To this MO question, Yemon Choi comments that
If $\tau$ is a faithful normal trace on a von Neumann algebra $M$, then IIRC $\tau(|x|)$ is equal to the supremum of $|\tau(xy)|$ as $y$ runs over all elements in unit ball of $M$.
Recall $|x|=(x^*x)^{1/2}$. Yemon then comments that:
It's in Dixmier's book on von Neumann algebras. I don't have a copy to hand but I once had to cite this fact and if I got it right, then it is in Section V.2 of his book (French version, but presumably also the English translation). I suspect it might originally be due to I. Segal back in the 1950s?
At the moment I am away from a library and have been unable to find a reference online... I am also not sure what norm is on $M$ in this framework... could it be the 1-norm again?
Therefore if you can't find an accessible reference for the following, is there a short easy proof around of:
Where $\tau$ is a faithful normal trace on a von Neumann algebra $M$, we have that $$\|x\|_1:=\tau(|x|)=\sup_{y\,:\,\|y\|_1\leq1}|\tau(xy)|.$$
I am looking at Theorem 2.4.16 in Murphy and am wondering is this what Yemon is thinking of...
The equality as written in the question is false. For instance you can take $M=\mathbb C^2$, with the trace $\tau(x_1,x_2)=x_1+x_2$, and $x=(1,1/2)$. If $\tau(|y|)=1$, this means that $y=(y_1,y_2)$ with $|y_1|+|y_2|=1$. Then $$ |\tau(xy)|=|y_1+y_2/2|\leq|y_1|+|y_2|/2=1-|y_2|/2\leq1, $$ so $\|x\|_1=3/2$ is never achieved by using elements of trace 1.
The equality that does hold is $(2)$ below (with operator norm instead of 1-norm). The key observation is the inequality $$\tag{1} |\tau(xy)|\leq \|x\|\,\tau(|y|). $$ (proof of this at the end).
Now, for a fixed $x\in M$, write the polar decomposition $x=u|x|$; then $|x|=u^*x$, with $\|u\|\leq1$. This and $(1)$ show that $$\tag{2} \|x\|_1=\sup\{|\tau(xy)|:\ y\in M,\ \|y\|\leq1\}. $$
For $y$ with $\tau(|y|)=1$, the inequality $(1)$ shows that $|\tau(xy)|\leq\|x\|$. For $\varepsilon>0$, let $p$ be the spectral projection of $|x|$ corresponding to the interval $[\|x\|-\varepsilon,\|x\|]$. Let $y=pu^*/\tau(p)$. Then $$ \tau(xy)\,\tau(p)=\tau(xpu^*)=\tau(u^*xp)=\tau(|x|\,p)\geq (\|x\|-\varepsilon)\,\tau(p), $$ so $\tau(xy)\geq\|x\|-\varepsilon$. Note that $\tau(|y|)=1$. We can do this for any $\varepsilon$, so $$\tag{3} \|x\|=\sup\{|\tau(xy)|:\ y\in M,\ \|y\|_1\leq1\}. $$ Note that $(2)$ and $(3)$ are precisely Hölder for $1,\infty$ and $\infty,1$.
$$\ $$
Proof of the inequality $(1)$. We prove first the inequality $$ |\tau(z)|\leq\tau(|z|). $$ Indeed, writing $z=u|z|$ for the polar decomposition of $z$ and using Cauchy-Schwarz, $$ |\tau(z)|^2=|\tau(u|z|^{1/2}|z|^{1/2})|^2\leq\tau(|z|^{1/2}u^*u|z|^{1/2})\tau(|z|) \leq\tau(|z|)^2. $$
Now, using that the square root is monotone, $$ |\tau(xy)|\leq\tau(|xy|)=\tau((y^*x^*xy)^{1/2})\leq\|x\|\,\tau((y^*y)^{1/2})=\|x\|\,\tau(|y|). $$