Let $C:[0,2\pi]\times [0,2\pi]\rightarrow[0,1/2]$ be given by $C(\theta_1,\theta_2)=\frac{1}{2}(\cos(\theta_2-\theta_1))^2$, $m=\min\{0, \alpha, \beta, \gamma, \alpha+\beta+\gamma\},$ and $M=\max\{0, \alpha, \beta, \gamma, \alpha+\beta+\gamma\}.$

$\textbf{Question:}$ Do there exist real numbers $\alpha, \beta, \gamma \in \textbf{R}$, such that the inequality

$$m\leq \alpha C(\theta_1,\theta_2)+\beta C(\theta_2,\theta_3)+ \gamma C(\theta_1,\theta_3)\ \leq M $$

is $\textbf{violated}$? If not, prove this.

$\textbf{My attempt:}$ I wrote a python script that checks the inequality for many values of $\alpha,\beta,\gamma\in[-5,5]$ and $\theta_1,\theta_2\in[0,2\pi]$. However, the inequality was never violated. I am trying to prove this result. I noticed that, if $\alpha,\beta,\gamma$ are all positive, then $m=0$ and $M=\alpha+\beta+\gamma$. Since $0\leq\frac{1}{2}(\cos(x))^2\leq\frac{1}{2}$ we find that the inequality is always satisfied in this case. A similar thing can be done for when $\alpha,\beta,\gamma$ are all negative. I am now stuck. Can this problem be solved by checking all the differecnt cases? (for example the case were only $\alpha\leq 0,$ or maybe the case were $\alpha+\beta\geq\gamma$) If yes, I don't see which cases I could check. If no, is there another method to solve this? Lastly, maybe I can somehow use some triangle inequality for the cosine?

Any help/hints are welcome.

(For the interested, the problem arose in an attempt to disprove the existence of hidden variables in quantum mechanics.)

  • 1
    $\begingroup$ I think that the symmetry of the problem's set-up would let you get away with checking only the case in which $\alpha$ is negative but $\beta, \gamma$ are positive. $\endgroup$ – Connor Harris May 16 '17 at 14:01
  • $\begingroup$ @ConnorHarris This seems to be true. Since if I switch the roles for $\gamma$ and $\alpha$, and then switch the roles for $\theta_1$ and $\theta_2$, I retrieve the same inequality. Do you have an idea how to prove the inequality for $\alpha\leq 0$ and $\gamma, \beta \geq 0$? $\endgroup$ – Meneer-Beer May 16 '17 at 14:08

There's actually no need for trigonometry. If $\alpha \leq 0$ and $\beta, \gamma \geq 0$, then $m = \alpha$ and $M = \max \{\beta, \gamma, \alpha + \beta + \gamma\}$. Let's denote $A = \alpha C(\theta_1, \theta_2) + \beta C(\theta_2, \theta_3) + \gamma C(\theta_3, \theta_1)$. It's clear that $\alpha C(\theta_1, \theta_2) \geq \alpha /2 > \alpha$, and the other terms of $A$ are positive, so the lower bound $m$ holds. Similarly, $A \leq \beta C(\theta_2, \theta_3) + \gamma C(\theta_3, \theta_1) \leq \frac{\beta + \gamma}{2} \leq \max \{\beta, \gamma\} \leq M$. The other cases in which $\alpha, \beta, \gamma$ do not all have the same sign follow from symmetry of the setup under permutation of the variables $\alpha, \beta, \gamma$ and under $(\alpha, \beta, \gamma) \mapsto (-\alpha, -\beta, -\gamma)$.

  • $\begingroup$ I think you mean $m=\alpha$, and $M=max\{\gamma, \beta, \alpha+\beta+\gamma\}$, or am I wrong? Also, thank you, I think you solved it. I will make sure all other cases can be proved under permutation, and then accept your answer. $\endgroup$ – Meneer-Beer May 16 '17 at 14:59
  • $\begingroup$ Oh, of course you're right. Will fix. $\endgroup$ – Connor Harris May 16 '17 at 15:03

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