0
$\begingroup$

i'm reading Numerical Optimization.

In page 45, Newton's method: "Since $\nabla ^{2} f\left( x^{*}\right)$ is nonsingular, there is a radius r > 0 such that $\Vert \nabla ^{2} f^{-1}_{k}\Vert \leqslant 2\Vert \nabla ^{2} f\left( x^{*}\right)^{-1}\Vert$ for all $x_{k}$ with $\Vert x_{k} -x^{*}\Vert \leqslant r$". I don't understand why, can anyone please explain for me.(Sorry for not typing all the criteria because it is too long and relevant to other section )

$\endgroup$
5
  • $\begingroup$ Your link is broken. However, it appears that you're referring to the proof of theorem 3.5 in Nocedal and Wright's Numerical Optimization, 2nd ed. The theorem has the hyothesis that $\nabla^{2} f(x)$ is Lipschitz continuous in a neighborhood of $x^{*}$. You need that hypothesis here. $\endgroup$
    – Brian Borchers
    Commented Sep 24, 2019 at 4:13
  • $\begingroup$ @BrianBorchers Thank you for the reply. From your hint, i see that r here imply the use of Lipschitz condition, which is $\Vert \nabla ^{2} f( x_{k}) -\nabla ^{2} f( x_{0})\Vert \leqslant \Vert x_{k} -x^{*}\Vert $. But i still don't see how can i apply it $\endgroup$
    – hkab
    Commented Sep 24, 2019 at 16:10
  • $\begingroup$ No, Lipschitz continuit of the Hessian means that there is a constant $L$ such that $\| \nabla^{2}f(x_{k})-\nabla^{2}(f(x^{*}) \| \leq L \| x_{k}-x^{*} \|$. Furthermore, when $x^{*}$ is sufficently far from any point of singularity, the inverse of the Hessian is also continuous. $\endgroup$
    – Brian Borchers
    Commented Sep 24, 2019 at 16:14
  • $\begingroup$ Sorry i mistyped the Lipschitz condition, from my poor algebra skills, i was trying to do something like this (i don't know why i can't use mathjax here). I still not know how can i derive the coefficient 2. Or maybe me approach is wrong. $\endgroup$
    – hkab
    Commented Sep 25, 2019 at 11:58
  • $\begingroup$ this question is actually duplicated, anyone run into this question should take a look at this $\endgroup$
    – hkab
    Commented Sep 28, 2019 at 1:30

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .