Question

Q. Is d(x, y) =√(│x-y│) a metric space?(solve it)

Answer 1

Answer on Question #71146 – Math – Real Analysis

Question

Is $d(x, y) = \sqrt{|x - y|}$ a metric?

Solution

First let us recall the definition of the metric

(see https://en.wikipedia.org/wiki/Metric_(mathematics)#Definition).

Let us assume that $x, y \in \mathbb{R}$. Now let us check all four conditions from definition.

1. $d(x, y) \geq 0$. Since $|x - y| \geq 0$ due to the property of absolute value, and the square root is also nonnegative, function we conclude that the condition 1 is satisfied.

2. $d(x, y) = 0 \Leftrightarrow \sqrt{|x - y|} = 0 \Leftrightarrow |x - y| = 0 \Leftrightarrow x = y$, so the second condition is satisfied.

3. $d(x, y) = \sqrt{|x - y|} = \sqrt{|y - x|} = d(y, x)$, and the condition of symmetry is also satisfied.

4. $d(x, z) \leq d(x, y) + d(y, z)$. Let us start from the well-known triangle inequality

(see https://en.wikipedia.org/wiki/Triangle_inequality#Example_norms)

$|a + b| \leq |a| + |b|$. Let us put $a := x - y, b := y - z \Rightarrow a + b = x - z$. Then we get:

$$|x - z| \leq |x - y| + |y - z|.$$

The last inequality is easily transformed to

$$|x - z| \leq |x - y| + |y - z| + 2\sqrt{|x - y| \cdot |y - z|} \Leftrightarrow |x - z| \leq \left(\sqrt{|x - y|} + \sqrt{|y - z|}\right)^2 \Leftrightarrow$$

$\Leftrightarrow \sqrt{|x - z|} \leq \sqrt{|x - y|} + \sqrt{|y - z|}$, so we obtained the required inequality

$d(x, z) \leq d(x, y) + d(y, z)$, and the last condition is also satisfied.

We see that all four conditions from definition are satisfied so $d(x, y) = \sqrt{|x - y|}$ is a metric.

**Answer**: Yes.

Answer provided by https://www.AssignmentExpert.com