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prove triangular inequality for $$ d(x,y)= \frac{||x-y||}{1+||x-y||}$$ that is $d(x,y) \leq d(x,z)+d(z,y)$ ofcourse ||.|| is a norm and has properties of norms

this usually works $$ \begin{aligned} d(x,y) = &\frac{||x-z+z-y||}{1+||x-y||} \\ & \leq \frac{||x-z|| +|| z-y||}{1+||x-y||} = \frac{||x-z||}{1+||x-y||} +\frac{||z-y||}{1+|x-y|} \\& \vdots \\ & =\frac{||x-z||}{1+||x-z||} +\frac{||z-y||}{1+|z-y|}=d(x,z)+d(z,y) \end{aligned}$$ therefore $d(x,y) \leq d(x,z)+d(z,y)$

but this dosent work. the next thing to do for mylself is try to go the other way start with right side of $\leq $ and get the other side. thiniking need to use schartz inequality. if this was on a quiz or test it would be taking too much time for me

second way attempt (I think it is closer) $$ \begin{aligned} d(x,z)+d(z,y) &= \frac{||x-z ||}{1+|x-z||}+\frac{||z-y ||}{1+||z-y||} \\ & = \frac{||x-z ||}{1+|x-z||} * \frac{1+||z-y||}{1+||z-y||} +\frac{||z-y ||}{1+||z-y||} * \frac{1+|x-z||}{1+|x-z||} \\&=\frac{||x-z||+||x-z|||z-y||}{1+||z-y||+||x-z||+||x-z|||z-y||} +\frac{||z-y||+||z-y||||x-z||}{1+||z-y||+||x-z||+||x-z|||z-y||} \\&=\frac{||x-z||+||x-z|||z-y||+_||z-y||+||z-y||||x-z||}{1+||z-y||+||x-z||+||x-z|||z-y||} \\&= \frac{2||x-z|| ||z-x||}{1+||z-y|| ||x-z||} \\& \vdots (missing steps, ) \geq \frac{||x-y|||}{1+||x-y||}=d(x,y) \end{aligned}$$ stuck, im going to be stubborn and still try to finish this proof later hopefully and think is done diff from the "duplicate"excersises (but I just glanced at them). after some thinking, The duplicate does answer the question way elegantly and I did a stupid mistake on my second attempt on second to last line.

Tiger Blood
  • 1,940

4 Answers4

2

For $d(x,y)\ne 0$, we have $$\frac{1}{{d\left( {x,y} \right)}} = \frac{{1 + \left\| {x - y} \right\|}} {{\left\| {x - y} \right\|}} = 1 + \frac{1}{{\left\| {x - y} \right\|}} \geqslant 1 + \frac{1}{{\left\| {x - z} \right\| + \left\| {z - y} \right\|}} = \frac{{1 + \left\| {x - z} \right\| + \left\| {z - y} \right\|}} {{\left\| {x - z} \right\| + \left\| {z - y} \right\|}}.$$ Thus $$\begin{gathered} d\left( {x,y} \right) \leqslant \frac{{\left\| {x - z} \right\| + \left\| {z - y} \right\|}} {{1 + \left\| {x - z} \right\| + \left\| {z - y} \right\|}} = \frac{{\left\| {x - z} \right\|}} {{1 + \left\| {x - z} \right\| + \left\| {z - y} \right\|}} + \frac{{\left\| {z - y} \right\|}} {{1 + \left\| {x - z} \right\| + \left\| {z - y} \right\|}} \hfill \\ \leqslant \frac{{\left\| {x - z} \right\|}} {{1 + \left\| {x - z} \right\|}} + \frac{{\left\| {z - y} \right\|}} {{1 + \left\| {z - y} \right\|}} = d\left( {x,z} \right) + d\left( {z,y} \right). \hfill \\ \end{gathered} $$

Baily
  • 1,539
1

Notice that the function $a\mapsto a/(1+a)$ is increasing and use that $d'(x,y)=||x-y||$ is also a metric (satisfies triangle inequality).

syzygy
  • 731
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Changing the metric $d$ to $F(d)$ preserves the properties of a metric if:

  • $F(0) = 0$
  • $F(d) > 0$ when $d > 0$
  • $F(a+b) \leq F(a) + F(b)$ for all $a,b \geq 0$

This is implied by, and in practice is equivalent to, $F$ being an increasing concave function. It is possible to artificially construct examples of $F$ that are increasing, subadditive and not concave.

$F(d)=\frac{d}{1+d}$ is the most famous instance of an increasing convex function that preserves metrics, because it has the simplest formula for an example that is bounded. Having an increasing function that transforms any metric to a bounded metric shows that boundedness is not a topologically invariant property of metric spaces.

zyx
  • 35,436
0

Besides other great answers and references, I tried to follow your thoughts to complete the proof.

\begin{aligned} d(x,z)+d(z,y) &= \frac{||x-z ||}{1+|x-z||}+\frac{||z-y ||}{1+||z-y||} \\ & = \frac{||x-z ||}{1+|x-z||} * \frac{1+||z-y||}{1+||z-y||} +\frac{||z-y ||}{1+||z-y||} * \frac{1+|x-z||}{1+|x-z||} \\&=\frac{||x-z||+||x-z|||z-y||}{(1+||x-z||) (1+||z-y||)} +\frac{||z-y||+||z-y||||x-z||}{(1+||x-z||) (1+||z-y||)} \\&=\frac{||x-z||+||x-z|||z-y||+_||z-y||+||z-y||||x-z||}{(1+||x-z||) (1+||z-y||)} \\& \textrm{Above are the same as you provide.} \\& = \frac{2||x-z|| ||z-y|| + ||x-z|| + ||z-y||}{(1+||x-z||) (1+||z-y||)} \\& \geq \frac{||x-z|| ||z-y|| + ||x-z|| + ||z-y||}{(1+||x-z||) (1+||z-y||)} \\& \textrm{ Notice the inequality holds becuase of removal of one ||x-z||||z-y|| term.} \\& = 1 - \frac{1}{(1+||x-z||) (1+||z-y||)} \\& \geq 1 - \frac{1}{(1+||x-y||)} \textrm{ since triangular inequality of d'(x, y) = ||x-y|| holds} \\& = d(x,y) \end{aligned}

Tai
  • 101