I want to show that $M=\{x\in \mathbb{R}^n: x^TQx\leq(a^Tx)^2, a^Tx\geq0\}$ where $Q$ is positive definite is a convex cone. I know a theorem that says $S\subseteq \mathbb{R}^n$ is convex cone if and only if:
- $\forall x,y \in S ,x+y\in S$
- $\forall x \in S, \forall \lambda \geq0, \lambda x\in S$
I've shown the second condition but I got stuck to show the first inequality for $x+y$.
As $\boldsymbol{Q}$ is positive definite, we can compute its square root matrix $\boldsymbol{D}$ defined by $$\boldsymbol{D}^T\boldsymbol{D}=\boldsymbol{Q}$$ Then, $S=\{\boldsymbol{x}\in \mathbb{R}^n: \boldsymbol{x}^T\boldsymbol{Qx}\le (\boldsymbol{a}^T\boldsymbol{x})^2, \boldsymbol{a}^T\boldsymbol{x}\geq0\}$
$\forall \boldsymbol{x,y} \in S$, it's obvious that $\boldsymbol{a}^T(\boldsymbol{x}+\boldsymbol{y})\geq0$. So, it suffices to prove that $$(\boldsymbol{x+y})^T\boldsymbol{Q}(\boldsymbol{x+y}) \le (\boldsymbol{a}^T(\boldsymbol{x+y}))^2 \tag{1}$$
The LHS of $(1)$ is equal to $$ \begin{align} LHS(1) &= (\boldsymbol{x+y})^T\boldsymbol{Q}(\boldsymbol{x+y}) \\ &= \boldsymbol{x}^T\boldsymbol{Q}\boldsymbol{x} +\boldsymbol{y}^T\boldsymbol{Q}\boldsymbol{y} +2\boldsymbol{x}^T\boldsymbol{Q}\boldsymbol{y} \le (\boldsymbol{a}^T\boldsymbol{x})^2+(\boldsymbol{a}^T\boldsymbol{y})^2 +2\boldsymbol{x}^T\boldsymbol{D}^T\boldsymbol{D}\boldsymbol{y} \end{align} $$
So, $(1)$ holds true if $(\boldsymbol{Dx})^T\boldsymbol{Dy} \le (\boldsymbol{a}^T\boldsymbol{x})(\boldsymbol{a}^T\boldsymbol{y}) \tag{2}$
Let denote $\boldsymbol{Dx}=(x_1,...,x_n)^T$ and $\boldsymbol{Dy}=(y_1,...,y_n)^T$, then the LHS of $(2)$ is equal to \begin{align} LHS(2) &= (\boldsymbol{Dx})^T\boldsymbol{Dy}\\ &= \sum_{i=1}^n x_iy_i\\ &\le \sqrt{\sum_{i=1}^n x_i^2}\sqrt{\sum_{i=1}^n y_i^2} = \sqrt{\left((\boldsymbol{Dx})^T(\boldsymbol{Dx})\right)}\sqrt{\left((\boldsymbol{Dy})^T(\boldsymbol{Dy})\right)} \le (\boldsymbol{a}^T\boldsymbol{x})(\boldsymbol{a}^T\boldsymbol{y})\\ \end{align}
We can conclude that $S$ is a convex cone. Q.E.D