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 $0< \theta <1$ for which $‖J(x)−J(y)‖ ≤ \theta / \mu$ for all $x,y \in N$,show that $\{x_k\}$ converges at least linearly to $x^∗$ whenever $x_0 \in N$.
I have a solution which relies on $\{x_k \}$ being in $N$ for every $k$, but I cannot show that this is the case. Could you please help me with this problem? Thank you very much.
My Solution Suppose that for some $k \ge 0$ we have $x_k \in N$. 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^*|| \cdot \frac {\theta}{ \mu} = \theta||x_k - x^*||. \end{align*}
So assuming $x_k \in N$ for every $k$, this argument shows that $\{x_k \}$ converges linearly.
If $N$ was a ball centered at $x^*$, I think this argument would also prove that $x_k \in N$ for every $k$, because assuming $x_k \in N$, we have shown that $x_{k+1}$ is closer to $x^*$ than $x_k$ is. But unfortunately $N$ is not a ball, it is a general convex neighborhood.
P.S. I would be very grateful if you could also tell me whether my proof is correct or not. Thank you very much.