Quadratic convergence is the hallmark of Newton's Method for root-solving. I'm looking for a result that implies the Newton result that looks like this:

Theorem : Let $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ satisfy the following conditions:

  1. There is a point $\bar{x}\in\mathbb{R}^n$ such that $f(\bar{x})= \bar{x}$ (ie. $\bar{x}$ is a fixed point of $f$)
  2. The Jacobian matrix $f^\prime (x)$ is Lipschitz in a neighborhood of $\bar{x}$.
  3. The matrix $f^\prime (\bar{x})$ is nilpotent, that is, all of its eigenvalues are zero.

Then there is a neighborhood of $\bar{x}$ where all points iterate under $f$ to $\bar{x}$ quadratically.

Can anyone tell me where to find such a result?


1 Answer 1


I don't know where else to find this result, but it is not hard to prove. Indeed, without loss of generality $\bar x=0$. So, we have $f(0)=0$, $M^n=0$ for $M:=f'(0)$, and $\|f'(x)-M\|\le2L\|x\|$ for some real $L\ge0$, some norms on $\mathbb R^{n\times n}$ and $\mathbb R^n$ (both denoted here by $\|\cdot\|$), some real $h>0$, and all $x\in B_h:=\{x\in\mathbb R^n\colon\|x\|\le h\}$. So, for $x\in B_h$ and $$g(x):=f(x)-Mx\tag{1}$$ one has $$\|g(x)\|=\|f(x)-f(0)-Mx\|=\Big\|\int_0^1(f'(tx)-M)x\,dt\Big\| $$ $$ \le\int_0^1\|f'(tx)-M\|\,\|x\|\,dt\le\int_0^1 2Lt\|x\|\,\|x\|\,dt=L\|x\|^2, $$ whence $\|f(x)\|\le\|M\|\|x\|+L\|x\|^2\le(\|M\|+Lh)\|x\|$. In fact, letting $\|M\|$ be the (say) Euclidean operator norm of the Jordan form of the nilpotent matrix $M$, we may assume that $\|M\|\le1$, so that $\|f(x)\|\le(1+Lh)\|x\|$ for $x\in B_h$. Hence, $x_j\in B_h$ for all $x\in B_{h_1}$ and $j=0,\dots,n-1$, where $h_1:=h/(1+Lh)^{n-1}$, and, for each $x$, the sequence $(x_j)$ is defined recursively by the formulas $x_0:=x$ and $x_{j+1}:=f(x_j)$.

By $(1)$, $f(x)=Mx+g(x)$. So, in view of the condition $M^n=0$, by induction, for any natural $N\ge n-1$ one obtains $$x_N=M^Nx+M^{N-1}g(x)+\dots+Mg(x_{N-2})+g(x_{N-1})$$ $$ =M^{n-1}g(x_{N-n})+\dots+Mg(x_{N-2})+g(x_{N-1}) $$ for $x\in B_{h_2}$, where $h_2:=\min[h_1,\frac1{nL}]$;
here we use the estimate $$\|x_N\|\le\|M^{n-1}g(x_{N-n})+\dots+Mg(x_{N-2})+g(x_{N-1})\|$$ $$ \le\|g(x_{N-n})\|+\dots+\|g(x_{N-2})\|+\|g(x_{N-1})\| $$ $$ \le L\|x_{N-n}\|^2+\dots+L\|x_{N-1}\|^2\le nLh_2^2\le h_2 $$ for $x\in B_{h_2}$. In particular, it follows that
$$\|x_N\| \le L(\|x_{N-n}\|^2+\dots+\|x_{N-1}\|^2) $$ for $x\in B_{h_2}$, which yields the desired quadratic convergence.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.