Successive approximation (Banach's fixed point theorem)

Question

Let (X,d) be a complete metric space and $T\colon X\to X$. Moreover it exists a $n\in\mathbb{N}$, so that the n-th power of $T$ is $q$-contractive. Yesterday my task was to If $T^n$ is $q$-contractive, $T$ exactly has one fixed point. Today my task is the following.

Consider any given $x_0\in X$. Let $\overline{x}\in X$ be the one and only fixed point of $T$ (from the linked task). Show that the sequence of successive approximation $$ x_m:=T^m(x_0) $$ converges to $\overline{x}$ and that there is the estimation $$ d(x_m,\overline{x})\leq q^k\max\limits_{j\in\left\{1,\ldots,n-1\right\}}\left\{d(T^j(x_0),\overline{x})\right\}, $$ whereat $m\in\mathbb{N}$ fullfills $$ kn\leq m <(k+1)n~\text{for }k\in\mathbb{N}_0:=\mathbb{N}\cup\left\{0\right\}. $$

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I only tried to show the convergence yet.

My idea is to assume w.l.o.g. that $n\leq m$ and then using Banach's fixed point theorem, which says that for $j\in\mathbb{N}_0$ and any starting point $x_0\in X$ it is $$ x_{j+1}:=T^n(x_j)\to \overline{x}.~~~~(*) $$ So I thought of the following: $$ x_m=T^m(x_0)=T^n(T^{m-n}(x_0))=T^n(x_{m-n}) $$

and for $m\to\infty$ it is $T^n(x_{m-n})=x_{m-n+1}\to\overline{x}$ because of $(*)$.

Is that okay?

Answer 1

Let $x_m = T^m (x_0)$. Let $\hat{x}$ be the unique fixed point.

Define the sequences $y_k^0 = x_{kn+0}$, $y_k^1 = x_{kn+1}$,all the way to $y_k^{n-1} = x_{kn+n-1}$.

We see that $y_{k+1}^i = T^n(y_k^i)$ for $i=0,...,n-1$, and so we have $y_k^i \to \hat{x}$ for all $i$. It follows that $x_m \to \hat{x}$.

As an aside, since $d(y_k^i ,\hat{x}) \le q^k d( y_0^i , \hat{x} )$, it is straightforward to show the other estimate.

Answer 2

Here is my idea for a proof of the estimation.

From the above construction of the series $y^i, i=0,\ldots,n-1$ and the $q$-contractivity of $T^n$ it follows that $$ d(y_k^i,\overline{x})=d(T^{kn}(y_0^i),T^n(\overline{x}))=d((T^n)^k(y_0^i),T^n(\overline{x}))\\=d((\underbrace{T^n\circ\ldots\circ T^n}_{k-\text{times}})(y_0^i),T^n(\overline{x}))\\\leq q~d((T^n)^{k-1}(y_0^i),\overline{x})=q~d((T^n)^{k-1}(y_0^i),T^n(\overline{x}))\\\leq q^2~d((T^n)^{k-2}(y_0^i),\overline{x})\leq\ldots\\\leq q^k~d(y_0^i,\overline{x}). $$ Because of $x_m=y_k^i$ for a $i\in\left\{0,\ldots,n-1\right\}$ and a $k\in\mathbb{N}_0$ for any $m\in\mathbb{N}_0$ and $y_0^i=T^i(x_0)$ it follows $$ d(x_m,\overline{x})=d(y_k^i,\overline{x})\leq q^k~d(y_0^i,\overline{x})=q^k~d(T^i(x_0),\overline{x})\\\leq q^k~\max\limits_{j\in\left\{0,\ldots,n-1\right\}}\left\{d(T^j(x_0),\overline{x})\right\}\\\leq q^k~\max\limits_{j\in\left\{1,\ldots,n-1\right\}}\left\{d(T^j(x_0),\overline{x})\right\}. $$

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Would be great to hear if this makes sense.

With greetings

math12