Uniform lower bound of positive-definite Hessian on unit sphere

Question

Let $f:\mathbb{R}^{n}\to\mathbb{R}$ be twice continuously differentiable with positive-definite Hessian (denoted by $\nabla^{2}f$), i.e., for all $x,y\in\mathbb{R}^{n}$, we have $y^{\top}\nabla^{2}f(x)y>0$. Is it possible to find some $\mu>0$ such that \begin{equation} y^{\top}\nabla^{2}f(x)y\geq\mu\lVert y\rVert^{2} \end{equation} holds for all $x\in B_{1}(0)$ and $y\in\mathbb{R}^{n}$? Here, $B_{1}(0)$ denotes the (closed) unit sphere.

Answer 1

$B_1(0)$ is compact and the minimal eigenvalue of $\nabla^2f$ depends continuously on $x$. (See this MSE Q&A for details & references.) Let $\mu$ be the minimal eigenvalue of $\nabla^2f$, attained at some $x\in B_1(0)$. If $\mu=0$ then $\nabla^2f(x)$ is singular, and hence fails to be positive definite. So $\mu>0$.