3
$\begingroup$

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 $d(x,y)\le d(x,z)+d(z,y), \forall x,y,z \in X$.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Arthur Aug 17 at 10:16
  • 1
    $\begingroup$ Hint: $d'(x, y) =\frac{d(x, y)}{1+d(x, y)}$ is also a metric if $d$ is. $\endgroup$ – Berci Aug 17 at 10:27
1
$\begingroup$

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 $$ f(a+b) = \frac{a}{a+b} f(a+b) + \frac{b}{a+b}f(a+b) \le $$ $$ \le f\big(\frac{a}{a+b}\cdot (a+b)\big) + f\big(\frac{b}{a+b}\cdot (a+b)\big) = $$ $$ = f(a) + f(b). $$

$\endgroup$
1
$\begingroup$

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 $|\cdot|$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.