A document I am reading on Von-Neumann algebras (VNA) asserts that it follows from Cauchy-Schwarz that if $M$ is a VNA, and $w$ is a positive linear functional on M that is merely norm continuous, then $|w(xp)|\leq w(px^*xp)^{1/2}w(p)^{1/2}$. I'm afraid that the author might need to be considering $|w(pxp)|$ instead because the pseudo-inner product that is relevant here is $\langle x, y\rangle=w(x^*y)$ not $\langle x, y\rangle=w(xy)$
Can someone either confirm my problem or solve it by deriving that inequality please? Thanks.
Just figured I'd add an example showing that the statement (as written) is indeed generally false.
Take $M$ to be the von Neumann algebra of $2 \times 2$ matrices over the complex numbers (with the usual addition, multiplication, and involution), and let $h = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, and define $\omega: M \to \mathbb{C}$ by $$ \omega(m) = \operatorname{trace}(h m), \qquad m \in M, $$ where $\operatorname{trace}$ is the usual (non-normalized) trace.
The map $\omega$ is evidently linear, and it is easily seen to be positive: in fact, whenever $h$ is a positive element of $M$, a short calculation shows that $\operatorname{trace} (h m^* m) = \operatorname{trace} ((m h^{1/2})^* (m h^{1/2})) \geq 0$ for all $m \in M$.
With $p = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $x = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$, short calculations show that $$ h x p = h p x^* x p = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}, \qquad h p = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} $$ so that $|\omega(xp)| = 1$ is larger than $\omega(p x^* x p)^{1/2} \omega(p)^{1/2} = 1 \cdot (1/2)^{1/2}= (1/2)^{1/2}$.
The inequality does hold if one additionally assumes that $\omega$ has the trace property $$ \omega(ab) = \omega(ba), \qquad a, b \in M, $$ because then $\omega(xp) = \omega(xpp) = \omega(pxp)$ and the usual Cauchy-Schwarz argument applies.