Let $M$ be a von Neumann algebra acting on a Hilbert space $H$ and we denote $M_+$ by the set of all positive elements of $M$. Let us now consider a map $\phi:M_+ \to [0,\infty]$ which satisfies the following conditions:
- $\phi(tx)=t\phi(x)$ for $t \ge 0,~x \in M_+$;
- $\phi(x+y)=\phi(x)+\phi(y)$ for $x,y \in M_+$.
Now assume that the set $N=\{x\in M: \phi(x^*x)<\infty\}$ is dense in $M$ with respect to ultraweak topology. Consider the set $U=\{x\in M_+:\phi(x)<\infty\}$. I want to show that $U$ is dense in $M_+$ with respect to ultraweak topology. Recall that, The ultraweak topology on $B(H)$ is the Hausdorff locally convex topology on $B(H)$ generated by the semi-norms
$$B(H) \to \mathbb R^+,~~u \mapsto |\text{Tr}(uv)|,~~v\in L^1(H),$$
where $L^1(H)$ denotes the set of trace class operators on $H$.
My approach: Let $x \in M_+$. Then $x=y^*y$ for some $y\in M.$ Since $N$ is ultraweakly dense in $M$, so there exists a net $(a_l)$ in $N$ such that $a_l \xrightarrow{\text{ultraweak}} y$. Since involution is continuous under ultraweak topology we have $a_l^* \xrightarrow{\text{ultraweak}} y^*.$ Also $a_l^*a_l \in N$ since $N$ is a left ideal, but joint multiplication is not continuous under ultraweak topology. Now I am unable to solve this. Please help me. Thank you for your time and effort.
Claim: $N \cap N^*$ is strongly dense in $M$.
Proof: We observe that as $N$ is a left ideal, $N \cap N^*$ is a $*$-subalgebra of $M$. Thus, take $A = \overline{N \cap N^*}^{||\cdot||}$, which is a $C^*$-subalgebra of $M$. $A$ then admits an approximate unit $\{a_\lambda\}_{\lambda}$. Since $a_\lambda$ is an increasing net of positive elements with norm bounded by 1, $a_\lambda \rightarrow p$ for some positive $p \in M$ strongly. We see that $xp = x$ for all $x \in A$.
Now, for any $x \in N$, we see that $x^*x \in N \cap N^* \subseteq A$. Thus, $x^*xp = x^*x$, so $x^*x(1 - p) = 0$ and then $x(1 - p) = 0$. Since $N$ is ultraweakly dense, we may take a net $x_\mu$ converging ultraweakly to 1. Then $1 - p = \lim_\mu x_\mu(1 - p) = 0$, i.e., $p = 1$. Now, since $N \cap N^*$ is norm-dense in $A$ and $A \ni a_\lambda = a_\lambda^{1/2}a_{\lambda}^{1/2} \rightarrow 1$ strongly, by choosing elements of $N \cap N^*$ close in norm to $a_\lambda^{1/2}$ and using the fact that $N \cap N^*$ is a $*$-subalgebra, we see that there exists $b_\eta \in (N \cap N^*)_+$ of uniformly bounded norms s.t. $b_\eta \rightarrow 1$ strongly.
For any $x \in M$ and $\eta$, we claim that $b_\eta x b_\eta \in N \cap N^*$. Indeed, $(b_\eta x b_\eta)^*(b_\eta x b_\eta) \leq ||b_\eta||^2||x||^2b_\eta^2$. As $b_\eta \in N$, $\phi(b_\eta^2) < \infty$, so $b_\eta x b_\eta \in N$. As $(b_\eta x b_\eta)^* = b_\eta x^* b_\eta$, the same argument shows that $b_\eta x b_\eta \in N^*$, which proves the claim. But then as multiplication is jointly continuous under the strong operator topology restricted to bounded sets, we see that $b_\eta x b_\eta \rightarrow x$ strongly. Since $x \in M$ is arbitrary, this proves the claim. QED
Now, as $N \cap N^*$ is strongly dense in $M$ and as $N \cap N^*$ is a $*$-subalgebra, we can apply Kaplansky density theorem to see that $(N \cap N^*)_1$ is dense in $(M)_1$ under the ultrastrong$^*$ topology. At this point, we can apply your argument to finish the proof as, again, multiplication is jointly continuous under the ultrastrong$^*$ topology restricted to bounded sets.