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.