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Thanks. It follows like this. \begin{align*} q\left(\frac{1}{2} \textbf{x} + \left(1 - \frac{1}{2}\right) (-\textbf{x})\right) = q(0) = c <& \frac{1}{2} q(\textbf{x}) + \left(1 - \frac{1}{2}\right) q(-\textbf{x}) = \textbf{x}^T G \textbf{x} + c\\ \textbf{x}^T G\textbf{x} >& 0 \end{align*} This...

Consider the quadratic function $\displaystyle q(\textbf{x}) = \frac{1}{2} \textbf{x}^T G \textbf{x} + \textbf{d}^T \textbf{x} + c$ on $\mathbb{R}^n$, where $\textbf{d}$ is a constant $n \times 1$ vector, $G$ is a constant $n \times n$ symmetric matrix and $c$ is a scalar. The gradient is...

Thank you I like Serena! I think I managed to simplify that. \begin{align*} & \theta x_1^2 + (1 - \theta) y_1^2 - (\theta x_1 + (1-\theta) y_1)^2 \\ =& \theta x_1^2 + (1-\theta)y_1^2 - ( \theta^2 x_1^2 + 2 (\theta - \theta^2) x_1 y_1 + (1-2\theta + \theta^2) y_1^2)\\ =& (\theta - \theta^2)...

I am stuck at the inequality proof of this convext set problem. $\Omega = \{ \textbf{x} \in \mathbb{R}^2 | x_1^2 - x_2 \leq 6 \}$ The set should be a convex set, meaning for $\textbf{x}, \textbf{y} \in \mathbb{R}^2$ and $\theta \in [0,1]$, $\theta \textbf{x} + (1-\theta)\textbf{y}$ also belong...