# A theorem about a tracial state in von Neumann algebra

I am reading a book about C*-algebra. There is a quotation below.

Let $M$ be a von Neumann algebra with a faithful normal tracial state $\tau$ and let $1_{M}\in N\subset M$ be von Neumann subalgebra. Then, for each $a\in N$, we define $\tau_{a}\in N_{\ast}$ by $\tau_{a}(y)=\tau(ya)~$(here, $N_{\ast}$ is the Banach space of normal linear functionals on $N$). Then, $||\tau_{a}||=\tau(|a|)$ (here, $|a|=(a^{\ast}a)^{1/2}$). Also note that $\{\tau_{a}: a\in N\}$ is a norm-dense linear subspace in $N_{\ast}$.(If, it were not dense, we could find $0\neq n\in N$ such that $\tau_{a}(n)=0$ for all $a\in N$, which is impossible since $\tau$ is faithful.)

I have two question about the quotation:

1. For the equation $||\tau_{a}||=\tau(|a|)$, through using polar decomposition of $a$ and the proposition of tracial state, we can prove $||\tau_{a}||\leq\tau(|a|)$. But how to prove $||\tau_{a}||\geq\tau(|a|)$?

2. How to explain $\{\tau_{a}: a\in N\}$ is a norm-dense linear subspace in $N_{\ast}$?

1. The trace satisfies the following inequality: $$|\tau(ya)|\leq\|y\|\,\tau(|a|).$$ This implies directly that $\|\tau_a\|\leq\tau(|a|)$. For the reverse inequality, if $a=v|a|$ is the polar decomposition, then $|a|=v^*a$ and so $\tau(|a|)=\tau(v^*a)\leq\|\tau_a\|$.
2. Suppose that $X=\{\tau_a:\ a\in N\}$ is not norm-dense in $N_*$. Note that $X$ is a vector space; in particular, it is convex. By Hahn-Banach, we can find nonzero $x\in(N_*)^*=N$ such that $X\subset\ker x$. This means that $0=x(\tau_a)=\tau_a(x)$ for all $a\in N$. In particular we can take $a=x^*$, so $\tau(x^*x)=0$. As the trace is faithful, $x=0$, a contradiction.
• I have prove that $||\tau_{a}|| \leq \tau(|a|)$. But how to prove $||\tau_{a}|| \geq \tau(|a|)$? Mar 10, 2014 at 2:05
• I suppose that, in the polar decomposition $a=v|a|$, $v$ is just a partial isometry (not isometry), then, how to get $|a|=v^{\ast}a$? And does $||v^{\ast}a|| \leq 1$? Then, you use the definition of norm of the operator to get $\tau(v^{\ast}a) \leq ||\tau_{a}||$? Mar 10, 2014 at 15:06
• In the polar decomposition of $a$, $v^*v$ is the projection onto the range of $|a|$. So $v^*a=v^*v|a|=|a|$. The fact that $v$ is a partial isometry implies $\|v\|=1$. The number $\|v^*a\|$ plays no role in the argument. And yes, $\tau(v^*a)\leq\|\tau_a\|$ by definition of the norm of $\tau_a$. Mar 10, 2014 at 15:10