Proving $d_1$ is a metric.

Question

If $d$ is a metric on a set X, then $d_1 = \frac{d(x,y)}{1+d(x,y)}$ is also a metric. I have proved the other conditions of being a metric except the triangle inequality. Please help!!

Answer 1

We have $$\frac{{1 + d\left( {x,y} \right)}}{{d\left( {x,y} \right)}} = 1 + \frac{1}{{d\left( {x,y} \right)}} \geqslant 1 + \frac{1}{{d\left( {x,z} \right) + d\left( {z,y} \right)}} = \frac{{1 + d\left( {x,z} \right) + d\left( {z,y} \right)}}{{d\left( {x,z} \right) + d\left( {z,y} \right)}}.$$ So $$\begin{gathered} \frac{{d\left( {x,y} \right)}}{{1 + d\left( {x,y} \right)}} \leqslant \frac{{d\left( {x,z} \right) + d\left( {z,y} \right)}}{{1 + d\left( {x,z} \right) + d\left( {z,y} \right)}} = \frac{{d\left( {x,z} \right)}} {{1 + d\left( {x,z} \right) + d\left( {z,y} \right)}} + \frac{{d\left( {z,y} \right)}} {{1 + d\left( {x,z} \right) + d\left( {z,y} \right)}} \hfill \\ \leqslant \frac{{d\left( {x,z} \right)}} {{1 + d\left( {x,z} \right)}} + \frac{{d\left( {z,y} \right)}} {{1 + d\left( {z,y} \right)}} \hfill \\ \end{gathered} $$

Answer 2

Hint: Use the fact that the function $f(x) = \frac{x}{1+x}$ is increasing and that d is a metric and satisfies triangle inequality!!

Answer 3

Hint: More generally, if $f:[0,\infty)\to[0,\infty)$ is an increasing, subadditive function such that $f(x)=0\Leftrightarrow x=0$, then $d'(x,y):=f(d(x,y))$ is a metric on $X$.

The case that interests you is when $f(x)=\frac{x}{1+x}$, but one could use the general result with $f(x):=\sqrt{x}$ and $f(x):=\min\{1,x\}$ in order to prove that $d(x,y):=\sqrt{|x-y|}$ and $d(x,y):=\min\{1,|x-y|\}$ are metrics on $\mathbb{R}$, for example.

Answer 4

Try as an aside to prove the following lemma, if $0 \leq a \leq b$ then \begin{equation} \frac{a}{1+a} \leq \frac{b}{1+b} \end{equation}