Hello and thanks for read.
We have the problem $\min_{c(x)=0} f(x)$ with $c:\mathbb{R}^n\rightarrow \mathbb{R}^m$. If we use Newton's method for solve the problem we get a succession $(x_k,\lambda_k)_{k\in\mathbb{N}}$ such as for the last $k$ it satisfies the stop condition $||\nabla l_x(x_k,\lambda_k)||_\infty<\epsilon$ and $||c(x)||_\infty<\epsilon$. We define $(x_*,\lambda_*)$ as the last approximation to the minimum and for all $k$ $$y_k=\frac{||\nabla l_x(x_k,\lambda_k)||_\infty+||c(x_k)||_\infty}{||\nabla l_x(x_1,\lambda_1)||_\infty+|c(x_1)||_\infty}$$ where $l(x,\lambda)$ is the lagrange function.
The question is: if $y_k$ have a quadratic convergence, can we imply that $(x_k,\lambda_k)$ have a quadratic convergence?.
I think that is necessary have conditions over $f$ and $c$, I interest in the case $f$ and $c_i'$ linear, $i=1,...,m$ .
I try to prove that exists $M_0,M_1>0$ such as for all $k$ $M_0||(x_k,\lambda_k)-(x_*,\lambda_*)||\le y_k\le M_1||(x_k,\lambda_k)-(x_*,\lambda_*)||$ because is that happened we will have $$M_0||(x_{k+1},\lambda_{k+1})-(x_*,\lambda_*)||\le y_{k+1} \le Cy_{k}^2\le CM_1^2||(x_k,\lambda_k)-(x_*,\lambda_*)||^2$$ but I dont know is this is a good way.
Thanks for help!.
