# Does weak convergence imply strong convergence in a von Neumann Algebra

Let $H$ be an Hilbert space, and $A$ a von Neumann subalgebra of $B(H)$.

It is made abundantly clear that this is equivalent to: $A$ is a s-o closed *-subalgebra, and also $A$ is w-o closed *-subalgebra.

What is not clear (to me), is whether the actual occurences of convergence are the same, namely:

letting $a_i, a\in A$, does

$a_i \longrightarrow a$ (weakly) imply

$a_i \longrightarrow a$ (strongly).

Is this even true in when restricting attention to the unit ball of $A$?

-

## 2 Answers

Let $H=\ell^2(\mathbb Z)$ with standard orthonormal basis $\{e_n\}_{n\in\mathbb Z}$, and $U$ the right shift on $H$, sending $e_n$ to $e_{n+1}$. Then $U^n\to 0$ in the weak operator topology as $n\to \infty$, but not in the strong operator topology. (Because $U$ is unitary, this could also be considered a commutative example.)

In general, topologies are determined by their convergent nets; sequences are not always enough, and it is possible for convergence of sequences in two different topologies to be the same. But for the case of the two topologies in question, in the infinite dimensional case their convergent sequences are not the same.

-

This is not true even in the commutative case. The classical counterexample is the system of Rademacher functions (normalised for the uniform norm) in $L^\infty$ of the unit interval.

`

-