I am self-studying so I hope anyone who can understand this topic kindly help me. I have problems with the following question.
(Q 2.10, "convex optimization" by stephen boyd.)
Let $C \subseteq \mathbb{R}^n$ be the solution set of a quadratic inequality, $$C=\{x\in{\mathbb{R}^n}|x^T Ax+b^Tx+c\leq0\}.$$ a) Show that C is convex if A is positive semidefinite.
b) Show that intersection of C and the hyperplane defined by $g^Tx+h=0$ (where $g \neq 0$) is convex if $A+\lambda gg^T \geq0$ for some $\lambda \in \mathbb{R}$.
My questions are:
Is it possible to solve (a) and (b) by showing $\theta x_1+ (1-\theta )x_2 \in C$, where $x_1,x_2 \in C$? If yes, how?
For (a), I have seen a proof by showing that the intersection of $C$ with an arbitrary line $\{\hat x+tv|t\in\mathbb{R}\}$ is convex. This approach ultimately leads to a solution of $D=\{\hat x+tv|\alpha t^2+\beta t+\gamma\leq 0\}$, where $\alpha,\beta,\gamma$ are expressed in terms of $x$. It says that $D$ is convex if $\alpha\geq0$. Nevertheless, I don't understand why this is the case. It seems that such statement comes from the fact that one is taking the second derivative of the quadratic inequality w.r.t. $t$ and therefore $\alpha\geq0$ implies convexity. But shouldn't we test the convexity in terms of $x$? Please correct my misunderstanding.
- I am very curious how $\lambda gg^T \geq0$ is obtained. That seems very arbitrary to me though I am aware that $g^Tv=0$.
- Is it acceptable to define $f(x)=x^T Ax+b^Tx+c$, and suggest that $\frac{\partial^2 f}{\partial x^2}>0$ implies convexity?
P.S.: I'm aware there is an explanation posted in mathstackexchange to the problem Q 2.10. Nevertheless, I am looking for answers to my questions.