I'm very confused on the following two-part question. Part a) requires just some small clarification, I think, but I'm a bit more lost on part b).
We have $C \subseteq \mathbb{R}^n$ defined by $$C := \left\{ x \in \mathbb{R}^n : x^TAx + b^Tx + c \leq 0 \right\}$$ where $A \in \mathbb{R}^{n \times n}$ is symmetric, $b \in \mathbb{R}^n$, and $c \in \mathbb{R}$.
Part a) Show that $C$ is convex if $A \succeq 0 $ (element-wise inequality).
Part b) Show that the intersection of $C$ and the hyperplane defined by $g^T x + h = 0,\, g\neq0$ is convex if $A + \psi gg^T \succeq 0$ for some $\psi \in \mathbb{R}$.
Part a)
So, I used the fact that a set is convex if its intersection with any line is also convex. So with $L = \{x:x = x_0 + \lambda v, x_0\in\mathbb{R}^n, v\in\mathbb{R}^n, \lambda \in \mathbb{R}\}$ I obtain
$S = L\cap C = \{x_o + \lambda v : \alpha\lambda^2 + \beta\lambda + \gamma \leq 0\}$ with
$\begin{align} \alpha &= v^TAv \\ \beta &= 2x_0^TAv + b^Tv \\ \gamma &= x_0^TAx_0+b^Tx_0+c \end{align}$
The solution manual states that the set $S$ is convex for $\alpha \geq 0$. I'm not sure why this is the case. Is it because $x_0 + \lambda v$ can have a positive region at its maximum if $\alpha < 0$ and will have negative regions around it?
It's clear to me that $\alpha \geq 0$ means that $v^TAv \geq 0$, which holds for all $v$ only if $A \succeq 0$. I'm just not sure on why $\alpha$ must be non-negative.
Part b)
I've looked at the solution manual, and I understand the symbolic manipulation that happened, but I don't know why the statements from which the derivation occurred were made and what their geometric meaning is.
Let $H = \{x: g^Tx + h = 0\}$. We denote, $\alpha$, $\beta$, and $\gamma$ as in the solution of part (a) and, in addition:
$\begin{align} \delta&= g^Tv \\ \epsilon&= g^Tx_0+h \end{align}$
Without loss of generality we can assume that $x_0 \in H$. (Why is this okay? Furthermore, this assumption forces $\epsilon = 0$ so I don't understand why the author felt the need to include $\epsilon$ in the first place.)
$C\cap H\cap L = \{x_o+\lambda v : \alpha\lambda^2 + \beta\lambda + \gamma \leq 0, \, \underset{\text{what is this?}}{ \delta\lambda=0 } \}$
I don't understand how the constraint involving $\delta$ was introduced; it obviously has something to do with the hyperplane, but I don't understand what.
From here I follow the arguments that lead to $A + \psi gg^T \succeq 0$.