Let $H$ be a separable, complex Hilbert space and let $\mathcal{B}(H)$ denote the algebra of bounded linear operators on $H$. Let $T \in \mathcal{B}(H)$. We define $$ A = \{ p(T,T^*) : p \in \mathbb{C}[z_1,z_2] \}.$$ $A$ is the subalgebra generated by $T,T^*$ - or the $*$-algebra generated by $T$. If we denote by $\overline{A}^\sigma$ its weak (operator) closure, I would like to know whether can the equality $$\overline{A}^\sigma = \mathcal{B}(H)$$ holds.
Thank you very much.
Yes, the unilateral shift $S$ on $l^2(\mathbb{N})$ generates $B(l^2(\mathbb{N}))$ as a von Neumann algebra. This is a consequence of the double commutant theorem and the fact that the only bounded operators which commute with both $S$ and $S^*$ are scalars.
(To see this, suppose $T$ commutes with both $S$ and $S^*$. Then $S^*Te_0 = TS^*e_0 = 0$ so $Te_0$ must be a scalar multiple of $e_0$, say $Te_0 = \lambda e_0$. Then $Te_n = TS^ne_0 = S^nTe_0 = \lambda e_n$ for all $n$, and therefore $T = \lambda I$.)
Yes, just take the unilateral shift on $H=\ell^2({\bf N})$. The norm closed (${\rm C}^*$-) algebra generated by this shift and its adjoint is the Toeplitz algebra, which contains all compact operators on $H$. Hence the WOT closure of the original $*$-algebra contains the WOT closure of $K(H)$, which is all of $B(H)$.
