Cauchy Sequence Contractive Condition

Question

Show that the sequence $\left(x_n\right)$ satisfies the cauchy criterion $x_1=1$ and $x_{n+1}=\frac{1}{2+x_n^2}\:\forall \:n\ge 1$

As per the hint for this problem, they just follow the regular route of $\left|x_{n+1}-x_n\right|$ after proving by mathematical induction that all terms in this sequence are greater than 0 and finally get to this point $\left|x_{n+1}-x_n\right|=\frac{\left|x_n^2-x_{n-1}^2\right|}{\left(2+x_n^2\right)\left(2+x_{n-1}^2\right)}\le \frac{\left|x_n-x_{n-1}\right|\left|x_n+x_{n-1}\right|}{4}$ After this the solution uses this inequality further $\left|x_n+x_{n-1}\right|\le 2$. Apparently, there is something known as the contractive condition for cauchy sequence. It would help if anyone could see how this inequality $\left|x_n+x_{n-1}\right|\le 2$ and further how the contractive condition is equivalent to the formal definition of cauchy.

Answer 1

It is clear that $x_n \geq 0$ and $x_{n+1}\leq \frac 1{2+0}=\frac 1 2$. So $|x_n+x_{n-1}|\leq |x_n|+|x_{n-1}|\leq \frac 1 2+\frac 1 2=1$.

This gives $|x_{n+1}-x_n|\leq \frac{|x_n-x_{n-1}||x_n+x_{n-1}|} 4\leq \frac{|x_n-x_{n-1}|} 4$. Iteration of this gives $|x_{n+1}-x_n|\leq |x_2-x_1| 4^{-(n-1)}$. Now $|x_m-x_n|\leq |x_m-x_{m-1}|+ |x_{m-1}-x_{m-2}|+\cdots +|x_{n+1}-x_n|$ for $m>n$. Can you take it from here?

Answer 2

We can prove the following property:

If $\,f:M\to M\,$ is a contraction mapping on a metric space $\,(M,d)\,$ and if $\,x_1\in M\,,\,$ $x_{n+1}=f\left(x_n\right),\;\forall n\in\Bbb N\,,\,$ then the sequence $\big\{x_n\big\}_{\!n\in\Bbb N}$ is a Cauchy sequence.

Proof:

Since $\,f:M\to M\,$ is a contraction mapping, there exists $\,k\in(0,1)\,$ such that $\;d\big(f(x),f(y)\big)\leqslant k\,d(x,y)\;$ for any $\,x,y\in M\,.\quad\color{blue}{(1)}$

For any $\,i\in\Bbb N\,$ it results that

$\begin{align}d\left(x_i,x_{i+1}\right)&=d\big(f(x_{i-1}),f(x_i)\big)\underset{(1)}{\leqslant}k\,d(x_{i-1},x_i)=\\[3pt]&=k\,d\big(f(x_{i-2}),f(x_{i-1})\big)\underset{(1)}{\leqslant}k^2d(x_{i-2},x_{i-1})=\\&\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\[3pt]&=k^{i-2}d\big(f(x_1),f(x_2)\big)\underset{(1)}{\leqslant}k^{i-1}d(x_1,x_2)\;.\quad\color{blue}{(2)}\end{align}$

For any $\,\varepsilon>0\,$ there exists $\,n_\varepsilon\!\!\in\!\Bbb N\,\land\, n_\varepsilon\!\!>\!\log_k\!\!\left[\!\dfrac{(1-k)\varepsilon}{1\!+\!d(x_1,x_2)}\!\right]$
such that, for any $\,n,m\in\Bbb N\,\land\,n,m>n_\varepsilon\,,\;$ it results that

$\begin{align}d(x_n,x_m)&\leqslant\begin{cases}\sum\limits_{i=n}^{m-1}d(x_i,x_{i+1})\underset{(2)}{\leqslant}\sum\limits_{i=n}^{m-1}k^{i-1}d(x_1,x_2)&\text{if }n<m\\[2pt]\;\,0&\text{if }n=m\\[2pt]\sum\limits_{i=m}^{n-1}d(x_i,x_{i+1})\underset{(2)}{\leqslant}\sum\limits_{i=m}^{n-1}k^{i-1}d(x_1,x_2)&\text{if }n>m\end{cases}\\[7pt]&=\begin{cases}k^{n-1}d(x_1,x_2)\!\!\sum\limits_{i=0}^{m-n-1}\!\!k^i\leqslant\dfrac{k^{n-1}d(x_1,x_2)}{1-k}&\text{if }n<m\\[1pt]\;\,0&\text{if }n=m\\k^{m-1}d(x_1,x_2)\!\!\sum\limits_{i=0}^{n-m-1}\!\!k^i\leqslant\dfrac{k^{m-1}d(x_1,x_2)}{1-k}&\text{if }n>m\end{cases}\\[7pt]&\leqslant\dfrac{k^{\min\{n,m\}-1}d(x_1,x_2)}{1-k}\leqslant\dfrac{k^{n_\varepsilon}d(x_1,x_2)}{1-k}\leqslant\\[3pt]&\leqslant k^{\log_k\left[\frac{(1-k)\varepsilon}{1+d(x_1,x_2)}\right]}\!\cdot\!\dfrac{d(x_1,x_2)}{1-k}=\\[3pt]&=\frac{(1-k)\varepsilon}{1+d(x_1,x_2)}\!\cdot\!\dfrac{d(x_1,x_2)}{1-k}<\varepsilon\,.\end{align}$

So we have proved that for any $\,\varepsilon>0\,$ there exists $\,n_\varepsilon\!\in\Bbb N\,$ such that for any $\,n,m\in\Bbb N\,\land\,n,m>n_\varepsilon\,$ it results that $\,d(x_n,x_m)<\varepsilon\;,\;\;$ that is ,

$\big\{x_n\big\}_{\!n\in\Bbb N}$ is a Cauchy sequence.

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$\;$
In particular, if $\,(M,d)\,$ is the metric space $\,\Bbb R\,$ with the Euclidean metric $\,d_{\Bbb R}(x,y)=|x-y|\;\,\forall x,y\in\Bbb R\,$ and $\,f:\Bbb R\to\Bbb R\,$ is the function defined as $\;f(x)=\dfrac1{2+x^2}$
for any $\,x\in\Bbb R\,,\,$ then
$f\,$ is a contraction mapping, indeed $\,\exists\,k\!=\!\dfrac14\!\in\!(0,1)$ such that

$\begin{align}d_{\Bbb R}\big(f(x),f(y)\big)&=\big|f(x)-f(y)\big|=\left|\dfrac1{2+x^2}-\dfrac1{2+y^2}\right|=\\[3pt]&=\left|\dfrac{y^2-x^2}{\left(2+x^2\right)\!\left(2+y^2\right)}\right|=\dfrac{|y+x||y-x|}{\left(2+x^2\right)\!\left(2+y^2\right)}\leqslant\\[3pt]&\leqslant\dfrac{|x|+|y|}{2\!\left(x^2+y^2+2\right)}\,\big|x-y\big|\leqslant\\[3pt]&\leqslant\dfrac{\frac12\left(x^2+1\right)+\frac12\left(y^2+1\right)}{2\!\left(x^2+y^2+2\right)}\,\big|x-y\big|=\\[3pt]&=\dfrac{\frac12\left(x^2+y^2+2\right)}{2\!\left(x^2+y^2+2\right)}\,\big|x-y\big|=\dfrac14\big|x-y\big|=\\[3pt]&=k\,d_{\Bbb R}(x,y)\quad\text{ for any }x,y\in\Bbb R\,.\end{align}$

Consequently, by applying the property that I previously proved, it follows that the sequence $\,\big\{x_n\big\}_{\!n\in\Bbb N}\,,\,$ obtained by $\,x_1=1\in\Bbb R\,,\,$ $x_{n+1}=f\left(x_n\right)=\dfrac1{2+x_n^2}\,,\;\forall n\in\Bbb N\;,$
is a Cauchy sequence.