# If $d(x,y)$ is a metric, then $\frac{d(x,y)}{1 + d(x,y)}$ is also a metric

Let $$(X,d)$$ be a metric space and for $$x,y \in X$$ define $$d_b(x,y) = \dfrac{d(x,y)}{1 + d(x,y)}$$ a) show that $$d_b$$ is a metric on $$X$$

Hint: consider the derivative of $$f(t)$$ = $$\dfrac{t}{1+t}$$

b) show that $$d$$ and $$d_b$$ are equivalent metrics.

c) let $$(X,d)$$ be $$(\mathbb{R}, |\cdot|)$$ Show that there exists no $$M>0$$ such that $$|x-y|$$ $$\leq$$ $$Md_b$$ $$(x,y)$$ for all $$x,y$$ $$\in$$ $$\mathbb{R}$$

I have calculated $$f(t)$$ and $$f'(t)$$ and from this I know $$f(t)$$ is an increasing function as $$f'(t)$$ is strictly positive. But I don't know where to go from here or how to do parts b) and c)

• I don't believe we can use the derivative to show the triangle inequality, since the original problem does not specify that X = R. I vote for the answer of ippiki-ookami. Dec 13, 2021 at 19:50

a) Separation and symmetry are clear. For the triangular inequality, there is a bit more work. Let $$f(t):=\frac{t}{1+t}\qquad f'(t)=\frac{1}{(1+t)^2}.$$ Since this function increases on $[0,+\infty)$, the triangular inequality of $d$ yieds $$d_b(x,y)=f(d(x,y))\leq f(d(x,z)+d(z,y))=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)}$$ $$\leq f(d(x,z))+f(d(z,y))=d_b(x,z)+d_b(z,y).$$

b) You need to show that a sequence converges to $x$ for $d$ if and only if it converges to $x$ for $d_B$.

Assume first that $d(x_n,x)\rightarrow 0$. Then $$d_B(x_n,x)=\frac{d(x_n,x)}{1+d(x_n,x)}\leq d(x_n,x)$$ so $d_B(x_n,x)$ tends to $0$.

Now if $d_B(x_n,x)$ tends to $0$, $d(x_n,x)$ is bounded by some $M>0$. Indeed, assume for a contradiction that $d(x_n,x)$ is unbounded. So there exists a subsequence $d(x_{n_k},x)$ which tends to $\pm\infty$. Then $d_B(x_{n_k},x)$ must tend to $1$. Contradiction.

Now $$\frac{d(x_n,x)}{1+M}\leq \frac{d(x_n,x)}{1+d(x_n,x)}=d_B(x_n,x).$$ So $d(x_n,x)$ tends to $0$.

c) Observe that $d_B$ is bounded while $|x-y|$ is unbounded. So such a minoration is impossible.

• Im still a little confused with c) and how i can use that fact to prove that no M>0 exists Feb 20, 2013 at 17:57
• @Mathsstudent147 Assume such an $M>0$ exists. Then consider $x=n\geq 0$ and $y=0$. You would have $n\leq M n/(1+n)\leq M$ for all $n$. A contradiction. Feb 20, 2013 at 17:59
• Hello can i ask why So there exists a subsequence which tends to infinity then dB(xn ,x) tends to 1 ?? Apr 1, 2017 at 12:35

Separation, identity of indiscernibles and symmetry are immediate by the properties of $d$. For the triangular inequality, here is a simpler step-by-step proof without having to consider derivatives:

$$d_b(x,z) = \frac{d(x,z)}{1 + d(x,z)} = \frac{1 + d(x,z) - 1}{1 + d(x,z)} = 1 - \frac{1}{1 + d(x,z)}$$

$$\leq 1 - \frac{1}{1 + d(x,y) + d(y,z)} = \frac{1 + d(x,y) + d(y,z) - 1}{1 + d(x,y) + d(y,z)}$$

$$= \frac{d(x,y)}{1 + d(x,y) + d(y,z)} + \frac{d(y,z)}{1 + d(x,y) + d(y,z)}$$

$$\leq \frac{d(x,y)}{1 + d(x,y)} + \frac{d(y,z)}{1 + d(y,z)} = d_b(x,y) + d_b(y,z)$$

where the first inequality is from the sub-additivity of $d$, and the second inequality is from the fact that reducing a positive denominator can only increase a positive quotient.

• Great answer! Thanks a lot Sep 26, 2021 at 19:16

you know that $d$ is a metric so $$d_b=\frac{d(x,y)}{1+d(x,y)}=1-\frac{1}{1+d(x,y)}$$ as $d(x,y)$ is greater equal zero, you have the positiv definit here, and the symmetrie. The triangle inequality should be shown similar.

for c ) take that $d_B$ is bounded (1 is a bound), but $|x-y|$ is not.

• My turn to be picky. The first two properties are trivial, but for the triangular inequality, there is some work to do. Then note that there are several notions of equivalence for norms. The one you mention is Lipschitz-equivalence, and it is not true here (take $d(x,y)=|x-y|$ on $\mathbb{R}$, one is unbounded, the other one is bounded, that's the point of question c). But they are topologically equivalent, which means that they define the same topology. Feb 20, 2013 at 16:18
• @julien thanks cause I already thaugt that they are not equivalent. we did only have the lipschitz equivalence. Feb 20, 2013 at 16:20
• That's the other way around. They are equivalent (topologically), but not Lipschitz equivalent. Feb 20, 2013 at 16:23
• oh sry my english is bad. I will correct it Feb 20, 2013 at 16:58
• I was figurring out how to proof triangle inequality, but I see now you already linked it. Feb 20, 2013 at 17:05

[Tried to generalise part (a), along the lines of Julien's proof, and David Ullrich's proof here]

Obs: Let $${ (X, d) }$$ be a metric space. Suppose $${ \varphi : \mathbb{R} _{\geq 0} \to \mathbb{R} _{\geq 0} }$$ is continuous, $${ \varphi(0) = 0 },$$ and $${ \varphi (t) \gt 0, \varphi ' (t) \geq 0, \varphi '' (t) \leq 0 }$$ for $${ t \in (0, \infty) }.$$
Then $${ \rho (x,y) := \varphi (d (x,y)) }$$ is also a metric on $$X.$$
Pf: Firstly, $${ \varphi (d(x,y)) = 0 \iff d(x,y) = 0 \iff x=y. }$$
Let $${ x,y,z \in X }.$$ Triangle inequality $${ \varphi(d(x,y)) + \varphi(d(y,z)) \geq \varphi(d(x,z)) }$$ also holds : Notice $${ \varphi }$$ is increasing on $${ [0, \infty) }$$ and $${ \varphi ' }$$ is decreasing on $${ (0, \infty) }.$$ From $${ d(x,y) + d(y,z) \geq d(x, z) },$$ we have $${ \varphi (d(x,y) + d(y,z)) \geq \varphi(d(x,z)) }.$$ So it suffices to show $${ \varphi (d(x,y)) + \varphi(d(y,z)) \geq \varphi( d(x,y) + d(y,z) ) }.$$
It suffices to show $${ \varphi(a) + \varphi(b) \geq \varphi(a+b) }$$ for all $${ a,b \in \mathbb{R} _{\geq 0} }.$$ Rephrasing, it suffices to show that for every $${ a \in \mathbb{R} _{\geq 0} },$$ $${ \phi _a (t) := \varphi (a) + \varphi (t) - \varphi(a+t) }$$ is $${ \geq 0 }$$ on $${ [0, \infty) }.$$
Let $${ a \in \mathbb{R} _{\geq 0} }.$$ We have $${ \phi _a (0) = 0 },$$ and $${ \phi ' _a (t) }$$ $${ = \varphi'(t) - \varphi'(a+t) \geq 0 }$$ for $${ t \in (0, \infty) }.$$ So $${ \phi _a }$$ is $${ \geq 0 }$$ on $${ [0, \infty) },$$ as needed.

So if $${ d }$$ is a metric on set $${ X },$$ so are $${ d _1 (x,y) = \frac{d(x,y)}{1+d(x,y)} ,}$$ $${ d _2 (x,y) = 1 - e ^{ - d (x,y) } },$$ $${ d _3 (x,y) = \log (1+d(x,y)) },$$ etc.

Edit If further $${ \varphi ' (t) \gt 0 }$$ for $${ t \in (0, \infty) },$$ the identity map $${ (X, d) \overset{\text{id}}{\to} (X, \rho) }$$ is a homeomorphism, because $${ \varphi }$$ has a continuous inverse $${ \varphi ^{-1} : \mathbb{R} _{\geq 0} \to \mathbb{R} _{\geq 0} }$$ and now $${ (x _n \to x \text{ in } (X,d) ) }$$ $${ \iff ( \lim _{n \to \infty} d(x _n, x) = 0 ) }$$ $${ \iff ( \lim _{n \to \infty} \varphi(d(x _n, x)) = 0 ) }$$ $${ \iff (x _n \to x \text{ in } (X, \rho) ) }$$ as needed.