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where $(X,d)$ is a metric and $x,y \in X$.

I know we need to show:

  • non-negativity: $d(x,y)\geq$ 0
  • $d(x,y)=0$ if and only if $x=y$
  • symmetry: $d(x,y)=d(y,x)$
  • $d(x,z)\leq d(x,y) + d(y,z)$

I think we need to consider the derivative of $f(t)=\frac{t}{1+t}$ however I am not sure what to do next.

Levi
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3 Answers3

19

If $(X, d)$ is any metric space, show that another metric on $X$ is defined by $$\overline d(x, y) = \frac{d(x, y)}{1 + d(x, y)}$$.

Proof:We must first show that $\overline d$ is a metric on $X$.

  1. Observe that $\displaystyle \overline d(x, y) = \frac{d(x, y)}{1 + d(x, y)}$ and $d(x, y) \geq 0$ since it is a metric of a metric space. It follows that $ \frac{d(x, y)}{1 + d(x, y)} \geq 0$ and thus Axiom 1 is satisfied for $\overline d$.
  2. Suppose $\overline d (x, y) = 0$. Then $\overline d(x, y) = \frac{d(x, y)}{1 + d(x, y)}$ which implies $d(x, y) = 0$. Since $d$ is a metric on $X$, by Axiom 2 it follows that $x = y$. Conversely suppose that $x = y$. Then $\overline d(x, y) = \frac{d(x, y)}{1 + d(x, y)} = \frac{0}{1 + 0} = \frac01 = 0$ and thus Axiom 2 is satisfied for $\overline d$.
  3. Observe that $\overline d(x, y) = \frac{d(x, y)}{1 + d(x, y)} = \frac{d(y, x)}{1 + d(y, x)} = \overline d(y, x)$ and thus Axiom 3 is satisfied for $\overline d$.
  4. Observe that given $d(z, w) \geq d(x, y)$ it follows that \begin{align*} d(x, y) &\leq d(z, w)\\ d(x, y) + d(x, y)d(z, w) &\leq d(z, w) + d(z, w)d(x, y)\\ d(x, y)(1 + d(z, w))&\leq d(z, w)(1 + d(x, y))\\ \frac{d(x, y)}{1 + d(x, y)} &\leq \frac{d(z, w)}{1 + d(z, w)} \end{align*} Now observe that $d(x, y) \leq d(x, z) + d(z, y)$ since $d$ is a metric. It immediately follows that \begin{align*} \overline d(x, y)&= \frac{d(x, y)}{1 + d(x, y)}\\ &\leq \frac{d(x, z) + d(z, y)}{1 + d(x, z) + d(z, y)}\\ &= \frac{d(x, z)}{1 + d(x, z) + d(z, y)} + \frac{d(z, y)}{1 + d(x, y) + d(z, y)}\\ &\leq \frac{d(x, z)}{1 + d(x, z)} + \frac{d(z, y)}{1 + d(y, z)}\\ &= \overline d(x, z) + \overline d(z, y)\\ \end{align*} Thus Axiom 4 is satisfied for $\overline d$.
Robert Cardona
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  • In axiom $4$, the first inequality does not seem to hold. How do you justify it? – T. Eskin Feb 21 '13 at 23:05
  • I believe I fixed the problem. It is greater than or equal to. It implies the second line of the last equation array. It shows that it is monotone increasing. – Robert Cardona Feb 21 '13 at 23:18
  • Ah, yes. You could use the fact there that the function $f$ is increasing. This works just fine then. – T. Eskin Feb 21 '13 at 23:40
  • You n.1 is wrong. you have $a/(1+a)$, and $a>0$, then $a/(1+a)<a$ because $a$ is less than $1+a$. Anyway, that point is as easy as: $a/(a+1)>0$ because $a>0$ because $a$ forms a metric space. Being $a=d(x,y)$ – MyUserIsThis Feb 21 '13 at 23:50
  • @MyUserIsThis Thanks! Your right! I fixed it! – Robert Cardona Feb 21 '13 at 23:58
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As you noted: the function $f(t)=\frac{t}{1+t}$ can be differentiated in $(-1,\infty)$, and you get $f '(t)=\frac{1}{(1+t)^{2}}>0$, which shows that $f$ is monotonically increasing.

I think the only non-trivial axiom here is the triangle inequality, so I will demonstrate how it follows by this observation. Whenever I use the fact that $f$ is monotonically increasing, I denote it by $(*)$.

Take $x,y,z\in X$. If $d(x,z)\leq d(x,y)$, then \begin{equation*} d_{b}(x,z)=\frac{d(x,z)}{1+d(x,z)}\overset{(*)}{\leq}\frac{d(x,y)}{1+d(x,y)}=d_{b}(x,y)\leq d_{b}(x,y)+d_{b}(y,z). \end{equation*} Similarly, if $d(x,z)\leq d(y,z)$ then $d_{b}(x,z)\leq d_{b}(x,y)+d_{b}(y,z)$.

So we may assume that $d(x,z)>d(x,y)$ and $d(x,z)>d(y,z)$. I denote this condition by $(**)$. Hence \begin{align*} d_{b}(x,z)&=\frac{d(x,z)}{1+d(x,z)}\leq\frac{d(x,y)+d(y,z)}{1+d(x,z)}=\frac{d(x,y)}{1+d(x,z)}+\frac{d(y,z)}{1+d(x,z)} \\ &\overset{(**)}{\leq}\frac{d(x,y)}{1+d(x,y)}+\frac{d(y,z)}{1+d(y,z)} =d_{b}(x,y)+d_{b}(y,z). \end{align*} So in any case, $d_{b}(x,z)\leq d_{b}(x,y)+d_{b}(y,z)$.

T. Eskin
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3

For any particular $x,y$ you know $d(x,y)\geq 0$. So define $a=d(x,y)$ and note that your new metric becomes

$$\frac a {1+a}$$

Since $d(x,y)$ is a metric by assumption, $a$ is not dependent on the ordering of $x$ and $y$ and so this is trivially symmetric. You then want to prove that $\frac{a}{1+a}=0\iff a=0$ and that $\forall a\ \frac{a}{1+a}\geq0$ (remember that $a=d(x,y)\geq0$ by assumption).

The triangle inequality is trickier. First show that $\frac a{1+a}$ is monotonically increasing. Denoting $a=d(x,z), b=d(x,y), c=d(y,z)$, you want to show that $a\leq b+c\implies \frac a{1+a}\leq \frac b{1+b}+\frac c{1+c}$. Since $\frac a{1+a}$ is monotonically increasing, we have that $a<b\implies \frac a{1+a}<\frac b{1+b}$, which you'll need.

Robert Mastragostino
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    Monotonicity doesn't seem to be sufficient. $f\left(t\right) = \frac{t}{1+t}$ must be shown to be subadditive. – S.B. Feb 21 '13 at 22:38
  • @S.B. I didn't mean to say that monotonicity would be sufficient, only necessary as part of the proof. Maybe it's just me, but I considered the rest of it self-explanatory once that stage of the setup is reached. – Robert Mastragostino Feb 22 '13 at 05:45