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I have a theorem to prove for numerical mathematics regarding a fixed-point iteration (henceforth FPI).

Problem statement:
Given $p\geq 2$ and $\Phi \in C^{p} [a,b]$ for some $a,b\in\mathbb{R}$ is a FPI-Function ($x^{(\nu+1)}=\Phi(x^{(\nu)})$). Let $\tilde{x}\in(a,b)$ be a fixed point for the FPI s.t. $\Phi^{(j)}(\tilde{x})=0\quad j=1,...,p-1$ but $\Phi^{(p)}(\tilde{x})\neq 0$. We have to show that the order of convergence of the FPI is not greater than $p$.
Relevant definition (order of convergence):
A FPI is said to converge in Order $\alpha$ if $\exists C\in\mathbb{R} >0$ s.t. for all $k\in\mathbb{N}$ holds: $|x^{(k+1)}-\tilde{x}| \leq C |x^{(k)}-\tilde{x}|^\alpha$
Own solution:
The proof of convergence of order at least $p$ is done by taylor expansion of $\Phi$ around $\tilde{x}$ with lagrange remainder $\frac{\Phi^{(p)}(\xi(x, \tilde{x}))}{p!}(x-\tilde{x})^p$ then setting $x=x^{(k)}$ and then taking either the supremum/maximum of $\Phi^{(p)}$ over $[x_k,\tilde{x}]$ or taking limit $k\rightarrow\infty$.
Now to prove that the order is not higher than $p$ I assume it is (proof by contradiction), that is $\exists C>0$ s.t. the definition holds with $\alpha=p+1$.
Then I use the Taylor expansion from the first part of the proof to get: $|\frac{\Phi^{(p)}(\xi(x^{(k)}, \tilde{x}))}{p!}||x^{(k)}-\tilde{x}|^p \leq C |x^{(k)}-\tilde{x}|^{p+1}$ (where inequality holds by assumption)
Seeing as the differences are positive and nonzero I can get rid of $|x^{(k)}-\tilde{x}|^p$ on both sides.

Now taking the limit $\lim_{k\rightarrow\infty}$ on both sides of the inequality I obtain: $0\neq |\Phi^{(p)}(\tilde{x})/p!| = \lim_{k\rightarrow\infty} |\Phi^{(p)}(\xi(x^{(k)},\tilde{x}))/p!| \leq C \lim_{k\rightarrow\infty} | x^{(k)} -\tilde{x}| = 0$
which is a contradiction - as desired. Therefore QED.

Question
My problem is that I am not sure whether this last part is sound.

Intuitively I'd say yes, because $\xi \in [x^{(k)},\tilde{x}]$ and therefore tends to $\tilde{x}$ if k tends to infinity (contraction of interval to a point). But experience has shown that what is intuitively clear to me mostly needs to be proven.

Therefore the question: Is the proof above valid or am I missing something fundamental (any other misses or things to be more precise about are welcome, too).

Thanks in advance

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1 Answer 1

up vote 0 down vote accepted

Yes, as $x^{(k)} \to x$ then $\xi^{(k)} := \xi(x^{(k)}, \tilde{x}) \to \tilde{x}$ by your argument. Then, since $\Phi^{(p)}$ is continuous (by assumption), it follows that $\Phi^{(p)}(\xi^{(k)}) \to \Phi^{(p)}(\tilde{x})$, as $k \to \infty$.

The next inequality feels better if you write $\ldots \leq C \liminf_{k \to \infty} |x^{(k)} - \tilde{x}|$, because when you estimate from above inside the limit, it may be by something that does not converge; but in your case it does, of course.

I hope I have understood and answered your question well.

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Thanks for the feedback! I like math, but am not confident at proofs therefore the question. Let's see if this holds up in the exercises. –  Nox Jan 14 at 0:31

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