Bell type inequalities - von Neumann Algebras

Question

I have a problem with proving part (4) of the following Lemma 2.1. It concerns Bell-type Inequalities. I attach the relevant definitions the lemma and the suggested proof (It begins with "Since..."). Is that proof correct?

Answer 1

The first assertion is that $\|T\|\leq\sqrt2$. When two selfadjoints $S,T$ commute, we have $|ST|=(|S|^2|T|^2)^{1/2}=|S|\,|T|$. Then, since everything is selfadjoint (even after taking sums and products) and commutes, \begin{align} \|A_1(B_1+B_2)+A_2(B_1-B_2)\|^2&=\|\,[A_1(B_1+B_2)+A_2(B_1-B_2)]^2\,\|\\ &=\|\,|A_1(B_1+B_2)+A_2(B_1-B_2)|^2\,\|\\ &\leq\|\,(|A_1(B_1+B_2)|+|A_2(B_1-B_2)|)^2\,\|\\ &\leq\|\,(|A_1|\,|B_1+B_2|+|A_2|\,|B_1-B_2|)^2\,\|\\ &\leq\|\,(|B_1+B_2|+|B_1-B_2|)^2\,\|\\ &\leq\|\,(B_1+B_2)^2+(B_1-B_2)^2+2|B_1+B_2|\,|B_1-B_2|\,\|\\ &=\|2B_1^2+2B_2^2+2|B_1^2-B_2^2|\,\|\\ &\leq\|2I+2I+2I\|=6. \end{align} Then $\|T\|\leq\tfrac12\,\sqrt6=\tfrac{\sqrt3}{\sqrt2}\leq\sqrt2$.

For the estimate, \begin{align} \phi(T)&=\phi(T)-\psi(T)+\psi(T)=(\phi-\psi)(T)+\psi(T)\\ &\leq\|\phi-\psi\|\,\|T\|+\psi(T) \leq\sqrt2\,\|\phi-\psi\|+\psi(T)\\ &\leq\sqrt2\|\phi-\psi\|+\beta(\psi,\mathscr A,\mathscr B). \end{align} As $\phi(T)\leq \sqrt2\|\phi-\psi\|+\beta(\psi,\mathscr A,\mathscr B)$ for all $T$, $$ \beta(\phi,\mathscr A,\mathscr B)\leq\sqrt2\|\phi-\psi\|+\beta(\psi,\mathscr A,\mathscr B), $$ so $$ \beta(\phi,\mathscr A,\mathscr B)-\beta(\psi,\mathscr A,\mathscr B)\leq\sqrt2\|\phi-\psi\|. $$ As the roles can be reversed, $$ |\beta(\phi,\mathscr A,\mathscr B)-\beta(\psi,\mathscr A,\mathscr B)|\leq\sqrt2\|\phi-\psi\|. $$