I'm doing part (a) problem 2.10 in textbook Convex Optimization by Boyd and Vandenberghe:
Let $C \subseteq \mathbf{R}^{n}$ be the solution set of a quadratic inequality,
$$ C=\left\{x \in \mathbb{R}^{n} \mid x^{T} A x+b^{T} x+c \leq 0\right\} \\ $$
with $A \in \mathbb{S}^{n}$, $b \in \mathbb{R}^{n}$, and $c \in \mathbb{R}$.
(a) Show that $C$ is convex if $A \succeq 0$.
In my attempt, I show that $C$ is convex without using the symmetry of $A$. Could you please verify if it's correct? I'm not sure why the authors include the assumption that $A \in \mathbb S^n$?
Let $f(x) = x^T A x + b^T x +c$. Then $\nabla^2 f(x) = A+A^T$ and $C = \{x \in \mathbb{R}^{n} \mid f(x) \le 0 \}$. Because $A\succeq 0$, so is $A^T\succeq 0$. As such, $\nabla^2 f(x) \succeq 0$. Finally, $f$ is convex. We need to prove $C$ is convex.
Let $x, y \in C$ and $\theta \in [0,1]$. Because $f$ is convex, $f(\theta x + (1-\theta)y) \le \theta f(x) + (1-\theta) f(y) \le 0$ as $f(x), f(y) \le 0$ and $\theta, 1-\theta \ge 0$. It follows that $\theta x + (1-\theta)y \in C$ and thus $C$ is convex.
