Problem description
Let $\mathcal{C} = \{X \in \mathbb{R}^n \mid g(X) \leq 0\}$ with $g(X)$ a convex function. Suppose I need to solve the feasibility problem, for a given $r>0$ $$ \exists ^?X \in \mathcal{C} \cap \{ X\in \mathbb{R}^n \mid X^T \cdot X \geq r^2\}$$
My attempt
I need to solve the following optimization problem \begin{align} \min_{x \in \mathbb{R}^n} &\quad g(x) \\\text{s.t}&\quad x^T\cdot x \geq r^2 \end{align} This unfortunately is a minimization of a convex function over a concave domain. However, consider the following matrix: $$ Q = \begin{bmatrix} A &B\\ B^T &C\end{bmatrix} \in \mathbb{R}^{(n+m)\times(n+m)}$$ with $C \in \mathbb{R}^{m\times m}$, $B \in \mathbb{R}^{n\times m}$ and $A \in \mathbb{R}^{n\times n}$. Then according to Schur's complement theorem one has: $$ C \succeq 0 \implies \left( Q \succeq0 \iff A - B \cdot C^{-1} \cdot B^T \succeq 0\right)$$ Therefore, because $\mathcal{I} \succeq 0$ \begin{align} r^2 - X^T\cdot X \geq 0 \iff X^T\cdot X \leq r^2 \iff \begin{bmatrix} r^2 &X^T\\ X &\mathcal{I}\end{bmatrix} \succeq 0 \end{align} Therefore $$ \begin{bmatrix} r^2 &X^T\\ X &\mathcal{I}\end{bmatrix} \nsucceq 0 \Rightarrow r^2 - X^T\cdot X < 0 $$ Hence a sufficient (but not necessary) condition for $X^T\cdot X \geq r^2$ is $\begin{bmatrix} r^2 &X^T\\ X &\mathcal{I}\end{bmatrix} \preceq 0$, which is convex in $X$. It is obtained: \begin{align} \min_{x \in \mathbb{R}^n} &\quad g(x) \\ \text{s.t} &\quad \begin{bmatrix} r^2 &X^T\\ X &\mathcal{I}\end{bmatrix} \preceq 0 \end{align}
Question
Is this correct? I am afraid while the necessary condition being the matrix not to be positive definite, asking for it to be negative definite is too much! Are there any better solution?