# Why is this inequality about norm of the inverse of a nonsingular matrix holds?

Nocedal and Wright, in their book Numerical Optimization, Version 2, just under (3.32) say:

Since $\nabla^2 f(x^*)$ is nonsingular, there is a radius $r>0$ such that $\Vert \nabla^2f(x_k)^{-1}\Vert \leq 2 \Vert \nabla^2f(x^*)^{-1} \Vert$ for all $x_k$ with $\Vert x_k - x^* \Vert \leq r$.

where $x^*$ is the minimizer (the position at which function $f$ attains its minimum), $\nabla^2f(x^*)$ is the Hessian matrix (matrix of second partial derivatives) at $x^*$, $\nabla^2f(x_k)$ is the Hessian matrix at $x_k$, and $k$ is the current step number.

Can you prove that statement? Why is the norm of inverse of a nonsingular, positive definite matrix has this property?

• The inequality looks weird. It's just $\Vert A \Vert \le 2 \Vert A \Vert$ Commented Dec 15, 2017 at 17:21
• Yes, @JiaqiLi. I think the RHS should be at $x^*$, not at $x_k$, but the questioner should fix it themselves before we go assuming it. Commented Dec 15, 2017 at 17:41
• @JiaqiLi and JonathanZ, you guys are right. I fixed the typo the question. Commented Dec 16, 2017 at 16:53
• If that's the case, then does my answer below work for you? If it's correct you can accept it by clicking on the check mark. If it doesn't apply I'm curious as to why. Commented Dec 16, 2017 at 22:53

I don't understand all the notation, but any positive-valued continuous function has the property that you can find a neighborhood around a point where the function is bounded by twice its value at that point (you apply the $\epsilon$-$\delta$ version of continuity with $\epsilon$ equal to the value of the function at the point). Can we use this here for the function $x \mapsto \|(\nabla^2 f)(x)^{-1}\|$?