# 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)

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.