Say $f:\mathbb{R}^3\to[0,\infty)$ and suppose $f$ is radial, i.e. $f=f(r)$ where $r=|x|$. Suppose we know that $f''(r)\ge c>0$. Then is it necessarily the case that the hessian of $f$ is a uniformly positive definite matrix, $v^T(\nabla\nabla f) v\ge c'|v|^2>0$ for all $v\in\mathbb{R}^3-\{0\}$?
This is clearly the case when $f=\frac{|x|^2}{2}$ because then $f''(r)=1$ and $\nabla\nabla f=I$. Just not sure if it's true in general.