I am working on the following problem:
Consider using Newton’s method to find $x \in \mathbb{R}^n$ such that $F(x) = 0,$ for $F:D \subseteq \mathbb{R}^n \to \mathbb{R}^n$, where $D$ is open and convex and $F$ is continuously differentiable on $D$. Suppose $x^∗$ satisfies $F(x∗) = 0$; and $J(x)$, the Jacobian of $F$ evaluated at $x$ satisfies $‖J^{−1}(x)‖\le \mu$ for some number $\mu > 0$ for all $x$ in a convex neighborhood $N \subseteq D$ that contains $x^∗$.
Assuming there is a constant $v$ such that $‖J(y)−J(x)‖≤v‖y−x‖$ for each $x,y \in N$, show that $\{x_k \}$ converges quadratically to $x^∗$ whenever $x_0 \in N.$
I have been able to solve the problem, but only by imposing two additional assumptions: $N$ is a ball centered at $x^*$, and $||x_0 - x^*|| < \frac 1M$ where $M = \frac 12 \mu v.$
\begin{align*} \\ \end{align*}
My Partial Solution:
Suppose that for some $k \ge 0$ we have $x_k \in N$, and that for $m = 0, 1, 2, ..., k-1$ we have $||x_{m+1} - x^*|| \le M ||x_m - x^*||^2$ where $M = \frac 12 v \mu$.
Let $h = x_k - x^*$ and define $$G(t) = F(x^* + th)$$ for $t \in [0, 1]$. Since $x^*, x_k \in N$ and $N$ is convex, $x^* + th \in N$ for every $t$. Thus $G$ is differentiable on $[0, 1]$ with $G'(t) = J(x^* + th)h$.
\begin{align*} ||x_{k+1} - x^*|| &= ||x_k - J^{-1}(x_k) F(x_k) - x^*||\\ \\ & \le ||J^{-1}(x_k)|| \cdot ||J(x_k)(x_k - x^*) - F(x_k)|| \\ \\ &\le \mu || J(x_k)(x_k - x^*) - (F(x_k) - F(x^*) )||\\ \\ & = \mu|| G'(1) - (G(1) - G(0))||\\ \\ & = \mu|| \int_0^1 G'(1) dt - \int_0^1 G'(t) dt||\\ \\ & \le \mu \int_0^1 ||G'(1) - G'(t)|| dt \\ \\ & \le \mu ||x_k - x^*|| \int_0^1 ||J(1) - J(x^* + th)|| dt\\ \\ & \le \mu ||x_k - x^*|| \int_0^1 t v ||h|| dt \\ & \le \mu ||x_k - x^*||^2 \frac 12 v \\ \\ & = M ||x_k - x^*||^2. \end{align*}
Let $y_m = ||x_m - x^*||$. We would like to prove that $y_m \to 0$. We have
\begin{align*} y_1 &\le My_0^2\\ y_2 &\le My_1^2 \le M^3 y_0^4\\ y_3 &\le My_2^2 \le M^7 y_0^8\\ &\vdots \\ y_m &\le M^{2^{m}-1} y_0^{2^m} = \frac 1M \left(M y_0\right)^{2^m} \end{align*}
Now if $y_0 < \frac 1M$ (assumed in my additional hypothesis), then $$y_{m+1} < \frac 1M (1)^{2m+2} = \frac 1M = y_0$$ This means that $x_{k}$ is closer to $x^*$ than $x_{k+1}$ is, and if $N$ is a ball (my other additional hypothesis) then $x_{k+1}$ is also in $N$.
Finally, $y_0 < 1M \implies \frac 1M \left(M y_0\right)^{2^m} \to 0$, so $y_m \to 0$ as desired.