[0,1]$). In particular $T_k$, with $k$ the characteristic of $[0,1/2]$ (one can simply verify this directly, $T_kT_gf=kgf=gkf=T_gT_kf$). Then $T_kh=0$. So $h$ is not cyclic for $A'$.