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Prove that $$ \left\{x=(x_1,x_2) \in \mathbb R^2 \mid \cos(x_1 + x_2) \ge \frac{\sqrt{2}}{2}, x_1^2 + x_2^2 \le \frac{\pi^2}{4}\right\}$$ is convex.

How should I do this? Hessian is made out of $-\sin(x_1 + x_2)$ so I can't determine if it's a positive definite matrix...And besides, there are two functions making the set, so does that mean I have to make two Hessians and both have to be positive definite for the set to be convex?

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  • $\begingroup$ what does $cos(x_1,x_2)$ mean? $\endgroup$ – mathemagician Dec 15 '13 at 23:47
  • $\begingroup$ That was supposed to be a plus ;) $\endgroup$ – khernik Dec 15 '13 at 23:51
  • $\begingroup$ I was wondering since when cos takes two arguments. $\endgroup$ – mathemagician Dec 15 '13 at 23:52
  • $\begingroup$ @khernik What is the question? $\endgroup$ – lcv Dec 15 '13 at 23:56
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    $\begingroup$ Is the second set a disc or what ? $\endgroup$ – derivative Dec 16 '13 at 0:04
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Hint: Since $|x_i|\le\pi/2$, the condition $\cos(x_1+x_2)$ is equivalent to $-\pi/4\le x_1+x_2\le\pi/4$, that is $x_2\ge-x_1-\pi/4$ and $x_2\le-x_1+\pi/4$. Now you may use the fact that the intersection of convex sets is convex.

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