Generalized triangle inequality in b-metric spaces

[ozkan12]

How is the generalized triangle inequality in b-metric spaces ? I find something...But I wonder your opinion...Thank you for your attention...

Especially if you write for n,m>0 m>n $d({x}_{n},{x}_{m})$$\le$..... I will be happy...

 

[Fernando Revilla]

How is the generalized triangle inequality in b-metric spaces ? I find something...But I wonder your opinion...Thank you for your attention... Especially if you write for n,m>0 m>n $d({x}_{n},{x}_{m})$$\le$..... I will be happy...

Suppose $m=n+3,$ and $d$ a $b-$metric. Then you'll easily prove
$$d(x_n,x_{n+3})\le\ldots\le sd(x_n,x_{n+1})+s^2d(x_{n+1},x_{n+2})+s^2d(x_{n+2},x_{n+3}).$$ But $s\ge 1\Rightarrow s^3\ge s^2,$ so we can write
$$d(x_n,x_{n+3})\le\ldots\le sd(x_n,x_{n+1})+s^2d(x_{n+1},x_{n+2})+s^3d(x_{n+2},x_{n+3}).$$ Now, you can find a simple formula for a generic $m>n,$ easily proved by inducction.

 

[ozkan12]

Dear professor

I find this...İs this true ?

$d\left({x}_{n},{x}_{m}\right)$$\le$$sd\left({x}_{n},{x}_{n+1}\right)$+${s}^{2}d\left({x}_{n+1},{x}_{n+2}\right)$+...+${s}^{n-m}d\left({x}_{m-1},{x}_{m}\right)$

 

[Fernando Revilla]

Dear professor I find this...İs this true ?
$d\left({x}_{n},{x}_{m}\right)$$\le$$sd\left({x}_{n},{x}_{n+1}\right)$+${s}^{2}d\left({x}_{n+1},{x}_{n+2}\right)$+...+${s}^{n-m}d\left({x}_{m-1},{x}_{m}\right)$

It should be (I suppose your $n-m$ is a typo): $$d\left({x}_{n},{x}_{m}\right)\le sd\left({x}_{n},{x}_{n+1}\right)+{s}^{2}d\left({x}_{n+1},{x}_{n+2}\right)+\cdots+{s}^{m-n}d\left({x}_{m-1},{x}_{m}\right).$$