Generalized triangle inequality in b-metric spaces

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SUMMARY

The discussion focuses on the generalized triangle inequality in b-metric spaces, specifically examining the relationship between distances defined by a b-metric. The participants derive a formula for distances between points indexed by n and m, where m > n, demonstrating that the distance can be expressed as a sum of scaled distances between consecutive points. The formula is confirmed to be valid through induction, establishing a clear relationship for any m > n. The key inequality presented is: $$d(x_n,x_m) \le sd(x_n,x_{n+1}) + s^2d(x_{n+1},x_{n+2}) + \cdots + s^{m-n}d(x_{m-1},x_m).$$

PREREQUISITES
  • Understanding of b-metric spaces
  • Familiarity with mathematical induction
  • Knowledge of distance functions and inequalities
  • Basic algebraic manipulation of inequalities
NEXT STEPS
  • Explore the properties of b-metric spaces in detail
  • Study mathematical induction techniques for proving inequalities
  • Investigate applications of generalized triangle inequalities in various mathematical contexts
  • Learn about other types of metric spaces and their properties
USEFUL FOR

Mathematicians, researchers in metric space theory, and students studying advanced geometry or analysis will benefit from this discussion.

ozkan12
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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...
 
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ozkan12 said:
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.
 
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)$
 
ozkan12 said:
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): $$d\left({x}_{n},{x}_{m}\right)\le sd\left({x}_{n},{x}_{n+1}\right)+{s}^{2}d\left({x}_{n+1},{x}_{n+2}\right)+\cdots+{s}^{m-n}d\left({x}_{m-1},{x}_{m}\right).$$
 

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