I am not sure if this question is too easy for mathoverflow - please tell me to remove this question if it is too e before any downvotes. I have asked this on MSE (Link), but it received only a few comments and no answers.

**Note: the screenshot at the bottom is where my question comes from.**

This question is quite different from other versions of conditions of convergence of Newton iteration. For example, Kantorovich theorem.

I am now analysing the Newton-Raphson iteration in general Banach spaces $E,F$. Let $x_0\in E$, and let $f:B_t(x_0)\to F$ be a differentiable function. ($B$ denotes an open ball with radius $t$.) $L(E,F)$ is the set of linear mapping from $E$ to $F$.

By definition, $f$ is differentiable at $x$ with derivative $Df_x\in L(E,F)$ (which is a linear functional from $E$ to $F$) if $\exists r(h),f(x+h)=f(x)+Df_x(h)+r(h)$, where $r(h)/\|h\|\to 0$ as $h\to 0$.

To make it simple, I assume that there exist $s>0$ such that

- $\|f(x_0)\|\leq t/(2s)$
- If $x,y\in B_t(x_0)$ then $\|Df_x-Df_y\|\leq 1/(2s)$
- $\forall x\in B_t(x_0),\exists J_x\in L(F,E)$ such that $J_xDf_x=Df_xJ_x=I_E$ and $\|J_x\|\leq s$.

Now let's work on the iteration. Let's **fix** $x\in B_t(x_0)$. Set $x_n=x_{n-1}-J_x(f(x_{n-1}))$. In real analysis course, we often take $x=x_{n-1}$, but here I have to fix $x$ to be anything in $B_t(x_0)$. Just assume for a moment that $\forall x\in B_t(x_0)$. I will explain why later.

Firstly I have to show that $x_n$ converges. Now I can use the inequality $$ \|f(a)-f(b)-T(a-b)\|\leq \|a-b\|\sup_{c\in [a,b]} \|Df_c-T\|, $$ where $[a,b]$ is the line segment joining $a,b$, and $T\in L(E,F)$.

To use this inequality, we define $g(y)=J_x(f(y))$, so $x_n=x_{n-1}-g(x_{n-1})$, and $Dg_y=J_xDf_y$.(**The reason why I cannot set $x=x_{n-1}$ is that** if I do it that way, then $g(y)=J_y(f(y))$, and I cannot find the derivative of $g$ in this case.) Since $x$ is fixed, we can assume there is NO $x$ dependence in $g$. Therefore,
$$
\|x_{n+1}-x_{n}\|=\|f(x_{n})-f(x_{n-1})-(x_{n}-x_{n-1})\|\\ \leq \|x_{n}-x_{n-1}\|\sup_{c\in [x_n,x_{n-1}]} \|Dg_c-I\|\\=\|x_{n}-x_{n-1}\|\sup_{c\in [x_n,x_{n-1}]} \|J_xDf_c-J_xDf_x\|\\ \leq \|x_{n}-x_{n-1}\|\|J_x\|\|Df_c-Df_x\|\\ \leq \frac{1}{2} \|x_{n}-x_{n-1}\|.
$$
Also,
$$
\|x_1-x_0\|=\|J_x(f(x_0))\|\leq t/2
$$
The conclusion is $\|x_n-x_{n-1}\|\leq t/2^n$.

**My question: is it really OK to let $x$ be anything fixed in $B_t(x_0)$? Does that really work? If it is wrong, how can I fix it?**

To prove that $f(x_n)$ converges to zero, I feel that I should prove something like $\|f(x_n)\|\leq t/(2^{n+1}s)$(Suggested in a book of real analysis). I try to start by considering this: $$ \|f(x_n)\|\leq \|Df_x\|\|x_{n+1}-x_n\| $$ but it goes nowhere. From $\|J_x\|\leq s$ we cannot obtain an upper bound on $Df_x$.

**So how can I prove $\|f(x_n)\|\leq t/(2^{n+1}s)$?**

**It should be clear that $x_n$ is a Cauchy sequence - but it might not converge into $B_t(x_0)$ - is that a problem?**

It is a long question, so if I have made mistakes please point it out.

Please look at the following screenshot if the above is not clear.

Source of my problem: *A course in mathematical analysis* (screenshot)

Here is a theorem of Kantorovich which is related but not the same.