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

Question

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)

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

[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.

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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.