Does realism imply locality or vice versa?

[stevendaryl]

where
$C_\lambda(\alpha, \alpha', \beta, \beta') \equiv X_A(\alpha, \lambda) X_B(\beta, \lambda) + X_A(\alpha', \lambda) X_B(\beta, \lambda) + X_A(\alpha, \lambda) X_B(\beta', \lambda) - X_A(\alpha', \lambda) X_B(\beta', \lambda)$

It might not be obvious, but since each of the Xs are between +1 and -1, it follows that $C_\lambda(\alpha, \alpha', \beta, \beta')$ must lie in the range $[-2, +2]$.

Letting

• $A \equiv X_A(\alpha, \lambda)$
• $A' \equiv X_A(\alpha', \lambda)$
• $B \equiv X_B(\beta, \lambda)$
• $B' \equiv X_B(\beta', \lambda)$
• $F(A,A', B, B') \equiv AB + A'B + AB' -A'B'$

We want to show that $-2 \leq F(A,A',B,B') \leq +2$, under the assumption that all 4 variables lie in the range $[-1, +1]$.

Let $A_{max}, B_{max}, A'_{max}, B'_{max}$ be a choice of the variables that maximizes $F$. Then:

• Either $A_{max} = \pm 1$ or $\frac{\partial F}{\partial A}|_{A = A_{max}, B=B_{max}, A'=A'_{max}, B' = B'_{max}} = 0$
• Either $A'_{max} = \pm 1$ or $\frac{\partial F}{\partial A'}|_{A = A_{max}, B=B_{max}, A'=A'_{max}, B' = B'_{max}} = 0$

Suppose $\frac{\partial F}{\partial A} = 0$. That implies that $B_{max} + B'_{max} = 0$. So in this case: $F(A_{max}, A'_{max}, B_{max}, B'_{max}) = A'_{max} (B_{max} - B'_{max}) \leq 2$

Suppose $\frac{\partial F}{\partial A'} = 0$. That implies that $B_{max} - B'_{max} = 0$. So in this case: $F(A_{max}, A'_{max}, B_{max}, B'_{max}) = A_{max} (B_{max} + B'_{max}) \leq 2$

So we conclude that if $F > 2$, it must be when $A_{max} = \pm 1$ and $A'_{max} = \pm 1$. We have two cases: they have the same sign, or they have opposite signs.

Suppose $A_{max} = \pm 1$ and $A'_{max} = +A_{max}$. Then $F(A_{max}, A'_{max}, B_{max}, B'_{max}) = \pm (B_{max} + B'_{max}) \pm (B_{max} - B'_{max}) = \pm 2 B_{max} \leq 2$

Suppose $A_{max} = \pm 1$ and $A'_{max} = -A_{max}$. Then $F(A_{max}, A'_{max}, B_{max}, B'_{max}) = \pm (B_{max} + B'_{max}) \mp (B_{max} - B'_{max}) = \pm 2 B'_{max} \leq 2$

So in all cases, $F(A,A',B, B') \leq 2$. We can similarly prove $-2 \leq F(A,A',B,B')$. So we conclude:

$-2 \leq F(A,A',B,B') \leq +2$