Triangle inequality in b-metric spaces

[ozkan12]

Let $X$ be a non-empty set and let $s\ge1$ be a given real number. A function $d:$ X $\times$ X$\to$ ${R}^{+}$ , is called a b-metric provided that, for all x,y,z $\in$ X,

1) d(x,y)=0 iff x=y,
2)d(x,y)=d(y,x),
3)d(x,z)$\le$s[d(x,y)+d(y,z)].

A pair (X,d) is called b-metric space. İt is clear that definition of b-metric space is a extension of usual metric space.

İn attachment, I didnt prove triangle inequality, please help me...thank you for your attention :)

View attachment 4626

 

[Fernando Revilla]

Hint: prove that for all \(\displaystyle x,y,z\in l_p\) with \(\displaystyle (0<p<1),\) we have \(\displaystyle d(x,z)\le 2^{1/p}\left[d(x,y)+d(y,z)\right].\)

 

[ozkan12]

yes we get this...But I didnt prove...Please can you write this and send me ? thank you for your attention :)

 

[Fernando Revilla]

yes we get this...But I didnt prove...Please can you write this and send me ? thank you for your attention :)

Using the following results you'll easily prove it.

1. If $0<p<1$ then $(a+b)^p\le a^p+b^p$ for all $a,b\ge 0.$

2. If $0<p<1$ then $f(t)=t^{1/p}$ is a convex funtion on $(0.+\infty)$ so $$\left(\frac{1}{2}a+\frac{1}{2}b\right)^{1/p}\le \frac{1}{2}a^{1/p}+\frac{1}{2}b^{1/p}$$ As a consequence $$(a+b)^{1/p}\le 2^{1/p}(a^{1/p}+b^{1/p})\quad \forall a,b \ge 0.$$

 

[ozkan12]

Thank you so much, But I didnt understand this...

 

[Fernando Revilla]

Thank you so much, But I didnt understand this...

Let us see, 15:11 - 14:54 are equivalent to 17 minutes. You can't say that you don't understand, you need some more effort. Show some work and I'll help you.

 

[ozkan12]

Dear professor,

I worked on this topic...But I didn't come to consequence...:)

 

[Fernando Revilla]

Dear professor,

I worked on this topic...But I didn't come to consequence...:)

I am going to solve your question (only for this time) but you should understand that this is not the best way for learning.

Suppose $x=(x_n),$ $y=(y_n),$ $z=(z_n)$ are elements of $l_p$ with $0<p<1.$ Then, using $$\left|x_n-z_n\right|=\left|(x_n-y_n)+(y_n-z_n)\right| \le \left|x_n-y_n\right|+\left|y_n-z_n\right|,$$
$$d(x,z)=\left(\sum_{n=1}^{\infty}\left|x_n-z_n\right|^p\right)^{1/p}\le \left(\sum_{n=1}^{\infty}\left(\left|x_n-y_n\right|+\left|y_n-z_n\right|\right)^p\right)^{1/p}.$$ Using $(a+b)^p\le a^p+b^p$ for every $a,b\ge 0,$
$$d(x,z)\le \left(\sum_{n=1}^{\infty}\left(\left|x_n-y_n\right|^p+\left|y_n-z_n\right|^p\right)\right)^{1/p}$$ $$=\left(\sum_{n=1}^{\infty}\left|x_n-y_n\right|^p+\sum_{n=1}^{\infty}\left|y_n-z_n\right|^p\right)^{1/p}.$$
Using $(a+b)^{1/p}\le 2^{1/p}(a^{1/p}+b^{1/p})\quad \forall a,b \ge 0,$ $$d(x,z)\le 2^{1/p}\left(\left(\sum_{n=1}^{\infty}\left|x_n-y_n\right|^p\right)^{1/p}+\left(\sum_{n=1}^{\infty}\left|y_n-z_n\right|^p\right)^{1/p}\right)$$
$$=d(x,y)+d(y,z).$$ That is, $d$ is an $s=2^{1/p}-$ metric.

 

[ozkan12]

Dear professor

I carried out this process...But I think this is false :) İndeed, this true :) Thank you so much...But in first post

How we get as a consequence

(a+b)^1/p ≤ 2^1/p (a^1/p+b^1/p )∀a,b≥0.

Thank you for your help dear professor, best wishes :)

 

[Fernando Revilla]

How we get as a consequence (a+b)^1/p ≤ 2^1/p (a^1/p+b^1/p )∀a,b≥0.

$$\left(\frac{1}{2}a+\frac{1}{2}b\right)^{1/p}\le \frac{1}{2}a^{1/p}+\frac{1}{2}b^{1/p}$$ $$\Rightarrow \frac{1}{2^{1/p}}(a+b)^{1/p}\le \frac{1}{2}\left(a^{1/p}+b^{1/p}\right)\le a^{1/p}+b^{1/p}$$ $$\Rightarrow (a+b)^{1/p}\le 2^{1/p}(a^{1/p}+b^{1/p}).$$

 

[ozkan12]

Dear professor,

Thank you for your attention and your help...But in

⇒1/2^1/p(a+b)^1/p≤1^2(a^1/p+b^1/p)≤a^1/p+b^1/p in second part I think 1/2 must be 1/2^1/p... İs it true ?

 

[Fernando Revilla]

İs it true ?

No, it isn't.