Let $\phi$ be a normal linear functional on a von Neumann algebra $M$. Define $L=\{x\in M:\phi(x^*x)=0\}$.Show that $L$ is $WOT$ closed in $M$. I have been trying to show that $L$ is $SOT$ closed and then use the fact that $L$ is convex to conclude that $L$ is indeed $WOT$ closed in $M$, but to no avail. Thanks for any help.
$Definition$- $\phi$ is said to be a normal linear functional on a von Neumann algebra $M$ if $\phi(x)\geq 0$ $\forall x\geq 0$ and $\phi$ is $\sigma-weakly$ continuous on $M$. I can only show that $L$ is $\sigma-weakly$ closed.Can somebody please help me to show that $L$ is actually $WOT$ closed. First note that using the $Cauchy$ $Schwarz$ $Inequality$ we have $x \in L\Leftrightarrow\phi(x^*y)=0$ for each $y$ $\in$ $M$. Let $x_\alpha\longrightarrow x$ $\sigma-weakly$ where $x_\alpha$ is a net in $L$ and $x \in M$.$\Rightarrow x_\alpha^*y\longrightarrow x^*y$ $\sigma-weakly$ for each $y \in M$ $\Rightarrow \phi(x_\alpha^*y)\longrightarrow \phi(x^*y)$ for each $y \in M$.But $\phi(x_\alpha^*y)=0$ $\forall \alpha$ and for each $y \in M$ $\Rightarrow \phi(x^*y)=0$ for each $y \in M$ $\Rightarrow x \in L$ $\Rightarrow$ $L$ is $\sigma-weakly$ closed in $M$.
The trick I know for this uses the multiplicative domain. The thing is to show that $$\tag{1} L=\{x\in M:\ \phi(yx)=0\,\ \forall y\in M\}. $$ The multiplicative domain of $\phi$ is the set where equality holds in the Kadison-Schwarz inequality, i.e. $$ R_\phi=\{x:\in M:\ \phi(x)^*\phi(x)=\phi(x^*x)\}. $$ It can be shown (there are proofs in Paulsen's book and in Ozawa's QWEP paper) that $$ R_\phi=\{x\in M:\ \phi(yx)=\phi(y)\phi(x)\,\ \forall y\in M\}. $$ Now $(1)$ follows easily, since by the Kadison-Schwarz inequality (note that $\phi$ is completely positive because it is a functional), for $x\in L$ we have $$ 0\leq\phi(x)^*\phi(x)\leq\phi(x^*x)=0, $$ so $x\in R_\phi$.