Proof that a function is a b-metric

Question

I'm trying to prove that this function is a b-metric. I can't prove the triangle inequality. Here's the definition of a b-metric space.bmetric definition. Here's the example. I wanna prove the last inequality.example Any idea would be helpful. My teacher told me that this is a basic and easy prove but I spent more than 2 hours and still can't understand how they got s=5/2. (Sorry for my english). Thank you!

Kavi Rama Murthy · Answer 1 · 2019-06-18T00:35:18.987

0

edited Jun 18 '19 at 00:35

answered Jun 18 '19 at 00:10

Kavi Rama Murthy

311,013

Thank you for this hint. I still can't understand how can I show that. I tried to use $|x-y|\leq |x|+|y|$ but still can't prove. I got something like $d(m,n)+d(n,p)\leq d(m,p)+2|1/n|$ – ursuv2 Jun 18 '19 at 00:33
@ursuv2 You have to deal only with the case when $d(m,n)=|\frac 1 n -\frac 1 m|$ and $d(n,p)=|\frac 1 n -\frac 1 p|$. In this case carefully look at the definition and prove that we must have $d(m,p)=|\frac 1 p -\frac 1 m|$ The inequality becomes obvious from this. – Kavi Rama Murthy Jun 18 '19 at 04:46

score 0 · Accepted Answer · answered Jun 18 '19 at 03:00

HINT.-What you need is that for all $m,n,p$ in $\mathbb N\cup\{\infty\}$ $$\frac{d(m,p)}{d(m,n)+d(n,p)}\le\frac52\tag1$$ You don't need to prove that in $(1)$ the maximum of the $LHS$ is equal to $\dfrac52$ (in fact, it seems that never $LHS=RHS$).

The maximum of the numerator is $5$ so $(m,p)=(2r+1,2s+1)$ or $(m,p)=(2r+1,\infty)$ (or $(\infty,2s+1)$ of course) and the minimum of the denominator is easily seen to be $2+2=4$ because of the taken values of $(m,p)$ (note that taking $m=n$ we would have $0+5\gt4$).

On the other hand one has

$$\left|\frac1m-\frac1p\right|=\left|\left(\frac1m-\frac1n\right)+\left(\frac1n-\frac1p\right)\right|$$ and the triangular inequality would work to discard this possibility.

Finally one has $$\frac{d(m,p)}{d(m,n)+d(n,p)}\le\frac54\le\frac52$$

Proof that a function is a b-metric

2 Answers2