MHB Best wishes :)Question: How do we use supremum in the proof of Cauchy sequence?

[ozkan12]

Let $\left(E,d\right)$ be a complete metric space, and $T,S:E\to E$ two mappings such that for all $x,y\in E$,

$d\left(Tx,Sy\right)\le M\left(x,y\right)-\varphi\left(M\left(x,y\right)\right)$,

where $\varphi:[0,\infty)\to [0,\infty)$ is a lower semicontinuous function with $\varphi\left(t\right)>0$ for $t\in(0,\infty)$ and $\varphi(0)=0$ $M(x,y)=max\left\{d(x,y), d(Tx,x),d(Sy,y),\frac{1}{2}\left[d(y,Tx)+d(x,Sy)\right]\right\}$

Proof of cauchy sequence

We will show that $\left\{{x}_{n}\right\}$ is Cauchy sequence. Let

${c}_{n}=sup\left\{d\left({x}_{j},{x}_{k}\right):j,k\ge n\right\}$. Then ${c}_{n}$ is decreasing. İf $\lim_{{n}\to{\infty}}{c}_{n}=0$, then we are done. Assume that $\lim_{{n}\to{\infty}}{c}_{n}=c>0$. Choose $\varepsilon<\frac{c}{8}$ small enough and select $N$ such that for all $n\ge N$.

$d\left({x}_{n+1},{x}_{n}\right)<\varepsilon$ and ${c}_{n}<c+\varepsilon$. By the definition of ${c}_{N+1}$, there exists $m,n\ge N+1$ such that $d\left({x}_{m},{x}_{n}\right)>{c}_{n}-\varepsilon \ge c-\varepsilon$. Replace ${x}_{m}$ by ${x}_{m+1}$ if necessary. We may assume that $m$ is even, $n$ is odd, and $d\left({x}_{m},{x}_{n}\right)>c-2\varepsilon$. Then $d\left({x}_{m-1},{x}_{n-1}\right)>c-4\varepsilon$, and $d\left({x}_{m},{x}_{n}\right)=d\left(T{x}_{m-1},S{x}_{n-1}\right)\le M\left({x}_{m-1},{x}_{n-1}\right)-\varphi\left(M\left({x}_{m-1},{x}_{n-1}\right)\right)$
$\le max\left\{d\left({x}_{m-1},{x}_{n-1}\right),\varepsilon,\varepsilon,\frac{1}{2}\left[d\left({x}_{n-1},{x}_{m}\right)+d\left({x}_{m-1},{x}_{n}\right)\right]\right\}-\varphi\left(\frac{c}{2}\right)$. We have proved that ${c}_{N}-\varphi\left(\frac{c}{2}\right)>{c}_{N+1}$ (if $\varepsilon$ is small enough). This is impossible. Thus, we must have $c=0$

Question: How ${c}_{N}-\varphi\left(\frac{c}{2}\right)>{c}_{N+1}$ is condratiction ? I didnt understand...Please can you help me ? Thank you for your attention...Best wishes :)

 

[Opalg]

Let $\left(E,d\right)$ be a complete metric space, and $T,S:E\to E$ two mappings such that for all $x,y\in E$,

$d\left(Tx,Sy\right)\le M\left(x,y\right)-\varphi\left(M\left(x,y\right)\right)$,

where $\varphi:[0,\infty)\to [0,\infty)$ is a lower semicontinuous function with $\varphi\left(t\right)>0$ for $t\in(0,\infty)$ and $\varphi(0)=0$ $M(x,y)=max\left\{d(x,y), d(Tx,x),d(Sy,y),\frac{1}{2}\left[d(y,Tx)+d(x,Sy)\right]\right\}$

Proof of cauchy sequence

We will show that $\left\{{x}_{n}\right\}$ is Cauchy sequence. Let

${c}_{n}=sup\left\{d\left({x}_{j},{x}_{k}\right):j,k\ge n\right\}$. Then ${c}_{n}$ is decreasing. İf $\lim_{{n}\to{\infty}}{c}_{n}=0$, then we are done. Assume that $\lim_{{n}\to{\infty}}{c}_{n}=c>0$. Choose $\varepsilon<\frac{c}{8}$ small enough and select $N$ such that for all $n\ge N$.

$d\left({x}_{n+1},{x}_{n}\right)<\varepsilon$ and ${c}_{n}<c+\varepsilon$. By the definition of ${c}_{N+1}$, there exists $m,n\ge N+1$ such that $d\left({x}_{m},{x}_{n}\right)>{c}_{n}-\varepsilon \ge c-\varepsilon$. Replace ${x}_{m}$ by ${x}_{m+1}$ if necessary. We may assume that $m$ is even, $n$ is odd, and $d\left({x}_{m},{x}_{n}\right)>c-2\varepsilon$. Then $d\left({x}_{m-1},{x}_{n-1}\right)>c-4\varepsilon$, and $d\left({x}_{m},{x}_{n}\right)=d\left(T{x}_{m-1},S{x}_{n-1}\right)\le M\left({x}_{m-1},{x}_{n-1}\right)-\varphi\left(M\left({x}_{m-1},{x}_{n-1}\right)\right)$
$\le max\left\{d\left({x}_{m-1},{x}_{n-1}\right),\varepsilon,\varepsilon,\frac{1}{2}\left[d\left({x}_{n-1},{x}_{m}\right)+d\left({x}_{m-1},{x}_{n}\right)\right]\right\}-\varphi\left(\frac{c}{2}\right)$. We have proved that ${c}_{N}-\varphi\left(\frac{c}{2}\right)>{c}_{N+1}$ (if $\varepsilon$ is small enough). This is impossible. Thus, we must have $c=0$

Question: How ${c}_{N}-\varphi\left(\frac{c}{2}\right)>{c}_{N+1}$ is condratiction ? I didnt understand...Please can you help me ? Thank you for your attention...Best wishes :)

It is hard to make sense of this because you have not said how the sequence $\{x_n\}$ is defined.

 

[ozkan12]

oh, excuse me dear opalg,

${x}_{n}=T{x}_{n-1}={T}^{n}{x}_{0}$. ${x}_{0}$ is arbitrary point in X metric space

 

[Opalg]

oh, excuse me dear opalg,

${x}_{n}=T{x}_{n-1}={T}^{n}{x}_{0}$. ${x}_{0}$ is arbitrary point in X metric space

Looking at the proof more carefully, I think that that definition is wrong. It looks as though the correct definition should be ${x}_{n}=T{x}_{n-1}$ if $n$ is even, and ${x}_{n}=S{x}_{n-1}$ if $n$ is odd.

I still fail to understand how this proof works, because it starts by claiming that $d(x_{n+1},x_n)$ can be made less than $\varepsilon$ for all sufficiently large $n$. I cannot see anything that might justify that.

However, if the proof is correct, it appears to show that ${c}_{N}-\varphi\!\left(\frac{c}{2}\right)>{c}_{N+1}$ for all sufficiently large $N$ (if $\varepsilon$ is small enough). This would mean that $${c}_{N+1} < {c}_{N}-\varphi\!\left(\tfrac{c}{2}\right),$$ $${c}_{N+2} < {c}_{N+1}-\varphi\!\left(\tfrac{c}{2}\right) < {c}_{N}-2\varphi\!\left(\tfrac{c}{2}\right),$$ $$\vdots$$ $${c}_{N+k} < {c}_{N}-k\varphi\!\left(\tfrac{c}{2}\right),$$ and for $k$ large enough the right side of that last inequality would become negative. That is clearly impossible, because ${c}_{N+k}$ has to be positive.

 

[ozkan12]

Dear professor Opalg,

Why right side of last inequality would become negative for "k" large enough ? I didnt understand...Can you explain..? Thank you for your attention :)

 

[Opalg]

Why right side of last inequality would become negative for "k" large enough ? I didnt understand...Can you explain..? Thank you for your attention :)

Try to work this one out for yourself. If you take a positive quantity (${c}_{N}$) and you subtract larger and larger multiples of another positive quantity ($\varphi\!\left(\tfrac{c}{2}\right)$), then eventually the result will become negative. That's not graduate-level analysis, it's primary school arithmetic.

 

[ozkan12]

Thank you dear Opalg, I think so but to ensure, I ask this...Thank you sou much, thank you for your attention :)

 

[ozkan12]

Dear professor,

How we get

${C}_{N+2}\le{C}_{N+1}-\varphi\left(\frac{c}{2}\right)$
${C}_{N+3}\le{C}_{N+2}-\varphi\left(\frac{c}{2}\right)$
.
.
.

By iteration ? Or by repating same process ?

 

[ozkan12]

İn this proof, I have some troubles...

First of all,

in ${c}_{n}=sup\left\{d\left({x}_{j},{x}_{k}\right):j,k\ge n\right\}$ why we use supremum ?

Also, How we know that $\left\{d\left({x}_{j},{x}_{k}\right):j,k\ge n\right\}$ has a supremum ? This set is not bounded...

Thank you for your attention...