Proving that the sequence space is a metric space in $\mathbb{R}$

Question

I need to show that $(X, d)$ is a metric space:

$X$ is the sequence space where its elements are sequences of real numbers, and $d : X \times X \rightarrow \mathbb{R}$ where d is defined as:

$$d(x,y) = \sum_{j=1}^\infty\frac{1}{2^j} \frac{|x_j - y_j|}{1 + |x_j - y_j|}$$

I think that I need to prove the right $\frac{|x_j - y_j|}{1 + |x_j - y_j|}$ is montone increasing but I have no clue how to start with proving the triangle inequality for this expression. Any help is appreciated.

Answer 1

Hint: First, the function $f(x)=\frac{x}{1+x}$ is monotone increasing when $x \geq 0$, which can be proved by methods like taking derivative.

Second, since $f(x)=\frac{x}{1+x}$ is monotone increasing, prove the triangle inequality holds for each component, which is $\frac{|x_j - y_j|}{1 + |x_j - y_j|}+\frac{|y_j-z_j|}{1 + |y_j-z_j|} \geq \frac{|x_j-z_j|}{1 + |x_j-z_j|}$.

Third, prove the metric satisfies the triangle inequality.

Answer 2

Let $f(x) = \frac{x}{1 + x}$. We wish to show that, for $x, y \ge 0$, we have $f(x + y) \le f(x) + f(y)$ (i.e. the function is subadditive over $[0, \infty)$). I hope you can see how this would prove triangle inequality for the proposed metric.

Suppose $x, y \ge 0$. Consider \begin{align*} f(x + y) \le f(x) + f(y) &\iff \frac{x + y}{1 + x + y} \le \frac{x}{1 + x} + \frac{y}{1 + y} \\ &\iff 1 - \frac{1}{1 + x + y} \le 2 - \frac{1}{1 + x} - \frac{1}{1 + y} \\ &\iff \frac{1}{1 + x} + \frac{1}{1 + y} \le 1 + \frac{1}{1 + x + y} \\ &\iff (1 + x + y)(1 + y) + (1 + x + y)(1 + x) \le (2 + x + y)(1 + x)(1 + y) \\ &\iff 1 + x + y \le (1 + x)(1 + y) \\ &\iff xy \ge 0, \end{align*} which is true.

You can also use the representation $f(x) = 1 - \frac{1}{1 + x}$ to quickly conclude that $f$ increases over $[0, \infty)$.

Answer 3

The most important idea is here, which shows that

If $d: X \times X \to \mathbb{R}$ is a metric, then so is $d'(x,y) = \frac{d(x,y)}{1+d(x,y)}$.

So in our case we apply it to $d(x,y) = |x-y|$ on the reals. The quotient metric has the pleasant property that it's bounded by $1$.

Then our formula just becomes a weighted sum of metrics, all with weights $>0$ and the weights $\frac{1}{2^j}$ serve mostly to ensure that the series is absolutely convergent (as the $n$-th term is bounded above by $\frac{1}{2^n}$, so the series sum never grows above $1$ too).

Symmetry is clear and the triangle inequality just becomes an infinite sum of component triangle inequalities (because we know that the quotients form a metric by themselves). The sum can only be $0$ iff all terms are etc. So it's then easy to check the metric properties.

Answer 4

We know the function $ f:R\rightarrow R$ given by $f(x)=\frac {x}{1+x}$ is increasing on $[0,\infty)$ since $f'>0\ on\ [0,\infty)$.Hence $|x_n-y_n|≤|x_n-z_n|+|z_n-y_n|$ so that $\frac {|x_n-y_n|}{1+|x_n-y_n|}≤ \frac {|x_n-z_n|+|z_n-y_n|}{1+|x_n-z_n|+|z_n-y_n|}=\frac {|x_n-z_n|}{1+|x_n-z_n|+|z_n-y_n|} + \frac {|z_n-y_n|}{1+|x_n-z_n|+|z_n-y_n|}≤\frac{|x_n-z_n|}{1+|x_n-z_n|}+\frac {|z_n-y_n|}{1+|z_n-y_n|}$

Now multiply by $\frac{1}{2^n}$ both first and last expression and then take sum over $n$.