The solution of the exercise 2.10 from the textbook Convex Optimization by Boyd & Vandenberghe, they say that the set:
$\{\hat{x} + tv \space |\space \alpha t^2 + \beta t + \gamma \le 0\}$
$\space \space \space \space \space \space \space \space$ with $\alpha = v^T Av, \space \beta = b^Tv + 2\hat{x}^T Av, \space and \space \gamma = c + b^T \hat{x} + \hat{x}^TA \hat{x} $
is convex if $\alpha \ge 0$. This is true for any $v$, if $v^T Av \ge 0$ for all $v$, i.e, $A \succeq 0$
I don't understant why "(the set) is convex if $\alpha \ge 0$." and why "This is true for any $v$, if $v^T Av \ge 0$ for all $v$, i.e, $A \succeq 0$".
Does someone has an idea?
Best Answer
When $\alpha \ge 0$, the quadratic function is convex,
The set of $t$ where $\alpha t^2 +\beta t + \gamma \le 0$ can be written in the form of $p \le t \le q$ where $p$ and $q$ are the roots of the quadratic polynomial, $q \ge p$. The correponding $\hat{x}+tv$ is a line segment.
If $\alpha < 0$, the corresponding set is the union of $\{t:t > q\}$ and $\{t:t <p\} $, which is not convex.