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stevendaryl said:where
[itex]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)[/itex]
It might not be obvious, but since each of the Xs are between +1 and -1, it follows that [itex]C_\lambda(\alpha, \alpha', \beta, \beta')[/itex] must lie in the range [itex][-2, +2][/itex].
Letting
- [itex]A \equiv X_A(\alpha, \lambda)[/itex]
- [itex]A' \equiv X_A(\alpha', \lambda)[/itex]
- [itex]B \equiv X_B(\beta, \lambda)[/itex]
- [itex]B' \equiv X_B(\beta', \lambda)[/itex]
- [itex]F(A,A', B, B') \equiv AB + A'B + AB' -A'B'[/itex]
Let [itex]A_{max}, B_{max}, A'_{max}, B'_{max}[/itex] be a choice of the variables that maximizes [itex]F[/itex]. Then:
- Either [itex]A_{max} = \pm 1[/itex] or [itex]\frac{\partial F}{\partial A}|_{A = A_{max}, B=B_{max}, A'=A'_{max}, B' = B'_{max}} = 0[/itex]
- Either [itex]A'_{max} = \pm 1[/itex] or [itex]\frac{\partial F}{\partial A'}|_{A = A_{max}, B=B_{max}, A'=A'_{max}, B' = B'_{max}} = 0[/itex]
Suppose [itex]\frac{\partial F}{\partial A'} = 0[/itex]. That implies that [itex]B_{max} - B'_{max} = 0[/itex]. So in this case: [itex]F(A_{max}, A'_{max}, B_{max}, B'_{max}) = A_{max} (B_{max} + B'_{max}) \leq 2[/itex]
So we conclude that if [itex]F > 2[/itex], it must be when [itex]A_{max} = \pm 1[/itex] and [itex]A'_{max} = \pm 1[/itex]. We have two cases: they have the same sign, or they have opposite signs.
Suppose [itex]A_{max} = \pm 1[/itex] and [itex]A'_{max} = +A_{max}[/itex]. Then [itex]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[/itex]
Suppose [itex]A_{max} = \pm 1[/itex] and [itex]A'_{max} = -A_{max}[/itex]. Then [itex]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[/itex]
So in all cases, [itex]F(A,A',B, B') \leq 2[/itex]. We can similarly prove [itex]-2 \leq F(A,A',B,B')[/itex]. So we conclude:
[itex]-2 \leq F(A,A',B,B') \leq +2[/itex]