Bell type inequalities - von Neumann Algebras 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?




 A: 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, 
$$
|\beta(\phi,\mathscr A,\mathscr B)-\beta(\psi,\mathscr A,\mathscr B)|\leq\sqrt2\|\phi-\psi\|.
$$
