Boundedness of a set under perturbation

Question

Denote by $\mathbb{R}^{n\times n}_{S^+}$ the set of all real, symmetric and positive definite $(n\times n)$- matrices. Let $Q_0\in \mathbb{R}^{n\times n}_{S^+}$ and choose $\varepsilon>0 $ such that the closed ball $\bar B(Q_0,\varepsilon)\subset \mathbb{R}^{n\times n}_{S^+}$. My question is whether the set $$\{x\in \mathbb{R}^n: x^TQx\le 1, Q\in \bar B(Q_0,\varepsilon)\}$$ is compact? Thank you in advance.

Answer 1

The set $\bar B(Q_0,\epsilon)$ is compact, and then the minimum of $$ (x,Q) \mapsto x^TQx $$ for $\|x\|=1$ and $Q\in \bar B(Q_0,\epsilon)$ exists, and is positive. Lets call this minimum $\lambda$. In fact, it is the smallest possible eigenvalue for matrices in $\bar B(Q_0,\epsilon)$.

Then if $x^TQx\le 1$ for some $x$ and $Q\in \bar B(Q_0,\epsilon)$ it follows that $\|x\| \le \lambda^{-1}$. And the set in question is bounded. In addition, it is closed, as it can be written as an intersection of closed sets $$ \bigcap_{Q \in \bar B(Q_0,\epsilon)} \{ x: \ x^TQx\le 1\}, $$ so we have compactness.