$w^*$-limit of projections in von Neumann algebra

Question

Let $\mathcal M$ be a semi finite von Neumann algebra with a normal faithful semi finite trace $\tau$. Let $(e_i)_{I\in I}$ be a net of projections in the von Neumann algebra which converges to an element $u$ in $w^*$ topology. Is it true that for any $x\in L_1(\mathcal m),$ $\|xe_i-xu\|_1\to 0$? If you have any reference please point that to me.

Answer 1

No, this is not true. Even when $M$ is abelian and $\tau$ is a normal faithful tracial state, this need not be true. An easy counterexample can be constructed as follows. Denote by $\mu_0$ the probability measure on $\{0,1\}$ defined by $\mu_0(0) = \mu_0(1) = 1/2$. Then define $(X,\mu) = (X_0,\mu_0)^\mathbb{N}$. Define the projections $e_n \in L^\infty(X,\mu)$ by $e_n(x) = x_n$. Then, $e_n \to 1/2$ in the weak$^*$ topology, but $\|e_n - 1/2\|_1 = 1/2$ for all $n$.

Answer 2

$L^1(\mathcal M)$ is the predual of $\mathcal M$. So the weak*-topology on your von Neumann algebra $\mathcal M$ is the same as the weak topology induced by the pairing $L^1(\mathcal M)\times \mathcal M\to \mathbb C$. So your result holds true. No assumption are needed on the von Neumann algebra (e.g. doesn't need to have a trace), and no assumptions are needed on the elements $e_i$ (they don't need to be projections).

Edit: my answer is wrong. I had misread the question -- see Yemon Choi's comment.