# Prove that $d$ is a metric on $X$.

Let $$X$$ be a set of real sequences and $$d(x,y)=\sum_{k=1}^{\infty}\dfrac{1}{2^k}\cdot \dfrac{\vert a_k-b_k \vert}{1+\vert a_k-b_k\vert}$$ with $$x=(a_k)_{k \in \mathbb{N}}$$, $$y=(b_k)_{k \in \mathbb{N}} \in X$$.

Prove that $$d$$ is a metric on $$X$$.

I can easily checking $$d(x,y) \ge 0$$, $$d(x,y)=0 \Leftrightarrow x=y$$ and $$d(x,y)=d(y,x)$$.

I stuck at checking the convergence of this sequence and the condition $$d(x,y)\le d(x,z)+d(z,y), \forall x,y,z \in X$$.

• Convergence is easy to show by comparison. Each term in the sum is less than $\frac1{2^k}$. Triangle inequality is where the real work lies. – Arthur Aug 17 at 10:16
• Hint: $d'(x, y) =\frac{d(x, y)}{1+d(x, y)}$ is also a metric if $d$ is. – Berci Aug 17 at 10:27

Consider a function $$f(x) = \frac{x}{1+x}$$ for nonnegative $$x$$. We need to prove that it is subadditive $$f(a+b) \le f(a) + f(b).$$ It is a concave function (check it, using the second derivative). Note also that $$f(tx) = f(tx + (1-t)0) \ge tf(x) + (1-t)f(0) = tf(x).$$
That's why we have $$f(a+b) = \frac{a}{a+b} f(a+b) + \frac{b}{a+b}f(a+b) \le$$ $$\le f\big(\frac{a}{a+b}\cdot (a+b)\big) + f\big(\frac{b}{a+b}\cdot (a+b)\big) =$$ $$= f(a) + f(b).$$
Proposition: If $$(X , d)$$ is a metric space, then $$d^*(x , y) = \frac{d(x , y)}{1 + d(x , y)}$$ is a metric on $$X$$.
Assume and try to prove last result later. Now we just use it to solve your problem. Let assume that $$d$$ a metric the set $$X$$ of your context, and let $$x = {(x_n)}_{n \in \mathbb{N}}$$, $$y = {(y_n)}_{n \in \mathbb{N}}$$ and $$z = {(z_n)}_{n \in \mathbb{N}}$$ be three ''points'' in $$X$$, and let $$a_n = \frac{d(x_n , z_n)}{2^n} \qquad \mbox{and} \qquad b_n = \frac{d(x_n , y_n) + d(y_n , z_n)}{2^n}$$ be two sequences on $$[0 , \infty)$$. Clearly, $$a_n \leq b_n$$ for all $$n \in \mathbb{N}$$, therefore $$\sum_{n = 1}^{\infty} \frac{d(x_n , z_n)}{2^n} \leq \sum_{n = 1}^{\infty} \frac{d(x_n , y_n) + d(y_n , z_n)}{2^n}\tag{1}\mbox{.}$$ You are done if you know how to use last proposition (this last $$d$$ is the $$d^*$$ in the proposition. Then the both series in $$(1)$$ converge, as $$d$$ takes a value less than $$1$$ when we apply it two points in $$X$$), and if you know that $$d_1$$ is a metric (the metric on $$\mathbb{R}$$ induced by the norm $$|\cdot|$$).