# One particular application of the Cauchy Schwarz Inequality

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.

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What document? What are $x$ and $p$? Is $p$ a projection, or at least a positive element of $M$? –  Jonas Meyer May 19 '12 at 21:05
Oops sorry I had planned to say that, but forgot. The document is a paper document that I can't share, and is also not relevant I think but p is a projection in M, x is an element of M. –  Jeff May 19 '12 at 21:19

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.